[1]\fnmPing-Hsun \surChuang

\equalcont

The author contributed to the appendix.

[1]\orgdivDepartment of Mathematics, \orgnameNational Taiwan University, \orgaddress\cityTaipei, \postcode10617, \countryTaiwan

On the Periods of Twisted Moments of the Kloosterman Connection

[email protected] \fnmJeng-Daw \surYu [email protected] *

Abstract

This paper aims to study the Betti homology and de Rham cohomology of twisted symmetric powers of the Kloosterman connection of rank two on the torus. We compute the period pairing and, with respect to certain bases, interpret these associated period numbers in terms of the Bessel moments. Via the rational structures on Betti homology and de Rham cohomology, we prove the $\mathbb{Q}$ -linear and quadratic relations among these Bessel moments.

keywords:

Periods, Bessel moments, Kloosterman connection

pacs:

[

MSC Classification]32G20, 33C10, 34M35

Statements and Declarations

This research was supported by the grant 108-2115-M-002-003-MY2 of the Ministry of Science and Technology. The authors have no relevant financial or non-financial interests to disclose.

1 Introduction

Let $\mathbb{G}_{\mathrm{m},z}=\operatorname{Spec}\,(\mathbb{Q}[z,z^{-1}])$ be the algebraic torus over $\mathbb{Q}$ with variable $z$ , and similarly for the torus $\mathbb{G}_{\mathrm{m},t}$ with variable $t$ . Let ${\mathrm{Kl}}_{2}$ be the Kloosterman connection (of rank two) on $\mathbb{G}_{\mathrm{m},z}$ corresponding to the differential operator $(z\partial_{z})^{2}-z$ . (For details, see section 2.) In [1], in order to study the Hodge aspects of the symmetric powers $\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ , Fresán, Sabbah, and Yu consider the following settings. Let $\left[2\right]:\mathbb{G}_{\mathrm{m},t}\rightarrow\mathbb{G}_{\mathrm{m},z}$ be the double cover induced by the ring homomorphism $\mathbb{Q}[z,z^{-1}]\rightarrow\mathbb{Q}[t,t^{-1}]$ , given by $z\mapsto t^{2}$ . One obtains the pullback connection

\widetilde{{\mathrm{Kl}}}_{2}=\left[2\right]^{+}{\mathrm{Kl}}_{2}.

The structure of $\widetilde{{\mathrm{Kl}}}_{2}$ is much simpler since it is the restriction to $\mathbb{G}_{\mathrm{m}}$ of the Fourier transform of a regular holonomic module on the affine line. In addition, the symmetric power $\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ appears in the pushforward $\left[2\right]_{+}\operatorname{Sym}^{k}{\widetilde{\mathrm{Kl}}}_{2}$ naturally in the decomposition [1, p. 1662]

\left[2\right]_{+}\operatorname{Sym}^{k}{\widetilde{\mathrm{Kl}}}_{2}\cong\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\oplus\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2},

where $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}=\left(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},{\mathrm{d}}+\frac{{\mathrm{d}}z}{2z}\right)\otimes\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ . In [2], Fresán, Sabbah, and Yu compute the de Rham cohomology and Betti homology for $\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ . In this paper, we study the analogues for $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ .

1.1 Historical results and our results

Let $I_{0}(t)$ and $K_{0}(t)$ be modified Bessel functions. Define the Bessel moments

\displaystyle{\mathrm{IKM}}_{k}(a,b)=\int_{0}^{\infty}I_{0}(t)^{a}K_{0}(t)^{k-a}t^{b}{\mathrm{d}}t

(1)

provided that $0\leq a\leq k$ are non-negative integers, $b\in\mathbb{Z}$ , and the convergence of this integral. The particular Bessel moments of the form ${\mathrm{IKM}}_{a+b}(a,2c-1)$ appear in two-dimensional quantum field theory as Feynman integrals [3, 4, 5]. From a mathematical point of view, these moments are realized as period integrals of $\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ and $\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ . For the details, we refer to [2]. In that paper, Fresán, Sabbah, and Yu developed the Hodge theory on symmetric powers of the generalized Kloosterman connection $\mathrm{Kl}_{n+1}$ of rank $(n+1)$ .

Sum rule identities

In [4, (220)], the authors provide the following conjecture on the $\mathbb{Q}(\pi)$ linear relation of Bessel moments which is called the “sum rule” in their paper.

Conjecture 1.

For each pair of integers $\left(n,k\right)$ with $n\geq 2k\geq 2$ , the following combination of Bessel moments vanish

\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^{m}\binom{n}{2m}\pi^{n-2m}{\mathrm{IKM}}_{2n}(n-2m,n-2k)=0.

Later in [6, (1.5)], Zhou uses the Hilbert transformation to prove this conjecture. Moreover, he also proves a “sum rule”:

Formula 2.

For each pair of integers $\left(n,k\right)$ with $n-1\geq 2k\geq 2$ , the following conbination of Bessel moments vanish

\sum_{m=1}^{\lfloor(n+1)/2\rfloor}(-1)^{m}\binom{n}{2m-1}\pi^{n-2m+1}{\mathrm{IKM}}_{2n}(n-2m+1,n-2k-1)=0.

When the involved exponents of $t$ in (1) are odd, these two identities are both reproved by Fresán, Sabbah, and Yu [2]. The proof is to study the connection $\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ on $\mathbb{G}_{\mathrm{m}}$ whose period integrals are those Bessel moments $\mathrm{IKM}_{k}(a,b)$ with odd $b$ . In this paper, by studying the connection $\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ on $\mathbb{G}_{\mathrm{m}}$ , we provide proofs of these two identities involving even powers of $t$ using a similar approach in section 5. The key point to consider the twisted connection $\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ is that the period integrals of $\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ are those Bessel moments $\mathrm{IKM}_{k}(a,b)$ with even $b$ . For example, we have the following result:

Formula 3 (Corollary 28).

For $k=4r+4$ , a multiple of $4$ , one has

\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2i)=\begin{cases}(-1)^{r}\pi^{2r+2}{\mathrm{IKM}}_{k}(2r+2,2i)&\text{if $0\leq i\leq r$},\\ (-1)^{r}\pi^{2r+2}{\mathrm{IKM}}_{k}^{\mathrm{reg}}(2r+2,2i)&\text{if $r+1\leq i\leq\lfloor\frac{k-1}{2}\rfloor$}.\end{cases}

The notation $\mathrm{IKM}^{\mathrm{reg}}_{k}(2r+2,2i)$ above denotes the regularized Bessel moments (see Lemma 21). Roughly speaking, the regularized Bessel moments are obtained from those Bessel moments $\mathrm{IKM}_{k}(a,b)$ with parameters $k,a,b$ that makes $\mathrm{IKM}_{k}(a,b)$ diverge but minus their divergent asymptotic. Therefore, our sum rule generalizes the sum rules in [6, 4].

$\mathbb{Q}$ -dimension of Bessel moments

In [7], Zhou considers the $\mathbb{Q}$ -vector subspace spanned by the Bessel moments in $\mathbb{C}$ . This vector subspace is finite-dimensional due to the sum rule. Similarly, we have the following upper bound of the dimension.

Theorem 4 (Corollary 30).

For any $k$ and any $0\leq a\leq\lfloor(k-1)/2\rfloor$ , the dimension of the $\mathbb{Q}$ -vector space generated by the Bessel moments has an upper bound:

\dim{\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j\in\left\{0\right\}\cup\mathbb{N}\right\}\leq\lfloor(k+1)/2\rfloor.

For $k$ even, the dimension of the $\mathbb{Q}$ -vector space generated by the regularized Bessel moments has an upper bound:

\dim{\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}\left(k/2,2j\right)\mid j\in\left\{0\right\}\cup\mathbb{N}\right\}\leq\lfloor(k+1)/2\rfloor.

Note that our statement involves the regularized Bessel moments. This conclusion is a more general result than the one given by Zhou.

Quadratic relations of Bessel moments

In [8], the authors prove a general result of quadratic relations between periods given by a self-dual connection. We apply this result and obtain the quadratic relation between the Bessel moments. Under certain bases of cohomologies, let $B$ be the topological pairing matrix on Betti homology, $D$ be the Poincaré pairing matrix on de Rham cohomology, and $P,P_{\mathrm{c}}$ be the period pairing matrices between these two homology and cohomology. Then, the quadratic relations on these periods (Bessel moments) are given by

\displaystyle PD^{-1}P_{\mathrm{c}}=(-1)^{k}(2\pi\sqrt{-1})^{k+1}B.

(2)

The entries of the matrices $P$ and $P_{\mathrm{c}}$ consist of $\mathbb{Q}$ -linear combinations of Bessel moments and regularized Bessel moments, which are obtained in Section 5. Moreover, due to the rational structure of Betti homology and de Rham cohomology, the corresponding pairing matrices $D,B$ consist of rational numbers.

Note that the quadratic relations among Bessel moments does not depend on the choices of the bases. The effect of changing the bases of these homologies and cohomologies is just the conjugation of the matrices $P,D,P_{\mathrm{c}},B$ and thus on (2).

Determinants of Bessel moment matrix

Another interesting result is to compute the determinants of certain matrices consisting of Bessel moments. In [3, Conjectures 4 and 7], Broadhurst conjectures closed formulae of the determinants of the following two $r\times r$ matrices $\mathbf{M}_{r}$ and $\mathbf{N}_{r}$ involving the Bessel moments:

\mathbf{M}_{r}=\big{(}{\mathrm{IKM}}_{2r+1}(a,2b-1)\big{)}_{1\leq a,b\leq r},\quad\mathbf{N}_{r}=\big{(}{\mathrm{IKM}}_{2r+2}(a,2b-1)\big{)}_{1\leq a,b\leq r}.

Later, in [9], Zhou uses an analytic method to prove these two determinant formulae. Using a similar method as Zhou, we give explicit determinant formulae:

Formula 5 (Corollary 39).

For $r\geq 1$ , we have

	$\displaystyle\det\big{(}{\mathrm{IKM}}_{2r-1}(i-1,2j-2)\big{)}_{1\leq i,j\leq r}$	$\displaystyle=\sqrt{\pi}^{r(r+1)}\sqrt{2}^{r(r-3)}\prod_{a=1}^{r-1}\dfrac{a^{r-a}}{\sqrt{2a+1}^{2a+1}},$
	$\displaystyle\det\big{(}{\mathrm{IKM}}_{2r}(i-1,2j-2)\big{)}_{1\leq i,j\leq r}$	$\displaystyle=\dfrac{\sqrt{\pi}^{(r+1)^{2}}}{\Gamma\left(\frac{r+1}{2}\right)}\dfrac{1}{\sqrt{2}^{r(r+3)}}\prod_{a=1}^{r-1}\dfrac{(2a+1)^{r-a}}{(a+1)^{a+1}}.$

1.2 Approach

In [10], Bloch and Esnault study irregular connections on curves and provide the associated homology theory. Due to their results, we study the de Rham cohomology and Betti homology of $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ on $\mathbb{G}_{\mathrm{m}}$ and provide explicit bases in order to find the periods.

In section 2, we introduce the twisted $k$ -th symmetric power of the Kloosterman connection $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ , which is the main object in this paper. We discuss the rational structures on the de Rham cohomology and Betti homology of the connection. Moreover, since the connection is self-dual, we introduce its algebraic and topological self-pairings. These pairings will play an important role in our computations.

In section 3, we study the de Rham cohomology and the de Rham cohomology with compact support of $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ and write down certain elements in these two cohomologies. Next, we introduce the Poincaré pairing between them and compute the pairing with respect to the elements we have constructed. Using the dimension result of de Rham cohomology, along with the non-vanishing determinant of the Poincaré pairing, in Corollary 14, we conclude that the explicit elements in de Rham cohomology form bases.

We study parallelly the Betti homology of $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ in section 4. Since our ambient space is a non-compact space $\mathbb{C}^{\times}$ , we need to modify our Betti homology theory by allowing the chain to go to $0$ or $\infty$ . By controlling the growth behaviors of the horizontal sections, we study the moderate decay Betti homology and rapid decay Betti homology on $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ . Similarly, We first write down some elements in the moderate decay homology and rapid decay homology and compute their topological pairing explicitly. Moreover, by the duality of de Rham cohomology and Betti homology, the dimension of Betti homology is the same as the de Rham cohomology. Together with the topological pairing, we conclude that they are bases in Corollary 20.

Finally, in section 5, we compute the period pairing between the de Rham cohomologies and the Betti homologies and interpret them in terms of the Bessel moments. Note that our variety $\mathbb{G}_{\mathrm{m}}=\operatorname{Spec}\mathbb{Q}[z,z^{-1}]$ and the connection $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ are defined over $\mathbb{Q}$ and therefore, the de Rham cohomology and Betti homology are naturally endowed with a $\mathbb{Q}$ -vector space structure. From the dimension constraint of homologies, after computing the period pairing, we obtain the $\mathbb{Q}$ -linear relation of Bessel moments (Formula 3) and an upper bound of $\mathbb{Q}$ -dimension of space spanned by the Bessel moments (Theorem 4). In addition, the self duality of $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ gives quadratic relations between these Bessel moments (2).

In appendix A.1, we provide an accurate analysis of the symmetric powers of the modified Bessel differential operator. The first usage belongs to section 3, which enables us to determine the dimension of the de Rham cohomology $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ . The second usage belongs to appendix A.2, which allows us to analyze the leading term of the Vanhove operator. This helps us to obtain the determinant formula (Formula 5).

2 The Kloosterman connection and its twisted symmetric powers

In this section, we recall the definition and basic properties of Kloosterman connection and its symmetric powers from [1, 2]. Besides, we recall the twisted connection on $\mathbb{G}_{\mathrm{m}}$ obtained from the decomposition of the pushforward of trivial connection under the cyclic cover of $\mathbb{G}_{\mathrm{m}}$ . Combining these connections, we obtain the twisted symmetric powers of the Kloosterman connection. Moreover, since these connections are all self-dual, the duality induces the algebraic pairings on them and the topological pairings on the sheaves of horizontal sections.

2.1 Self-duality and pairing on $\mathrm{Kl}_{2}$

The connection ${\mathrm{Kl}}_{2}=(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}}v_{0}\oplus\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}}v_{1},\nabla)$ consists of a rank $2$ free sheaf on $\mathbb{G}_{\mathrm{m},z}=\operatorname{Spec}\mathbb{Q}[z,z^{-1}]$ with basis of sections $v_{0}$ and $v_{1}$ and the connection $\nabla$ on it given by

\displaystyle z\nabla\left(v_{0},v_{1}\right)=\left(v_{0},v_{1}\right)\begin{pmatrix}0&z\\ 1&0\end{pmatrix}{\mathrm{d}}z.

That is, $z\nabla v_{0}=v_{1}\mathrm{d}z$ and $\nabla v_{1}=v_{0}\mathrm{d}z$ . The connection $\mathrm{Kl}_{2}$ is self-dual in the sense that there exists an algebraic horizontal pairing $\langle\ ,\,\rangle_{\text{alg}}$ on it:

\left(\left\langle v_{i},v_{j}\right\rangle_{\mathrm{alg}}\right)_{0\leq i,j\leq 1}=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}

such that $\lambda:\mathrm{Kl}_{2}\rightarrow\mathrm{Kl}_{2}^{\vee}$ by $\left(v_{0},v_{1}\right)\mapsto\left(v_{1}^{\vee},-v_{0}^{\vee}\right)$ makes the following diagram commute.

Here ${\mathrm{Kl}}_{2}^{\vee}$ denotes the dual connection with the dual basis $\{v_{0}^{\vee},v_{1}^{\vee}\}$ .

Recall that the modified Bessel functions $I_{0}(t)$ and $K_{0}(t)$ satisfy the differential equation $((t\partial_{t})^{2}-t^{2})y=0$ and the Wronskian relation

I_{0}(t)K_{0}^{\prime}(t)-I_{0}^{\prime}(t)K_{0}(t)=\frac{-1}{t}.

(3)

Under the change of variable $z=\frac{t^{2}}{4}$ , the differential equation $((t\partial_{t})^{2}-t^{2})y=0$ becomes $4((z\partial_{z})^{2}-z)y=0$ . Define $A_{0},B_{0}$ be the fundamental solutions to the differential equation $((z\partial_{z})^{2}-z)y=0$ by rescaling the modified Bessel functions. In addition, define $A_{1},B_{1}$ by $z\partial_{z}$ differential of $A_{0},B_{0}$ :

	$\displaystyle A_{0}(z)=-2I_{0}(2\sqrt{z}),$	$\displaystyle\qquad A_{1}(z)=z\partial_{z}A_{0}(z);$
	$\displaystyle B_{0}(z)=2K_{0}(2\sqrt{z}),$	$\displaystyle\qquad B_{1}(z)=z\partial_{z}B_{0}(z).$

Here, the function $\sqrt{z}$ is taken to be the principal branch on the range $|\arg z\,|<\pi$ . For other $z$ , these functions are defined via the analytic continuation. Throughout this paper, the multivalued functions such as $z^{k/2}$ or $z^{-1/4}$ are all treated in this way without a mention. The functions $A_{0}(z)$ and $B_{0}(z)$ are annihilated by the operator $\left(z\partial_{z}\right)^{2}-z$ and real-valued on the ray $\mathbb{R}_{>0}$ . This gives

\partial_{z}A_{1}(z)=A_{0}(z),\qquad\partial_{z}B_{1}(z)=B_{0}(z).

Together with the Wronskian relation $A_{0}B_{1}-A_{1}B_{0}=2$ from (3), we obtain a basis of horizontal sections of $\nabla$ on ${\mathrm{Kl}}_{2}$

\displaystyle e_{0}=\frac{1}{2}(A_{0}v_{1}-A_{1}v_{0}),\qquad e_{1}=\frac{1}{2\pi\sqrt{-1}}(B_{0}v_{1}-B_{1}v_{0}).

(4)

Denote $\mathrm{Kl}_{2}^{\nabla}$ the local system of $\mathbb{Q}$ -vector space generated by $e_{0},e_{1}$ . There exists a topological pairing $\langle\ ,\,\rangle_{\mathrm{top}}=2\pi\sqrt{-1}\langle\ ,\,\rangle_{\mathrm{alg}}$ on $\mathrm{Kl}_{2}^{\nabla}$ :

\left(\left\langle e_{i},e_{j}\right\rangle_{\mathrm{top}}\right)_{0\leq i,j\leq 1}=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}.

2.2 Rational structures and pairings on $(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ .

Consider the double cover $\left[2\right]:\mathbb{G}_{\mathrm{m},t}\rightarrow\mathbb{G}_{\mathrm{m},z}$ induced by the ring homomorphism $\mathbb{Q}[z,z^{-1}]\rightarrow\mathbb{Q}[t,t^{-1}]$ , $z\mapsto t^{2}$ . Let $T=(\mathcal{O}_{\mathbb{G}_{\mathrm{m},t}},{\mathrm{d}})$ be the trivial connection on $\mathbb{G}_{\mathrm{m},t}$ . Via the ring homomorphism $\mathbb{Q}[z,z^{-1}]\rightarrow\mathbb{Q}[t,t^{-1}]$ , we view $\mathbb{Q}[t,t^{-1}]$ as the $\mathbb{Q}[z,z^{-1}]$ -module. Then, from the decomposition of $\mathbb{Q}[z,z^{-1}]$ -modules

\mathbb{Q}[t,t^{-1}]=\mathbb{Q}[z,z^{-1}]\oplus t\cdot\mathbb{Q}[z,z^{-1}],

the pushforward connection $\left[2\right]_{+}T$ decomposes into the direct sum

(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},{\mathrm{d}})\oplus\left(t\cdot\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},{\mathrm{d}}\right).

The second component $\left(t\cdot\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},{\mathrm{d}}\right)$ is isomorphic to $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ via the following diagram

(5)

The dual connection of $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ is given by $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}-\frac{1}{2}\frac{\mathrm{d}z}{z})$ , and the two are isomorphic via multiplication by $z$ , that is, the following diagram commutes.

This induces a perfect algebraic horizontal pairing $\langle\ ,\,\rangle_{\mathrm{alg}}$ on $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ given by

\langle 1,1\rangle_{\mathrm{alg}}=z.

On the other hand, the rational structure of the local system of horizontal sections of $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},t}},\mathrm{d})$ is generated by $1$ . Under the isomorphism (5), the rational structure of local system of horizontal sections of $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ is generated by $\frac{1}{t}=\frac{1}{\sqrt{z}}$ . Its dual connection $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}-\frac{1}{2}\frac{\mathrm{d}z}{z})$ has local system of horizontal sections generated by $\sqrt{z}$ . This induces a rational topological pairing $\langle\ ,\,\rangle_{\mathrm{top}}$ on $(\mathcal{O}_{\mathbb{G}_{\mathrm{m},z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})^{\nabla}$

\left\langle\frac{1}{\sqrt{z}},\frac{1}{\sqrt{z}}\right\rangle_{\mathrm{top}}=1.

