Counting Non-abelian Coverings of Algebraic Curve

Let $X$ be a smooth projective variety, and assume that there is a free action of $\mathbb{Z}/n\mathbb{Z}$ on $X$. Then the quotient $Y=X/(\mathbb{Z}/n\mathbb{Z})$ is also smooth and projective. Thus, $\pi:X\to Y$ is an etale cyclic covering of degree $n$.

The pushforward $\pi_{*}\mathcal{O}_{X}$ has a structure of $\mathcal{O}_{Y}$ algebra. Under the action of $\mathbb{Z}/n\mathbb{Z}$, $\pi_{*}\mathcal{O}_{X}$ decomposes into irreducible representations of $\mathbb{Z}/n\mathbb{Z}$. Namely, we have:

where $L\in Pic(Y)$ is a $n$-torsion line bundle, which means $\mathcal{L}^{\otimes n}\cong\mathcal{O}_{Y}$.

Conversely, take a $n$-torsion line bundle $\mathcal{L}$ over $Y$, we can construct such an $X$ explicitly: fixing an isomorphism $\mathcal{L}^{\otimes n}\cong\mathcal{O}_{Y}$, we can endow the sheaf $\bigoplus_{i=0}^{n-1}\mathcal{L}^{-i}$ with the structure of a $\mathcal{O}_{Y}$-algebra. Let $X=Spec(\bigoplus_{i=0}^{n-1}\mathcal{L}^{-i})$, then the $\mathcal{O}_{Y}$-structure map $\mathcal{O}_{Y}\to\bigoplus_{i=0}^{n-1}\mathcal{L}^{-i}$ gives a surjection $X\to Y$, which is an etale cyclic $n$-covering. Note that under this construction, in order to guarantee $X$ to be connected, we require $n$ to be minimum: i.e. $n=ord(\mathcal{L})$ in the group $Pic(Y)$. Moreover, if we replace $\mathcal{L}$ by any $\mathcal{L}^{k}$ with $gcd(k,n)=1$, the vairety $X^{\prime}$ we obtain is isomorphic to $X$ constructed above, because $\bigoplus_{i=0}^{n-1}\mathcal{L}^{-i}\cong\bigoplus_{i=0}^{n-1}\mathcal{L}^{-ki}$ as $\mathcal{O}_{Y}$-algebra. Therefore, if one wants to count $X$ up to isomorphism, such equvilance $\mathcal{L}\sim\mathcal{L}^{k}$ needs to be modulo out.

More geometrically, suppose we have a $n$-torsion line bundle $\mathcal{L}$ on $Y$, and the constant section $1\in\Gamma(Y,\mathcal{O}_{Y})$. We denote $\mathbb{L}$ the total space of $\mathcal{L}$, and we let $p:\mathbb{L}\to Y$ be the bundle projection. If $t\in\Gamma(\mathbb{L},p^{*}\mathcal{L})$ is the tautological section, the zero divisor of $p^{*}1-t^{n}$ defines $X$ in $\mathbb{L}$. $\pi=p|_{X}$ exhibits $X$ as an etale cyclic $n$-covering of $Y$. For such $X$, we can also verify that $\pi_{*}\mathcal{O}_{X}\cong\bigoplus_{i=0}^{n-1}\mathcal{L}^{-i}$. Under such description, let $K=\left\langle\sigma\right\rangle\cong\mathbb{Z}/n\mathbb{Z}$, and let $\chi$ denote the generating character of $K$, i.e. $\chi(\sigma)=\zeta_{n}$, a primitive $n$-th root of unity. Then $\mathcal{L}^{-i}$ is the eigenspace of $\chi^{i}$.

The above argument exhibits a bijection between $n$-torsion points in $Pic(Y)$ and etale cyclic $n$-covering of $Y$. In the case of a smooth projective curve $C$ of genus $g\geq 2$, $n$-torsion points in $Pic^{0}(C)\subset Pic(C)$ is just $J_{C}[n]$, where $J_{C}$ is the Jacobian of $C$. Over $\mathbb{C}$, we know that $J_{C}[n]\cong(\mathbb{Z}/n\mathbb{Z})^{\oplus 2g}$. If we denote $S\subset(\mathbb{Z}/n\mathbb{Z})^{\oplus 2g}$ as the primitive elements in $(\mathbb{Z}/n\mathbb{Z})^{\oplus 2g}$: i.e. $S=\{v\in(\mathbb{Z}/n\mathbb{Z})^{\oplus 2g},ord(v)=n\}$, then up to isomorphisms, the connected etale cyclic $n$-covering of $C$ is bijective to $S/(\mathbb{Z}/n\mathbb{Z})^{\times}$.

