A Microlocal Analysis of the Lévy Generator with Conjugate Points

Fix $u\in C^{\infty}(M)$ and $x\in M$. Write $\mathscr{A}=A_{1}+A_{2}$, where

	$\displaystyle(A_{1}u)(x)$	$\displaystyle=C_{n,\alpha}\operatorname{p{.}v.}\int_{T_{x}M\setminus\{0\}}\chi\left(|v|_{g}^{2}\right)\frac{u(\exp_{x}(v))-u(x)}{|v|_{g}^{n+2\alpha}}\,dT_{x}(v),$
	$\displaystyle(A_{2}u)(x)$	$\displaystyle=C_{n,\alpha}\int_{T_{x}M\setminus\{0\}}\left(1-\chi\left(|v|_{g}^{2}\right)\right)\frac{u(\exp_{x}(v))-u(x)}{|v|_{g}^{n+2\alpha}}\,dT_{x}(v).$

By our choice of $\chi$, the integrand of $A_{1}$ vanishes whenever $|v|_{g}^{2}>r_{\operatorname{inj}}^{2}/2$, so we can make the change of variables $y=\exp_{x}(v)$. Then

$$(A_{1}u)(x)=C_{n,\alpha}\operatorname{p{.}v.}\int_{M}\chi(d_{g}(x,y)^{2})\frac{u(y)-u(x)}{d_{g}(x,y)^{n+2\alpha}}J(x,y)\,dV_{g}(y),$$

where $d_{g}(x,y)$ is the Riemannian distance from $x$ to $y$, $J(x,y)$ is the Jacobian determinant of the map $y\mapsto\exp_{x}^{-1}(y)$, and $dV_{g}$ is the Riemannian density of $(M,g)$. Hence $\mathscr{A}_{2\alpha}:=A_{1}$ is a pseudodifferential operator of order $2\alpha$ on $M$.

For $A_{2}$, use polar coordinates to write

$$(A_{2}u)(x)=C_{n,\alpha}\int_{S_{x}M}\int_{0}^{\infty}\left(1-\chi\left(s^{2}\right)\right)\frac{u(\exp_{x}(sv))-u(x)}{s^{1+2\alpha}}\,ds\,dS_{x}(v),$$

where $dS_{x}$ is the Riemannian density of $S_{x}M$ induced by the metric $g|_{S_{x}M}$. Since $R_{a}$ preserves $C^{\infty}(SM)$ and $\pi_{*}$ integrates over the fibers of $\pi$, we can write

$$(A_{2}u)(x)=(\pi_{*}\circ R_{a}\circ\pi^{*}u)(x)-(\pi_{*}\circ R_{a}\circ\pi^{*}1)(x)u(x).$$

Let $\mathscr{A}_{0}$ be the operator which multiplies by the constant function $-(\pi_{*}\circ R_{a}\circ\pi^{*}1)$. Since $\mathscr{A}_{0}$ is a pseudodifferential operator of order $0$ on $M$, this completes the proof. ∎

As noted above, we can use a partition of unity to define the composition of $p_{*}$ and $(\psi\tilde{a})^{m}\circ\Phi^{*}$. Then by localizing and reducing to the case of two properly supported Fourier integral operators, the proof boils down to an analysis of the canonical relations.

Since $C_{\Phi^{*}}$ is the canonical relation of $(\psi\tilde{a})^{m}\circ\Phi^{*}$, Lemma 2.4 implies that

$$C_{R_{a,\psi}}=C_{p_{*}}\circ C_{\Phi^{*}}.$$

Therefore, by the clean intersection calculus, it is enough to prove the following:

1. (i)

   The intersection

   (2)

   $$C:=(C_{p_{*}}\times C_{\Phi^{*}})\cap\big{(}T^{*}SM\times\Delta\big{(}T^{*}(SM\times\mathbb{R})\big{)}\times T^{*}SM\big{)}$$

   is clean in the sense that $C$ is an embedded submanifold, and at every point $c\in C$ the tangent space $T_{c}C$ equals the intersection of the tangent spaces of the two manifolds being intersected.

2. (ii)

   The projection map $\pi_{C}:C\to T^{*}SM\times T^{*}SM$ is proper.

3. (iii)

   For every $(\xi,\tilde{\xi})\in T^{*}SM\times T^{*}SM$, the fiber $\pi_{C}^{-1}(\xi,\tilde{\xi})$ is connected.

If we are willing to work with local canonical relations, then point (iii) can be omitted. Point (iii) will hold in this case, but in later results it may not.