2.3 Algebraic and topological pairings on $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$

The $k$ -th symmetric product of $\mathrm{Kl}_{2}$ , $\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ , is a rank $k+1$ free sheaf over $\mathcal{O}_{\mathbb{G}_{\mathrm{m}}}$ with basis of sections

v_{0}^{a}v_{1}^{k-a}=\frac{1}{|\mathfrak{S}_{k}|}\sum_{\sigma\in\mathfrak{S}_{k}}\sigma\left(v_{0}^{\otimes a}\otimes v_{1}^{\otimes k-a}\right)\qquad a=0,1,\ldots,k,

where $\mathfrak{S}_{k}$ is the symmetric group on $k$ elements. It is endowed with the induced connection from $\left(\mathrm{Kl}_{2},\nabla\right)$ . After twisting with the connection $\left(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{{\mathrm{d}z}}{z}\right)$ , we define

\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}=\left(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},{\mathrm{d}}+\frac{1}{2}\frac{{\mathrm{d}}z}{z}\right)\otimes\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}.

The induced connection $\nabla$ on $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ is given by

\displaystyle\nabla v_{0}^{a}v_{1}^{k-a}

\displaystyle=(k-a)v_{0}^{a+1}v_{1}^{k-a-1}{\mathrm{d}}z+\dfrac{a}{z}v_{0}^{a-1}v_{1}^{k-a+1}{\mathrm{d}}z+\dfrac{1}{2z}v_{0}^{a}v_{1}^{k-a}{\mathrm{d}}z.

(6)

Note that $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ is the same sheaf as $\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ but endowed with a different connection.

Via the self-duality on ${\mathrm{Kl}}_{2}$ and on $\left(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z}\right)$ , we have the perfect algebraic pairing $\left\langle\ ,\ \right\rangle_{\mathrm{alg}}$ on $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ :

given by

\left\langle v_{0}^{k-a}v_{1}^{a},v_{0}^{k-b}v_{1}^{b}\right\rangle_{\mathrm{alg}}=z\delta_{k,a+b}(-1)^{a}\dfrac{a!b!}{k!}=(2\pi\sqrt{-1})^{k}\left\langle e_{0}^{k-a}e_{1}^{a},e_{0}^{k-b}e_{1}^{b}\right\rangle_{\mathrm{alg}}.

Indeed,

	$\displaystyle\left\langle v_{0}^{k-a}v_{1}^{a},v_{0}^{k-b}v_{1}^{b}\right\rangle_{\mathrm{alg}}$	$\displaystyle=\left\langle\frac{1}{\|\mathfrak{S}_{k}\|}\sum_{\sigma\in\mathfrak{S}_{k}}\sigma\left(v_{0}^{\otimes k-a}\otimes v_{1}^{\otimes a}\right),\frac{1}{\|\mathfrak{S}_{k}\|}\sum_{\tau\in\mathfrak{S}_{k}}\tau\left(v_{0}^{\otimes k-b}\otimes v_{1}^{\otimes b}\right)\right\rangle_{\mathrm{alg}}$
		$\displaystyle=\frac{1}{(k!)^{2}}\sum_{\sigma,\tau\in\mathfrak{S}_{k}}\langle\sigma(v_{0}^{\otimes k-a}\otimes v_{1}^{\otimes a}),\tau(v_{0}^{\otimes k-b}\otimes v_{1}^{\otimes b})\rangle_{\mathrm{alg},\mathrm{Kl}_{2}}\cdot\langle 1,1\rangle_{\mathrm{alg},(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})}$
		$\displaystyle=\frac{1}{(k!)^{2}}(\delta_{k-a,b}k!a!b!(-1)^{a})\cdot z=z\delta_{k,a+b}(-1)^{a}\frac{a!b!}{k!}.$

By the definition of $e_{0},e_{1}$ in (4) and Wronskian relation $A_{0}B_{1}-A_{1}B_{0}=2$ , similar computation shows the formula for the algebraic pairing $\langle e_{0}^{k-a}e_{1}^{a},e_{0}^{k-b}e_{1}^{b}\rangle_{\mathrm{alg}}$ .

The local system $(\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2})^{\nabla}$ is a $\mathbb{Q}$ -vector space generated by the horizontal sections

\dfrac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}=\dfrac{1}{\sqrt{z}}\sum_{\sigma\in\mathfrak{S}_{k}}\sigma\big{(}e_{0}^{\otimes a}\otimes e_{1}^{\otimes k-a}\big{)},\quad a=0,1,\cdots,k,

(7)

which are the products of the horizontal sections of the connections $\left(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{{\mathrm{d}}z}{z}\right)$ and $\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ . The topological pairing $\langle\ ,\,\rangle$ on $\mathrm{Kl}_{2}^{\nabla}$ induces a topological pairing on $(\operatorname{Sym}^{k}\mathrm{Kl}_{2})^{\nabla}$ and thus on $(\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2})^{\nabla}$ :

where $\mathbb{Q}$ on the right hand side is the constant sheaf associated with the field $\mathbb{Q}$ on $\mathbb{G}_{\mathrm{m}}$ . This pairing reads

\left<\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a},\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}\right>_{\mathrm{top}}=\delta_{k,a+b}(-1)^{k-a}\dfrac{a!b!}{k!}

by the similar computation as above.

3 The de Rham cohomology

In this section, we study the de Rham cohomology of the twisted Kloosterman connection $H^{1}_{\mathrm{dR}}(\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2})$ and its dual, compact support de Rham cohomology $H^{1}_{\mathrm{dR,c}}(\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2})$ . We will write down certain elements in these cohomologies explicitly and compute the Poincaré pairing between these elements. Finally, we conclude the bases of these two cohomologies in the end of this section.

3.1 Dimension of $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$

Proposition 6.

For the connection $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ on $\mathbb{G}_{\mathrm{m}}$ , we have

\dim H^{1}_{\mathrm{dR}}\Big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\Big{)}=\left\lfloor\dfrac{k+1}{2}\right\rfloor.

Proof.

In [11, Lemma 2.9.13], we have the following formula.

Lemma 7.

On $\mathbb{G}_{\mathrm{m}}$ with parameter $z$ , let $\mathcal{D}=\mathcal{O}_{\mathbb{G}_{\mathrm{m}}}[\partial_{z}]$ be the ring of all differential operators on $\mathbb{G}_{\mathrm{m}}$ . Write $\theta_{z}=z\partial_{z}$ . For a non-zero element $L\in\mathcal{D}$ , write $L$ into a finite sum of the form $\sum_{i}z^{i}P_{i}(\theta_{z})$ , where $P_{i}(x)\in\mathbb{Q}[x]$ . Define integers $a,b$ by

a_{L}:=\max\left\{i\mid P_{i}\neq 0\right\};\ \ \ \ b_{L}:=\min\left\{i\mid P_{i}\neq 0\right\}.

Then the Euler characteristic of the $\mathcal{D}$ -module $\mathcal{D}/\mathcal{D}L$ is given by $\chi\left(\mathbb{G}_{\mathrm{m}},\mathcal{D}/\mathcal{D}L\right)=-\left(a_{L}-b_{L}\right)$ .

In this proof, we will follow the notations as in this lemma. Now, the differential operator on $\mathbb{G}_{\mathrm{m}}$ associated with the connection ${\mathrm{Kl}}_{2}$ is given by $\theta_{z}^{2}-z$ which annihilates $v_{0}$ and has fundamental solutions $A_{0}(z)$ and $B_{0}(z)$ . Then, the differential operator for $\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ is given by the $k$ -th symmetric power of $\theta_{z}^{2}-z$ , i.e., the differential operator annihilates $v_{0}^{k}$ and has fundamental solutions $A_{0}^{i}B_{0}^{k-i}$ for $i=0,\ldots,k$ . Denote this operator by $\widetilde{L}_{k+1}\in\mathcal{D}$ . For $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ , the corresponding differential operator reads $\frac{1}{\sqrt{z}}\widetilde{L}_{k+1}\sqrt{z}=:L$ since the solution is now given by $\frac{1}{\sqrt{z}}A_{0}^{i}B_{0}^{k-i}$ for $i=0,1,\ldots,k$ .

Recall in subsection 2.1, $L_{2}=(t\partial_{t})^{2}-t^{2}$ is the differential operator annihilates $I_{0}(t)$ and $K_{0}(t)$ . Write $L_{k+1}$ to be the $k$ -th symmetric power of $(t\partial_{t})^{2}-t^{2}$ . That is, $L_{k+1}$ annihilates $I_{0}^{a}(t)K_{0}^{k-a}(t)$ for $a=0,\ldots,k$ . As discussed in subsection 2.1, the change of variable $z=\frac{t^{2}}{4}$ sends $L_{k+1}$ to $\widetilde{L}_{k+1}$ . By Proposition 37, we have that $a_{L_{k+1}}=2\lfloor\frac{k+1}{2}\rfloor$ , $b_{L_{k+1}}=0$ . Therefore, by the degree $2$ change of variable $z=\frac{t^{2}}{4}$ , we conclude $a_{\widetilde{L}_{k+1}}=\lfloor\frac{k+1}{2}\rfloor$ , $b_{\widetilde{L}_{k+1}}=0$ .

Using the fact that $\frac{1}{\sqrt{z}}\theta_{z}\sqrt{z}=\theta_{z}+\frac{1}{2}$ , we have $\frac{1}{\sqrt{z}}\widetilde{L}_{k+1}\sqrt{z}=\sum z^{i}P_{i}\left(\theta_{z}+\frac{1}{2}\right)$ whenever $\widetilde{L}_{k+1}=\sum z^{i}P_{i}(\theta_{z})$ . This shows $a_{L}=a_{\widetilde{L}_{k+1}}$ and $b_{L}=b_{\widetilde{L}_{k+1}}$ . Therefore, by Lemma 7, we have

\chi\left(\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\right)=\chi\left(\mathbb{G}_{\mathrm{m}},\mathcal{D}/\mathcal{D}L\right)=-\left\lfloor\frac{k+1}{2}\right\rfloor.

Similar to the behavior of $I_{0}$ and $K_{0}$ [12, §10.30(i)], $A_{0}$ is holomorphic at $0$ and has exponential growth near infinity, and $B_{0}$ has a log pole at $0$ . These imply all of the solutions $\frac{1}{\sqrt{z}}A_{0}^{i}B_{0}^{k-i}$ are not algebraic solutions, and thus $H^{0}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}=0$ . Hence, combining the fact that $H^{2}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}=0$ by Artin vanishing theorem, we conclude that $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ has dimension $\left\lfloor\frac{k+1}{2}\right\rfloor$ . ∎

Remark 8.

In [11, Lemma 2.9.13], Katz provides the proof of lemma 7 only in the case $\mathbb{G}_{\mathrm{m},\mathbb{C}}$ which is over $\mathbb{C}$ . Yet, the same proof still works in our situation $\mathbb{G}_{\mathrm{m},\mathbb{Q}}$ which is over $\mathbb{Q}$ .

3.2 Compactly supported de Rham cohomology

Write $k^{\prime}=\lfloor\frac{k-1}{2}\rfloor$ . Consider the $k^{\prime}+1$ elements $\left\{v_{0}^{k}z^{i}\frac{{\mathrm{d}z}}{z}\right\}_{i=0}^{k^{\prime}}$ in $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ . We will prove these elements form a $\mathbb{Q}$ -basis. (See Corollary 14.) The Poincaré dual of the de Rham cohomology is the de Rham cohomology with compact support. An element in the de Rham cohomology with compact support $H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ is represented by a triple $(\xi,\eta,\omega)$ , where $\omega\in H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ and $\xi,\eta$ are formal solutions to $\nabla\xi=\nabla\eta=\omega$ at $0$ and $\infty$ respectively (See [8, Corollary 3.5]). The solutions are provided by the following lemma.

Lemma 9.

Suppose that $k\equiv 0,1,3\bmod{4}$ . For $0\leq i\leq k^{\prime}$ , there exists $(\xi_{i},\eta_{i})\in\big{(}\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}_{\widehat{0}}\oplus\big{(}\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}_{\widehat{\infty}}$ such that $\nabla\xi_{i}=\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z}$ .

On the other hand, let $k\equiv 2\bmod{4}$ , say $k=4r+2$ . For $0\leq i\leq k^{\prime}$ with $i\neq r$ , there exists $(\xi_{i},\eta_{i})\in\big{(}\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}_{\widehat{0}}\oplus\big{(}\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}_{\widehat{\infty}}$ such that

\nabla\xi_{i}=\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{{\mathrm{d}}z}{z},

where $\gamma_{k,n}\in\mathbb{Q}$ are the coefficients in the asymptotic expansion of $(-A_{0}\left(z\right)B_{0}\left(z\right))^{k/2}$ given by (11) below.

Proof.

Near $0$ , we want to find

\xi_{i}=\sum_{a=0}^{k}\xi_{i,a}(z)v_{0}^{a}v_{1}^{k-a}\in\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}

such that $\nabla\xi_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}$ . Using the connection formula (6) on $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$ , we need to solve:

	$\displaystyle\frac{{\mathrm{d}}}{\mathrm{d}z}\xi_{i,k}(z)+(k-1)\xi_{i,k-1}(z)+\frac{1}{2z}\xi_{i,k}(z)$	$\displaystyle=z^{i-1},$
	$\displaystyle\frac{{\mathrm{d}}}{\mathrm{d}z}\xi_{i,a}(z)+(k-a+1)\xi_{i,a-1}(z)+\frac{a+1}{z}\xi_{i,a+1}(z)+\frac{1}{2z}\xi_{i,a}(z)$	$\displaystyle=0{\text{ for }}a=1,2,\cdots,k-1,$
	$\displaystyle\frac{{\mathrm{d}}}{\mathrm{d}z}\xi_{i,0}(z)+\frac{1}{z}\xi_{i,1}(z)+\frac{1}{2z}\xi_{i,0}(z)$	$\displaystyle=0.$

Write $\xi_{i,a}=\sum_{n=0}^{\infty}\xi_{i,a,n}z^{n}$ . We solve $\xi_{i,a,n}$ recursively on $n$ . Suppose that we have solved $\xi_{i,a,j}$ for $j<n$ . Compare the coefficient of $z^{n-1}$ of the above system of equations and get

\left(\begin{array}[]{ccccc}&&&&n+\frac{1}{2}\\ &&&n+\frac{1}{2}&k\\ &&n+\frac{1}{2}&k-1&\\ &\iddots&\iddots&&\\ n+\frac{1}{2}&1&&&\end{array}\right)\left(\begin{array}[]{c}\xi_{i,0,n}\\ \xi_{i,1,n}\\ \xi_{i,2,n}\\ \vdots\\ \xi_{i,k,n}\end{array}\right)={\text{lower order combinations}}.

Since the first square matrix is invertible, $\xi_{i,a,n}$ is determined uniquely. Thus, we find $\xi_{i}\in\displaystyle\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}$ such that $\nabla\xi_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}$ . In $k\equiv 2\bmod{4}$ case, we only need to replace $\xi_{i}$ by $\xi_{i}-\gamma_{k,i-r}\xi_{r}$ .

Next, we turn to investigate the formal solutions at $\infty$ using horizontal frames. We have the modified Bessel functions have the asymptotic expansions at $\frac{1}{t}$ [13, §7.23]

$\displaystyle I_{0}(t)$	$\displaystyle\sim e^{t}\frac{1}{\sqrt{2\pi t}}\sum_{n=0}^{\infty}\frac{((2n-1)!!)^{2}}{2^{3n}n!}\frac{1}{t^{n}},$	$\displaystyle\|\arg t\,\|<\frac{1}{2}\pi$	(8)
$\displaystyle K_{0}(t)$	$\displaystyle\sim e^{-t}\sqrt{\frac{\pi}{2t}}\sum_{n=0}^{\infty}(-1)^{n}\frac{((2n-1)!!)^{2}}{2^{3n}n!}\frac{1}{t^{n}},$	$\displaystyle\|\arg t\,\|<\frac{3}{2}\pi$	(9)
$\displaystyle I_{0}(t)K_{0}(t)$	$\displaystyle\sim\frac{1}{2t}\sum_{n=0}^{\infty}\frac{((2n-1)!!)^{3}}{2^{3n}n!}\frac{1}{t^{2n}}.$		(10)

Here, the notation $n!!$ is the double factorial of a positive integer $n$ defined by

n!!=\prod_{k=0}^{\lceil\frac{n}{2}\rceil-1}(n-2k).

Let $w=\frac{1}{z}$ be the local coordinate at $z=\infty$ . For $k$ even, by the last asymptotic expansion, we have

\displaystyle(-A_{0}(z)B_{0}(z))^{k/2}\sim w^{1/4}\sum_{n=0}^{\infty}\gamma_{k,n}w^{n},

(11)

where $\gamma_{k,0}=1$ and $\gamma_{k,n}>0$ for all $n>0$ . For convenience, we set $\gamma_{k,j}=0$ for all $j<0$ .

Following the notation in (4), let us set $\overline{e}_{1}=\pi\sqrt{-1}e_{1}$ . Then $\frac{1}{\sqrt{z}}e_{0}^{a}\overline{e_{1}}^{k-a}$ are horizontal sections. Using the Wronskian relation $A_{0}B_{1}-A_{1}B_{0}=2$ , we have $v_{0}=B_{0}e_{0}-A_{0}\overline{e}_{1}$ . Then, we obtain

\displaystyle v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z}

\displaystyle=-\sum_{a=0}^{k}\binom{k}{a}\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}\mathrm{d}w.

To solve the formal solution $\eta_{i}$ of $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z}$ , We first solve $\eta_{i,a}$ for each $i,a$ such that

\displaystyle\mathrm{d}\eta_{i,a}=\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}.

Then, $\eta_{i}=-\sum_{a=0}^{k}\binom{k}{a}\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}$ is the desired solution. Moreover, since the function $\eta_{i,a}$ have $w^{k/4}$ and the exponential factor (see (12) below), we need to justify that $\eta_{i}$ lies in $\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket w\rrbracket v_{0}^{a}v_{1}^{k-a}$ (not just in $\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket w^{1/4},e^{-1/\sqrt{w}}\rrbracket v_{0}^{a}v_{1}^{k-a}$ ).

Near $\infty$ , we have the expansion

\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}=\begin{cases}\sqrt{\pi}^{k-2a}e^{-2(k-2a)/\sqrt{w}}w^{k/4-i-3/2}\cdot F_{i,a},&a\neq\frac{k}{2}\\ w^{k/4-i-3/2}\left(\sum_{n=0}^{\infty}\frac{\left(\left(2n-1\right)!!\right)^{3}}{2^{5n}n!}w^{n}\right)^{k/2},&a=\frac{k}{2}\end{cases}

where $F_{i,a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$ . When $a\neq\frac{k}{2}$ , we can find an antiderivative $\eta_{i,a}$ of $\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}$ with the expansion

\displaystyle\eta_{i,a}=\frac{\sqrt{\pi}^{k-2a}}{k-2a}e^{-2(k-2a)/\sqrt{w}}w^{k/4-i}\cdot G_{i,a}

(12)

for some $G_{i,a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$ . We analyze $\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}$ . Write $e_{0}^{k-a}\overline{e}_{1}^{a}$ back to the expression in basis $v_{0}^{b}v_{1}^{k-b}$ :

	$\displaystyle e_{0}^{k-a}\overline{e}_{1}^{a}$	$\displaystyle=2^{-k}(A_{0}v_{1}-A_{1}v_{0})^{k-a}\cdot(B_{0}v_{1}-B_{1}v_{0})^{a}$
		$\displaystyle=2^{-k}e^{2(k-a)\sqrt{z}}\frac{1}{\sqrt{\pi}^{k-a}}(F_{1}v_{0}-F_{2}v_{1})^{k-a}\cdot e^{-2a\sqrt{z}}\sqrt{\pi}^{a}(G_{1}v_{1}-G_{2}v_{0})^{a}$
		$\displaystyle=2^{-k}e^{2(k-2a)\sqrt{z}}\sqrt{\pi}^{2a-k}(F_{1}v_{0}-F_{2}v_{1})^{k-a}(G_{1}v_{1}-G_{2}v_{0})^{a},$

where $F_{1},F_{2},G_{1},G_{2}\in z^{1/4}\mathbb{Q}\llbracket z^{-1/4}\rrbracket$ . Thus,

\displaystyle\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}

\displaystyle=\frac{2^{-k}}{k-2a}w^{k/4-i+1/2}G_{i,a}(F_{1}v_{0}-F_{2}v_{1})^{k-a}(G_{1}v_{1}-G_{2}v_{0})^{a},

where $F_{1},F_{2},G_{1},G_{2}\in z^{1/4}\mathbb{Q}\llbracket z^{-1/4}\rrbracket$ . We conclude that the desired $\eta_{i}=-\sum_{a=0}^{k}\binom{k}{a}\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}$ has no exponential factor as a combination of monomials $v_{0}^{k-b}v_{1}^{b}$ , that is, $\eta_{i}$ lies in $\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket w^{1/4}\rrbracket v_{0}^{a}v_{1}^{k-a}$ .