The idea of the first method is using similarity between matrices. To begin with, recall that the Jacobian $J_{D}=H^{0}(D,\omega_{D})^{*}/H_{1}(D,\mathbb{Z})$. Since $\sigma$ is an automorphism of $J_{D}$, it is also an isomorphism of the lattice $H_{1}(D,\mathbb{Z})\cong\mathbb{Z}^{\oplus 2h}$. In this way, $\sigma$ becomes an element in $GL_{2h}(\mathbb{Z})$, satisfying $\sigma^{n}=id$. Since we know the eigenspace decomposition of $H^{0}(D,\omega_{D})$ under the action of $\sigma$, we can easily deduce what matrix is $\sigma$ similar to over $\mathbb{C}$. To see this, the complexification of $\sigma$, namely $\sigma\otimes\mathbb{C}$, acts on $H_{1}(D,\mathbb{Z})\otimes\mathbb{C}$, which is $H_{1}(D,\mathbb{C})$. Since $H^{1}(D,\mathbb{C})=H^{1,0}(D)\oplus H^{0,1}(D)$, taking dual, we obtain $H_{1}(D,\mathbb{C})\cong H^{0}(D,\omega_{D})\oplus H^{0}(D,\omega_{D})^{*}$. Further recall that: $H^{0}(D,\omega_{D})\cong H^{0}(C,\omega_{C})\oplus\bigoplus_{i=1}^{n-1}H^{0}(C,\omega_{C}\otimes\mathcal{L}^{-i})$ gives the eigenspace decomposition of the action of $\sigma$, we obtain:

If we denote $J=diag(\zeta,\zeta^{2},\cdots,\zeta^{n-1})$, then $\sigma$ is similar to the matrix $I_{2g}\oplus(J\otimes I_{2g-2})$ over $\mathbb{C}$.

Define the $(n-1)\times(n-1)$ matrix $G=$: $\begin{pmatrix}0&\cdots&0&-1\\ 1&\cdots&0&-1\\ \vdots&\ddots&\vdots&\vdots\\ 0&\cdots&1&-1\end{pmatrix}$ It is easy to see $G$ is similar to $J$ over $\mathbb{C}$. Thus, $\sigma$ is similar to $I_{2g}\oplus(G\otimes I_{2g-2})$ over $\mathbb{C}$. Note that both $\sigma$ and $I_{2g}\oplus(G\otimes I_{2g-2})$ over $\mathbb{C}$ is defined over $\mathbb{Z}$, thus over $\mathbb{Q}$, we deduce that $\sigma$ and $I_{2g}\oplus(G\otimes I_{2g-2})$ is similar over $\mathbb{Q}$. The difficulty here we cannot obtain similarity over $\mathbb{Z}$. To make the argument work, we must require $gcd(m,n)=1$ in the following.

The goal is to obtain the similarity type of $\sigma$ after reduction modulo $m$. Once we get that information, we will know how $\sigma$ acts on $\frac{1}{m}H_{1}(D,\mathbb{Z})/H_{1}(D,\mathbb{Z})$, which is just the $m$-torsion points in $J_{D}$. We achieve our goal in several steps:

Step 1:
Take any prime $p|m$, consider the similarity type of $\sigma(p):=\sigma$ mod $p$. Here we view $\sigma(p)$ as an element in $GL_{2h}(\mathbb{F}_{p})$. Because $\sigma$ is similar to $I_{2g}\oplus(G\otimes I_{2g-2})$ over $\mathbb{Q}$, they have the same characteristic polynomial and minimal polynomial over $\mathbb{Q}$. Therefore, the characteristic polynomial of $\sigma$ equals to $(x-1)^{2g}(1+x+\cdots+x^{n-1})^{2g-2}$, and the minimal polynomial of $\sigma$ equals to $x^{n}-1$. The characteristic polynomial of $\sigma(p)$ is just $(x-1)^{2g}(1+x+\cdots+x^{n-1})^{2g-2}$ mod $p$, and the minimal polynomial of $\sigma(p)$, say $q(x)$, must dividing $x^{n}-1$ mod $p$. Since $gcd(p,n)=1$, $x^{n}-1$ has distinct roots over $\overline{\mathbb{F}}_{p}$. Assume $q(x)\neq x^{n}-1$ mod $p$, then some factors of $x^{n}-1$ must be left out by $q(x)$. As a result, such factors cannot appear in the characteristic polynomial $(x-1)^{2g}(1+x+\cdots+x^{n-1})^{2g-2}$ mod $p$. However, $(x-1)^{2g}(1+x+\cdots+x^{n-1})^{2g-2}$ mod $p$ has the same factor with $x^{n}-1$ mod $p$, leading to a contradiction. Thus, $q(x)=x^{n}-1$ mod $p$.