Consider the map

$$G:T^{*}SM\times\mathbb{R}\to T^{*}(SM\times\mathbb{R})$$

defined by

$$G(\xi,s)=dp|_{(\pi_{T^{*}SM}(\xi),s)}^{t}\,\xi.$$

If we include the base points, then

(3)

$$G(\pi_{T^{*}SM}(\xi),\xi,s)=\big{(}(\pi_{T^{*}SM}(\xi),s),(\xi,0)\big{)},$$

so $G$ is a smooth embedding. The injectivity of $G$ implies that $\pi_{C}$ is injective. Hence point (iii) holds, and assuming points (i) and (ii) are true the excess of the intersection (2) is zero.

To begin proving point (i), let $Z$ be the smooth rank-$(2n-1)$ subbundle of $T^{*}(SM\times\mathbb{R})$ whose fiber over each point $(v,s)\in SM\times\mathbb{R}$ is

$$Z_{(v,s)}:=\operatorname{Range}\left(d\Phi|_{(v,s)}^{t}\right).$$

In other words, $Z_{(v,s)}$ is the subspace of $T_{(v,s)}^{*}(SM\times\mathbb{R})$ which is conormal to the kernel of $d\Phi|_{(v,s)}$. To see that $Z$ is a smooth subbundle, just use the rank theorem to write

$$\Phi\left(y^{1},\dots,y^{2n}\right)=\left(y^{1},\dots,y^{2n-1}\right)$$

locally, and note that $\left(dy^{1},\dots,dy^{2n-1}\right)$ is a smooth local frame for $Z$. Because $d\Phi|_{(v,s)}^{t}$ is a linear isomorphism from $T_{\Phi(v,s)}^{*}SM$ to $Z_{(v,s)}$, we can define a smooth map

(4)

$$d\Phi^{-t}:Z\to T^{*}SM$$

whose restriction to each fiber $Z_{(v,s)}$ is the inverse $(d\Phi|_{(v,s)}^{t})^{-1}:Z_{(v,s)}\to T_{\Phi(v,s)}^{*}SM$.

The domain of our smooth parametrization of $C$ will be the set

$$\mathcal{O}:=\operatorname{Range}(G)\cap Z.$$

We claim that $\mathcal{O}$ is an embedded submanifold of dimension $4n-2$. Let

(5)

$$q:Z\to\mathbb{R}$$

be the restriction to $Z$ of the projection map

$$T^{*}(SM\times\mathbb{R})\ni\big{(}(v,s),(\xi,\sigma)\big{)}\mapsto\sigma.$$

Since $\mathcal{O}$ equals the level set $q^{-1}(0)$, the claim is true if $dq|_{\zeta}$ is nonzero for all $\zeta\in Z$. Let $\pi_{V}(\zeta)=(v,s)$. Then $\zeta=d\Phi|_{(v,s)}^{t}\,\theta$ for some $\theta\in T_{\Phi(v,s)}^{*}SM$. Choose slice coordinates for $SM$ near $v$ and near $\Phi(v,s)$, and fix natural coordinates for $T^{*}SM$ associated with the latter. Then locally we can write $\Phi=\left(\Phi^{1},\dots,\Phi^{2n-1}\right)$ and $\theta=(\theta_{1},\dots,\theta_{2n-1})$. Since $\Phi(v,\cdot)$ is a unit-speed geodesic, we may suppose without loss of generality that

$$\left.\frac{d}{d\tilde{s}}\right|_{\tilde{s}=s}\Phi^{1}(v,\tilde{s})\neq 0.$$

Define a curve $\beta=\left(\beta^{1},\dots,\beta^{2n}\right)$ in $T^{*}SM$ as the composition of $d\Phi|_{(v,s)}^{t}$ and the curve

$$\mathbb{R}\ni\tau\mapsto(\theta_{1}+\tau,\theta_{2},\dots,\theta_{2n-1})\in T_{\Phi(v,s)}^{*}SM.$$

Then $\beta$ is a smooth curve in $Z_{(v,s)}$ such that $\beta(0)=\zeta$. Moreover,

(6)

$$\left.\frac{d}{d\tau}\right|_{\tau=0}(q\circ\beta)(\tau)=\left.\frac{d}{d\tau}\right|_{\tau=0}\beta^{2n}(\tau)=\left.\frac{d}{d\tilde{s}}\right|_{\tilde{s}=s}\Phi^{1}(v,\tilde{s})\neq 0.$$

Therefore, as claimed, $\mathcal{O}$ is an embedded submanifold of dimension $4n-2$.