Next, we will prove $\eta_{i}$ lies in $\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket w\rrbracket v_{0}^{a}v_{1}^{k-a}$ by showing $\eta_{i}$ is invariant under the Galois group action. Let $\sigma:w^{1/4}\mapsto\sqrt{-1}w^{1/4}$ be the generator of the Galois group of the extension $\mathbb{C}(w^{1/4})$ of $\mathbb{C}(w)$ . From the monodromy action [12, 10.34.5] of $I_{0},K_{0}$ , the $\sigma$ action on $A_{i},B_{i}$ is given by

\displaystyle\sigma\left(A_{j},B_{j}\right)

\displaystyle=\left(\frac{1}{\pi\sqrt{-1}}B_{j},-\pi\sqrt{-1}A_{j}\right){\text{ for }}j=0,1,

and thus on $e_{0},\overline{e}_{1}$ by

\displaystyle\sigma\left(e_{0},\overline{e}_{1}\right)=\left(\frac{1}{\pi\sqrt{-1}}\overline{e}_{1},-\pi\sqrt{-1}e_{0}\right);

\displaystyle\quad\sigma\left(e_{0}^{k-a}\overline{e}_{1}^{a}\right)=(\sqrt{-1})^{-k}\pi^{2a-k}e_{0}^{a}\overline{e}_{1}^{k-a}.

Moreover, we have

\sigma\left(\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}\right)=\eta_{i,k-a}\frac{1}{\sqrt{z}}e_{0}^{a}\overline{e}_{1}^{k-a}.

Hence, when $k\equiv 1,3\bmod{4}$ , the element $\eta_{i}$ is fixed by $\sigma$ and

\displaystyle\eta_{i}=-\sum_{a=0}^{k}\binom{k}{a}\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}\in\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket w\rrbracket v_{0}^{a}v_{1}^{k-a}.

This gives $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}.$

When $k=4r+4$ and $a=2r+2$ , the exponents of $w$ of the expansion of $\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}$ are in $\frac{1}{2}+\mathbb{Z}$ and one takes

\eta_{i,2r+2}\sim\frac{w^{r-i+1/2}}{r-i+1/2}G_{i}

where $G_{i}\in 1+w\mathbb{Q}\llbracket w\rrbracket$ . More precisely, we have

G_{i}=1+\sum_{n=1}^{\infty}\frac{r-i+1/2}{r-i+1/2+n}\gamma_{k,n}w^{n}.

Moreover, $\eta_{i,2r+2}\frac{1}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+2}$ has no exponential factor as a combination of monomials $v_{0}^{k-b}v_{1}^{b}$ and is invariant under $\sigma$ . Hence, when $k\equiv 0\bmod{4}$ , we take an element

\eta_{i}=-\sum_{a=0}^{k}\binom{k}{a}\eta_{i,a}\frac{1}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}\in\displaystyle\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}

This gives $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}$ .

Now, suppose that $k=4r+2$ , a positive integer congruent to $2$ modulo $4$ , and $a=2r+1$ . Using the expansion (11), we have the residue:

\operatorname{Res}_{w}\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}=\gamma_{k,i-r},

which vanishes if and only if $i\leq r-1$ . Therefore, for $i\geq r$ , there exists

\eta_{i,2r+1}\sim\frac{1}{r-i}w^{r-i}\cdot H_{i}

such that

\mathrm{d}\eta_{i,2r+1}=\left(w^{-i-3/2}-\gamma_{k,i-r}w^{-r-3/2}\right)(-A_{0}B_{0})^{2r+1}\mathrm{d}w

where $H_{i}\in 1+w\mathbb{Q}\llbracket w\rrbracket$ . Also, $\eta_{i,2r+1}\frac{1}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1}$ is invariant under $\sigma$ . Moreover, $\eta_{i,2r+1}\frac{1}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1}$ has no exponential factor as a combination of monomials $v_{0}^{k-b}v_{1}^{b}$ . Thus, we have

\displaystyle v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{\mathrm{d}z}{z}

\displaystyle=\nabla\left(-\sum_{\begin{subarray}{c}a=0\\ a\neq k/2\end{subarray}}^{k}\binom{k}{a}\frac{\eta_{i,a}-\gamma_{k,i-r}\eta_{r,a}}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}-\binom{k}{k/2}\frac{\eta_{i,2r+1}}{\sqrt{z}}e_{0}^{2r+1}\overline{e}_{1}^{2r+1}\right)

and hence we find an element $\eta_{i}$ in $\displaystyle\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}$ such that $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{{\mathrm{d}}z}{z}$ . ∎

Now, we define some elements in the de Rham cohomology and the de Rham cohomology with compact support. In next subsection, we will prove that these elements form bases of the corresponding cohomology spaces (see Corollary 14).

Definition 10.

In the de Rham cohomology $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ , the classes $\omega_{k,i}$ are given as follows.

1.

When $k\equiv 0,1,3\bmod{4}$ , define the $k^{\prime}+1$ elements:

$\omega_{k,i}=v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z}{\text{ for }}i=0,1,2,\cdots,k^{\prime}.$

When $k\equiv 2\bmod{4}$ , write $k=4r+2$ and define the $k^{\prime}$ elements:

\omega_{k,i}=\begin{cases}v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z},&0\leq i\leq r-1;\\ v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{{\mathrm{d}}z}{z},&r+1\leq i\leq 2r,\end{cases}

where $\gamma_{k,n}\in\mathbb{Q}$ are the coefficients in the asymptotic expansion of $(-A_{0}\left(z\right)B_{0}\left(z\right))^{k/2}$ given by (11) above.

From the Lemma 9, we define the elements in the compactly supported de Rham cohomology.

Definition 11.

We define certain elements in the compactly supported de Rham cohomology $H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ as follows.

1.

When $k\equiv 0,1,3\bmod{4}$ , define $k^{\prime}+1$ elements

$\widetilde{\omega}_{k,i}=(\xi_{i},\eta_{i},\omega_{k,i}){\text{ for }}0\leq i\leq k^{\prime},$

where $\nabla\xi_{i}=\nabla\eta_{i}=\omega_{k,i}$ .

When $k\equiv 2\bmod{4}$ , write $k=4r+2$ and define $k^{\prime}$ elements

\widetilde{\omega}_{k,i}=(\xi_{i},\eta_{i},\omega_{k,i}){\text{ for }}0\leq i\leq r-1{\text{ and }}r+1\leq i\leq k^{\prime},

where $\nabla\xi_{i}=\nabla\eta_{i}=\omega_{k,i}$ .

In the case that $k\equiv 2\bmod{4}$ , write $k=4r+2$ and further define

\widehat{m}_{2r+1}=\left(0,2^{k}\frac{1}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1},0\right)\in H^{1}_{\mathrm{dR,c}}\Big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\Big{)}.

Here, $\overline{e}_{1}:=\pi\sqrt{-1}e_{1}$ and $e_{0},e_{1}$ are horizontal sections of $\mathrm{Kl}_{2}$ defined in (4).

Remark 12.

The pair of the formal solutions $(\xi_{i},\eta_{i})$ is unique except in the case that there are solutions $(\xi,\eta)$ to $\nabla\xi=\nabla\eta=0$ . The latter happens only when $k\equiv 2\bmod{4}$ . In this circumstance, we fix the choice of $(\xi_{i},\eta_{i})$ to be the one constructed in the proof of Lemma 9. These expressions will be used in the computations of Poincaré pairing and period pairing in the rest of this paper.

Further, we define the middle part de Rham cohomology, $H^{1}_{\mathrm{mid}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ , to be the image of the projection $H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}\rightarrow H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ , $(\xi,\eta,\omega)\mapsto\omega$ . We therefore have

	$\displaystyle\omega_{k,i}\in H^{1}_{\mathrm{mid}}\Big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\Big{)}$	$\displaystyle{\text{ for }}0\leq i\leq k^{\prime}{\text{ when }}k\equiv 0,1,3\bmod{4};$
	$\displaystyle\omega_{k,i}\in H^{1}_{\mathrm{mid}}\Big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\Big{)}$	$\displaystyle{\text{ for }}0\leq i\leq k^{\prime},\,i\neq r{\text{ when }}k=4r+2.$

We may regard $H^{1}_{\mathrm{mid}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ as a quotient of $H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ containing the class of elements $\widetilde{\omega}_{k,i}$ .

3.3 Poincaré pairing

We have the following Poincaré pairing between the de Rham cohomology and the compactly supported de Rham cohomology. Recall the algebraic pairing $\left\langle\ ,\ \right\rangle_{\mathrm{alg}}$ is introduced in section 2.3.

Here, a one-form $\eta$ occurs in $\langle\widehat{m},\eta\rangle_{\mathrm{alg}}$ . This algebraic pairing means $\langle\widehat{m},f\rangle_{\mathrm{alg}}\mathrm{d}z$ whenever $\eta=f\mathrm{d}z$ . The notation $\mathbb{Q}(-k-1)$ is the $(k+1)$ -time tensor product of the Tate structures $\mathbb{Q}(-1)$ . As a vector space, $\mathbb{Q}(-k-1)$ is nothing but $\mathbb{Q}$ . Here, in consideration of Hodge filtrations, we use $\mathbb{Q}(-k-1)$ instead of $\mathbb{Q}$ to indicate the Hodge filtrations on both sides respect the Poincaré pairing. Note that the Poincaré pairing induces on the middle part de Rham cohomology which we still call it $\langle\ ,\ \rangle_{\mathrm{Poin}}$ :

Proposition 13.

Under the notation as in Definition 11, for $j\geq 0$ , we have the Poincaré pairing

\displaystyle\left\langle\widetilde{\omega}_{k,i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{Poin}}

\displaystyle=\begin{cases}0&{\text{if }}i+j\leq k^{\prime}-1,\ k:{\text{arbitrary,}}\\ (-2)^{k^{\prime}}\frac{k^{\prime}!}{k!!}&{\text{if }}i+j=k^{\prime},\ k\equiv 1,3\bmod{4},\\ \frac{\binom{k}{k/2}}{2^{k}(r-i+1/2)}&{\text{if }}i+j=k^{\prime},\ k\equiv 0\bmod{4},\ k=4r+4,\\ -\frac{\binom{k}{k/2}}{2^{k}(r-i)}&{\text{if }}i+j=k^{\prime},\ k\equiv 2\bmod{4},\ k=4r+2.\end{cases}

Moreover, if $k=4r+2$ , we have

\left\langle\widehat{m}_{2r+1},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{Poin}}=\begin{cases}0&{\text{if }}j<r,\\ \gamma_{k,j-r}&{\text{if }}j\geq r.\end{cases}

In particular, the Poincaré pairing matrix between $k^{\prime}+1$ elements in Definition 11 and $k^{\prime}+1$ elements $\left\{v_{0}^{k}z^{i}\frac{{\mathrm{d}z}}{z}\right\}_{i=0}^{k^{\prime}}$ in $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ is non-degenerate.

Proof.

In this proof, we will follow the notations as in the proof of Lemma 9. We first discuss the residue at $z=0$ . For any $k$ and $0\leq i,j\leq k^{\prime}$ , we compute

	$\displaystyle\operatorname{Res}_{z}\left\langle\xi_{i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$	$\displaystyle=\sum_{a=0}^{k}\operatorname{Res}_{z}\left\langle\xi_{i,a}v_{0}^{a}v_{1}^{k-a},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$
		$\displaystyle=\operatorname{Res}_{z}\left\langle\xi_{i,0}v_{1}^{k},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$
		$\displaystyle=\operatorname{Res}_{z}\left((-1)^{k}\xi_{i,0}z^{j}\right)=0$

where $\xi_{i,0}\in\mathbb{Q}\llbracket z\rrbracket$ .

Next, we discuss the residue at $z=\infty$ . When $k\equiv 1,3\bmod{4}$ and for any $0\leq i,j\leq k^{\prime}$ , we compute

	$\displaystyle\operatorname{Res}_{w}\left\langle\eta_{i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$	$\displaystyle=-\sum_{a=0}^{k}\binom{k}{a}\operatorname{Res}_{w}\left\langle\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$
		$\displaystyle=\sum_{a,b=0}^{k}\binom{k}{a}\binom{k}{b}\operatorname{Res}_{w}\left\langle\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}},\frac{(-A_{0})^{b}B_{0}^{k-b}}{w^{j+3/2}}\frac{e_{0}^{k-b}\overline{e}_{1}^{b}}{\sqrt{z}}\mathrm{d}w\right\rangle_{\mathrm{alg}}$
		$\displaystyle=\frac{1}{2^{k}}\sum_{a=0}^{k}(-1)^{a}\binom{k}{a}\operatorname{Res}_{w}\left(\eta_{i,a}\frac{1}{w^{j+3/2}}(-A_{0})^{k-a}B_{0}^{a}{\mathrm{d}}w\right)$
		$\displaystyle=\frac{1}{2^{k}}\sum_{a=0}^{k}\frac{(-1)^{a}\binom{k}{a}}{k-2a}\operatorname{Res}_{w}\left(w^{(k-1)/2-i-j-1}G_{i,a}F_{j,k-a}{\mathrm{d}}w\right)$
		$\displaystyle=\begin{cases}0&{\text{ if }}i+j\leq k^{\prime}-1,\\ (-2)^{k^{\prime}}\frac{k^{\prime}!}{k!!}&{\text{ if }}i+j=k^{\prime}.\end{cases}$

where $G_{i,a},F_{j,k-a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$ and the last equality follows from [2, lemma 3.18].

When $k\equiv 0\bmod{4}$ , write $k=4r+4$ . For any $0\leq i,j\leq k^{\prime}$ , we compute

	$\displaystyle\operatorname{Res}_{w}\left\langle\eta_{i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$	$\displaystyle=\frac{1}{2^{k}}\sum_{\begin{subarray}{c}a=0\\ a\neq\frac{k}{2}\end{subarray}}^{k}\frac{(-1)^{a}\binom{k}{a}}{k-2a}\operatorname{Res}_{w}\left(w^{(k-1)/2-i-j-1}G_{i,a}F_{j,k-a}{\mathrm{d}}w\right)$
		$\displaystyle\qquad+\frac{(-1)^{k/2}}{2^{k}}\binom{k}{k/2}\operatorname{Res}_{w}\left(\eta_{i,2r+2}\frac{1}{w^{j+3/2}}(-A_{0}B_{0})^{2r+2}{\mathrm{d}}w\right)$
		$\displaystyle=\frac{1}{2^{k}}\sum_{\begin{subarray}{c}a=0\\ a\neq\frac{k}{2}\end{subarray}}^{k}\frac{(-1)^{a}\binom{k}{a}}{k-2a}\operatorname{Res}_{w}\left(w^{(k-1)/2-i-j-1}G_{i,a}F_{j,k-a}{\mathrm{d}}w\right)$
		$\displaystyle\qquad+\frac{\binom{k}{k/2}}{2^{k}(r-i+1/2)}\operatorname{Res}_{w}\left(w^{k^{\prime}-i-j-1}\cdot G_{i}F_{2r+2}{\mathrm{d}}w\right)$
		$\displaystyle=\begin{cases}0&{\text{ if }}i+j\leq k^{\prime}-1,\\ \frac{\binom{k}{k/2}}{2^{k}(r-i+1/2)}&{\text{ if }}i+j=k^{\prime}.\end{cases}$

where $G_{i,a},F_{j,k-a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$ , $G_{i}\in 1+w\mathbb{Q}\llbracket w\rrbracket$ and $F_{2r+2}=\left(\sum_{n=0}^{\infty}\frac{\left(\left(2n-1\right)!!\right)^{3}}{2^{5n}n!}w^{n}\right)^{2r+2}$ .

When $k\equiv 2\bmod{4}$ , the computation is similar to the case $k\equiv 0\bmod{4}$ .

Finally, we compute

	$\displaystyle\operatorname{Res}_{w}\left\langle\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}$	$\displaystyle=-\sum_{b=0}^{k}\binom{k}{b}\operatorname{Res}_{w}\left\langle\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},\frac{(-A_{0})^{b}B_{0}^{k-b}}{w^{j+3/2}}\frac{e_{0}^{k-b}\overline{e}_{1}^{b}}{\sqrt{z}}{\mathrm{d}}w\right\rangle_{\mathrm{alg}}$
		$\displaystyle=-\binom{k}{k/2}\operatorname{Res}_{w}\left\langle\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},\frac{(-A_{0}B_{0})^{2r+1}}{w^{j+3/2}}\frac{(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}}{\mathrm{d}}w\right\rangle_{\mathrm{alg}}$
		$\displaystyle=(-1)^{2r+2}\operatorname{Res}_{w}\left(\frac{(-A_{0}B_{0})^{2r+1}}{w^{j+3/2}}{\mathrm{d}}w\right)$
		$\displaystyle=\operatorname{Res}_{w}\left(w^{k/4-j-3/2}\sum_{n=0}^{\infty}\gamma_{k,n}w^{n}{\mathrm{d}}w\right)$
		$\displaystyle=\begin{cases}0&{\text{if }}j<r,\\ \gamma_{k,j-r}&{\text{if }}j\geq r.\end{cases}$

Combining these residues, we obtain this proposition. ∎

Corollary 14 (Bases in de Rham side).

Let $k$ be a positive integer.

1.

$H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ has basis $\left\{v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\}_{j=0}^{k^{\prime}}$ .

$H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ has basis

\begin{cases}\left\{\widetilde{\omega}_{k,j}\right\}_{j=0}^{k^{\prime}}&{\text{ if }}k\equiv 0,1,3\bmod{4},\\ \left\{\widetilde{\omega}_{k,j}\right\}_{j=0}^{r-1}\cup\left\{\widetilde{\omega}_{k,j}\right\}_{j=r+1}^{k^{\prime}}\cup\left\{\widehat{m}_{2r+1}\right\}&{\text{ if }}k\equiv 2\bmod{4}{\text{ with }}k=4r+2.\end{cases}

$H_{{\mathrm{dR,mid}}}^{1}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ has basis

\begin{cases}\left\{\omega_{k,i}\right\}_{i=0}^{k^{\prime}}&{\text{ if }}k\equiv 0,1,3\bmod{4},\\ \left\{\omega_{k,i}\right\}_{i=0}^{r-1}\cup\left\{\omega_{k,i}\right\}_{i=r+1}^{k^{\prime}}&{\text{ if }}k\equiv 2\bmod{4}{\text{ with }}k=4r+2.\end{cases}

Proof.

Putting the dimension result in Proposition 6, the non-vanishing determinant of the Poincaré pairing matrix in Proposition 13 together, we obtain this corollary, thanks to the following simple observation in linear algebra. ∎

Fact 15.

Let $V$ and $W$ be two $n$ -dimensional vector spaces over a field $F$ . Suppose that $\langle\ ,\ \rangle:V\times W\rightarrow F$ is a bilinear pairing. If $\{v_{1},\ldots,v_{n}\}\subseteq V$ and $\{w_{1},\ldots,w_{n}\}\subseteq W$ are subsets of vectors such that the matrix

\big{(}\langle v_{i},w_{j}\rangle\big{)}_{i,j=1\ldots n}\in M_{n}(F)

is invertible, then $\{v_{1},\ldots,v_{n}\}$ is a basis of $V$ and $\{w_{1},\ldots,w_{n}\}$ is a basis of $W$ .

4 The local system and the associated homology

In this section, we study the rapid decay homology and moderate decay homology of the local system $(\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2})^{\nabla}$ . We write down the explicit cycles in these homologies and compute their Betti intersection pairing. In the end, we finish this section by concluding the bases of these two homologies.

In order to write down the cycles in the homology, we need to understand the monodromy action of the horizontal sections of $\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2}$ . Recall $\{e_{0},e_{1}\}$ is the basis of the local system ${\mathrm{Kl}}_{2}^{\nabla}$ defined in (4). From [12, 10.25(ii)], the modified Bessel function $I_{0}(t)$ is entire. On the other hand, $K_{0}(t)$ extends analytically to a multivalued function on $\mathbb{C}^{\times}$ satisfying the monodromy $K_{0}(e^{\pi\sqrt{-1}}t)=K_{0}(t)-\pi\sqrt{-1}I_{0}(t)$ from [12, 10.34]. This implies $e_{0},e_{1}$ undergo the monodromy action $T:\left(e_{0},e_{1}\right)\mapsto\left(e_{0},e_{1}+e_{0}\right)$ near $0$ . Then the basis in (7) of the local system $(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}$ satisfies $T:\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}\mapsto\frac{-1}{\sqrt{z}}e_{0}^{a}(e_{1}+e_{0})^{k-a}$ near $0$ .

4.1 Rapid decay cycles

Write $k^{\prime}=\left\lfloor\frac{k-1}{2}\right\rfloor$ . Denote the chains on $\mathbb{C}^{\times}$ :

	$\displaystyle\sigma_{0}$	$\displaystyle={\text{the unit circle}},{\text{ starting at }}1{\text{ and oriented counterclockwise}};$
	$\displaystyle\sigma_{+}$	$\displaystyle=\text{the interval $\left[1,\infty\right)$},{\text{ starting at }}1{\text{ toward }}+\infty.$

By the asymptotic expansion (8), (9), the horizontal sections $\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}$ decay exponentially along $\sigma_{+}$ for $a=0,1,\ldots,k^{\prime}$ . We have the following lemma describing some elements in the rapid decay homology.

Lemma 16.

For $0\leq b\leq k^{\prime}$ , the elements

\displaystyle\delta_{b}

\displaystyle=\sigma_{+}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\dfrac{1}{2}\sigma_{0}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}+\sum_{n=1}^{k-b}d_{k-b}(n)\sigma_{0}^{2n}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b},

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are rapid decay cycles in $H^{\mathrm{rd}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ , where $d_{n}(i)$ are real numbers satisfying

\displaystyle\sum_{i=1}^{n}d_{n}(i)\left(2i\right)^{m}

\displaystyle=-\dfrac{1}{2},{\text{ for }}m=1,2,\cdots,n.

In fact, by Cramer’s rule, one can write $d_{n}(i)=\frac{(-1)^{i}}{n!2^{n+1}}\binom{n}{i}\frac{\left(2n-1\right)!!}{2i-1}$ uniquely.

Proof.

We need to prove that $d_{n}(i)$ makes $\delta_{b}$ into a cycle, that is, $\partial\delta_{b}=0$ . The boundaries of chains $\sigma_{+}$ and $\sigma_{0}$ in $\delta_{b}$ support at the point $1\in\mathbb{C}^{\times}$ . It suffices to check that the coefficient of $1\in\mathbb{C}^{\times}$ in $\partial\delta_{b}$ is $0$ . Indeed, considering the monodromy action $T$ describe above, a direct computation shows the coefficient of $1\in\mathbb{C}^{\times}$ in $\partial\delta_{b}$ is

		$\displaystyle\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\dfrac{1}{2}\left(\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}+\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+e_{0})^{k-b}\right)$
		$\displaystyle\hskip 40.0pt+\sum_{n=1}^{k-b}d_{k-b}(n)\left(\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+2ne_{0})^{k-b}\right)$
	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\dfrac{1}{2}\left(\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}+\sum_{j=0}^{k-b}\binom{k-b}{j}\frac{1}{\sqrt{z}}e_{0}^{b+j}e_{1}^{k-b-j}\right)$
		$\displaystyle\hskip 40.0pt+\sum_{n=1}^{k-b}d_{k-b}(n)\left(\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\sum_{m=0}^{k-b}\binom{k-b}{m}(2n)^{m}\frac{1}{\sqrt{z}}e_{0}^{b+m}e_{1}^{k-b-m}\right)$
	$\displaystyle=$	$\displaystyle-\dfrac{1}{2}\sum_{j=1}^{k-b}\binom{k-b}{j}\frac{1}{\sqrt{z}}e_{0}^{b+j}e_{1}^{k-b-j}-\sum_{m=1}^{k-b}\binom{k-b}{m}\sum_{n=1}^{k-b}d_{k-b}(n)(2n)^{m}\frac{1}{\sqrt{z}}e_{0}^{b+m}e_{1}^{k-b-m}$
	$\displaystyle=$	$\displaystyle\sum_{j=1}^{k-b}\binom{k-b}{j}\left(-\dfrac{1}{2}-\sum_{n=1}^{k-b}d_{k-b}(n)(2n)^{j}\right)\frac{1}{\sqrt{z}}e_{0}^{b+j}e_{1}^{k-b-j}=0,$

where the last equality is the assumption on real numbers $d_{n}(i)$ . ∎

From this lemma, we have $k^{\prime}+1$ elements $\{\delta_{b}\}_{b=0}^{k^{\prime}}$ in the rapid decay homology $H^{\mathrm{rd}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ . At the end of this section, we will prove these elements form a basis (see Corollary 20).

4.2 Moderate decay cycles

Define one more chain

\displaystyle\mathbb{R}_{+}

\displaystyle=\text{the half line $\left[0,\infty\right)$, starting at $0$ toward $+\infty$}.

By [12, §10.30(i)], the modified Bessel function $K_{0}\left(t\right)$ has log pole at $0$ , so the horizontal sections $\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}$ decay moderately along $\mathbb{R}_{+}$ near $0$ for $a=0,1,\ldots,\lfloor\frac{k}{2}\rfloor$ . Moreover, by the expression (10), $(I_{0}K_{0})^{a}$ decay polynomially along $\mathbb{R}_{+}$ near $\infty$ . Then, we define the moderate decay cycles in $H^{\mathrm{mod}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$

\displaystyle\gamma_{a}=\mathbb{R}_{+}\otimes\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a},{\text{ for }}a=0,1,2,\cdots,\left\lfloor\dfrac{k}{2}\right\rfloor.

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They are indeed a cycle. The proof is the same as the above lemma by taking the homotopy as the radius of $\sigma_{0}$ tends to $0$ and $\sigma_{+}$ tends to $\mathbb{R}_{+}$ . Since a rapid decay cycle is a moderate decay cycle as well, we have the natural map

This natural map sends $\delta_{b}$ to $\gamma_{b}$ for $b=0,1,\cdots,k^{\prime}$ by the homotopy argument. The following lemma shows when $k\equiv 2\bmod{4}$ , $\displaystyle\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\delta_{2j}$ belongs to the kernel of this map.

Lemma 17.

In $H^{\mathrm{mod}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ , one has

	$\displaystyle\sum_{j=0}^{k/4}\binom{k/2}{2j}\gamma_{2j}=0{\text{ if }}k\equiv 0\bmod{4};$
	$\displaystyle\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\gamma_{2j}=0{\text{ if }}k\equiv 2\bmod{4}.$

Proof.

Let $\rho:\left\{\left(x,y\right)\in\mathbb{R}^{2}\mid 0<x,y,x+y<1\right\}\rightarrow\mathbb{C}$ be the open simplicial $2$ -chain

\rho\left(x,y\right)=\tan\dfrac{\pi\left(x+y\right)}{2}\exp\left(4\sqrt{-1}\tan^{-1}\dfrac{y}{x}\right)

that covers $\mathbb{C}$ once. If $k$ is even, by the asymptotic expansion (10), the singular chain

\Delta=\rho\otimes\left(\frac{1}{\sqrt{z}}(e_{1}-e_{0})^{k/2}e_{1}^{k/2}\right)

has moderate growth. The boundary of $\rho$ consists of two positive real lines $\mathbb{R}_{+}$ . From the monodromy action $T:(e_{0},e_{1})\mapsto(e_{0},e_{1}+e_{0})$ , one computes $\partial\Delta$ :

	$\displaystyle\partial\Delta$	$\displaystyle=\mathbb{R}_{+}\otimes\left(\frac{1}{\sqrt{z}}(e_{1}-e_{0})^{k/2}e_{1}^{k/2}\right)+\mathbb{R}_{+}\otimes\left(\frac{1}{\sqrt{z}}e_{1}^{k/2}(e_{0}+e_{1})^{k/2}\right)$
		$\displaystyle=\sum_{i=0}^{k/2}(-1)^{i}\binom{k/2}{i}\mathbb{R}_{+}\otimes\frac{1}{\sqrt{z}}e_{0}^{i}e_{1}^{k-i}+\sum_{i=0}^{k/2}\binom{k/2}{i}\mathbb{R}_{+}\otimes\frac{1}{\sqrt{z}}e_{0}^{i}e_{1}^{k-i}$
		$\displaystyle=\sum_{i=0}^{k/2}(1+(-1)^{i})\binom{k/2}{i}\gamma_{i}.$

When $k\equiv 0\bmod{4}$ , this reads

\frac{1}{2}\partial\Delta=\sum_{j=0}^{k/4}\binom{k/2}{2j}\gamma_{2j}.

Thus, $\sum_{j=0}^{k/4}\binom{k/2}{2j}\gamma_{2j}$ is homologous to zero in $H_{1}^{\mathrm{mod}}(\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}\mathrm{Kl}_{2})^{\nabla})$ . The case when $k\equiv 2\bmod{4}$ is similar. ∎

Here, we have written down the $1+\lfloor\frac{k}{2}\rfloor$ elements $\{\gamma_{a}\}_{a=0}^{\lfloor k/2\rfloor}$ in the moderate decay homology $H^{\mathrm{mod}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ . At the end of this section, we will prove that these elements form a basis modulo the linear relation given in the above lemma (see Corollary 20).

Similar to the middle part de Rham cohomology in the previous section, we define the middle part Betti homology $H_{1}^{{\mathrm{mid}}}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ to be the image of $H_{1}^{\mathrm{rd}}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ in $H_{1}^{\mathrm{mod}}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ . More precisely, we have

	$\displaystyle\gamma_{i}\in H_{1}^{{\mathrm{mid}}}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$	$\displaystyle{\text{ for }}0\leq i\leq k^{\prime}{\text{ when }}k\equiv 0,1,3\bmod{4};$
	$\displaystyle\gamma_{i}\in H_{1}^{{\mathrm{mid}}}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$	$\displaystyle{\text{ for }}1\leq i\leq k^{\prime}{\text{ when }}k\equiv 2\bmod{4}.$

Also, we may regard $H_{1}^{{\mathrm{mid}}}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ as the quotient of $H^{\mathrm{rd}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ containing the class of elements $\delta_{b}$ . At the end of this section, we will prove these elements form a basis (see Corollary 20).

4.3 Betti intersection pairing

We use the topological pairing $\left\langle\ ,\ \right\rangle_{\mathrm{top}}$ introduced in section 2.3 to define the Betti intersection pairing

Here, we need to find representatives of $\delta=\sum\sigma_{i}\otimes s_{\sigma_{i}}$ and $\gamma=\sum\tau_{j}\otimes s_{\tau_{j}}$ in their homology classes respectively such that any two chains $\sigma_{i}$ and $\tau_{j}$ intersect transversally for all $i,j$ . Then, for each pair $(i,j)$ , $\sigma_{i}\cap\tau_{j}$ consists of only finitely many topological intersection points. The sum over $\sigma_{i}\cap\tau_{j}$ is then the sum of the topological pairings of the corresponding sections at each intersection point. Note that the Betti intersection pairing induces on the middle part Betti homology which we still call it $\langle\ ,\ \rangle_{\mathrm{Betti}}$ :

To compute the topological pairing with respect to the elements we had written down, we need to introduce the Euler numbers and Euler polynomials. The Euler polynomials $E_{n}(x)$ are given by the following power series, and we define the numbers $E_{n}$ for $n\geq 0$ as in [14],

\sum_{n=0}^{\infty}E_{n}(x)\frac{z^{n}}{n!}=\frac{2}{e^{z}+1}e^{xz},\ \ E_{n}=E_{n}(0).

The first few $E_{n}$ are

$E_{0}$	$E_{1}$	$E_{2}$	$E_{3}$	$E_{4}$	$E_{5}$	$E_{6}$
$1$	$-1/2$	$0$	$1/4$	$0$	$-1/2$	$0$

We have the inversion formula for Euler polynomials,

x^{n}=E_{n}(x)+\dfrac{1}{2}\sum_{k=0}^{n-1}\binom{n}{k}E_{k}(x).

Evaluating at $x=0$ , we get

\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}E_{k}=-2E_{n}.

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Proposition 18.

We have the Betti intersection pairing

\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}=(-1)^{a}\frac{\binom{k-b}{a}}{\binom{k}{a}}\frac{-1}{2}E_{k-a-b}=\frac{(-1)^{a+1}}{2}\frac{(k-a)!(k-b)!}{k!}\frac{E_{k-a-b}}{(k-a-b)!}

for $b=0,\cdots,k^{\prime}$ and $a=0,\cdots,\left\lfloor\frac{k}{2}\right\rfloor$ .

Proof.

Fix some $-\pi<\theta_{0}<0$ and let $x_{0}=\exp(\sqrt{-1}\theta_{0})$ . To compute the pairing $\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}$ , we move the ray $\sigma_{+}$ by adding the scalar $(x_{0}-1)$ and let the circle $\sigma_{0}$ start at $x_{0}$ . Then the component $\sigma_{0}^{j}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}$ in the deformed $\delta_{b}$ meets $\gamma_{a}$ topologically $j$ times at the same point $+1\in\mathbb{C}^{\times}$ . At the $i$ -th intersection, the factor $\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}$ becomes $(-1)^{i-1}\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+(i-1)e_{0})^{k-b}$ and we have

\left\langle(-1)^{i-1}\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+(i-1)e_{0})^{k-b},\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}=(-1)^{i-1}(i-1)^{k-a-b}(-1)^{a}\dfrac{\binom{k-b}{a}}{\binom{k}{a}}.

By adding these contributions, we obtain

\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}=\dfrac{\binom{k-b}{a}}{\binom{k}{a}}(-1)^{a}\sum_{n=1}^{k-b}d_{k-b}(n)T_{k-a-b}(2n),

where

T_{n}(k)=\sum_{\ell=0}^{k-1}(-1)^{\ell}\ell^{n}=-1+2^{n}-\cdots+(-1)^{k-1}(k-1)^{n}.

Kim [14] gave the following relation for $T_{n}(k)$ :

T_{n}(k)=\dfrac{(-1)^{k+1}}{2}\sum_{\ell=0}^{n-1}\binom{n}{\ell}E_{\ell}k^{n-\ell}+\dfrac{E_{n}}{2}\left(1+(-1)^{k+1}\right).

Now, we have the following computation

	$\displaystyle\sum_{n=1}^{k-b}d_{k-b}(n)T_{k-a-b}(2n)$	$\displaystyle=\sum_{n=1}^{k-b}d_{k-b}(n)\left[\dfrac{-1}{2}\sum_{\ell=0}^{k-a-b-1}\binom{k-a-b}{\ell}E_{\ell}(2n)^{k-a-b-\ell}\right]$
		$\displaystyle=\dfrac{-1}{2}\sum_{\ell=0}^{k-a-b-1}\binom{k-a-b}{\ell}E_{\ell}\sum_{n=1}^{k-b}d_{k-b}(n)(2n)^{k-a-b-\ell}$
		$\displaystyle=\dfrac{1}{4}\displaystyle\sum_{\ell=0}^{k-a-b-1}\binom{k-a-b}{\ell}E_{\ell}$
		$\displaystyle=\dfrac{-1}{2}E_{k-a-b}$

where the last equality follows from (15). ∎

Consider the $(k^{\prime}+1)\times(k^{\prime}+1)$ pairing matrix

B_{k}=\begin{cases}\left(\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}\right)_{\ 0\leq b\leq k^{\prime},\ 0\leq a\leq\lfloor\frac{k}{2}\rfloor}&{\text{ if }}k{\text{ is odd,}}\\ \left(\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}\right)_{0\leq b\leq k^{\prime},\ 1\leq a\leq\frac{k}{2}}&{\text{ if }}k{\text{ is even.}}\end{cases}

By Proposition 18, when $k$ is even, we have

B_{k}=\left(\begin{array}[]{ccc}\frac{(-1)^{2}}{2}\frac{(k-1)!k!}{k!}\frac{E_{k-1}}{(k-1)!}&\cdots&\frac{(-1)^{k/2+1}}{2}\frac{(k/2)!k!}{k!}\frac{E_{k/2}}{(k/2)!}\\ \vdots&\ddots&\vdots\\ \frac{(-1)^{2}}{2}\frac{(k-1)!(k/2+1)!}{k!}\frac{E_{k/2}}{(k/2)!}&\cdots&\frac{(-1)^{k/2+1}}{2}\frac{(k/2)!(k/2+1)!}{k!}\frac{E_{1}}{\left(1\right)!}\end{array}\right)

and that

B_{k-1}=\left(\begin{array}[]{ccc}\frac{(-1)}{2}\frac{(k-1)!(k-1)!}{(k-1)!}\frac{E_{k-1}}{(k-1)!}&\cdots&\frac{(-1)^{k/2}}{2}\frac{(k/2)!(k-1)!}{(k-1)!}\frac{E_{k/2}}{(k/2)!}\\ \vdots&\ddots&\vdots\\ \frac{(-1)}{2}\frac{(k-1)!(k/2)!}{(k-1)!}\frac{E_{k/2}}{(k/2)!}&\cdots&\frac{(-1)^{k/2}}{2}\frac{(k/2)!(k/2)!}{(k-1)!}\frac{E_{1}}{\left(1\right)!}\end{array}\right).

Then we obtain the relation

B_{k}=-\frac{1}{k}\operatorname{diag}(k,k-1,\cdots,k/2+1)\cdot B_{k-1}.

(16)

Thus, $B_{k}$ and $B_{k-1}$ have the same rank whenever $k$ is even. Moreover, we may compute the determinant of $B_{k}$ explicitly as given in the following proposition.

Proposition 19.

The determinant of $B_{k}$ is given by the following.

1.

When $k$ is odd, we have

$\displaystyle\det B_{k}$ $\displaystyle=2^{-k-1}\prod_{a=1}^{k^{\prime}}a^{k^{\prime}+1-2a}(2a+1)^{k^{\prime}-2a}.$

When $k$ is even, we have

\displaystyle\det B_{k}

\displaystyle=(-1)^{(k^{\prime}+1)(k^{\prime}+3)}2^{-k}\prod_{a=1}^{k^{\prime}}(a+1)^{k^{\prime}-2a-1}(2a+1)^{k^{\prime}+1-2a}.

In particular, they are all non-vanishing.

Proof.

Set $\mathcal{E}_{2n-1}=(-1)^{n}2^{2n-1}E_{2n-1}$ . Apply the result [15, Eq. H12] in the following computations.

When $k=2k^{\prime}+1$ is odd, we have

	$\displaystyle\det B_{k}$	$\displaystyle=\frac{(-1)^{(k^{\prime}+1)(k^{\prime}+2)/2}}{(2\cdot k!)^{k^{\prime}+1}}\left[\prod_{i=k^{\prime}+1}^{k}i!\right]^{2}\det\left(\begin{array}[]{ccc}\frac{E_{k}}{k!}&\cdots&\frac{E_{k^{\prime}+1}}{(k^{\prime}+1)!}\\ \vdots&\ddots&\vdots\\ \frac{E_{k^{\prime}+1}}{(k^{\prime}+1)!}&\cdots&\frac{E_{1}}{\left(1\right)!}\end{array}\right)$
		$\displaystyle=\frac{1}{(2\cdot k!)^{k^{\prime}+1}}\left[\prod_{i=k^{\prime}+1}^{k}i!\right]^{2}\frac{1}{2^{(k^{\prime}+1)^{2}}}\det\left(\begin{array}[]{ccc}\frac{\mathcal{E}_{k}}{k!}&\cdots&\frac{\mathcal{E}_{k^{\prime}+1}}{(k^{\prime}+1)!}\\ \vdots&\ddots&\vdots\\ \frac{\mathcal{E}_{k^{\prime}+1}}{(k^{\prime}+1)!}&\cdots&\frac{\mathcal{E}_{1}}{\left(1\right)!}\end{array}\right)$
		$\displaystyle=\frac{1}{2^{(k^{\prime}+1)(k^{\prime}+2)}(k!)^{k^{\prime}+1}}\left[\prod_{i=k^{\prime}+1}^{k}i!\right]^{2}2^{k^{\prime 2}}\frac{k^{\prime}!}{k!}\prod_{j=1}^{k^{\prime}}\frac{(j-1)!^{2}}{(2j-1)!^{2}}$
		$\displaystyle=\frac{1}{2^{k+1}}\prod_{a=1}^{k^{\prime}}a^{k^{\prime}+1-2a}(2a+1)^{k^{\prime}-2a}.$

When $k=2k^{\prime}+2$ is even, we have

	$\displaystyle\det B_{k}$	$\displaystyle=\frac{(-1)^{(k^{\prime}+1)(k^{\prime}+4)/2}}{(2\cdot k!)^{k^{\prime}+1}}\left[\prod_{i=k^{\prime}+1}^{k-1}i!\left(i+1\right)!\right]\det\left(\begin{array}[]{ccc}\frac{E_{k-1}}{(k-1)!}&\cdots&\frac{E_{k^{\prime}+1}}{(k^{\prime}+1)!}\\ \vdots&\ddots&\vdots\\ \frac{E_{k^{\prime}+1}}{(k^{\prime}+1)!}&\cdots&\frac{E_{1}}{\left(1\right)!}\end{array}\right)$
		$\displaystyle=\frac{(-1)^{(k^{\prime}+1)(k^{\prime}+4)/2}}{(2\cdot k!)^{k^{\prime}+1}}\left[\prod_{i=k^{\prime}+1}^{k-1}i!\left(i+1\right)!\right]$
		$\displaystyle\hskip 100.0pt\cdot\frac{(\sqrt{-1})^{(k^{\prime}+1)(k^{\prime}+2)}}{2^{(k^{\prime}+1)^{2}}}\det\left(\begin{array}[]{ccc}\frac{\mathcal{E}_{k-1}}{(k-1)!}&\cdots&\frac{\mathcal{E}_{k^{\prime}+1}}{(k^{\prime}+1)!}\\ \vdots&\ddots&\vdots\\ \frac{\mathcal{E}_{k^{\prime}+1}}{(k^{\prime}+1)!}&\cdots&\frac{\mathcal{E}_{1}}{\left(1\right)!}\end{array}\right)$
		$\displaystyle=\frac{(-1)^{(k^{\prime}+1)(k^{\prime}+3)}}{2^{(k^{\prime}+1)(k^{\prime}+2)}(k!)^{k^{\prime}+1}}\left[\prod_{i=k^{\prime}+1}^{k-1}i!\left(i+1\right)!\right]2^{k^{\prime 2}}\frac{k^{\prime}!}{(k-1)!}\prod_{j=1}^{k^{\prime}}\frac{(j-1)!^{2}}{(2j-1)!^{2}}$
		$\displaystyle=\frac{(-1)^{(k^{\prime}+1)(k^{\prime}+3)}}{2^{k}}\prod_{a=1}^{k^{\prime}}(a+1)^{k^{\prime}-2a-1}(2a+1)^{k^{\prime}+1-2a}.\qed$

Finally, before we conclude the basis of Betti homologies, we need to introduce the period pairings here. However, the details of the pairings will be given in the next section. By [2, Corollary 2.11], there exist two perfect pairings

Here, the notation $V_{\mathbb{C}}$ means $V\otimes_{\mathbb{Q}}\mathbb{C}$ . For the next corollary, we just need to use the fact that these pairings are perfect. In the next section, we will compute these two pairings explicitly.

Corollary 20.

The natural map

sending $\delta_{b}$ to $\gamma_{b}$ is an isomorphism when $k\equiv 0,1,3\bmod 4$ and has a one-dimensional kernel when $k\equiv 2\bmod 4$ . Moreover, we find the following.

1.

$H^{\mathrm{rd}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ has basis $\{\delta_{b}\}_{b=0}^{k^{\prime}}$ .

$H^{\mathrm{mod}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ has basis

\begin{cases}\left\{\gamma_{a}\right\}_{a=0}^{k^{\prime}}&{\text{ if }}k{\text{ is odd;}}\\ \left\{\gamma_{a}\right\}_{a=1}^{k/2}&{\text{ if }}k{\text{ is even.}}\end{cases}

$H^{\mathrm{mid}}_{1}\big{(}\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla}\big{)}$ has basis

\begin{cases}\left\{\gamma_{a}\right\}_{a=0}^{k^{\prime}}&{\text{ if }}k\equiv 0,1,3\bmod{4};\\ \left\{\gamma_{a}\right\}_{a=1}^{k^{\prime}}&{\text{ if }}k\equiv 2\bmod{4}.\\ \end{cases}

Proof.

From the perfect period pairings, the dimension of rapid decay homology and moderate decay homology are both $k^{\prime}+1$ by Proposition 6. Then, by the Fact 15 and the non-vanishing determinant of $B_{k}$ in Proposition 19, we conclude $1.$ and $2.$ This also shows the natural map which sends $\delta_{b}$ to $\gamma_{b}$ for $b=0,\ldots,k^{\prime}$ is an isomorphism when $k\equiv 1,3\bmod 4$ . When $k\equiv 2\bmod{4}$ , Lemma 17 describes the one-dimensional kernel of the natural map. Moreover, $B_{k}$ has full rank $k/2$ when $k$ is even by the relation (16). Hence, we conclude that the natural map is an isomorphism when $k\equiv 0\bmod{4}$ . ∎

5 Twisted moments as periods

In this section, we compute the period pairing of the basis of de Rham cohomology and Betti homology in Corollary 14 and Corollary 20. Also, we interpret these periods as the Bessel moments and regularized Bessel moments.

5.1 Bessel moments and regularized Bessel moments

The Bessel moments are defined by

\displaystyle{\mathrm{IKM}}_{k}(a,b)=\int_{0}^{\infty}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}\mathrm{d}t.

provided the convergence of the integral, that is, for non-negative integers $k,a,b$ satisfying $a\leq k^{\prime}$ , $b\geq 0$ or $a=\frac{k}{2}$ , $0\leq b<k^{\prime}$ . The justification is given in the following lemma. Moreover, if $a=\frac{k}{2}$ and $b\geq k^{\prime}$ , by analyzing the singular integral, we could define the regularized Bessel moments $\mathrm{IKM}_{k}^{\mathrm{reg}}\left(\frac{k}{2},b\right)$ by subtracting the singular part of the integral. The precise definition is also given in the following lemma.

Lemma 21.

The integral expression of Bessel moments

{\mathrm{IKM}}_{k}(a,b)=\int_{0}^{\infty}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t

converges for non-negative integers $k,a,b$ satisfying $a\leq k^{\prime}$ , $b\geq 0$ or $a=\frac{k}{2}$ , $0\leq b<k^{\prime}$ . Moreover, in the case that $k$ is even, $a=\frac{k}{2}$ , and $b\geq k^{\prime}$ with $b$ even, the following two limits exist for $2j\geq k^{\prime}$ :

	$\displaystyle{\mathrm{IKM}}^{\mathrm{reg}}_{k}\left(\frac{k}{2},2j\right)$	$\displaystyle:=\lim_{t\rightarrow\infty}\left(\int_{0}^{t}(I_{0}K_{0})^{2r+2}s^{2j}{\mathrm{d}}s-\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}t^{2m+1}}{2^{k-2j+2m}(2m+1)}\right)\ \ {\text{if }}k=4r+4,$
	$\displaystyle{\mathrm{IKM}}^{\mathrm{reg}}_{k}\left(\frac{k}{2},2j\right)$	$\displaystyle:=\lim_{\begin{subarray}{c}t\rightarrow\infty\\ \varepsilon\rightarrow 0^{+}\end{subarray}}\left(\int_{\varepsilon}^{t}(I_{0}K_{0})^{2r+1}s^{2j}{\mathrm{d}}s-\frac{2\gamma_{k,j-r}}{2^{k-2j}}\int_{\varepsilon}^{t}\frac{\mathrm{d}s}{s}-\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}t^{2m+2}}{2^{k-2j+2m+1}(2m+2)}\right)$
		$\displaystyle\hskip 284.52756pt{\text{if }}k=4r+2.$

Proof.

Near $0$ , by [12, §10.30(i)] we have the asymptotics

	$\displaystyle I_{0}(t)$	$\displaystyle=1+O(t^{2});$		(17)
	$\displaystyle K_{0}(t)$	$\displaystyle=-\left(\gamma+\log\frac{t}{2}\right)+O(t^{2}\log t),$		(18)

where $\gamma$ is the Euler constant. Then, the integral $\int_{0}^{1}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t$ converges for all $0\leq a\leq\frac{k}{2}$ and any $b\geq 0$ .

Near $\infty$ , from (8) and (9), when $0\leq a\leq k^{\prime}$ , $I_{0}^{a}(t)K_{0}^{k-a}(t)$ decays exponentially and hence the integral $\int_{1}^{\infty}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t$ converges.

When $k$ is even and $a=\frac{k}{2}$ , near $\infty$ , by (11), we have the asymptotic expansion

\left(I_{0}K_{0}\right)^{k/2}t^{2j}=\frac{1}{2^{k/2}}\sum_{n=0}^{\infty}\gamma_{k,n}4^{n}t^{-2n-k/2+2j}.

Taking integration, we have

\int_{\varepsilon}^{t}(I_{0}K_{0})^{k/2}s^{2j}\mathrm{d}s=\frac{1}{2^{k/2}}\sum_{n=0}^{\infty}\gamma_{k,n}4^{n}\int_{\varepsilon}^{t}s^{-2n-k/2+2j}\mathrm{d}s.

Using the fact that $\int_{1}^{\infty}t^{\alpha}{\mathrm{d}}t$ converges if and only if $\alpha<-1$ and $\int_{0}^{1}t^{\alpha}{\mathrm{d}}t$ converges if and only if $\alpha>-1$ , the divergent part of the integral $\int_{\varepsilon}^{t}\left(I_{0}K_{0}\right)^{k/2}s^{2j}{\mathrm{d}}s$ as $t\rightarrow\infty$ , $\varepsilon\rightarrow 0^{+}$ is

	$\displaystyle\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}}{2^{k-2j+2m}}\frac{t^{2m+1}}{\left(2m+1\right)}$	$\displaystyle{\text{ if }}k=4r+4$		(19)
	$\displaystyle\frac{2\gamma_{k,j-r}}{2^{k-2j}}\int_{\varepsilon}^{t}\frac{\mathrm{d}s}{s}+\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}}{2^{k-2j+2m+1}}\frac{t^{2m+2}}{(2m+2)}$	$\displaystyle{\text{ if }}k=4r+2$		(20)

Hence, after subtracting the divergent part of the integral, we conclude that the limits $\mathrm{IKM}_{k}^{\mathrm{reg}}(\frac{k}{2},2j)$ exist. ∎

Remark 22.

For $a=\frac{k}{2}$ and $b\geq k^{\prime}$ with odd $b$ , the integral $\int_{0}^{\infty}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t$ also diverges. We may similarly define the regularized Bessel moments in this case. See [2, Definitions 6.1 and 6.4].

5.2 Period pairing and compactly supported period pairing

By [2, Corollary 2.11], there exist the following two perfect pairings. The period pairing is defined to be

\left\langle\sigma\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b},\omega\right\rangle_{\mathrm{per}}=\int_{\sigma}\frac{1}{\sqrt{z}}\left\langle e_{0}^{b}e_{1}^{k-b},\omega\right\rangle_{\mathrm{top}}.

Here, the notation $V_{\mathbb{C}}$ means $V\otimes_{\mathbb{Q}}\mathbb{C}$ . There is a one-form $\omega$ occurs in $\langle e_{0}^{b}e_{1}^{k-b},\omega\rangle_{\mathrm{top}}$ . This topological pairing means $\langle e_{0}^{b}e_{1}^{k-b},f\rangle_{\mathrm{top}}\mathrm{d}z$ whenever $\omega=f\mathrm{d}z$ . That is, we take the pairing $\langle\ ,\ \rangle_{\mathrm{top}}$ only on the coefficients. Note that the period pairing induces on the middle part Betti homology $H_{1}^{\mathrm{mid}}(\mathbb{G}_{\mathrm{m}},(\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})^{\nabla})_{\mathbb{C}}$ and middle part de Rham cohomology $H^{1}_{\mathrm{mid}}(\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2})_{\mathbb{C}}$ by the restriction:

Moreover, the compactly supported period pairing is defined to be

\displaystyle\left\langle(\xi,\eta,\omega),\sigma\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{per,c}}

\displaystyle=\int_{\sigma}\frac{1}{\sqrt{z}}\left\langle\omega,e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{top}}-\frac{1}{\sqrt{z}}\left\langle\eta,e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{top}}+\frac{1}{\sqrt{z}}\left\langle\xi,e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{top}}.

Remark 23.

Note that the order of homology and cohomology in these two pairing are different. This is because we want to write down the matrix expression of quadratic relation (22) preventing the transpose notation.

Proposition 24.

The period pairing of the rapid decay cycle $\delta_{b}$ in (13) and the de Rham cohomology class $\omega_{k,j}$ in Definition 10 is given by

\left\langle\delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}=(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}{\mathrm{IKM}}_{k}(b,2j)

for $0\leq b\leq k^{\prime}$ and $0\leq j\leq k^{\prime}$ .

Proof.

Denote $\varepsilon\sigma_{0}$ to be the scaling of the chain $\sigma_{0}$ , that is, $\varepsilon\sigma_{0}$ is a chain of a circle of radius $\varepsilon$ . Similarly, denote $\varepsilon\sigma_{+}$ to be the chain of the ray $[\varepsilon,\infty)$ . Then, since $\sigma_{0}$ and $\sigma_{+}$ are homotopy to $\varepsilon\sigma_{0}$ and $\sigma_{+}$ respectively, we may replace $\sigma_{0}$ and $\sigma_{+}$ in $\delta_{b}$ by $\varepsilon\sigma_{0}$ and $\varepsilon\sigma_{+}$ respectively in the following computation. We compute

	$\displaystyle\left\langle\delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}$	$\displaystyle=\int_{\varepsilon\left(\sigma_{+}-\frac{1}{2}\sigma_{0}+\sum_{n=1}^{k-b}d_{k-b}(n)\sigma_{0}^{2n}\right)}\frac{1}{\sqrt{z}}\left\langle e_{0}^{b}e_{1}^{k-b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{top}}$
		$\displaystyle=(-1)^{k-b}(\pi\sqrt{-1})^{b}\int_{\varepsilon\left(\sigma_{+}-\frac{1}{2}\sigma_{0}+\sum_{n=1}^{k-b}d_{k-b}(n)\sigma_{0}^{2n}\right)}\sqrt{z}(-A_{0})^{b}B_{0}^{k-b}z^{j-1}\mathrm{d}z$
		$\displaystyle=(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k}\int_{\varepsilon\left(\sigma_{+}-\frac{1}{2}\sigma_{0}+\sum_{n=1}^{k-b}d_{k-b}(n)\sigma_{0}^{2n}\right)}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z$
		$\displaystyle=(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k}\int_{\varepsilon}^{\infty}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z$
		$\displaystyle\hskip 20.0pt-\frac{1}{2}(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k}\int_{\varepsilon\sigma_{0}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z$
		$\displaystyle\hskip 20.0pt+\sum_{n=1}^{k-b}d_{k-b}(n)(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k}\int_{\varepsilon\sigma_{0}^{2n}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z.$

Changing the coordinate by $z=\frac{t^{2}}{4}$ , the first term becomes

(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}\int_{2\sqrt{\varepsilon}}^{\infty}I_{0}(t)^{b}K_{0}(t)^{k-b}t^{2j}\mathrm{d}t.

When $\varepsilon\rightarrow 0^{+}$ , this term tends to $(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}{\mathrm{IKM}}_{k}(b,2j)$ .

For the other two terms, $\int_{\varepsilon\sigma_{0}^{p}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z$ tends to zero as $\varepsilon\rightarrow 0^{+}$ for the following reason. As $s\rightarrow 0^{+}$ , we have the asymptotic expansions (17) and (18). Then, as $\varepsilon\rightarrow 0^{+}$ for all $j\geq 0$ , we have the estimate

	$\displaystyle\left\|\int_{\varepsilon\sigma_{0}^{p}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z\right\|$	$\displaystyle\leq\int_{\varepsilon\sigma_{0}^{p}}\left\|z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\right\|\left\|\mathrm{d}z\right\|$
		$\displaystyle\leq\int_{0}^{2\pi p}\varepsilon^{j-1}\sqrt{\varepsilon}\left\|I_{0}\left(2\sqrt{\varepsilon e^{i\theta}}\right)^{b}K_{0}\left(2\sqrt{\varepsilon e^{i\theta}}\right)^{k-b}\right\|\varepsilon{\mathrm{d}}\theta$
		$\displaystyle\leq\varepsilon^{j}\sqrt{\varepsilon}\int_{0}^{2\pi p}\left\|\gamma+\log\sqrt{\varepsilon e^{i\theta}}\right\|^{k-b}{\mathrm{d}}\theta$
		$\displaystyle=\varepsilon^{j}\sqrt{\varepsilon}\int_{0}^{2\pi p}\left\|\gamma+\log\sqrt{\varepsilon}+\frac{1}{2}i\theta\right\|^{k-b}{\mathrm{d}}\theta\rightarrow 0.\qed$

Proposition 25.

The compactly supported period pairing of the compactly supported de Rham cohomology $\widetilde{\omega}_{k,j}$ in Definition 11 and moderate decay cycle $\gamma_{a}$ in (14) is given by

\left\langle\widetilde{\omega}_{k,j},\gamma_{a}\right\rangle_{\mathrm{per,c}}=2^{k-2j}(-1)^{k-a}(\pi\sqrt{-1})^{a}\cdot{\mathrm{IKM}},

where $0\leq a\leq\lfloor k/2\rfloor,0\leq j\leq k^{\prime}$ with $j\neq r$ if $k\equiv 2\bmod{4}$ , and

{\mathrm{IKM}}=\begin{cases}{\mathrm{IKM}}^{\text{reg}}_{k}(a,2j)&\text{if $4\mid k,a=k/2,r+1\leq j\leq k^{\prime}$},\\ {\mathrm{IKM}}_{k}(a,2j)-\gamma_{k,j-k^{\prime}/2}2^{2j-k^{\prime}}{\mathrm{IKM}}_{k}(a,k^{\prime})&\text{if $4\mid(k+2),0\leq a\leq k^{\prime},r+1\leq j\leq k^{\prime}$},\\ {\mathrm{IKM}}^{\text{reg}}_{k}(a,2j)-\gamma_{k,j-k^{\prime}/2}2^{2j-k^{\prime}}{\mathrm{IKM}}^{\text{reg}}_{k}(a,k^{\prime})&\text{if $4\mid(k+2),a=k^{\prime}+1,r+1\leq j\leq k^{\prime}$},\\ {\mathrm{IKM}}_{k}(a,2j)&\text{otherwise}.\end{cases}

Moreover, when $k=4r+2$ , we have

\left\langle\widehat{m}_{2r+1},\gamma_{a}\right\rangle_{\mathrm{per,c}}=\delta_{a,2r+1}(\pi\sqrt{-1})^{a}2^{k}\frac{1}{\binom{k}{k/2}}.

Proof.

When $k\equiv 1,3\bmod{4}$ . We compute the compactly supported period pairing

	$\displaystyle\left\langle\left(\xi_{j},\eta_{j},\omega_{k,j}\right),\gamma_{a}\right\rangle_{\mathrm{per,c}}$	$\displaystyle=\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}z^{j}\frac{{\mathrm{d}}z}{z}-\frac{1}{\sqrt{z}}\left\langle-\sum_{c=0}^{k}\binom{k}{c}\frac{\eta_{j,c}}{\sqrt{z}}e_{0}^{k-c}\overline{e}_{1}^{c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle\hskip 20.0pt+\frac{1}{\sqrt{z}}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle=\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle\sum_{c=0}^{k}\binom{k}{c}\left(-A_{0}(z)\right)^{c}B_{0}(z)^{k-c}e_{0}^{k-c}\overline{e}_{1}^{c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}z^{j-1}{\mathrm{d}}z$
		$\displaystyle\hskip 20.0pt+(\pi\sqrt{-1})^{a}(-1)^{a}\eta_{j,a}+\frac{1}{\sqrt{z}}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle=(-1)^{a}(\pi\sqrt{-1})^{a}\int_{\mathbb{R}_{+}}\left(-A_{0}(z)\right)^{a}B_{0}(z)^{k-a}\sqrt{z}\,z^{j-1}{\mathrm{d}}z$
		$\displaystyle\hskip 20.0pt+(-1)^{a}(\pi\sqrt{-1})^{a}\eta_{j,a}+\frac{1}{\sqrt{z}}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle=(-1)^{a}(\pi\sqrt{-1})^{a}2^{k-2j}\int_{\mathbb{R}_{+}}I_{0}(s)^{a}K_{0}(s)^{k-a}s^{2j}{\mathrm{d}}s+(-1)^{a}(\pi\sqrt{-1})^{a}\eta_{j,a}$
		$\displaystyle\hskip 20.0pt+\frac{2}{s}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$

where the last equality follows by the change of variable $z=\frac{s^{2}}{4}$ . The first term converges by Lemma 21. Since $k>2a$ , by (12), the second term tends to zero as $s\rightarrow\infty$ . The third term tends to zero as $s\rightarrow 0$ since all $\xi_{j,c}\in\mathbb{Q}\llbracket\frac{s^{2}}{4}\rrbracket$ and the topological pairing gives a factor $\frac{s^{2}}{4}$ .

When $k\equiv 0\bmod{4}$ , write $k=4r+4$ . We compute the compactly supported period pairing

	$\displaystyle\left\langle\left(\xi_{j},\eta_{j},\omega_{k,j}\right),\gamma_{a}\right\rangle_{\mathrm{per,c}}$	$\displaystyle=\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}z^{j}\frac{{\mathrm{d}}z}{z}-\frac{1}{\sqrt{z}}\left\langle-\sum_{c=0}^{k}\binom{k}{c}\frac{\eta_{j,c}}{\sqrt{z}}e_{0}^{k-c}\overline{e}_{1}^{c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle\hskip 20.0pt+\frac{1}{\sqrt{z}}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle=2^{k-2j}(-1)^{a}(\pi\sqrt{-1})^{a}\int_{\mathbb{R}_{+}}I_{0}(s)^{a}K_{0}(s)^{k-a}s^{2j}{\mathrm{d}}s$
		$\displaystyle\hskip 20.0pt+(-1)^{a}(\pi\sqrt{-1})^{a}\eta_{j,a}+\frac{2}{s}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$

where the last equality is the change of variable $z=\frac{s^{2}}{4}$ . The third term tends to zero as $s\rightarrow 0$ since all $\xi_{j,c}\in\mathbb{Q}\llbracket\frac{s^{2}}{4}\rrbracket$ and the topological pairing gives a factor $\frac{s^{2}}{4}$ . By the same argument above, when $a=0,1,\cdots,\frac{k-2}{2}=k^{\prime}$ , that is, $k>2a$ , we have that the first term converges and the second term tends to zero as $s\rightarrow\infty$ .

Now, we turn to analyze the case that $a=\frac{k}{2}$ . The pairing becomes

\left\langle\left(\xi_{j},\eta_{j},\omega_{k,j}\right),\gamma_{a}\right\rangle_{\mathrm{per,c}}=2^{k-2j}(-\pi\sqrt{-1})^{a}\int_{0}^{s}I_{0}(t)^{a}K_{0}(t)^{k-a}s^{2j}{\mathrm{d}}s+(-\pi\sqrt{-1})^{a}\eta_{j,2r+2}

This term converges as $s\rightarrow\infty$ for the following reason:

The singular part of the integral $\left(I_{0}K_{0}\right)^{2r+2}s^{2j}$ is given by (19) and $\eta_{j,2r+2}$ has expansion

\eta_{j,2r+2}\sim\frac{2^{2r-2j+1}}{r-j+1/2}s^{2j-2r-1}\cdot G_{i}\sim 2^{2r-2j+1}\sum_{n=0}^{\infty}\frac{2^{2n}\gamma_{k,n}}{r-j+1/2+n}s^{2j-2r-1-2n}

Thus, both of the singular terms cancel.

When $k\equiv 2\bmod{4}$ , write $k=4r+2$ . Recall from Definition 11 the elements $\widetilde{\omega}_{k,j}$ and $\widehat{m}_{2r+1}$ in $H^{1}_{{\mathrm{dR,c}}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ . If we use the convention that $\gamma_{k,p}=0$ whenever $p<0$ , we rewrite

	$\displaystyle\widetilde{\omega}_{k,i}$	$\displaystyle=\left(\xi_{i},\eta_{i},\omega_{k,i}\right)$
		$\displaystyle=\left(\sum_{a=0}^{k}\xi_{i,a}(z)v_{0}^{a}v_{1}^{k-a}-\gamma_{k,i-r}\sum_{a=0}^{k}\xi_{r,a}(z)v_{0}^{a}v_{1}^{k-a},\right.$
		$\displaystyle\hskip 20.0pt-\sum_{\begin{subarray}{c}a=0\\ a\neq k/2\end{subarray}}^{k}\binom{k}{a}\frac{\eta_{i,a}-\gamma_{k,i-r}\eta_{r,a}}{\sqrt{z}}e_{0}^{k-a}\overline{e}_{1}^{a}-\binom{k}{k/2}\frac{\eta_{i,2r+1}}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1},$
		$\displaystyle\hskip 20.0pt\left.v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{\mathrm{d}z}{z}\right)$

In the pairing $\left\langle\widetilde{\omega}_{k,j},\gamma_{a}\right\rangle_{\mathrm{per,c}}$ , the third term

\frac{2}{s}\left\langle\sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}

tends to zero as $s\rightarrow 0$ since all $\xi_{j,c}\in\mathbb{C}\llbracket\frac{s^{2}}{4}\rrbracket$ and the topological pairing gives a factor $\frac{s^{2}}{4}$ . The other two terms are equal to

	$\displaystyle\left(\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}z^{j}\frac{{\mathrm{d}}z}{z}-\gamma_{k,j-r}\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}z^{r}\frac{{\mathrm{d}}z}{z}\right)$
	$\displaystyle-\frac{1}{\sqrt{z}}\left\langle-\sum_{\begin{subarray}{c}b=0\\ b\neq k/2\end{subarray}}^{k}\binom{k}{b}\frac{\eta_{j,b}-\gamma_{k,j-r}\eta_{r,b}}{\sqrt{z}}e_{0}^{k-b}\overline{e}_{1}^{b}-\binom{k}{k/2}\frac{\eta_{j,2r+1}}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
	$\displaystyle=(-1)^{a}(\pi\sqrt{-1})^{a}\left(2^{k-2j}\int_{\mathbb{R}_{+}}I_{0}(s)^{a}K_{0}(s)^{k-a}s^{2j}{\mathrm{d}}s-2^{k-2r}\gamma_{k,j-r}\int_{\mathbb{R}_{+}}I_{0}(s)^{a}K_{0}(s)^{k-a}s^{2r}{\mathrm{d}}s\right)$
	$\displaystyle\hskip 20.0pt+\frac{1}{z}\left\langle\sum_{\begin{subarray}{c}b=0\\ b\neq k/2\end{subarray}}^{k}\binom{k}{b}\left(\eta_{j,b}-\gamma_{k,j-r}\eta_{r,b}\right)e_{0}^{k-b}\overline{e}_{1}^{b}+\binom{k}{k/2}\eta_{j,2r+1}(e_{0}\overline{e}_{1})^{2r+1},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$

We analyze the convergence of these terms. When $2a<k$ or $j<r$ , the integral $\int_{0}^{t}I_{0}(s)^{a}K_{0}(s)^{k-a}s^{2j}{\mathrm{d}}s$ converges as $t\rightarrow\infty$ by Lemma 21. The second term is equal to

\displaystyle(-1)^{a}(\pi\sqrt{-1})^{a}\left(\eta_{j,a}-\gamma_{k,j-r}\eta_{r,a}\right).

By the expansion of $\eta_{i,a}$ :

\eta_{i,a}\sim\frac{\sqrt{\pi}^{k-2a}}{k-2a}e^{-(k-2a)s}\left(\frac{4}{s^{2}}\right)^{k/4-i}\cdot G_{i,a},

where $G_{i,a}\in 1+\frac{2}{s}\mathbb{Q}\llbracket\frac{2}{s}\rrbracket$ , this term tends to $0$ as $s\rightarrow\infty$ .

When $a=\frac{k}{2}$ and $j\geq r$ , the integral $\int_{0}^{t}I_{0}(s)^{a}K_{0}(s)^{k-a}s^{2j}{\mathrm{d}}s$ has the singular part (20). The second term is equal to

(-1)^{2r+1}(\pi\sqrt{-1})^{2r+1}\eta_{j,2r+1}=(-1)^{2r+1}(\pi\sqrt{-1})^{2r+1}\frac{1}{r-j}\left(\frac{4}{s^{2}}\right)^{r-j}\cdot H_{i}

where $H_{i}\in 1+\frac{4}{s^{2}}\mathbb{Q}\llbracket\frac{4}{s^{2}}\rrbracket$ . Thus, the singular part of this term is

(-1)^{2r+1}(\pi\sqrt{-1})^{2r+1}\sum_{n=1}^{j-r}\frac{-\gamma_{k,j-r-n}}{n}\left(\frac{4}{s^{2}}\right)^{-n}.

In consequence, the singular parts cancel.

Finally, for $a=0,1,\cdots,\frac{k}{2}$ , we have

	$\displaystyle\left\langle\left(0,\frac{2^{k}}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1},0\right),\gamma_{a}\right\rangle_{\mathrm{per,c}}$	$\displaystyle=-\frac{2^{k}}{\sqrt{z}}\left\langle\frac{1}{\sqrt{z}}(e_{0}\overline{e}_{1})^{2r+1},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}$
		$\displaystyle=\delta_{a,2r+1}(\pi\sqrt{-1})^{a}2^{k}\frac{1}{\binom{k}{k/2}}.\qed$

Corollary 26.

The period matrix of the period pairing with respective to the bases $\left\{\delta_{b}\right\}_{b=0}^{k^{\prime}}$ of $H_{1}^{\mathrm{rd}}$ and $\left\{\omega_{k,j}\right\}_{j=0}^{k^{\prime}}$ of $H^{1}_{\mathrm{dR}}$ is $P=(P_{bj})$ , where

P_{bj}=\left\langle\delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}=(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}{\mathrm{IKM}}_{k}(b,2j)

for $0\leq b\leq k^{\prime}$ and $0\leq j\leq k^{\prime}$ . Moreover, $P$ is invertible.

Remark 27 (determinant of the period matrix).

In fact,

\displaystyle\det P

\displaystyle=(\pi\sqrt{-1})^{k^{\prime}(k^{\prime}+1)/2}(-1)^{(2k-k^{\prime})(k^{\prime}+1)/2}2^{(k-k^{\prime})(k^{\prime}+1)}\det\left({\mathrm{IKM}}_{k}(b,2j)\right),

where

\displaystyle\det\left({\mathrm{IKM}}_{k}(b,2j)\right)=\begin{cases}\det\left(M_{k^{\prime}+1}\right)&{\text{if }}k{\text{ is odd;}}\\ \det\left(N_{k^{\prime}+1}\right)&{\text{if }}k{\text{ is even.}}\\ \end{cases}

The definition of $M_{r}$ and $N_{r}$ are given in Appendix A.2 and their determinants are given in Corollary 39 explicitly.

5.3 $\mathbb{Q}$ -linear and quadratic relations on Bessel moments

We have now developed all the tools and computations to see the wonderful results in $\mathbb{Q}$ -linear and quadratic relations on Bessel moments.

Corollary 28.

For $k=4r+4$ ,

\displaystyle\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2i)=\begin{cases}(-1)^{r}\pi^{2r+2}{\mathrm{IKM}}_{k}(2r+2,2i)&\text{if $0\leq i\leq r$},\\ (-1)^{r}\pi^{2r+2}{\mathrm{IKM}}_{k}^{\mathrm{reg}}(2r+2,2i)&\text{if $r+1\leq i\leq k^{\prime}$}.\end{cases}

For $k=4r+2$ ,

\displaystyle\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2i)=\begin{cases}0&\text{if $0\leq i\leq r-1$},\\ \gamma_{k,i-r}2^{2i-2r}\displaystyle\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2r)&\text{if $r+1\leq i\leq 2r$}.\end{cases}

(21)

Proof.

By Lemma 17, we know that

	$\displaystyle\sum_{j=0}^{k/4}\binom{k/2}{2j}\gamma_{2j}$	$\displaystyle=0{\text{ if }}k\equiv 0\bmod{4};$
	$\displaystyle\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\gamma_{2j}$	$\displaystyle=0{\text{ if }}k\equiv 2\bmod{4}.$

Then take the pairing with $\widetilde{\omega}_{k,i}$ in the compactly supported de Rham cohomology. Combining with the result of Proposition 25, we obtain the desired algebraic relation. ∎

Remark 29.

The above linear algebraic relations for $i$ in the range $0\leq i\leq r$ , under the name sum rule identities, are previously proved by analytic method in [6] (see [6, (1.3)] for $k\equiv 2\bmod{4}$ and [6, (1.5)] for $k\equiv 0\bmod{4}$ ).

Corollary 30.

For any $k$ and any $0\leq a\leq k^{\prime}$ , the dimension of the $\mathbb{Q}$ -vector space generated by the Bessel moments has an upper bound:

\dim{\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j\in\left\{0\right\}\cup\mathbb{N}\right\}\leq k^{\prime}+1.

If $k$ is even, the dimension of the $\mathbb{Q}$ -vector space generated by the regularized Bessel moments has an upper bound:

\dim{\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}(k/2,2j)\mid j\in\left\{0\right\}\cup\mathbb{N}\right\}\leq k^{\prime}+1.

Here when $0\leq j\leq\left\lfloor\frac{k-1}{4}\right\rfloor=r$ , we do not need to regularize the Bessel moments, that is, ${\mathrm{IKM}}^{\mathrm{reg}}_{k}(k/2,2j)={\mathrm{IKM}}_{k}(k/2,2j)$ (see Lemma 21).

Proof.

We know that the dimensions of $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ and $H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ are $k^{\prime}+1$ .

For each integer $s>k^{\prime}$ , since $\left\{v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\}_{j=0,\cdots,k^{\prime}}$ form a basis of $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ , we may express $v_{0}^{k}z^{s}\frac{{\mathrm{d}}z}{z}$ as the $\mathbb{Q}$ -linear combination of the basis. Then after we take the period pairing between $v_{0}^{k}z^{s}\frac{{\mathrm{d}}z}{z}$ and the rapid decay cycle $\delta_{a}$ (see Proposition 24), the $\mathbb{Q}$ -linear relation becomes a $\mathbb{Q}$ -linear relation for the Bessel moments

\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j=0,\cdots,k^{\prime}\right\}\cup\left\{{\mathrm{IKM}}_{k}(a,2s)\right\}.

If $k$ is even, similarly, when $s>k^{\prime}$ , express $\widetilde{\omega}_{k,s}\in H^{1}_{\mathrm{dR,c}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$ as the $\mathbb{Q}$ -linear combination of the basis describe in Corollary 14. Then taking the compactly supported period pairing (see Proposition 25), the $\mathbb{Q}$ -linear equivalence become the $\mathbb{Q}$ -linear relation for the regularized Bessel moments

\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j=0,\cdots,r-1\right\}\cup\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}(a,2j)\mid j=r,\cdots,k^{\prime}\right\}\cup\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}(a,2s)\right\}.\qed

Remark 31.

In [16], Borwein and Salvy provide a recurrence to find out the $\mathbb{Q}$ -linear combination for Bessel moments by analyzing the symmetric power of the modified Bessel differential operator. Moreover, Zhou proves a similar result in [7] for the $\mathbb{Q}$ -linear dependence for Bessel moments ${\mathrm{IKM}}_{k}(a,2j-1)$ . Our result is parallel to Zhou’s result.

Proposition 32.

With respect to the bases of $H^{{\mathrm{rd}}}_{1}$ , $H^{{\mathrm{mod}}}_{1}$ , $H_{\mathrm{dR}}^{1}$ , and $H_{\mathrm{dR,c}}^{1}$ described in Corollaries 14, 20, we form the pairing matrices:

1.

$B$ , the Betti intersection pairing matrix between $H^{{\mathrm{rd}}}_{1}$ and $H^{{\mathrm{mod}}}_{1}$ in Proposition 18.
2.

$D$ , the Poincaré pairing matrix between $H_{\mathrm{dR,c}}^{1}$ and $H_{\mathrm{dR}}^{1}$ in Proposition 13.
3.

$P$ , the period pairing matrix between $H^{{\mathrm{rd}}}_{1}$ and $H_{\mathrm{dR}}^{1}$ in Proposition 24.
4.

$P_{\mathrm{c}}$ , the period pairing matrix between $H_{\mathrm{dR,c}}^{1}$ and $H^{{\mathrm{mod}}}_{1}$ in Proposition 25.
5.

$B_{\mathrm{mid}}$ , the Betti pairing matrix on $H^{\mathrm{mid}}_{1}$ .
6.

$D_{\mathrm{mid}}$ , the Poincaré pairing matrix on $H_{\mathrm{mid}}^{1}$ .
7.

$P_{\mathrm{mid}}$ , the period pairing matrix between $H^{\mathrm{mid}}_{1}$ and $H_{\mathrm{mid}}^{1}$ .¹¹1 $B,D,P,P_{\mathrm{c}}$ are square matrices of size $k^{\prime}+1$ and that $B_{\mathrm{mid}},D_{\mathrm{mid}},P_{\mathrm{mid}}$ are of size $k^{\prime}+1-\delta_{4\mathbb{Z}+2,k}$ . When $k\equiv 0,1,3\bmod{4}$ , we have $B=B_{\mathrm{mid}}$ , $D=D_{\mathrm{mid}}$ , and $P_{\mathrm{mid}}=P=P_{\mathrm{c}}^{t}$ .

Then we have the algebraic quadratic relations

	$\displaystyle PD^{-1}P_{\mathrm{c}}$	$\displaystyle=(-1)^{k}(2\pi\sqrt{-1})^{k+1}B,$		(22)
	$\displaystyle P_{\mathrm{mid}}D^{-1}_{\mathrm{mid}}P_{\mathrm{mid}}^{t}$	$\displaystyle=(-1)^{k}(2\pi\sqrt{-1})^{k+1}B_{\mathrm{mid}}.$		(23)

Proof.

This quadratic relation is a general phenomenon on periods of meromorphic flat connection on complex manifolds. We refer to [8, Corollaries 2.14, 2.16] for more details. ∎

From this proposition, when $k\equiv 0,1,3\bmod{4}$ , we see the Bessel moments have quadratic relation given by (22). On the other hand, when $k\equiv 2\bmod{4}$ , the relation involves some combination of Bessel moments and regularized Bessel moments in the matrix $P_{\mathrm{c}}$ . In the following discussion, we provide another expression of this relation, and we will see the pure quadratic relation involving only Bessel moments.

When $k\equiv 2\bmod{4}$ , write $k=4r+2$ and define two $(k^{\prime}+1)\times k^{\prime}$ matrices with rational coefficients:

R_{k}=\left(\begin{array}[]{cccc}I_{r}&&0&\\ 0&-\gamma_{k,1}&\cdots&-\gamma_{k,k^{\prime}-r}\\ 0&&I_{r}&\end{array}\right),\quad L_{k}=\left(\begin{array}[]{ccccccc}0&-\binom{k/2}{2}&0&-\binom{k/2}{4}&\cdots&0&-\binom{k/2}{k^{\prime}}\\ &&&I_{k^{\prime}}&\end{array}\right).

By the linear relations (21) in Corollary 28, we have

PR_{k}=L_{k}P_{\mathrm{mid}}.

Also, $P_{\mathrm{mid}}$ is obtained by deleting the first row of $L_{k}P_{\mathrm{mid}}$ . Set $\widetilde{B}=L_{k}B_{\mathrm{mid}}L_{k}^{t}$ and $\widetilde{D}=R_{k}D^{-1}_{\mathrm{mid}}R_{k}^{t}$ which are square matrices of size $k^{\prime}+1$ with rational coefficients. Then $B_{\mathrm{mid}}$ is obtained by deleting the first row and column from $\widetilde{B}$ . Therefore, the quadratic relation (23) (involving linear combinations of Bessel moments) now becomes

P\widetilde{D}P^{t}=(-1)^{k}(2\pi\sqrt{-1})^{k+1}\widetilde{B}

(involving pure Bessel moments).

Remark 33.

The matrices $\widetilde{B}$ and $\widetilde{D}$ in the above expression are singular because of the linear relations (21) in Corollary 28. This expression is equivalent to the middle part quadratic relation (23) together with linear relations (21).

Proposition 34.

When $k=4r+2$ , the middle part period matrix is a $k^{\prime}\times k^{\prime}$ matrix given by

P_{\mathrm{mid}}=\left(\left\langle\delta_{b},\omega_{k,i}\right\rangle_{\mathrm{per}}\right)_{b=1,\cdots,k^{\prime},\ i=0,\cdots,\hat{r},\cdots,k^{\prime}}.

The determinant of this matrix $P_{\mathrm{mid}}$ is given by

\det P_{\mathrm{mid}}=\pi^{r(k+1)}\sqrt{-1}^{r(k^{\prime}-1)}\frac{2^{r(2r+1)}}{r!}\prod_{a=1}^{k^{\prime}}\frac{(2a+1)^{k^{\prime}+1-a}}{(a+1)^{a+1}}.

Proof.

The matrix $P_{\mathrm{mid}}$ appears in the upper left of the compactly supported period pairing matrix $P_{\mathrm{c}}$ . Just take determinant on (22) and then use the results of Propositions 13, 19, and Remark 27. ∎

Appendix A The Bessel operator and determinants of Bessel moments

A.1 Symmetric power of the modified Bessel differential operator

Consider the Weyl algebra $\mathbb{Q}\langle t,\partial_{t}\rangle$ consisting of ordinary differential operators. Write $\theta=t\partial_{t}$ . The modified Bessel differential operator is an element in the subalgebra $\mathbb{Q}\langle t^{2},\theta\rangle$ given by $L_{2}=\theta^{2}-t^{2}$ . The corresponding solutions are the modified Bessel functions $I_{0}(t)$ and $K_{0}(t)$ . The $n$ -th symmetric power $L_{n+1}\in\mathbb{Q}\left\langle\theta,t^{2}\right\rangle$ of $L_{2}$ has order $n+1$ and the corresponding solutions are $I_{0}^{a}(t)K_{0}^{n-a}(t)$ for $0\leq a\leq n$ . By [17, 16], the operator $L_{n+1}=L_{n+1,n}$ can be obtained by the recurrence relation as follows:

\displaystyle\begin{split}L_{0,n}&=1,\\ L_{1,n}&=\theta,\\ L_{k+1,n}&=\theta L_{k,n}-t^{2}k\left(n+1-k\right)L_{k-1,n},\ 1\leq k\leq n.\end{split}

(24)

Here we provide two more concrete results about the operator $L_{n+1}$ .

Put the degree on $\mathbb{Q}\langle t,\theta\rangle$ as $\deg t=\deg\theta=1$ . The associated graded ring $\operatorname{gr}\mathbb{Q}\langle t,\theta\rangle=\mathbb{Q}[\overline{t},\overline{\theta}]$ is a polynomial ring where $\overline{t}$ and $\overline{\theta}$ are the images of $t$ and $\theta$ , respectively.

Proposition 35.

The image of $L_{n+1}$ in $\mathbb{Q}[\overline{t},\overline{\theta}]$ is the polynomial

\overline{L}_{n+1}(\overline{t},\overline{\theta})=\begin{cases}\prod_{i=1}^{r}\left(\overline{\theta}^{2}-(2i-1)^{2}\overline{t}^{2}\right)&\text{if $n+1=2r$ is even},\\ \overline{\theta}\prod_{i=1}^{r}\left(\overline{\theta}^{2}-(2i)^{2}\overline{t}^{2}\right)&\text{if $n+1=2r+1$ is odd}.\end{cases}

(25)

Proof.

Taking the images in $\mathbb{Q}[\overline{t},\overline{\theta}]$ of the relation (24), we obtain $\overline{L}_{n+1}=\overline{L}_{n+1,n}$ satisfying

	$\displaystyle\overline{L}_{0,n}$	$\displaystyle=1,\quad\overline{L}_{1,n}=\overline{\theta},$
	$\displaystyle\overline{L}_{k+1,n}$	$\displaystyle=\overline{\theta}\,\overline{L}_{k,n}-t^{2}k\left(n+1-k\right)\overline{L}_{k-1,n},\ 1\leq k\leq n.$

The formula (25) is then a consequence of the following combinatorics lemma. ∎

Lemma 36.

For any $m\in\mathbb{N}$ , set the recurrence for $\lambda_{n,m}(x),n\in\mathbb{N}\cup\{0\}$ ,

	$\displaystyle\lambda_{0,m}$	$\displaystyle=1,\quad\lambda_{1,m}=x,$
	$\displaystyle\lambda_{k+1,m}$	$\displaystyle=x\lambda_{k,m}-k\left(m+1-k\right)\lambda_{k-1,m},\quad k\geq 1.$

Then we have

\displaystyle\lambda_{m+1,m}(x)=\begin{cases}\prod_{i=1}^{r}\left(x^{2}-(2i-1)^{2}\right),&m+1=2r,\\ x\prod_{i=1}^{r}\left(x^{2}-(2i)^{2}\right),&m+1=2r+1.\end{cases}

(26)

Proof.

Notice that $\lambda_{i,m}$ is a monic integral polynomial of degree $i$ for any $m$ . Consider the formal generating function²²2This generating function satisfies the differential equation $-y^{4}f^{\prime\prime}(y)-(2y^{3}-my^{3})f^{\prime}(y)+(1-xy+my^{2})f(y)=1.$ :

f_{m,x}(y)=\sum_{i=0}^{\infty}\lambda_{i,m}(x)y^{i}.

An induction on $i$ immediately yields the relation $f_{m,x-1}\left(y\right)+f_{m,x+1}\left(y\right)=2f_{m-1,x}\left(y\right)$ for any $m$ and $x$ .³³3Equality also holds when viewed as the solution of the corresponding differential equations. In other words, $\lambda_{i,m}(x-1)+\lambda_{i,m}(x+1)=2\lambda_{i,m-1}(x)$ for all $i$ . Therefore we obtain

\lambda_{m+1,m}(x-1)+\lambda_{m+1,m}(x+1)=2\lambda_{m+1,m-1}(x)=2x\lambda_{m,m-1}(x)

by the recurrence. Thus, since $\lambda_{m+1,m}(x)$ is a monic polynomial of degree $m+1$ , it is uniquely determined by the above functional equation when the polynomial $\lambda_{m,m-1}(x)$ is given. Hence, by the induction, it suffices to show that

	$\displaystyle\prod_{i=1}^{r}\left((x-1)^{2}-(2i-1)^{2}\right)+\prod_{i=1}^{r}\left((x+1)^{2}-(2i-1)^{2}\right)$	$\displaystyle=2x^{2}\prod_{i=1}^{r-1}\left(x^{2}-(2i)^{2}\right);$
	$\displaystyle(x-1)\prod_{i=1}^{r}\left((x-1)^{2}-(2i)^{2}\right)+(x+1)\prod_{i=1}^{r}\left((x+1)^{2}-(2i)^{2}\right)$	$\displaystyle=2x\prod_{i=1}^{r}\left(x^{2}-(2i-1)^{2}\right),$

which are straightforward to verify. ∎

Proposition 37.

Write $L_{n+1}$ into the form $\sum_{i}t^{i}P_{i}\left(\theta\right)$ , where $P_{i}(x)\in\mathbb{Q}\left[x\right]$ . Define the integers $a,b$ by

a=\max\left\{i\mid P_{i}\neq 0\right\};\quad b=\min\left\{i\mid P_{i}\neq 0\right\}.

Then we have $a=2\left\lfloor\frac{n+1}{2}\right\rfloor$ and $b=0$ .

Proof.

By the recurrence (24), if we set $\deg t=1$ and $\deg\theta=0$ , we easily see that $L_{j}$ has degree $2\left\lfloor\frac{j}{2}\right\rfloor$ by the interchanging relation $\theta t=t+t\theta$ . Thus, we have $a=2\left\lfloor\frac{n+1}{2}\right\rfloor$ . On the other hand, if we set $\deg t=0$ and $\deg\theta=1$ , we see that the leading term of $L_{k+1}$ is given by $\theta^{k+1}$ . Therefore, we conclude that $b=0$ by Proposition 35. ∎

A.2 Two-scale Bessel moments

From now on, we take for granted the properties of modified Bessel functions $I_{0}(t),K_{0}(t)$ in the treatise [13].

Recall the Bessel moments ${\mathrm{IKM}}_{k}(a,b)$ given in section 5.1. For $r\in\mathbb{Z}_{\geq 1}$ , define the two $r\times r$ matrices

M_{r}=\begin{pmatrix}{\mathrm{IKM}}_{2r-1}(i-1,2j-2)\end{pmatrix}_{1\leq i,j\leq r},\quad N_{r}=\begin{pmatrix}{\mathrm{IKM}}_{2r}(i-1,2j-2)\end{pmatrix}_{1\leq i,j\leq r}.

We aim to determine the two scalars $\det M_{r}$ , $\det N_{r}$ adapting the inductive methods explored by Zhou [9].

For the initial values, we have [13, §13.21, Eq.(8)]

M_{1}=\int_{0}^{\infty}K_{0}(t)\,{\mathrm{d}}t=\frac{\pi}{2},

(27)

and, by [13, §13.72], one has

	$\displaystyle N_{1}=\int_{0}^{\infty}K_{0}^{2}(t)\,{\mathrm{d}}t$	$\displaystyle=\frac{1}{2}\int_{0}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2t\cosh x\cosh y}\,{\mathrm{d}}x{\mathrm{d}}y{\mathrm{d}}t$
		$\displaystyle=\frac{1}{4}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{{\mathrm{d}}x{\mathrm{d}}y}{\cosh x\cosh y}$
		$\displaystyle=\frac{\pi^{2}}{4}.$

For $r\in\mathbb{Z}_{\geq 0}$ , let $\omega_{2r+1}(x)$ be the Wronskian of the $(2r+1)$ functions $f_{i}(x)$

f_{i}(x)=\begin{cases}\int_{0}^{\infty}I_{0}(xt)I^{i-1}_{0}(t)K_{0}^{2r-i+1}(t)\,{\mathrm{d}}t,&1\leq i\leq r,\\ \int_{0}^{\infty}K_{0}(xt)I^{i-r-1}_{0}(t)K_{0}^{3r-i+1}(t)\,{\mathrm{d}}t,&r<i\leq 2r+1.\end{cases}

(28)

The functions $f_{i}$ are well-defined and analytic on the interval $(0,2)$ and hence so is $\omega_{2r+1}$ . In particular, $\omega_{1}(x)=\int_{0}^{\infty}K_{0}(xt)\,{\mathrm{d}}t=\frac{\pi}{2x}$ by (27).

For $r\in\mathbb{Z}_{\geq 1}$ , let $\omega_{2r}(x)$ be the Wronskian of the $2r$ functions $g_{i}(x)$ where

g_{i}(x)=\begin{cases}\int_{0}^{\infty}I_{0}(xt)I^{i-1}_{0}(t)K_{0}^{2r-i}(t)\,{\mathrm{d}}t,&1\leq i\leq r,\\ \int_{0}^{\infty}K_{0}(xt)I^{i-r-1}_{0}(t)K_{0}^{3r-i}(t)\,{\mathrm{d}}t,&r<i\leq 2r.\end{cases}

All entries in the Wronskian matrix are well-defined analytic functions on the interval $(0,1)$ and so is $\omega_{2r}(x)$ .

Proposition 38.

The determinant $\omega_{k}(x)$ and its evaluation at $x=1$ are given by the following formulae:

For $r\in\mathbb{Z}_{\geq 1}$ ,

	$\displaystyle\omega_{2r+1}(x)$	$\displaystyle=\frac{(-1)^{\frac{r(r+1)}{2}}}{2}\left[\frac{1}{x^{2}}\prod_{i=1}^{r}\frac{(2i)^{2}}{(2i)^{2}-x^{2}}\right]^{\frac{2r+1}{2}}\Gamma\Big{(}\frac{r+1}{2}\Big{)}^{2}\big{(}\det N_{r}\big{)}^{2},$
	$\displaystyle\omega_{2r+1}(1)$	$\displaystyle=(-1)^{r(r+1)/2}\det M_{r}\cdot\det M_{r+1}.$

For $r\in\mathbb{Z}_{\geq 2}$ ,

	$\displaystyle\omega_{2r}(x)$	$\displaystyle=(-1)^{\frac{r(r+1)}{2}}\left[\frac{1}{x}\prod_{i=1}^{r}\frac{(2i-1)^{2}}{(2i-1)^{2}-x^{2}}\right]^{r}\big{(}\det M_{r}\big{)}^{2},$
	$\displaystyle\lim_{x\to 1^{-}}2^{r}(1-x)^{r}\omega_{2r}(x)$	$\displaystyle=(-1)^{r(r+1)/2}(r-1)!\det N_{r-1}\cdot\det N_{r}.$

The above proposition leads to the recursive formulae

	$\displaystyle\det M_{r}\cdot\det M_{r+1}$	$\displaystyle=\frac{1}{2}\left[\frac{2^{r}r!\sqrt{2r+1}}{(2r+1)!!}\right]^{2r+1}\Gamma\Big{(}\frac{r+1}{2}\Big{)}^{2}\big{(}\det N_{r}\big{)}^{2}$	$\displaystyle(r\geq 1),$
	$\displaystyle\det N_{r-1}\cdot\det N_{r}$	$\displaystyle=\frac{2^{r}}{(r-1)!}\left[\frac{(2r-1)!!\sqrt{2r}}{2^{r}r!}\right]^{2r}\big{(}\det M_{r}\big{)}^{2}$	$\displaystyle(r\geq 2).$		(29)

With the initial data $M_{1}=\frac{\pi}{2},N_{1}=\frac{\pi^{2}}{4}$ and the relation

\Gamma\left(\frac{r}{2}\right)\Gamma\left(\frac{r+1}{2}\right)=\frac{\left(r-1\right)!}{2^{r-1}}\sqrt{\pi},

one immediately obtains the following results by induction.

Corollary 39.

For positive integers $r$ , we have

	$\displaystyle\det M_{r}$	$\displaystyle=\sqrt{\pi}^{r(r+1)}\sqrt{2}^{r(r-3)}\prod_{a=1}^{r-1}\frac{a^{r-a}}{\sqrt{2a+1}^{2a+1}},$
	$\displaystyle\det N_{r}$	$\displaystyle=\frac{1}{\Gamma\left(\frac{r+1}{2}\right)}\frac{\sqrt{\pi}^{(r+1)^{2}}}{\sqrt{2}^{r(r+3)}}\prod_{a=1}^{r-1}\frac{(2a+1)^{r-a}}{(a+1)^{a+1}}.$

In particular, the two scalars $\sqrt{(2r-1)!!}\,\pi^{-m_{r}}\det M_{r}$ and $\pi^{-n_{r}}\det N_{r}$ are positive rational numbers, where $m_{r}=\frac{r(r+1)}{2}$ and $n_{r}=\left\lfloor\frac{(r+1)^{2}}{2}\right\rfloor$ .

A.3 The Vanhove operators

The adjoint $L_{n+1}^{*}$ of $L_{n+1}$ is derived under the convolution $(t,\partial_{t})\mapsto(t,-\partial_{t})$ (so $\theta\mapsto-(\theta+1)$ ) and hence the leading term of the signed adjoint $\Lambda_{n+1}=(-1)^{n+1}L_{n+1}^{*}$ equals $\overline{L}_{n+1}(\overline{\theta},\overline{t})$ by Proposition 35. For $F(xt)=I_{0}(xt),K_{0}(xt)$ and $G(t)=I_{0}^{a}(t)K_{0}^{n-a}(t)$ , we have, by integration by parts,

\int_{0}^{\infty}(\Lambda_{n+1}F(xt))G(t)\,{\mathrm{d}}t=(-1)^{n+1}\int_{0}^{\infty}F(xt)(L_{n+1}G(t))\,{\mathrm{d}}t=0.

The Vanhove operator $V_{n+1}\in\mathbb{Q}\left\langle\partial_{x},x^{\pm 1}\right\rangle$ is of order $(n+1)$ such that $V_{n+1}F(xt)=\Lambda_{n+1}F(xt)$ . So one has $V_{n+1}f_{i}=0$ for $f_{i}(x)$ in (28) and consequently $\omega_{n+1}(x)$ satisfies a first order linear differential equation (See (30) below).

Lemma 40.

Let $\lambda_{n+1}(x)=\overline{L}_{n+1}(1,x^{-1})\in\mathbb{Q}[x^{-1}]$ of order $2\left\lfloor\frac{n+1}{2}\right\rfloor$ with respect to $x^{-1}$ . Let $\theta_{x}=x\partial_{x}$ . One has

	$\displaystyle V_{n+1}$	$\displaystyle=\lambda_{n+1}(x)\theta_{x}^{n+1}+(n+1)\left[\lambda_{n+1}(x)+\frac{x\lambda_{n+1}^{\prime}(x)}{2}\right]\theta_{x}^{n}+\delta_{1}$
		$\displaystyle=x^{n+1}\lambda_{n+1}(x)\partial_{x}^{n+1}+\frac{n+1}{2}x^{n}\left[(n+2)\lambda_{n+1}(x)+x\lambda_{n+1}^{\prime}(x)\right]\partial_{x}^{n}+\delta_{2}$

where $\delta_{1},\delta_{2}$ are of order at most $(n-1)$ with respect to $\partial_{x}$ in $\mathbb{Q}\langle\partial_{x},x^{\pm 1}\rangle$ .

Proof.

By Vanhove [18], there exists $\widetilde{L}_{n-1}\in\mathbb{Q}\langle\partial_{x},x^{\pm 1}\rangle$ of order $(n-1)$ such that

t\widetilde{L}_{n-1}F(xt)=\Lambda_{n+1}\frac{F(xt)}{t}.

The operator $\widetilde{L}_{n-1}$ is of the form ([9, Eq. (4.29)])

\widetilde{L}_{n-1}=x^{2}\lambda(x)\theta_{x}^{n-1}+x^{2}\left[2(n-1)\lambda(x)+\frac{n-1}{2}x\lambda^{\prime}(x)\right]\theta_{x}^{n-2}+\widetilde{\delta}

where $\widetilde{\delta}$ is of order at most $(n-3)$ with respect to $\partial_{x}$ in $\mathbb{Q}\langle\partial_{x},x^{\pm 1}\rangle$ ⁴⁴4Comparing $\widetilde{L}_{n-1}(\theta_{x})$ with Zhou’s Vanhove operator $\widetilde{L}_{n-1}(\theta_{u})$ , we set his variable $u=x^{2}$ and multiply $\widetilde{L}_{n-1}(\theta_{u})$ by $2^{n-1}$ ..

Set

\Delta_{n}(\theta_{t})=\Lambda_{n+1}(\theta_{t})-\Lambda_{n+1}(\theta_{t}-1).

Since $\theta_{t}\frac{1}{t}=\frac{1}{t}(\theta_{t}-1)$ in $\mathbb{Q}\left\langle\partial_{t},t\right\rangle$ , we have

	$\displaystyle\Lambda_{n+1}(\theta_{t})F(xt)$	$\displaystyle=t\Lambda_{n+1}(\theta_{t})\frac{F(xt)}{t}+\Delta_{n}(\theta_{t})F(xt)$
		$\displaystyle=\Big{[}t^{2}\widetilde{L}_{n-1}(\theta_{x})+\Delta_{n}(\theta_{t})\Big{]}F(xt).$

Since $t^{2}F(xt)=\frac{1}{x^{2}}\theta_{x}^{2}F(xt)$ , we have

	$\displaystyle t^{2}\widetilde{L}_{n-1}(\theta_{x})F(xt)$	$\displaystyle=\widetilde{L}_{n-1}(\theta_{x})\frac{1}{x^{2}}\theta_{x}^{2}F(xt)$
		$\displaystyle=\frac{1}{x^{2}}\widetilde{L}_{n-1}(\theta_{x}-2)\theta_{x}^{2}F(xt)$

and the differential operator reads

\lambda(x)\theta_{x}^{n+1}+\frac{n-1}{2}x\lambda^{\prime}(x)\theta_{x}^{n}+\delta_{3}

where $\delta_{3}$ is of order at most $(n-1)$ with respect to $\partial_{x}$ in $\mathbb{Q}\langle\partial_{x},x^{\pm 1}\rangle$ .

On the other hand, since $\theta_{t}F(xt)=\theta_{x}F(xt)$ and by Proposition 35, we have

\Delta_{n}(\theta_{t})F(xt)=\left[\Lambda_{n+1}\left(\theta\right)-\Lambda_{n+1}\left(\theta-1\right)\right]F\left(xt\right)=\left[\left((n+1)\lambda(x)+x\lambda^{\prime}(x)\right)\theta_{x}^{n}+\delta_{4}\right]F(xt)

where $\delta_{4}$ is of order at most $(n-1)$ with respect to $\partial_{x}$ in $\mathbb{Q}\langle\partial_{x},x^{-1}\rangle$ . Therefore the leading two terms of $V_{n+1}$ are determined. ∎

Rationality of $\omega_{n+1}(x)$

Lemma 40 yields

\omega_{n+1}^{\prime}(x)=-\frac{n+1}{2x}\left[(n+2)+\frac{x\lambda_{n+1}^{\prime}(x)}{\lambda_{n+1}(x)}\right]\omega_{n+1}(x).

(30)

Since $\omega_{n+1}(x)$ takes real values on $(0,1)$ , one obtains

\omega_{n+1}(x)=C_{n+1}\left[(-1)^{\left\lfloor\frac{n+1}{2}\right\rfloor}x^{n+2}\lambda_{n+1}(x)\right]^{-\frac{n+1}{2}}

for some real constant $C_{n+1}$ for each $n\in\mathbb{Z}_{\geq 0}$ . We shall determine $C_{n+1}$ by investigating the limiting behavior of $\omega_{n+1}(x)$ as $x\to 0^{+}$ .

A.4 Singularities of $\omega_{n+1}(x)$

For $F(xt)=I_{0}(xt)$ or $K_{0}(xt)$ , we have

\partial_{x}F(xt)=tF^{\prime}(xt),\quad\partial_{x}^{2}F(xt)=-\frac{t}{x}F^{\prime}(xt)+t^{2}F(xt).

So $\omega_{n+1}(x)$ coincides with the determinant of the matrix $\Omega_{n+1}(x)$ of size $(n+1)$ whose $(i,j)$ -entry is

\begin{cases}\int_{0}^{\infty}I_{0}(xt)I_{0}^{j-1}(t)K_{0}^{n-j+1}(t)t^{i-1}\,{\mathrm{d}}t,&1\leq j\leq\left\lfloor\frac{n+1}{2}\right\rfloor,i=1,3,\cdots,2\left\lfloor\frac{n}{2}\right\rfloor+1,\\ \int_{0}^{\infty}tI_{0}^{\prime}(xt)I_{0}^{j-1}(t)K_{0}^{n-j+1}(t)t^{i-2}\,{\mathrm{d}}t,&1\leq j\leq\left\lfloor\frac{n+1}{2}\right\rfloor,i=2,4,\cdots,2\left\lfloor\frac{n+1}{2}\right\rfloor,\\ \int_{0}^{\infty}K_{0}(xt)I_{0}^{j-r-1}(t)K_{0}^{n-j+r+1}(t)t^{i-1}\,{\mathrm{d}}t,&\left\lfloor\frac{n+1}{2}\right\rfloor<j\leq n+1,i=1,3,\cdots,2\left\lfloor\frac{n}{2}\right\rfloor+1,\\ \int_{0}^{\infty}tK_{0}^{\prime}(xt)I_{0}^{j-r-1}(t)K_{0}^{n-j+r+1}(t)t^{i-2}\,{\mathrm{d}}t,&\left\lfloor\frac{n+1}{2}\right\rfloor<j\leq n+1,i=2,4,\cdots,2\left\lfloor\frac{n+1}{2}\right\rfloor.\end{cases}

Properties of $I_{0}(t)$ and $K_{0}(t)$

We collect some properties of the modified Bessel functions $I_{0}(t)$ and $K_{0}(t)$ in order to obtain information of $\omega_{n+1}(x)$ as $x\to 0^{+},1^{-}$ .

The function $I_{0}(t)$ is entire and even; it is real and increasing on the half line $[1,\infty)$ . The function $K_{0}(t)$ has a logarithmic pole at $x=0$ ; it is real and decreasing on $(0,\infty)$ . On the half plane $\Re(t)>0$ , we have the asymptotic approximations

I_{0}(t)=\frac{e^{t}}{\sqrt{2\pi t}}\left(1+O\Big{(}\frac{1}{t}\Big{)}\right),\quad K_{0}(t)=\sqrt{\frac{\pi}{2t}}e^{-t}\left(1+O\Big{(}\frac{1}{t}\Big{)}\right)

as $t\to\infty$ . In particular, for a positive integer $a$ ,

[I_{0}(t)K_{0}(t)]^{a}-\frac{1}{(2t)^{a}}=O\Big{(}\frac{1}{t^{a+1}}\Big{)}

as $t\to\infty$ along the real line. One has the boundedness

\sup_{t>0}\frac{|tK^{\prime}_{0}(t)+1|}{t(1+|\log t|)}<\infty.

For $c\in\mathbb{Z}_{\geq 0}$ , one has the evaluation [13, §13.21, Eq.(8)]

\int_{0}^{\infty}K_{0}(t)t^{c}\,{\mathrm{d}}t=2^{c-1}\Gamma\Big{(}\frac{c+1}{2}\Big{)}^{2}.

Integrations

With the data collected above, we list some consequences for the integrals that appear in the entries of the matrix $\Omega_{n+1}(x)$ .

For $0\leq a<b$ and $c\geq 0$ , one obtains

	$\displaystyle\int_{0}^{\infty}K_{0}(xt)I_{0}^{a}(t)K_{0}^{b}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=O(\log x),$		(31)
	$\displaystyle\int_{0}^{\infty}I_{0}^{\prime}(xt)I_{0}^{a}(t)K_{0}^{b}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=O(x)$		(32)

and

	$\displaystyle\int_{0}^{\infty}tK^{\prime}_{0}(xt)I_{0}^{a}(t)K_{0}^{b}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=\frac{-1}{x}\left[\int_{0}^{\infty}I_{0}^{a}(t)K_{0}^{b}(t)t^{c}\,{\mathrm{d}}t-\int_{0}^{\infty}\big{(}xtK_{0}^{\prime}(xt)+1\big{)}I_{0}^{a}(t)K_{0}^{b}(t)t^{c}\,{\mathrm{d}}t\right]$
		$\displaystyle=\frac{-1}{x}\int_{0}^{\infty}I_{0}^{a}(t)K_{0}^{b}(t)t^{c}\,{\mathrm{d}}t+O(\log x)$		(33)

as $x\to 0^{+}$ . For $0\leq c\leq a$ and as $x\to 0^{+}$ , we thus have

	$\displaystyle\int_{0}^{\infty}K_{0}(xt)I_{0}^{a}(t)K_{0}^{a}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=O\Big{(}\int_{0}^{\infty}K_{0}(xt)\,{\mathrm{d}}t\Big{)}=O\Big{(}\frac{1}{x}\Big{)},$		(34)
	$\displaystyle\int_{0}^{\infty}tK_{0}^{\prime}(xt)I_{0}^{a}(t)K_{0}^{a}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=O\Big{(}\int_{0}^{\infty}tK_{0}^{\prime}(xt)\,{\mathrm{d}}t\Big{)}=O\Big{(}\frac{1}{x^{2}}\Big{)}.$		(35)

If $0\leq a<c$ and as $x\to 0^{+}$ , then

$\displaystyle\int_{0}^{\infty}K_{0}(xt)I_{0}^{a}(t)K_{0}^{a}(t)t^{c}{\mathrm{d}}t$	$\displaystyle=\int_{0}^{\infty}\frac{K_{0}(xt)t^{c-a}}{2^{a}}{\mathrm{d}}t+\int_{0}^{\infty}K_{0}(xt)\Big{[}I_{0}^{a}(t)K_{0}^{a}(t)-\frac{1}{(2t)^{a}}\Big{]}t^{c}{\mathrm{d}}t$
	$\displaystyle=\frac{2^{c-2a-1}}{x^{c-a+1}}\Gamma\Big{(}\frac{c-a+1}{2}\Big{)}^{2}+O\Big{(}\frac{1}{x^{c-a}}\Big{)},$	(36)
$\displaystyle\int_{0}^{\infty}tK_{0}^{\prime}(xt)I_{0}^{a}(t)K_{0}^{a}(t)t^{c}{\mathrm{d}}t$	$\displaystyle=\int_{0}^{\infty}\frac{tK_{0}^{\prime}(xt)t^{c-a}}{2^{a}}{\mathrm{d}}t+\int_{0}^{\infty}tK_{0}^{\prime}(xt)\Big{[}I_{0}^{a}(t)K_{0}^{a}(t)-\frac{1}{(2t)^{a}}\Big{]}t^{c}{\mathrm{d}}t$
	$\displaystyle=\frac{c-a+1}{2^{a}x}\int_{0}^{\infty}K_{0}(xt)t^{c-a}\,{\mathrm{d}}t+O\Big{(}\frac{1}{x^{2}}\Big{)}$
	$\displaystyle=O\Big{(}\frac{1}{x^{c-a+2}}\Big{)}.$	(37)

On the real line, we have [9, Lemma 4.5]

\lim_{x\to 1^{-}}\frac{\int_{0}^{\infty}I_{0}(xt)K_{0}(t)\,{\mathrm{d}}t}{-\log(1-x)}=\frac{1}{2}

and for $c\in\mathbb{Z}_{\geq 0}$ ,

\lim_{x\to 1^{-}}(1-x)^{c+1}\int_{0}^{\infty}I_{0}(xt)K_{0}(t)t^{c+1}\,{\mathrm{d}}t=\frac{c!}{2}=\lim_{x\to 1^{-}}(1-x)^{c+1}\int_{0}^{\infty}tI_{0}^{\prime}(xt)K_{0}(t)t^{c}\,{\mathrm{d}}t.

Therefore for $a\geq 1,a>c$ and $x\to 1^{-}$ , one has

$\displaystyle\int_{0}^{\infty}I_{0}(xt)I_{0}^{a-1}(t)K_{0}^{a}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=\int_{0}^{\infty}I_{0}(xt)K_{0}(t)\big{[}I_{0}^{a-1}(t)K_{0}^{a-1}(t)t^{c}\big{]}\,{\mathrm{d}}t$
	$\displaystyle=O\Big{(}\int_{0}^{\infty}I_{0}(xt)K_{0}(t)\,{\mathrm{d}}t\Big{)}$
	$\displaystyle=O\big{(}\log(1-x)\big{)},$	(38)
$\displaystyle\int_{0}^{\infty}tI_{0}^{\prime}(xt)I_{0}^{a-1}(t)K_{0}^{a}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle=\int_{0}^{\infty}tI_{0}^{\prime}(xt)K_{0}(t)\big{[}I_{0}^{a-1}(t)K_{0}^{a-1}(t)t^{c}\big{]}\,{\mathrm{d}}t$
	$\displaystyle=O\Big{(}\int_{0}^{\infty}tI^{\prime}_{0}(xt)K_{0}(t)\,{\mathrm{d}}t\Big{)}$
	$\displaystyle=O\Big{(}\frac{1}{1-x}\Big{)}.$	(39)

If $c\geq a\geq 1$ and $x\to 1^{-}$ , then

$\displaystyle\int_{0}^{\infty}I_{0}(xt)I_{0}^{a-1}(t)K_{0}^{a}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle\begin{aligned} =&\int_{0}^{\infty}\frac{I_{0}(xt)K_{0}(t)t^{c-a+1}}{2^{a-1}}\,{\mathrm{d}}t\\ &\hskip 20.0pt+\int_{0}^{\infty}I_{0}(xt)K_{0}(t)\Big{[}I_{0}^{a-1}(t)K_{0}^{a-1}(t)-\frac{1}{(2t)^{a-1}}\Big{]}t^{c}\,{\mathrm{d}}t\end{aligned}$
	$\displaystyle=\frac{(c-a)!}{2^{a}(1-x)^{c-a+1}}+o\Big{(}\frac{1}{(1-x)^{c-a+1}}\Big{)},$	(40)
$\displaystyle\int_{0}^{\infty}tI^{\prime}_{0}(xt)I_{0}^{a-1}(t)K_{0}^{a}(t)t^{c}\,{\mathrm{d}}t$	$\displaystyle\begin{aligned} =&\int_{0}^{\infty}\frac{tI^{\prime}_{0}(xt)K_{0}(t)t^{c-a+1}}{2^{a-1}}\,{\mathrm{d}}t\\ &\hskip 20.0pt+\int_{0}^{\infty}tI^{\prime}_{0}(xt)K_{0}(t)\Big{[}I_{0}^{a-1}(t)K_{0}^{a-1}(t)-\frac{1}{(2t)^{a-1}}\Big{]}t^{c}\,{\mathrm{d}}t\end{aligned}$
	$\displaystyle=\frac{(c-a+1)!}{2^{a}(1-x)^{c-a+2}}+o\Big{(}\frac{1}{(1-x)^{c-a+2}}\Big{)}.$	(41)

Notice that the error terms in the above two formulas are of class small $o$ ; it is needed in the investigation of the limit of $\omega_{2r}$ as $x\to 1^{-}$ below.

Evaluation of $\omega_{2r+1}(x)$ at $x=1$

All entries of $\Omega_{2r+1}(x)$ can be evaluated at $x=1$ . We move the $2i$ -th row to row $i$ in $\Omega_{2r+1}(1)$ for $1\leq i\leq r$ and then subtract the $(r+j+1)$ -st column from the $j$ -th column of the resulting matrix for $1\leq j\leq r$ . By (3) on the upper-left block, we obtain

\omega_{2r+1}(1)=(-1)^{\frac{r(r+1)}{2}}\det\begin{pmatrix}M_{r}&*\\ 0&M_{r+1}\end{pmatrix}.

Behavior of $\omega_{2r+1}(x)$ as $x\to 0^{+}$

Fix $r\geq 1$ . We move row $(2i-1)$ of $\Omega_{2r+1}(x)$ to row $i$ for $1\leq i\leq r$ , which creates a sign $(-1)^{r(r-1)/2}$ to the determinant $\omega_{2r+1}(x)$ . As $x\to 0^{+}$ , the resulting matrix decomposes into $(r,r,1)\times(r,r,1)$ blocks of the form

\begin{pmatrix}N_{r}+o(1)&O(\log x)&O\big{(}\frac{1}{x^{r}}\big{)}\\ O(x)&\frac{-1}{x}N_{r}+O(\log x)&O\big{(}\frac{1}{x^{r}}\big{)}\\ O(1)&O(\log x)&\frac{1}{2x^{r+1}}\Gamma\big{(}\frac{r+1}{2}\big{)}^{2}+O\big{(}\frac{1}{x^{r}}\big{)}\end{pmatrix}

by direct evaluation and (32) in the left three blocks, (31) and (33) in the middle, and (34), (35), (36) and (37) in the last column. The leading term of $\omega_{2r+1}(x)$ , which is of order $x^{-(2r+1)}$ , comes from the diagonal blocks and one gets

\lim_{x\to 0^{+}}x^{2r+1}\omega_{2r+1}(x)=(-1)^{\frac{r(r+1)}{2}}\frac{1}{2}\Gamma\Big{(}\frac{r+1}{2}\Big{)}^{2}{\det}^{2}N_{r}.

Behavior of $\omega_{2r}(x)$ as $x\to 1^{-}$

Fix $r\geq 2$ . We move $2i$ -th row of $\Omega_{2r}(x)$ to row $i$ for $i=1,2,\cdots,(r-1)$ and $r$ -th column to the last, which adds a sign $(-1)^{r(r+1)/2}$ to the determinant $\omega_{2r}(x)$ . We subtract $j$ -th column by $(r+j)$ -th for $j=1,2,\cdots,(r-1)$ . As $x\to 1^{-}$ , the resulting matrix decomposes into $(r-1,r,1)\times(r-1,r,1)$ blocks of the form

\begin{pmatrix}N_{r-1}+o(1)&O(1)&O\big{(}\frac{1}{(1-x)^{r-1}}\big{)}\\ 0&N_{r}+o(1)&O\big{(}\frac{1}{(1-x)^{r-1}}\big{)}\\ O(1)&O(1)&\frac{(r-1)!}{2^{r}(1-x)^{r}}+o\big{(}\frac{1}{(1-x)^{r}}\big{)}\end{pmatrix}

by (3) and direct evaluation in the left three blocks, direct evaluation in the middle, and (38), (39), (40) and (41) in the last column. The leading term of $\omega_{2r}(x)$ , which is of order $(1-x)^{-r}$ , comes from the diagonal blocks. It yields

\lim_{x\to 1^{-}}(1-x)^{r}\omega_{2r}(x)=(-1)^{\frac{r(r+1)}{2}}\frac{(r-1)!}{2^{r}}\det N_{r-1}\det N_{r}.

Behavior of $\omega_{2r}(x)$ as $x\to 0^{+}$

Fix $r\geq 2$ . We move row $(2i-1)$ of $\Omega_{2r}(x)$ to row $i$ for $1\leq i\leq r$ , which adds a sign $(-1)^{r(r-1)/2}$ to the determinant $\omega_{2r}(x)$ . As $x\to 0^{+}$ , the resulting matrix decomposes into four blocks of equal size of the form

\begin{pmatrix}M_{r}+o(1)&O(\log x)\\ O(x)&\frac{-1}{x}M_{r}+O(\log x)\end{pmatrix}

by direct evaluation and (32) in the left two blocks and (31) and (33) in the right. This leads to

\lim_{x\to 0^{+}}x^{r}\omega_{2r}(x)=(-1)^{\frac{r(r+1)}{2}}{\det}^{2}M_{r}.

Remark 41.

Proposition 38 indeed holds for $\omega_{2}(x)$ by the same analysis if we set $\det N_{0}=1$ ; it is consistent with the relation (29) for $r=1$ .

References

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	$\displaystyle\left\|\int_{\varepsilon\sigma_{0}^{p}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z\right\|$	$\displaystyle\leq\int_{\varepsilon\sigma_{0}^{p}}\left\|z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\right\|\left\|\mathrm{d}z\right\|$
		$\displaystyle\leq\int_{0}^{2\pi p}\varepsilon^{j-1}\sqrt{\varepsilon}\left\|I_{0}\left(2\sqrt{\varepsilon e^{i\theta}}\right)^{b}K_{0}\left(2\sqrt{\varepsilon e^{i\theta}}\right)^{k-b}\right\|\varepsilon{\mathrm{d}}\theta$
		$\displaystyle\leq\varepsilon^{j}\sqrt{\varepsilon}\int_{0}^{2\pi p}\left\|\gamma+\log\sqrt{\varepsilon e^{i\theta}}\right\|^{k-b}{\mathrm{d}}\theta$
		$\displaystyle=\varepsilon^{j}\sqrt{\varepsilon}\int_{0}^{2\pi p}\left\|\gamma+\log\sqrt{\varepsilon}+\frac{1}{2}i\theta\right\|^{k-b}{\mathrm{d}}\theta\rightarrow 0.\qed$

On the Periods of Twisted Moments of the Kloosterman Connection

Abstract

keywords:

pacs:

Statements and Declarations

1 Introduction

1.1 Historical results and our results

Sum rule identities

Conjecture 1.

Formula 2.

Formula 3 (Corollary 28).

ℚ\mathbb{Q}-dimension of Bessel moments

Theorem 4 (Corollary 30).

Quadratic relations of Bessel moments

Determinants of Bessel moment matrix

Formula 5 (Corollary 39).

1.2 Approach

2 The Kloosterman connection and its twisted symmetric powers

2.1 Self-duality and pairing on Kl2\mathrm{Kl}_{2}

2.2 Rational structures and pairings on (𝒪𝔾m,d+12​d​zz)(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z}).

2.3 Algebraic and topological pairings on z​Symk⁡Kl2\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}

3 The de Rham cohomology

3.1 Dimension of HdR1​(𝔾m,z​Symk⁡Kl2)H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}

Proposition 6.

Proof.

Lemma 7.

Remark 8.

3.2 Compactly supported de Rham cohomology

Lemma 9.

Proof.

Definition 10.

Definition 11.

Remark 12.

3.3 Poincaré pairing

Proposition 13.

Proof.

Corollary 14 (Bases in de Rham side).

Proof.

Fact 15.

4 The local system and the associated homology

4.1 Rapid decay cycles

Lemma 16.

Proof.

4.2 Moderate decay cycles

Lemma 17.

Proof.

4.3 Betti intersection pairing

Proposition 18.

Proof.

Proposition 19.

Proof.

Corollary 20.

Proof.

5 Twisted moments as periods

5.1 Bessel moments and regularized Bessel moments

Lemma 21.

Proof.

Remark 22.

5.2 Period pairing and compactly supported period pairing

Remark 23.

Proposition 24.

Proof.

Proposition 25.

Proof.

Corollary 26.

Remark 27 (determinant of the period matrix).

5.3 ℚ\mathbb{Q}-linear and quadratic relations on Bessel moments

Corollary 28.

Proof.

Remark 29.

Corollary 30.

Proof.

Remark 31.

Proposition 32.

Proof.

Remark 33.

Proposition 34.

Proof.

Appendix A The Bessel operator and determinants of Bessel moments

A.1 Symmetric power of the modified Bessel differential operator

$\mathbb{Q}$ -dimension of Bessel moments

2.1 Self-duality and pairing on $\mathrm{Kl}_{2}$

2.2 Rational structures and pairings on $(\mathcal{O}_{\mathbb{G}_{\mathrm{m}}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ .

2.3 Algebraic and topological pairings on $\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}$

3.1 Dimension of $H^{1}_{\mathrm{dR}}\big{(}\mathbb{G}_{\mathrm{m}},\sqrt{z}\operatorname{Sym}^{k}{\mathrm{Kl}}_{2}\big{)}$

5.3 $\mathbb{Q}$ -linear and quadratic relations on Bessel moments

Rationality of $\omega_{n+1}(x)$

A.4 Singularities of $\omega_{n+1}(x)$

Properties of $I_{0}(t)$ and $K_{0}(t)$

Evaluation of $\omega_{2r+1}(x)$ at $x=1$

Behavior of $\omega_{2r+1}(x)$ as $x\to 0^{+}$

Behavior of $\omega_{2r}(x)$ as $x\to 1^{-}$

Behavior of $\omega_{2r}(x)$ as $x\to 0^{+}$