Therefore, $\sigma(p)$ and $I_{2g}\oplus(G\otimes I_{2g-2})$ mod $p$ have the same characteristic and minimal polynomial. Note that the minimal polynomial $x^{n}-1$ does not have multiple root over $\overline{\mathbb{F}}_{p}$, they are both diagonalizable over $\overline{\mathbb{F}}_{p}$. Same characteristic polynomial gives the same multiplicity of eigenvalues. Thus, over $\overline{\mathbb{F}}_{p}$, $\sigma(p)$ is similar to $I_{2g}\oplus(G\otimes I_{2g-2})$. Notice both matrices are defined over $\mathbb{F}_{p}$, they are actually similar over $\mathbb{F}_{p}$.

Step 2:
In order to lift the similarity relation from over $\mathbb{F}_{p}$ to over $\mathbb{Z}/m\mathbb{Z}$, we need some results of J.Pomfret [Pom73]:

Lemma 4.2: Let $R$ be a finite local ring with maximal ideal $M$, and $R/M=\mathbb{F}_{p^{f}}=k$. Let $\alpha$, $\beta$ be elements of $GL_{n}(R)$ with $gcd(|\left\langle\alpha\right\rangle|,p)=gcd(|\left\langle\beta\right\rangle|,p)=1$ (where $|\left\langle x\right\rangle|$ means the order of $x$). Then $\alpha$ is similar to $\beta$ if and only if $\alpha$ is similar to $\beta$ modulo $M$.

If $R$ is a finite commutative ring with identity, then $R$ is uniquely isomorphic to a product of finite local rings: $R\cong\prod_{i=1}^{t}R_{i}$. Suppose $R_{i}$ is a finite local ring with maximal ideal $M_{i}$, and $R_{i}/M_{i}=k_{i}$, we have the epimorphisms:

Combining lemma 4.2 and (4.11), we get:

Lemma 4.3: Let $R$ be a finite commutative ring with identity and let the cardinality of $R$ be $m$. Two elements $\alpha$ and $\beta$ of $GL_{n}(R)$ satisfying $gcd(|\left\langle\alpha\right\rangle|,m)=gcd(|\left\langle\beta\right\rangle|,m)=1$ are similar if and only if they are similar over each residue field $k_{i}$.

In step 1, we see $\sigma(p)$ and $I_{2g}\oplus(G\otimes I_{2g-2})$ are similar over $\mathbb{F}_{p}$ for any $p|m$. Also note that $\mathbb{F}_{p}$’s are all the residue fields of $\mathbb{Z}/m\mathbb{Z}$. Applying lemma 4.3 to our case: $R=\mathbb{Z}/m\mathbb{Z}$, $\alpha=\sigma(m)$ ($=\sigma$ mod $m$), and $\beta=I_{2g}\oplus(G\otimes I_{2g-2})$, we see that $\sigma(m)$ is similar to $I_{2g}\oplus(G\otimes I_{2g-2})$ over $\mathbb{Z}/m\mathbb{Z}$.

Step 3:
Denote $A=G\otimes I_{2g-2}$, viewed as an element in $GL_{(2g-2)(n-1)}(\mathbb{Z}/m\mathbb{Z})$. The part $I_{2g}$ corresponds to the eigenspace with eigenvalue $1$. Since we only want the eigenspace with eigenvalue $\lambda\neq 1$, it is equivalent to count the eigen-directions that $A$ have acting on $(\mathbb{Z}/m\mathbb{Z})^{\oplus(2g-2)(n-1)}$.