Now we can use $\mathcal{O}$ to parametrize $C$. Indeed, if $G_{\xi}^{-1}$ is the inverse of (3) composed with the projection onto the $\xi$ component, then the map

$$P_{C}:\mathcal{O}\to T^{*}SM\times\mathcal{O}\times\mathcal{O}\times T^{*}SM$$

defined by

$$P_{C}(\zeta)=\left(G_{\xi}^{-1}(\zeta),\zeta,\zeta,d\Phi^{-t}\zeta\right)$$

is a smooth embedding, so $C$ is an embedded submanifold of dimension $4n-2$.

To finish proving point (i), fix $c\in C$. Since $T_{c}C$ is necessarily contained in the intersection of the tangent spaces of the manifolds on the right-hand side of (2), it is enough to show the reverse containment. Observe that $C_{p_{*}}\times C_{\Phi^{*}}$ can be parametrized by the map

$$P_{C_{p_{*}}\times C_{\Phi^{*}}}:T^{*}SM\times\mathbb{R}\times Z\to T^{*}SM\times\operatorname{Range}(G)\times Z\times T^{*}SM$$

defined by

$$P_{C_{p_{*}}\times C_{\Phi^{*}}}(\xi,s,\zeta)=\left(\xi,G(\xi,s),\zeta,d\Phi^{-t}\zeta\right).$$

Hence any vector $X\in T_{c}(C_{p_{*}}\times C_{\Phi^{*}})$ is the velocity of some smooth curve

$$\mathbb{R}\ni\tau\mapsto(\xi(\tau),s(\tau),\zeta(\tau))\in T^{*}SM\times\mathbb{R}\times Z,$$

meaning $P_{C_{p_{*}}\times C_{\Phi^{*}}}(\xi(0),s(0),\zeta(0))=c$ and

$$X=\left.\frac{d}{d\tau}\right|_{\tau=0}P_{C_{p_{*}}\times C_{\Phi^{*}}}(\xi(\tau),s(\tau),\zeta(\tau)).$$

The vector $X$ is also in $T_{c}\big{(}T^{*}SM\times\Delta\big{(}T^{*}(SM\times\mathbb{R})\big{)}\times T^{*}SM\big{)}$ if and only if

$$\left.\frac{d}{d\tau}\right|_{\tau=0}G(\xi(\tau),s(\tau))=\left.\frac{d}{d\tau}\right|_{\tau=0}\zeta(\tau).$$

Thus, in any local coordinates, $G(\xi,s)$ and $\zeta$ agree to first order at $\tau=0$. Then (3) implies that the same is true of $\xi$ and $G_{\xi}^{-1}(\zeta)$. Hence

	$\displaystyle X$	$\displaystyle=\left.\frac{d}{d\tau}\right|_{\tau=0}P_{C_{p_{*}}\times C_{\Phi^{*}}}(\xi(\tau),s(\tau),\zeta(\tau))$
		$\displaystyle=\left.\frac{d}{d\tau}\right|_{\tau=0}P_{C}(\zeta(\tau)),$

which means $X\in T_{c}C$. This completes the proof of point (i).

For point (ii), suppose we have a compact set $K\subset T^{*}SM\times T^{*}SM$. Then there exists a constant $\rho>0$ such that

$$K\subset S:=\left\{(\xi,\tilde{\xi})\in T^{*}SM\times T^{*}SM:|\xi|_{g}+|\tilde{\xi}|_{g}\leq\rho\right\}.$$

Hence $P_{C}^{-1}(\pi_{C}^{-1}(S))$ is precisely the set

$$\left\{\zeta\in\mathcal{O}:|G_{\xi}^{-1}(\zeta)|_{g}+|d\Phi^{-t}\zeta|_{g}\leq\rho\right\},$$

which is compact by continuity. Then $P_{C}^{-1}(\pi_{C}^{-1}(K))$ is compact, because it is a closed subset of the compact set $P_{C}^{-1}(\pi_{C}^{-1}(S))$. Since $P_{C}$ is a diffeomorphism onto $C$, this implies that $\pi_{C}$ is a proper map. Thus point (ii) holds.

Since points (i)-(iii) hold and the excess is zero, the map $\pi_{C}\circ P_{C}$ is a smooth embedding, so its image $C_{R_{a,\psi}}$ is an embedded submanifold. Since $p_{*}$ and $(\psi\tilde{a})^{m}\circ\Phi^{*}$ are Fourier integral operators of order $-1/4$, we conclude that $R_{a,\psi}\in\mathcal{I}^{-1/2}(SM\times SM,C_{R_{a,\psi}}^{\prime})$. ∎

Since $R_{a,\psi}$ equals $p_{*}\circ(\psi\tilde{a})^{m}\circ\Phi^{*}$, we know

$$\operatorname{WF}^{\prime}(R_{a,\psi})\subset C_{p_{*}}\circ\operatorname{WF}^{\prime}((\psi\tilde{a})^{m})\circ C_{\Phi^{*}}.$$

Because the Schwartz kernel of $(\psi\tilde{a})^{m}$ is smooth away from $\operatorname{supp}(\psi\tilde{a}\otimes\psi\tilde{a})$, and $a\in C^{\infty}(\mathbb{R})$ is supported away from $0$, we also know

	$\displaystyle\operatorname{WF}^{\prime}((\psi\tilde{a})^{m})\subset\Big{\{}\big{(}\hat{\xi},\sigma,\hat{\xi},\sigma\big{)}\in\Delta\big{(}T^{*}(SM\times\mathbb{R})\big{)}:$	$\displaystyle\ \pi_{T^{*}\mathbb{R}}(\sigma)\neq 0\ \text{and}$
		$\displaystyle\ \left(\pi_{T^{*}SM}(\hat{\xi}),\pi_{T^{*}\mathbb{R}}(\sigma)\right)\in\operatorname{supp}(\psi)\Big{\}}.$

Putting these two containments together yields the result. ∎

Using Theorem 3.1 and Lemma 2.4, one can check that $C_{L_{\psi}}$ equals $C_{R_{a,\psi}}\circ C_{\pi^{*}}$, so it suffices to show points (i)-(ii) in the proof of Theorem 3.1 hold with (2) replaced by

(7)

$$C:=(C_{R_{a,\psi}}\times C_{\pi^{*}})\cap(T^{*}SM\times\Delta(T^{*}SM)\times T^{*}M).$$

(Since we omit point (iii), in general $C_{L_{\psi}}$ will only be a local canonical relation.)

To begin proving point (i), let $V$ be the smooth rank-$n$ subbundle of $T^{*}SM$ whose fiber over each vector $v\in SM$ is

$$V_{v}:=\operatorname{Range}\left(d\pi|_{v}^{t}\right).$$

Similar to (4), we can define a smooth map

$$d\pi^{-t}:V\to T^{*}M$$

whose restriction to each fiber $V_{v}$ is the inverse $(d\pi|_{v}^{t})^{-1}:V_{v}\to T_{\pi(v)}^{*}M$. Let

$$\widetilde{Z}=\left(d\Phi^{-t}\right)^{-1}(V).$$

Since $d\Phi^{-t}:Z\to T^{*}SM$ is a smooth submersion and $V$ is an embedded codimension-$(n-1)$ submanifold of $T^{*}SM$, we know $\widetilde{Z}$ is an embedded codimension-$(n-1)$ submanifold of $Z$.

Using a similar argument as the proof of Theorem 3.1, we will show that

$$\widetilde{\mathcal{O}}:=\mathcal{O}\cap\widetilde{Z}$$

is an embedded submanifold of dimension $3n-1$. Let $\tilde{q}:\widetilde{Z}\to\mathbb{R}$ be the restriction to $\widetilde{Z}$ of the function $q$ defined in (5). Then $\widetilde{\mathcal{O}}$ equals the level set $\tilde{q}^{-1}(0)$, so it suffices to show that $d\tilde{q}|_{\tilde{\zeta}}$ is nonzero for all $\tilde{\zeta}\in\widetilde{Z}$. Let $\pi_{V}(\tilde{\zeta})=(v,s)$. Then for some $\tilde{\eta}\in T_{\gamma_{v}(s)}^{*}M$, we have

$$\tilde{\zeta}=d\Phi|_{(v,s)}^{t}\circ d\pi|_{\Phi(v,s)}^{t}\,\tilde{\eta}.$$

Choose slice coordinates for $SM$ near $\Phi(v,s)$, which yield corresponding coordinates for $M$ near $\gamma_{v}(s)$. Fix natural coordinates on $T^{*}M$. Then locally we can write $\tilde{\eta}=(\tilde{\eta}_{1},\dots,\tilde{\eta}_{n})$. Define a curve $\beta$ in $T^{*}SM$ as the composition of $d\Phi|_{(v,s)}^{t}\circ d\pi|_{\Phi(v,s)}^{t}$ and the curve

$$\mathbb{R}\ni\tau\mapsto(\tilde{\eta}_{1}+\tau,\tilde{\eta}_{2},\dots,\tilde{\eta}_{n})\in T_{\gamma_{v}(s)}^{*}M.$$

Then $\beta$ is a smooth curve in $\widetilde{Z}$ such that $\beta(0)=\tilde{\zeta}$. Moreover, similar to (6),

$$\left.\frac{d}{d\tau}\right|_{\tau=0}(\tilde{q}\circ\beta)(\tau)\neq 0.$$

This proves that $\widetilde{\mathcal{O}}$ is an embedded submanifold of dimension $3n-1$.