We need to understand $(\mathbb{Z}/m\mathbb{Z})^{\oplus(2g-2)(n-1)}$ as a $(\mathbb{Z}/m\mathbb{Z})[A]$ module. Since $A=G\otimes I_{2g-2}$, we only need to study the $(\mathbb{Z}/m\mathbb{Z})[G]$-mod structure on $(\mathbb{Z}/m\mathbb{Z})^{\oplus(n-1)}$, because as a $(\mathbb{Z}/m\mathbb{Z})[A]$ module, $(\mathbb{Z}/m\mathbb{Z})^{\oplus(2g-2)(n-1)}$ is just isomorphic to the direct sum of $(2g-2)$-many copies of $(\mathbb{Z}/m\mathbb{Z})^{\oplus(n-1)}$, with each summand carrying the same $(\mathbb{Z}/m\mathbb{Z})[G]$-mod structure. The minimal polynomial and the characteristic polynomial of $G$ both equals to $f(x)=1+x+x^{2}+\cdots+x^{n-1}$. Therefore, as a $(\mathbb{Z}/m\mathbb{Z})[G]$-mod, $(\mathbb{Z}/m\mathbb{Z})^{\oplus(n-1)}$ is isomorphic to $(\mathbb{Z}/m\mathbb{Z})[G]$.

Now we compute $(\mathbb{Z}/m\mathbb{Z})[G]$. We have the decomposition:

For a set of eigenvectors: $\{v_{p}\in(\mathbb{Z}/p^{e}\mathbb{Z})^{\oplus(n-1)}\}_{p|m}$, satisfying $Gv_{p}=\lambda_{p}v_{p}$ mod $p^{e}$, we can find a unique vector $v\in(\mathbb{Z}/m\mathbb{Z})^{\oplus(n-1)}$, satifying $Gv=\lambda v$, and $v\equiv v_{p}$ mod $p^{e}$, $\lambda\equiv\lambda_{p}$ mod $p^{e}$. In this way, we reduce the computation to $(\mathbb{Z}/p^{e}\mathbb{Z})[G]$.

Note that $(\mathbb{Z}/p^{e}\mathbb{Z})[G]\cong(\mathbb{Z}/p^{e}\mathbb{Z})[x]/f(x)$, and $f(x)=\prod_{d|n,d\neq 1}\Phi_{d}(x)$. It suffices for us to consider the behavior of decomposition of each $\Phi_{d}(x)$ mod $p^{e}$. We have the following lemma:

Lemma 4.4: For $d\nmid p-1$, $\Phi_{d}(x)$ has no degree one factor modulo $p^{e}$. For $d\mid p-1$, $\Phi_{d}(x)$ decompose completely into $\phi(d)$-many degree one factors modulo $p^{e}$.

Proof: Passing to the residue field of $\mathbb{Z}/p^{e}\mathbb{Z}$, which is just $\mathbb{F}_{p}$, we first the decomposition of $\Phi_{d}(x)$ in $\mathbb{F}_{p}[x]$. If $\Phi_{d}(x)$ mod $p^{e}$ has any linear term $(x-\lambda)$, then so does $\Phi_{d}(x)$ mod $p$, because $\lambda\in(\mathbb{Z}/p^{e}\mathbb{Z})^{\times}$ implies $\lambda\neq 0$ mod $p$. This shows that if $\Phi_{d}(x)$ mod $p$ does not have any linear factor, then so does $\Phi_{d}(x)$ mod $p^{e}$. The only case that $\Phi_{d}(x)$ has root over $\mathbb{F}_{p}$ is that: the order of $p$ in $\mathbb{Z}/d\mathbb{Z}$ is exactly one, i.e. $d\mid p-1$. (In this case, primitive $d$-th roots of unity will be fixed under Frobenius, thus lie in $\mathbb{F}_{p}$).
Assuming $d\mid p-1$, then $\Phi_{d}(x)$ splits completely in $\mathbb{F}_{p}[x]$:

Since $gcd(d,p)=1$, roots $a_{i}$ and $a_{j}$ are distinct. By Hensel’s lemma, we can lift the factorization into $\mathbb{Z}/p^{e}\mathbb{Z}$, namely:

with $\tilde{a}_{i}\in(\mathbb{Z}/p^{e}\mathbb{Z})^{\times}$ distinct. The lemma is proven.

With the aid of lemma 4.4, let us count how many linear factors can $f(x)$ factor out in $(\mathbb{Z}/p^{e}\mathbb{Z})[x]$. Only for those $d$, satisfying: $d\mid n$, $d\neq 1$, and $d\mid p-1$, $\Phi_{d}(x)$ will contribute $\phi(d)$-many linear factors. Therefore, the total number of linear factors in $(\mathbb{Z}/p^{e}\mathbb{Z})[x]/f(x)$ equals to:

Tensoring by $I_{2g-2}$, we pass from $(\mathbb{Z}/p^{e}\mathbb{Z})[G]$ to $(\mathbb{Z}/p^{e}\mathbb{Z})[A]$. By doing so, the one dimensional eigenspace with eigenvalue $\tilde{a}_{i}$ (corresponding to the linear factor $(x-\tilde{a}_{i})$) is enlarged into a $(2g-2)$-dimensional eigenspace still with eigenvalue $\tilde{a}_{i}$. Since the roots of $f(x)$ are distinct, all the eigenvectors come from the union of such $(2g-2)$-dimensional eigenspaces. We first count the number of primitive points in those eigenspaces:

Take a $N$ dimensional space, $V=(\mathbb{Z}/p^{e}\mathbb{Z})^{\oplus N}$, the primitive vectors (those elements of order $p^{e}$) in $V$ can be viewed as $(\mathbb{Z}/p^{e}\mathbb{Z})^{\oplus N}-(p\mathbb{Z}/p^{e}\mathbb{Z})^{\oplus N}$. Therefore, the number of primitive vectors in $V$ equals to $p^{eN}(1-p^{-N})$. Furthermore, suppose $\mathbb{Z}/m\mathbb{Z}\cong\prod_{p|m}\mathbb{Z}/p^{e}\mathbb{Z}$, and suppose $X$ is a $N$ dimensional space over $\mathbb{Z}/m\mathbb{Z}$, say $X=(\mathbb{Z}/m\mathbb{Z})^{\oplus N}$, then a vector $v\in X$ is primitive if and only if $v_{p}=v$ mod $p^{e}$ is primitive in $X_{p}=(\mathbb{Z}/p^{e}\mathbb{Z})^{\oplus N}$ for every $p\mid m$.

Combining all the arguments above together, we obtain the number of primitive eigenvector for $A$ in $(\mathbb{Z}/m\mathbb{Z})^{\oplus(2g-2)(n-1)}$ equals to:

Since what we need to count is the number of eigen-directions, we need to modulo thus primitive eigenvectors by scalars in $(\mathbb{Z}/m\mathbb{Z})^{\times}$. Thus, we need to divide (4.16) by $\phi(m)$:

Where $T_{g}(m,n)$ is just the number we are looking for. Reduce (4.17), we obtain:

Summing up, we get the following counting result:

Theorem 4.5: Let $C$ be a smooth projective curve of genus $g\geq 2$, and suppose $m$ and $n$ are coprime integers. For a given etale cyclic $n$-covering $\pi:D\to C$, up to isomorphism, there are $T_{g}(m,n)$-many curves $E$, such that $E\to D$ is an etale cyclic $m$-covering, and the composition $E\to C$ is Galois with non-abelian Galois group $\mathbb{Z}/m\mathbb{Z}\rtimes\mathbb{Z}/n\mathbb{Z}$.

Here we will give an alternative method to count the number of such curves $E$. Recall that the $m$-torsion points in $J_{D}[m]$ are nothing but $\frac{1}{m}H_{1}(D,\mathbb{Z})/H_{1}(D,\mathbb{Z})$. If we can spell out the action of $\sigma$ on $H_{1}(D,\mathbb{Z})$, we will get an explicit matrix representation of $\sigma\in GL_{2h}(\mathbb{Z})$, up to conjugation over $\mathbb{Z}$. That will be enough for us to compute the eigenvectors of $\sigma(m)=\sigma$ mod $m$. The key point is a topological lemma below:

Lemma 4.6: Suppose $\Sigma_{g}$ is a closed topological surface of genus $g\geq 2$, and $\Sigma_{h}$, where $h=n(g-1)+1$, is a normal covering surface of $\Sigma_{g}$: $\Sigma_{h}\to\Sigma_{g}$, with group of deck transformation $\mathbb{Z}/n\mathbb{Z}$. Then such covering is equivalent to the ’standard’ cyclic covering between surfaces, as illustrated by Figure 1 [MER03] and Figure 2 [LM18].

We first need to translate the word standard into more concrete definition. Let $\pi=\pi_{1}(\Sigma_{g})$ be the fundamental group of genus $g$ closed surface, and let $a_{1},b_{1},\cdots,a_{g},b_{g}$ be a basis of $\pi$ satisfying $\prod_{i=1}^{g}[a_{i},b_{i}]=e$. Then the standard $\mathbb{Z}/n\mathbb{Z}$-covering corresponds to the epimorphism $\pi\to\mathbb{Z}/n\mathbb{Z}\to 1$, sending $a_{1}$ to a generator of $\mathbb{Z}/n\mathbb{Z}$ and $b_{1},a_{2},b_{2},\cdots a_{g},b_{g}$ to $0$. In this case, we have a short exact sequence: