* Southwood & Kivelson (1993) Southwood, D. J., & Kivelson, M. G. 1993, Journal of Geophysical Research: Space Physics, 98, 9181, doi: https://doi.org/10.1029/92JA02837
  * Sudan (1963) Sudan, R. N. 1963, The Physics of Fluids, 6, 57, doi: 10.1063/1.1724508
  * Sudan (1965) —. 1965, The Physics of Fluids, 8, 153, doi: 10.1063/1.1761082
  * Westfold (1959) Westfold, K. C. 1959, ApJ, 130, 241, doi: 10.1086/146713
  * White (2022) White, C. J. 2022, ApJS, 262, 28, doi: 10.3847/1538-4365/ac77ef
  * White et al. (2019) White, C. J., Stone, J. M., & Quataert, E. 2019, ApJ, 874, 168, doi: 10.3847/1538-4357/ab0c0c


fixed, we have

(3.55) $\text{if
}\,\,a_{\lambda,\nu}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})\cdot
a_{\lambda,\nu^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime},\tilde{s}^{\prime})\neq
0,\\\ \text{then }\,\,|\tilde{y}|,\,|\tilde{y}^{\prime}|\leq
C_{\delta}\theta,\,\,\text{and }\,\,|\tilde{y}-\tilde{y}^{\prime}|\geq
c_{\delta}\theta.$

Additionally, if $\delta_{0}<1/8$ as in (2.6) is small enough, then for
sufficiently small $\theta$ we have

(3.56) $\text{if
}\,a_{\lambda,\nu}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})\neq
0\,\,\text{then
}\,|\delta-2\kappa_{\ell}^{c_{0}\theta}(\tilde{t}-\tilde{s})|\leq
4\delta_{0}\delta,\\\ \text{and }\,\,\text{if
}\,a_{\lambda,\nu^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{s}^{\prime})\neq
0\,\,\text{then
}\,|\delta-2\kappa_{\ell}^{c_{0}\theta}(\tilde{t}-\tilde{s}^{\prime})|\leq
4\delta_{0}\delta,$

###### Proof.

The first assertion in (3.55) about the size of $\tilde{y}$ and
$\tilde{y}^{\prime}$ follows trivially from (3.34), (3.39) and (3.40). To see
the assertion regarding the important separation of the $\tilde{y}$-variables,
recall that $\nu,\nu^{\prime}\in c_{0}\theta\cdot\mathbb{Z}^{2n-3}$, and by
(3.23), $|\nu-\nu^{\prime}|\in[\frac{1}{4}\theta,4^{n}\theta]$. Thus, if we
write $\nu=(c_{0}\theta j,c_{0}\theta\ell)$ and $\nu^{\prime}=(c_{0}\theta
j^{\prime},c_{0}\theta\ell^{\prime})$, we can divide into the following two
cases:

(i) $|j-j^{\prime}|\geq\frac{1}{8}$. In this case, the spatial parts,
$\overline{\gamma}_{j}$ and $\overline{\gamma}_{j^{\prime}}$ of the
Schrödinger curves $\iota_{\nu}$ and $\iota_{\nu^{\prime}}$ have angle
$\approx\theta$. By (3.32) if the product of the amplitudes in (3.55) is
nonzero then we must have in our original Fermi normal coordinates that, for a
fixed constant $C_{1}^{\prime}$,
$x\in{\mathcal{T}}_{C_{1}^{\prime}c_{0}\theta}(\overline{\gamma}_{j})\cap{\mathcal{T}}_{C_{1}^{\prime}c_{0}\theta}(\overline{\gamma}_{j^{\prime}})$,
$y\in{\mathcal{T}}_{C_{1}^{\prime}c_{0}\theta}(\overline{\gamma}_{j})$ and
$y^{\prime}={\mathcal{T}}_{C_{1}^{\prime}c_{0}\theta}(\overline{\gamma}_{j^{\prime}})$.
Here, of course, ${\mathcal{T}}_{r}(\overline{\gamma})$ denotes an $r$-tube
about $\overline{\gamma}$ in $M^{n-1}$. By (3.35) we must also have
$d_{g}(x,y),d_{g}(x,y^{\prime})\in[\delta-\delta_{0}\delta,\delta+\delta_{0}\delta]$
for our small $\delta_{0}>0$ if the product is nonzero. Since we are assuming
(3.23) the two tubes of width $\approx c_{0}\theta$ intersect at angle
$\approx\theta$ at $(x,t)$, which implies that in our original Fermi normal
coordinates
$|(y_{1},\dots,y_{n-2})-(y_{1}^{\prime},\dots,y_{n-2}^{\prime})|\approx\theta$
if the above product is nonzero and $c_{0}$ and $\theta$ are small. By (3.40),
this yields the assertion in (3.55) about the separation of $\tilde{y}$ and
$\tilde{y}^{\prime}$ under our assumption that $j\neq j^{\prime}$. Note that
the smaller $\delta$ becomes we have to choose $c_{0}$ to be correspondingly
small, but we are assuming here that $\delta$ is fixed (as we shall do later).

(ii) $|\ell-\ell^{\prime}|\geq\frac{1}{8}$. In this case we have
$|\ell-\ell^{\prime}|\approx 1$. Recall that in our Fermi normal coordinates,
we have

(3.57) $d_{g}(x,y)=|x_{n-1}-y_{n-1}|,\,\,\,\frac{\partial}{\partial
x_{j}}d_{g}(x,y),\,\,\frac{\partial}{\partial y_{j}}d_{g}(x,y)=0,\\\
\text{if}\,\,x=(0,\dots,0,x_{n-1}),\,\,y=(0,\dots,0,y_{n-1}),\text{and
}\,\,\,j=1,\dots,n-2.$

Also we know that by (3.34)

(3.58) $|(x_{1},\dots,x_{n-2})|+|(y_{1},\dots,y_{n-2})|\leq
C_{0}\theta,\quad\text{if }\,a_{\lambda,\nu}a_{\lambda,\nu^{\prime}}\neq 0.$

Since the function $d_{g}(x,y)$ is smooth when $d_{g}(x,y)\approx\delta$, by
(3.57), (3.58) and Taylor’s expansion, we have

$|d_{g}(x,y)-(x_{n-1}-y_{n-1})|,\,|d_{g}(x,y^{\prime})-(x_{n-1}-y_{n-1}^{\prime})|\leq
C_{\delta}\theta^{2},\quad\text{if
}\,a_{\lambda,\nu}a_{\lambda,\nu^{\prime}}\neq 0,$

since by (3.35) both of the amplitudes vanish if $x_{n-1}-y_{n-1}<0$. If we
let $\theta$ to be small enough, $C_{\delta}\theta^{2}$ is much smaller than
$c_{0}\theta$, consequently, if
$|(x_{n-1}-y_{n-1})+2\kappa_{\ell}^{c_{0}\theta}(t-s)|\geq
C^{\prime}_{0}c_{0}\theta$ with $C_{0}^{\prime}$ large enough we must have
that $|d_{g}(x,y)+2\kappa^{c_{0}\theta}_{\ell}(t-s)|\geq C_{0}c_{0}\theta$
with $C_{0}$ as in (3.33), which means that $a_{\nu}=0$ if
$|(x_{n-1}-y_{n-1})+2\kappa_{\ell}^{c_{0}\theta}(t-s)|\geq
C^{\prime}_{0}c_{0}\theta$ for this choice of $C^{\prime}_{0}$ (which is
independent of $c_{0}$). We similarly have $a_{\nu^{\prime}}=0$ if
$|(x_{n-1}-y^{\prime}_{n-1})+2\kappa_{\ell^{\prime}}^{c_{0}\theta}(t-s)|\geq
C^{\prime}_{0}c_{0}\theta$. By (3.39) this means that for a uniform constant
$C_{1}$ if $a_{\nu}\neq 0$ we must have $|\tilde{x}_{n-1}-\tilde{y}_{n-1}|\leq
C_{1}c_{0}\theta$, and if $a_{\nu^{\prime}}\neq 0$ we must have
$|(\tilde{x}_{n-1}-\tilde{y}_{n-1}^{\prime})-2(\kappa_{\ell}^{c_{0}\theta})^{-1}(\kappa^{c_{0}\theta}_{\ell^{\prime}}-\kappa^{c_{0}}_{\ell})(t-s)|\leq
C_{1}c_{0}\theta$. Since (3.33) and (3.35) imply that $-(t-s)\sim\delta$ on
the support of the amplitudes and thus
$|2(\kappa_{\ell}^{c_{0}\theta})^{-1}(\kappa^{c_{0}\theta}_{\ell}-\kappa^{c_{0}\theta}_{\ell^{\prime}})(t-s)|$
must be larger than a fixed multiple of $\theta$ if
$|\ell-\ell^{\prime}|\approx 1$ and $a_{\lambda,\nu}\cdot
a_{\lambda,\nu^{\prime}}\neq 0$. So, in this case, if $c_{0}$ is small enough,
we must have $|\tilde{y}_{n-1}-\tilde{y}_{n-1}^{\prime}|\approx\theta$ if
$a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}\neq 0$, which finishes the
proof of the first assertion regarding the separation of $\tilde{y}$ and
$\tilde{y}^{\prime}$ in (3.55) .

It remains to prove (3.56). Since we are assuming $|\tilde{x}|\leq
C_{0}\theta$, it follows from the first part of (3.55) that both of the
amplitudes in (3.56) will vanish if we do not have
$|\tilde{y}|,\,|\tilde{y}^{\prime}|=O(\theta)$ and hence

(3.59)
$a_{\lambda,\nu}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=a_{\lambda,\nu^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=0\quad\text{if
}\,\,\,|\tilde{x}_{n-1}-\tilde{y}_{n-1}|\geq C^{\prime}\theta,$

for some constant $C^{\prime}$. By (3.39)

$(\tilde{x}_{n-1}-\tilde{y}_{n-1},\tilde{t}-\tilde{s})=-(\kappa_{\ell}^{c_{0}\theta})^{-1}\bigl{(}x_{n-1}-y_{n-1}+2\kappa^{c_{0}\theta}_{\ell}(t-s),\,-(x_{n-1}-y_{n-1})-\kappa^{c_{0}\theta}_{\ell}(t-s))\bigr{)},$

and so $(\tilde{x}_{n-1}-\tilde{y}_{n-1})+(\tilde{t}-\tilde{s})=-(t-s)$. Since
$|\ell-\ell^{\prime}|\approx 1$ implies
$\kappa^{c_{0}\theta}_{\ell}-\kappa^{c_{0}\theta}_{\ell^{\prime}}=O(\theta)$,
by (3.33) and (3.59), we conclude that both amplitudes vanish if we do not
have

$|d_{g}(x,y)-2\kappa_{\ell}^{c_{0}\theta}(\tilde{t}-\tilde{s})|\leq
C^{\prime}\theta,$

for some uniform constant $C^{\prime}$. By the first part of (3.35) we obtain
(3.56) if $\theta$ is sufficiently small. ∎

Let us now prove the bilinear oscillatory integral estimates (3.53) which will
finish the proof of Proposition 3.3.

To prove (3.53), in addition to following the proof of [23, Theorem 1.3], we
shall also follow related arguments of two of us [4] which proved analogous
bilinear estimates in the $1+2$ dimensional setting (one lower dimension than
here) using the simpler classical bilinear oscillatory integral estimates
implicit in Hörmander [19]. Similar arguments were in the paper [5] by these
two authors.

Just as in [23] we first perform a parabolic scaling as in (3.49) and (3.51)
to be able to apply the main estimate, Theorem 1.1, in Lee [23]. So for small
$\lambda^{-1/8}\leq\theta\ll 1$, we let

(3.60)
$\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})=\theta^{-2}\tilde{\varphi}(\theta\tilde{x},\tilde{t};\theta\tilde{y},\tilde{s}),\quad\text{and
}\,\,\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime})=\theta^{-2}\tilde{\varphi}(\theta\tilde{x},\tilde{t};\theta\tilde{y}^{\prime},\tilde{s}^{\prime}).$

and corresponding amplitudes

(3.61)
$a^{\theta}_{\lambda,\nu}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=a_{\lambda,\nu}(\theta
x,\tilde{t};\theta\tilde{y},\tilde{s})\quad\text{and
}\,\,a^{\theta}_{\lambda,\nu^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=a_{\lambda,\nu}(\theta
x,\tilde{t};\theta\tilde{y}^{\prime},\tilde{s}^{\prime}).$

Then, as we noted before

$D^{\beta}_{\tilde{x},\tilde{t},\tilde{y}}a^{\theta}_{\lambda,\mu}=O_{\beta}(1),\,\mu=\nu,\nu^{\prime}\quad\text{and
}\,\,D^{\beta}_{\tilde{x},\tilde{t},\tilde{y}}\phi_{j}=O_{\beta}(1),\,\,\phi_{1}=\phi^{\theta}{\tilde{s}},\,\,\phi_{2}=\phi^{\theta}_{\tilde{s}^{\prime}}.$

By (3.55) and (3.56) we also have the key separation properties for small
enough $\theta$

(3.62) $\text{if
}\,\,a^{\theta}_{\lambda,\nu}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})a^{\theta}_{\lambda,\nu^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime},\tilde{s}^{\prime})\neq
0\\\ \text{then
}\,|\tilde{y}|,|\tilde{y}^{\prime}|=O(1),\,\,|\tilde{y}-\tilde{y}^{\prime}|\geq
c_{\delta}\,\,\text{and }\,\,|\tilde{s}-\tilde{s}^{\prime}|\leq
8\delta_{0}\delta,$

with $\delta$ and $\delta_{0}$ as in (2.6).

Additionally, by a simple scaling argument, our remaining task of the section,
(3.53), is equivalent to the following for small enough $\theta$:

(3.63)
$\bigl{\|}B^{\theta,\tilde{s},\tilde{s}^{\prime}}_{\lambda,\nu,\nu^{\prime}}(h_{1},h_{2})\bigr{\|}_{L^{q/2}_{t,x}(\\{|\tilde{x}_{0}|\leq
C_{1}\\}\times[-1,1])}\lesssim_{\varepsilon}\bigl{(}\lambda\theta^{2}\bigr{)}^{-\frac{2n}{q}+\varepsilon}\|h_{1}\|_{L^{2}_{x}}\|h_{2}\|_{L^{2}_{x}},\,q=\tfrac{2(n+2)}{n},$

where we have the scaled version of (3.52), i.e.,

(3.64)
$B^{\mu,\tilde{s},\tilde{s}^{\prime}}_{\lambda,\nu,\nu^{\prime}}(h_{1},h_{2})(x,t)=(\alpha_{m}(t))^{2}\times\\\
\iint
e^{i(\lambda\theta^{2})[\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})+\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime})]}a^{\theta}_{\lambda,\nu}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})a^{\theta}_{\lambda,\nu^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime},\tilde{s}^{\prime})\,h_{1}(\tilde{y})h_{2}(\tilde{y}^{\prime})\,d\tilde{y}d\tilde{y}^{\prime}.$

To prove this, let us see how we can use our earlier observation that (3.41)
and (3.50) implies that $\tilde{\varphi}$ vanishes to second order when
$(\tilde{x},\tilde{y})=(0,0)$ to see that the scaled phase functions in (3.60)
closely resemble Euclidean ones if $\theta$ is small which will allow us to
verify the hypotheses of Lee’s bilinear oscillatory integral theorem [23,
Theorem 1.3] if $\delta,\delta_{0}>0$ in (2.6) are fixed small enough.

To be more specific, let

(3.65)
$A(\tilde{t},\tilde{s})=\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s}),\,\,\,B(\tilde{t},\tilde{s})=\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{x}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s}),\\\
\text{and
}\,\,C(\tilde{t},\tilde{s})=\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{x}_{j}\partial\tilde{x}_{k}}(0,\tilde{t};0,\tilde{s}).$

Then the Taylor expansion about $(\tilde{x},\tilde{y})=(0,0)$ of
$\tilde{\varphi}$ is

(3.66)
$\tilde{\varphi}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=\tfrac{1}{2}\tilde{y}^{T}A(\tilde{t},\tilde{s})\tilde{y}+\tilde{x}^{T}B(\tilde{t},\tilde{s})\tilde{y}+\tfrac{1}{2}\tilde{x}^{T}C(\tilde{t},\tilde{s})\tilde{x}+r(\tilde{x},\tilde{t};\tilde{y},\tilde{s}),$

where $r(\tilde{x},\tilde{t};\tilde{y},\tilde{s})$ vanishes to third order at
$(\tilde{x},\tilde{y})=(0,0)$. So,

(3.67)
$D^{\beta}_{\tilde{x},\tilde{y},\tilde{t},\tilde{s}}r^{\theta}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=O(\theta),\quad\text{if
}\,\,r^{\theta}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=\theta^{-2}r(\theta\tilde{x},\tilde{t};\theta\tilde{y},\tilde{s}),$

which means that $r^{\theta}\to 0$ in the $C^{\infty}$ topology as $\theta\to
0$.

To use (3.66) we shall use parabolic scaling and the following lemma, whose
proof we postpone until the end of this subsection, which says that if
$\delta,\delta_{0}>0$ in (2.6) are small enough then the phase functions
$\phi_{\tilde{s}}$ and $\phi_{\tilde{s}^{\prime}}$ in (3.54) satisfy the
Carleson-Sjölin condition (see [29, §2.2.2] and [33]).

######  Lemma 3.7.

Let $A(\tilde{t},\tilde{s})$ and $B(\tilde{t},\tilde{s})$ be as in (3.65).
Then if $\delta,\delta_{0}>0$ in (2.6) are small enough

(3.68) $\det
B(\tilde{t},\tilde{s})=\det\frac{\partial^{2}\tilde{\varphi}(0,\tilde{t};0,\tilde{s})}{\partial\tilde{x}_{j}\partial\tilde{y}_{k}}\neq
0,\quad\text{if }\,\,a^{\theta}_{\lambda,\nu}\cdot
a^{\theta}_{\lambda,\nu^{\prime}}\neq 0.$

Furthermore, on the support of $a^{\theta}_{\lambda,\nu}\cdot
a^{\theta}_{\lambda,\nu^{\prime}}$,
$-(\tfrac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s}))^{-1}=-(\tfrac{\partial}{\partial\tilde{t}}\tfrac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s}))^{-1}$
is positive definite, i.e.,

(3.69)
$\xi^{t}\Bigl{(}-\frac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s})\Bigr{)}^{-1}\xi,\,\,\xi^{t}\Bigl{(}-\frac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s}^{\prime})\Bigr{)}^{-1}\xi\geq
c_{\delta}|\xi|^{2},\quad\text{if }\,a^{\theta}_{\lambda,\nu}\cdot
a^{\theta}_{\lambda,\nu^{\prime}}\neq 0,$

and also

(3.70)
$\Bigl{|}\frac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s})\xi\Bigr{|}\geq
c_{\delta}|\xi|,\,\Bigl{|}\frac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s}^{\prime})\xi\Bigr{|}\geq
c_{\delta}|\xi|\,\,\,\text{if }\,\,a^{\theta}_{\lambda,\nu}\cdot
a^{\theta}_{\lambda,\nu^{\prime}}\neq 0,$

Let us use (3.66) and (3.67) and this lemma to see that we can obtain our
remaining estimate (3.63) via Lee’s [23, Theorem 1.1]. As we shall see, it is
crucial for us that
$-\frac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s})$ is positive
definite.

Note that, in addition to the $\theta$ parameter, (3.63) also involves the
$(\tilde{s},\tilde{s}^{\prime})$ parameters. For simplicity, let us first see
how Lee’s result yields (3.63) in the case where these two parameters agree,
i.e. $\tilde{s}=\tilde{s}^{\prime}$. We then will argue that if $\delta_{0}$
in (2.6) and hence (3.55) is fixed small enough we can also handle the case
where $\tilde{s}\neq\tilde{s}^{\prime}$.

To do this we first note that the parabolic scaling in (3.67), which agrees
with that in (3.60), preserves the first three terms in the right of (3.66)
since they are quadratic. Also, in proving (3.64), we may subtract
$\tfrac{1}{2}\tilde{x}^{t}C(\tilde{t},\tilde{s})\tilde{x}$ from
$\phi^{\theta}_{\tilde{s}}$ and
$\tfrac{1}{2}\tilde{x}^{t}C(\tilde{t},\tilde{s}^{\prime})\tilde{x}$ from
$\phi^{\theta}_{\tilde{s}^{\prime}}$ as these quadratic terms do not involve
$\tilde{y}$. We point out that this trivial reduction also works if
$\tilde{s}\neq\tilde{s}^{\prime}$.

Next, we note that, by (3.68) and our temporary assumption that
$\tilde{s}=\tilde{s}^{\prime}$, after making a linear change of variables in
$\tilde{x}$ (depending on $\tilde{t},\tilde{s}$), we may reduce to the case
where $B(\tilde{t},\tilde{s})=I_{n-1}$, the $(n-1)\times(n-1)$ identity
matrix. This means for the case where $\tilde{s}=\tilde{s}^{\prime}$ we have
reduced matters to showing that (3.63) is valid where

(3.71)
$\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})=\langle\tilde{x},\tilde{y}\rangle+\frac{1}{2}\sum_{j,k=1}^{n-1}\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s})\tilde{y}_{j}\tilde{y}_{k}+\tilde{r}^{\theta}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})\\\
=\langle\tilde{x},\tilde{y}\rangle+\tilde{y}^{t}A(\tilde{t},\tilde{s})\tilde{y}+\tilde{r}^{\theta}(\tilde{x},\tilde{t};\tilde{y},\tilde{s}),$

with $\tilde{r}^{\theta}$ denoting $r^{\theta}$ rewritten in the new
$\tilde{x}$ variables coming from $B(\tilde{t},\tilde{s})$. In view of (3.68),
(3.67) remains valid for $\tilde{r}^{\theta}$. For later use, we note that if
we change variables according to $\tilde{s}$ as above, then for
$\tilde{s}^{\prime}$ near $\tilde{s}$ we have for

(3.72)
$B(\tilde{t},\tilde{s},\tilde{s}^{\prime})=(B(\tilde{t},\tilde{s}^{\prime}))^{t}\,((B(\tilde{t},\tilde{s})^{-1})^{t}=I_{n-1}+O(|\tilde{s}-\tilde{s}^{\prime}|),$
(3.73)
$\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t},\tilde{y})=\langle
x,B(\tilde{t},\tilde{s},\tilde{s}^{\prime})\tilde{y}\rangle+\frac{1}{2}\sum_{j,k=1}^{n-1}\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s}^{\prime})\tilde{y}_{j}\tilde{y}_{k}+\tilde{r}^{\theta}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})\\\
=\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})+O(|\tilde{s}-\tilde{s}^{\prime}|).$

We fix $\delta$ and $\delta_{0}$ in (2.6) so that the conclusions of Lemmas
3.5 and 3.7 are valid. We can also now fix also finally fix $c_{0}$ so that
the results of Lemma 3.5 and Lemma 3.6 are valid. If we only had to treat the
case where $\tilde{s}=\tilde{s}^{\prime}$ in (3.63) then the above choice of
$\delta_{0}$ would suffice; however, as we shall momentally see, to handle the
cases where $\tilde{s}\neq\tilde{s}^{\prime}$ we shall need to choose
$\delta_{0}$ small enough so that we can exploit the last part of (3.62).

Let us now verify that we can apply [23, Theorem 1.3] to obtain (3.63) for
sufficiently small $\theta$. This would complete the proof of Proposition 3.3.
We recall that we are assuming for now that $\tilde{s}=\tilde{s}^{\prime}$ and
that we have reduced to the case where $B(\tilde{t},\tilde{s})=I_{n-1}$ and
$C(\tilde{t},\tilde{s})=0$ in (3.66) and so

(3.74)
$\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})=\langle\tilde{x},\tilde{y}\rangle+\tfrac{1}{2}\tilde{y}^{t}A(\tilde{t},\tilde{s})\tilde{y}+\tilde{r}^{\theta}(\tilde{x},\tilde{t};\tilde{y},\tilde{s}),$

with $\tilde{r}^{\theta}$ satisfying the bounds in (3.67).

By (3.67) and (3.74) we have

(3.75)
$\frac{\partial\phi^{\theta}_{\tilde{s}}}{\partial\tilde{x}}(\tilde{x},\tilde{t};\tilde{y})=\tilde{y}+\frac{\partial\tilde{r}^{\theta}}{\partial\tilde{x}}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=\tilde{y}+\varepsilon(\theta,\tilde{x},\tilde{t},\tilde{s};\tilde{y}),$

where $\tilde{y}\to\varepsilon(\cdot;\,\tilde{y})$ and its derivatives are
$O(\theta)$. Thus, for small enough $\theta$, the inverse function also
satisfies

(3.76)
$\tilde{y}\to\bigl{(}\frac{\partial\phi^{\theta}_{\tilde{s}}}{\partial\tilde{x}}(\tilde{x},\tilde{t};\,\cdot\,)\bigr{)}^{-1}(\tilde{y})=\tilde{y}+\tilde{\varepsilon}(\theta,\tilde{x},\tilde{s};\tilde{y}),$

where

(3.77)
$D^{\beta}_{\tilde{y}}\tilde{\varepsilon}(\theta,\tilde{x},\tilde{t},\tilde{s};\tilde{y})=O_{\beta}(\theta).$

Define, in the notation of [23],

(3.78)
$q^{\theta}_{s}(\tilde{x},\tilde{t};\tilde{y})=\tfrac{\partial}{\partial\tilde{t}}\phi^{\theta}_{s}\bigl{(}\tilde{x},\tilde{t};\bigl{(}\tfrac{\partial\phi^{\theta}_{s}}{\partial\tilde{x}}(\tilde{x},\tilde{t};\,\cdot\,)\bigr{)}^{-1}(\tilde{y})\bigr{)}\\\
=\bigl{(}\tfrac{\partial}{\partial\tilde{t}}\phi^{\theta}_{s}\bigr{)}\bigl{(}\tilde{x},\tilde{t};\tilde{y}+\tilde{\varepsilon}(\theta,\tilde{x},\tilde{t},s;\tilde{y})\bigr{)},\quad
s=\tilde{s},\tilde{s}^{\prime},$

as well as

(3.79)
$\delta^{\theta}_{\tilde{s},\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})=\partial_{\tilde{y}}q^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\partial_{\tilde{x}}\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y}))-\partial_{\tilde{y}}q^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\partial_{\tilde{x}}\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime})).$

Even though we are assuming for the moment that $\tilde{s}=\tilde{s}^{\prime}$
these two quantities will be needed for $\tilde{s}\neq\tilde{s}^{\prime}$ as
well to be able to use [23, Theorem 1.1] to obtain (3.63).

Then [23, (1.4)], the conditions to ensure the bounds (3.63), are

(3.80)
$\bigl{|}\bigl{\langle}\partial^{2}_{\tilde{x}\tilde{y}}\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})\delta_{\tilde{s},\tilde{s}^{\prime}}^{\theta},\,\bigl{[}\,\partial^{2}_{\tilde{x},\tilde{y}}\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})\,\bigr{]}^{-1}\bigl{[}\,\partial^{2}_{\tilde{y}\tilde{y}}q^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\partial_{\tilde{x}}\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y}))\,\bigr{]}^{-1}\delta_{\tilde{s},\tilde{s}^{\prime}}^{\theta}\bigr{\rangle}\bigr{|}>0,\\\
\quad\delta_{\tilde{s},\tilde{s}^{\prime}}^{\theta}=\delta^{\theta}_{\tilde{s},\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime}),\quad\text{on
}\,\,\,\text{supp }(a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}),$

as well as

(3.81)
$\bigl{|}\bigl{\langle}\partial^{2}_{\tilde{x}\tilde{y}}\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime})\delta_{\tilde{s},\tilde{s}^{\prime}}^{\theta},\bigl{[}\partial^{2}_{\tilde{x},\tilde{y}}\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime})\,\bigr{]}^{-1}\bigl{[}\,\partial^{2}_{\tilde{y}\tilde{y}}q^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\partial_{\tilde{x}}\phi^{\theta}_{\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y}^{\prime}))\bigr{]}^{-1}\delta_{\tilde{s},\tilde{s}^{\prime}}^{\theta}\bigr{\rangle}\bigr{|}>0,\\\
\quad\delta_{\tilde{s},\tilde{s}^{\prime}}^{\theta}=\delta^{\theta}_{\tilde{s},\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime}),\quad\text{on
}\,\,\,\text{supp }(a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}).$

Note that by (3.67), (3.71), (3.93), (3.77) and (3.78) for small $\theta$ we
have

(3.82)
$\bigl{(}q^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})\bigr{)}^{-1}=\bigl{(}\tfrac{\partial
A}{\partial\tilde{t}}(\tilde{t},\tilde{s})\bigr{)}^{-1}+O(\theta),$

and also, by (3.75) and (3.76),

(3.83)
$\partial^{2}_{\tilde{x},\tilde{y}}\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})=I_{n-1}+O(\theta),$

as well as

(3.84)
$\bigl{(}\partial^{2}_{\tilde{x},\tilde{y}}\phi^{\theta}_{\tilde{s}}(\tilde{x},\tilde{t};\tilde{y})\bigr{)}^{-1}=I_{n-1}+O(\theta).$

By (3.71), (3.70), (3.78) and the separation condition in (3.55), if
$\tilde{s}=\tilde{s}^{\prime}$ we have

(3.85)
$|\delta^{\theta}_{\tilde{s},\tilde{s}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})|>0\quad\text{on
}\,\,\text{supp }(a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}),$

if $\theta$ is small enough. Thus, in this case the quantities inside the
absolute values in (3.80) and (3.81) equal

(3.86)
$\bigl{\langle}\delta^{\theta}_{\tilde{s},\tilde{s}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime}),\,\bigl{(}\tfrac{\partial
A}{\partial
t}(\tilde{t},s)\bigr{)}^{-1}\delta^{\theta}_{\tilde{s},\tilde{s}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})\bigr{\rangle}+O(\theta)\,\,\,\text{on
}\,\,\text{supp }(a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}),$

and, therefore, by (3.69) and (3.85) the conditions (3.80) and (3.81) are
valid for small enough $\theta$ when $\tilde{s}=\tilde{s}^{\prime}$. So, by
[23, Theorem 1.1], we obtain (3.63), we obtain (3.63) in this case.

If $\tilde{s}\neq\tilde{s}^{\prime}$ in (3.63), we must replace
$\delta^{\theta}_{\tilde{s},\tilde{s}}$ by
$\delta^{\theta}_{\tilde{s},\tilde{s}^{\prime}}$. In order to accommodate
this, we first need to use the fact that, by the last part of (3.55),

$\delta^{\theta}_{\tilde{s},\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})=\delta^{\theta}_{\tilde{s},\tilde{s}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})+O(|\tilde{s}-\tilde{s}^{\prime}|)\quad\text{on
}\,\text{supp }(a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}).$

Thus, by the last part of (3.62),

$\delta^{\theta}_{\tilde{s},\tilde{s}^{\prime}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})=\delta^{\theta}_{\tilde{s},\tilde{s}}(\tilde{x},\tilde{t};\tilde{y},\tilde{y}^{\prime})+O(\delta_{0})\quad\text{on
}\,\text{supp }(a_{\lambda,\nu}\cdot a_{\lambda,\nu^{\prime}}).$

This means that, if we replace $O(\theta)$ by $O(\theta+\delta_{0})$ in
(3.86), then the quanitity in (3.80) is of this form.

The other condition, (3.81) involves the phase function
$\phi^{\theta}_{\tilde{s}^{\prime}}$ and the associated
$q^{\theta}_{\tilde{s}^{\prime}}$. However if
$B=B(\tilde{t},\tilde{s},\tilde{s}^{\prime})$ is as in (3.72), then we have
the analog of (3.75) where we replace the first term in the right side of
(3.75) by $B\tilde{y}$ and the first term in the right side of (3.76) by
$B^{-1}\tilde{y}$. Also, of course $\tfrac{\partial
A}{\partial\tilde{t}}(\tilde{t},\tilde{s}^{\prime})=\tfrac{\partial
A}{\partial\tilde{t}}(\tilde{t},\tilde{s})+O(|\tilde{s}-\tilde{s}^{\prime}|)$.
Consequently, $q^{\theta}_{\tilde{s}^{\prime}}$ will agree with
$q^{\theta}_{\tilde{s}}$ when $a_{\lambda,\nu}\cdot
a_{\lambda,\nu^{\prime}}\neq 0$ up to a $O(|s-s^{\prime}|)=O(\delta_{0})$
error, and by (3.72) the analogs of (3.83) and (3.84) remain valid if
$\tilde{s}$ is replaced by $\tilde{s}^{\prime}$ if $O(\theta)$ there is
replaced by $O(\theta+\delta_{0})$. So, like (3.80), if we replace $O(\theta)$
by $O(\theta+\delta_{0})$ in (3.86), then the quantity in (3.81) is of this
form.

Thus, if $\delta_{0}$ in (2.6) is (finally) fixed small enough, and, as above,
$\theta$ is small enough we conclude that the condition (1.4) in [23] is
valid, which yields (3.63) and completes the proof of Proposition 3.3. ∎

###### Proof of Lemma 3.7.

Let us first prove (3.69) and (3.70) since they are slightly more difficult
than the other estimate, (3.68), in the lemma.

If we recall (3.50) we see that

(3.87)
$A(\tilde{t},\tilde{s})=\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s})=\frac{\partial^{2}\varphi}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s}),$

where $\varphi$ is as in (3.36).

By (3.43) we have

(3.88)
$\frac{\partial^{2}\varphi}{\partial\tilde{y}_{n-1}^{2}}(0,\tilde{t};0,\tilde{s})=\frac{(\kappa_{\ell}^{c_{0}\theta})^{2}}{2(\tilde{t}-\tilde{s})}.$

Additionally, by (3.38) and (3.39) we have

(3.89)
$\varphi(0,\tilde{t};\tilde{y},\tilde{s})=\frac{\big{[}d_{g}\bigl{(}(0,\dots,0,2\kappa_{\ell}^{c_{0}\theta}\tilde{t}),\,(\tilde{y}_{1},\dots,\tilde{y}_{n-2},2\kappa^{c_{0}\theta}_{\ell}\tilde{s}+\kappa_{\ell}^{c_{0}\theta}\tilde{y}_{n-1})\bigr{)}\bigr{]}^{2}}{4(\tilde{t}-\tilde{s}-\tilde{y}_{n-1})}.$

By (3.89) we have

(3.90)
$\frac{\partial^{2}\varphi}{\partial\tilde{y}_{j}\partial\tilde{y}_{n-1}}(0,\tilde{t};0,\tilde{s})\equiv
0,\quad\text{if }\,\,j=1,\dots,n-2.$

The remaining part of the Hessian in (3.87) is

(3.91)
$\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s})=\frac{\partial^{2}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}\frac{\bigl{[}d_{g}\bigl{(}(0,\dots,0,,2\kappa_{\ell}^{c_{0}\theta}\tilde{t}),(\tilde{y}_{1},\dots,2\kappa_{\ell}^{c_{0}\theta}\tilde{s}\bigr{)}\bigr{]}^{2}}{4(\tilde{t}-\tilde{s})},\\\
\text{when }\,\,\tilde{y}_{1}=\dots=\tilde{y}_{n-2}=0\,\,\,\text{and
}\,\,1\leq j,k\leq n-2.$

To compute this, we recall that the Schrödinger coordinates
$(\tilde{x},\tilde{t})$ in (3.39) and (3.40) come from the Fermi normal
coordinates $(x_{1},\dots,x_{n-1})$ about the spatial geodesic
$\overline{\gamma}_{j}$, and that in these coordinates
$\overline{\gamma_{j}}=(0,\dots,0,t)$ and on this geodesic the metric is
$\delta_{j,k}$ (rectangular) and the Christoffel symbols vanish there as well

As a result, in the Fermi normal coordinates, we must have that the full
Hessian of the square of the distance function satisfies

$\frac{\partial^{2}}{\partial y_{j}\partial
y_{k}}\bigl{[}d_{g}\bigl{(}(0,\dots,0,2\kappa_{\ell}^{c_{0}\theta}\tilde{t}),y\bigr{)}\bigr{]}^{2}=2I_{n-1},\quad\text{if
}\,y=(0,\dots,0,2\kappa_{\ell}^{c_{0}\theta}\tilde{t}).$

This along with (3.40) means that (3.91), the remaining piece of the Hessian
in (3.87), must be of the form

$\frac{\partial^{2}\varphi}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}=\frac{1}{2(\tilde{t}-\tilde{s})}\delta_{j,k}+O(1),\quad\text{if
}\,\,1\leq j,k\leq n-2.$

Note that since

$(\tilde{t},\tilde{y}_{1},\dots,\tilde{y}_{n-2})\to
d_{g}((0,\dots,0,2\kappa^{c_{0}\theta}_{\ell}\tilde{t}),(\tilde{y}_{1},\dots,\tilde{y}_{n-2},2\kappa^{c_{0}\theta}_{\ell}\tilde{s}^{\prime}))$

is smooth we also obtain

(3.92)
$\frac{\partial}{\partial\tilde{t}}\frac{\partial^{2}\varphi}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s})=\frac{-1}{2(\tilde{t}-\tilde{s})^{2}}\delta_{j,k}+O((\tilde{t}-\tilde{s})^{-1}),\,\,1\leq
j,k\leq n-2.$

Therefore, by (3.65), (3.87), (3.88), (3.90) and (3.92), we have

(3.93)
$-2(\tilde{t}-\tilde{s})^{2}\frac{\partial}{\partial\tilde{t}}A(\tilde{t},\tilde{s})=-2(\tilde{t}-\tilde{s})^{2}\frac{\partial}{\partial\tilde{t}}\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{y}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s})=J_{\kappa_{\ell}^{c_{0}\theta}}+O(|\tilde{t}-\tilde{s}|),\\\
\text{if }\quad J_{\kappa_{\ell}^{c_{0}\theta}}=\text{diag
}(1,\dots,1,(\kappa_{\ell}^{c_{0}\theta})^{2}).$

Note that $\kappa^{c_{0}\theta}_{\ell}\in[1/10,10]$. Therefore by (3.56) if
$\delta$ is fixed small enough in (2.6) and if $\delta_{0}$ there is smaller
than $1/8$, by (3.56), we have that $-(\partial
A(\tilde{t},\tilde{s})/\partial\tilde{t})^{-1}$ is positive definite on the
support of the amplitudes in (3.64). Thus, we obtain (3.69) and (3.70) for
some $c_{\delta}>0$. Indeed, one may take $c_{\delta}\sim\delta^{-2}$,

The proof of the other (3.68) is very similar. If we use (3.36) we see that
since, by (3.35) and (3.56), $d_{g}(x,y)\approx|t-s|\approx\delta$ on
$\text{supp }a^{\theta}_{\lambda,j}\cdot a^{\theta}_{\lambda,\nu^{\prime}}$,
we may assume that $|x|,\,|y|\leq C\delta$. We then have

$(d_{g}(x,y))^{2}=|x-y|^{2}+r(x,y),\quad\text{where
}\,r(x,y)=O\big{(}(|x|+|y|)|x-y|^{2}\big{)}\,\,\text{and }\,r\in C^{\infty}.$

Therefore, by (3.38), (3.50) and (3.39), if
$\bar{x}=(\tilde{x}_{1},\tilde{x}_{2},\dots,\tilde{x}_{n-2})$ and
$\bar{y}=(\tilde{y}_{1},\tilde{y}_{2},\dots,\tilde{y}_{n-2})$

(3.94)
$\tilde{\varphi}(\tilde{x},\tilde{t};\tilde{y},\tilde{s})=\frac{(\kappa_{\ell}^{c_{0}\theta})^{2}(\tilde{x}_{n-1}-\tilde{y}_{n-1})^{2}+|\bar{x}-\bar{y}|^{2}+r(x,y)}{4\big{(}\tilde{t}-\tilde{s}+(\tilde{x}_{n-1}-\tilde{y}_{n-1})\big{)}}-(\kappa_{\ell}^{c_{0}\theta})^{2}(\tilde{t}-\tilde{s}).$

Consequently, by the proof of (3.69) and (3.70) we have that for
$J_{\kappa_{\ell}^{c_{0}\theta}}$ as above

$B(\tilde{t},\tilde{s})=\frac{\partial^{2}\tilde{\varphi}}{\partial\tilde{x}_{j}\partial\tilde{y}_{k}}(0,\tilde{t};0,\tilde{s})=-\frac{1}{2(\tilde{t}-\tilde{s})}J_{\kappa_{\ell}^{c_{0}\theta}}+O(1),$

which yields (3.68) if $\delta$ is small enough. ∎

##  4\. Kernel estimates

In this section we finish up matters by proving the various kernel estimates
that we have utilized.

4.1. Basic kernel estimates on manifolds of nonpositive curvature

Let us prove the kernel estimates that we used on $A_{+}$.

######  Proposition 4.1.

Let $S_{\lambda}(x,t;y,s)$ denote the kernel

$\eta(t/T)\eta(s/T)\beta^{2}(P/\lambda)\bigl{(}e^{-i(t-s)\lambda^{-1}\Delta_{g}}\bigr{)}(x,y).$

Then if $M=M^{n-1}$ has nonpositive sectional curvatures and
$T=c_{0}\log\lambda$ with $c_{0}=c_{0}(M)>0$ sufficiently small, we have for
$\lambda\gg 1$

(4.1) $|S_{\lambda}(x,t;y,s)|\leq
C\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}}\exp(C_{M}|t-s|).$

To prove this we note that for fixed $t$ and $s$,
$\beta^{2}(P/\lambda)e^{-i(t-s)\lambda^{-1}\Delta_{g}}=\beta^{2}(P/\lambda)e^{i(t-s)\lambda^{-1}P^{2}}$
is the Fourier multiplier operator on $M^{n-1}$ with

(4.2)
$m(\lambda,t-s;\tau)=\beta^{2}(|\tau|/\lambda)e^{i(t-s)\lambda^{-1}\tau^{2}}.$

We have extended $m$ to be an even function of $\tau$ so that we can write

(4.3)
$\beta^{2}(P/\lambda)e^{-i(t-s)\lambda^{-1}\Delta_{g}}=(2\pi)^{-1}\int_{-\infty}^{\infty}\hat{m}(\lambda,t-s;r)\,\cos
r\sqrt{-\Delta_{g}}\,dr,$

where

(4.4) $\hat{m}(\lambda,t-s;r)=\int_{-\infty}^{\infty}e^{-i\tau
r}\beta^{2}(|\tau|/\lambda)\,e^{i(t-s)\lambda^{-1}\tau^{2}}\,d\tau.$

We note that, by a simple integration by parts argument,

(4.5)
$\partial^{k}_{r}\hat{m}(\lambda,t-s;r)=O(\lambda^{-N}(1+|r|)^{-N})\,\forall\,N,\,\,\text{if
}\,|t-s|\leq 1\,\,\text{and }\,\,\,|r|\geq C_{0},$

with $C_{0}$ sufficiently large. Similarly

(4.6)
$\partial^{k}_{r}\hat{m}(\lambda,t-s;r)=O(\lambda^{-N}(1+|r|)^{-N})\,\forall\,N,\\\
\text{if }\,|t-s|\in[2^{j-1},2^{j}],\,\,\text{and }\,\,|r|\geq
C_{0}2^{j},\,\,j=1,2,\dots,$

with $C_{0}$ fixed large enough. Since $\beta(|\tau|/\lambda)=0$ if
$|\tau|\notin[\lambda/4,2\lambda]$ one may take $C_{0}=100$, as we shall do.

To use this fix an even function $a\in C^{\infty}_{0}({\mathbb{R}})$
satisfying

$a(r)=1,\,\,|r|\leq 100\quad\text{and }\,\,a(r)=0\,\,\text{if }\,\,|r|\geq
200.$

Then by crude eigenfunction estimates and the Weyl formula, if we let

(4.7) $\tilde{S}_{\lambda,0}(x,t;y,s)=(2\pi)^{-1}\int
a(r)\hat{m}(\lambda,t-s,r)\cos rP\,dr$

we have

(4.8)
$\tilde{S}_{\lambda,0}(x,t;y,s)-\bigl{(}\beta^{2}(P/\lambda)e^{-i(t-s)\lambda^{-1}\Delta_{g}}\bigr{)}(x,y)=O(\lambda^{-N})\,\forall\,N\quad\text{if
}\,|t-s|\leq 1,$

and if

(4.9) $\tilde{S}_{\lambda,j}(x,t;y,s)=(2\pi)^{-1}\int
a(2^{-j}r)\hat{m}(\lambda,t-s,r)\cos rP\,dr$

we have

(4.10)
$\tilde{S}_{\lambda,j}(x,t;y,s)-\bigl{(}\beta^{2}(P/\lambda)e^{-i(t-s)\lambda^{-1}\Delta_{g}}\bigr{)}(x,y)=O(\lambda^{-N})\,\forall\,N\\\
\text{if }\,|t-s|\in[2^{j-1},2^{j}],\,\,j=1,2,\dots.$

Consequently, we would have (4.1) if we could show that

(4.11)
$|\tilde{S}_{\lambda,0}(x,t;y,s)|\leq\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}}\quad\text{when
}\,\,|t-s|\leq 1,$

as well as

(4.12)
$|\tilde{S}_{\lambda,j}(x,t;y,s)|\leq\lambda^{\frac{n-1}{2}}\exp(C2^{j}),\quad\text{if
}\,|t-s|\in[2^{j-1},2^{j}]\\\ \text{with }\,\,j=1,2,\dots\,\,\text{and
}\,\,\,2^{j}\leq c_{0}\log\lambda$

with $c_{0}=c_{0}(M)$ fixed small enough.

To prove (4.11) and (4.12) we shall argue as in Bérard [2] (see also [28,
§3.6]). Just as in [2], [6], [31] and other works we shall want to use the
Hadamard parametrix and the Cartan-Hadamard theorem to lift the calculations
that will be needed up to the universal cover $({\mathbb{R}}^{n-1},\tilde{g})$
of $(M^{n-1},g)$.

We therefore let $\\{\alpha\\}=\Gamma$ denote the group of deck
transfermations preserving the associated covering map
$\kappa:{\mathbb{R}}^{n-1}\to M^{n-1}$ coming from the exponential map at the
point in $M^{n-1}$ with coordinates $0$ in $\Omega$ in §4 above. The metric
$\tilde{g}$ on ${\mathbb{R}}^{n-1}$ is the pullback of the metric $g$ on
$M^{n-1}$ via $\kappa$. Choose a Dirichlet domain $D\simeq M^{n-1}$ for
$M^{n-1}$ centered at the lift of the point with coordinates $0$.

As in earlier works (see [28]) we recall that if $\tilde{x}$ denotes the lift
of $x\in M^{n-1}$ to $D$, then we have the following formula

(4.13) $(\cos tP)(x,y)=(\cos
t\sqrt{-\Delta_{g}})(x,y)=\sum_{\alpha\in\Gamma}\bigl{(}\cos
t\sqrt{-\Delta_{\tilde{g}}}\bigr{)}(\tilde{x},\alpha(\tilde{y})).$

As a result, if we set

(4.14) $K_{\lambda,0}(\tilde{x},t;\tilde{y},s)=(2\pi)^{-1}\int
a(r)\hat{m}(\lambda,t-s;r)\,\bigl{(}\cos
r\sqrt{-\Delta_{\tilde{g}}}\bigr{)}(\tilde{x},\tilde{y})\,dr,$

we have the formula

(4.15)
$\tilde{S}_{\lambda,0}(x,t;y,s)=\sum_{\alpha\in\Gamma}K_{\lambda,0}(\tilde{x},t;\alpha(\tilde{y}),s)).$

Similarly, if we set

(4.16) $K_{\lambda,j}(\tilde{x},t;\tilde{y},s)=(2\pi)^{-1}\int
a(2^{-j}r)\hat{m}(\lambda,t-s;r)\,\bigl{(}\cos
r\sqrt{-\Delta_{\tilde{g}}}\bigr{)}(\tilde{x},\tilde{y})\,dr,$

we have

(4.17)
$\tilde{S}_{\lambda,j}(x,t;y,s)=\sum_{\alpha\in\Gamma}K_{\lambda,j}(\tilde{x},t;\alpha(\tilde{y}),s).$

Also, by Huygen’s principle and the support properties of $a$, we have that

(4.18) $K_{\lambda,0}(\tilde{x},\tilde{y})=0\quad\text{if
}\,\,d_{\tilde{g}}(\tilde{x},\tilde{y})\geq C_{1},\,\,\,\text{and
}\,\,K_{\lambda,j}(\tilde{x},\tilde{y})=0\,\,\text{if
}\,\,d_{\tilde{g}}(\tilde{x},\tilde{y})\geq C_{1}2^{j}$

for a uniform constant $C_{1}$. Based on this, we conclude that the number of
non-zero summands in the right side of (4.15) is $O(1)$ since
$\alpha(D)\cap\alpha^{\prime}(D)=\emptyset$ if $\alpha\neq\alpha^{\prime}$.
Also, by simple volume estimates, the number of $\alpha\in\Gamma$ for which
$d_{\tilde{g}}(D,\alpha(D))\leq\mu$ is $O(\exp(C\mu))$ for a uniform constant
$C$ if $\mu=2^{j}$, $j=1,2,\dots$, and so the number of nonzero summands in
the right side of (4.16) is $O(\exp(C2^{j}))$. As a result, we would obtain
(4.11) if we could show that

(4.19) $|K_{\lambda,0}(\tilde{x},t;\tilde{y},s)|\leq
C\lambda^{\frac{n-1}{2}}|t-s|^{-\tfrac{n-1}{2}},\quad\text{if }\,\,|t-s|\leq
1,$

while (4.12) would follow from the estimate

(4.20) $|K_{\lambda,j}(\tilde{x},t;\tilde{y},s)|\leq
C\lambda^{\frac{n-1}{2}}\exp(C2^{j}),\\\ \text{if
}\,\,|t-s|\in[2^{j-1},2^{j}],\,\,j=1,2,\dots,\,2^{j}\leq c_{0}\log\lambda,$

with $c_{0}=c_{0}(M)$ sufficiently small.

To prove these two estimates, we can use the Hadamard parametrix for
$\partial_{r}^{2}-\Delta_{\tilde{g}}$ since $({\mathbb{R}}^{n-1},\tilde{g})$
is a Riemannian manifold without conjugate points, i.e., its injectivity
radius is infinite. Thus, we can use the Hadamard parametrix to write for
$\tilde{x}\in D$, $\tilde{y}\in{\mathbb{R}}^{n-1}$ and $|r|\geq c_{0}>0$

(4.21) $\bigl{(}\cos
r\sqrt{-\Delta}_{\tilde{g}}\bigr{)}(\tilde{x},\tilde{y})=\sum_{\nu=0}^{N}w_{\nu}(\tilde{x},\tilde{y})W_{\nu}(r,\tilde{x},\tilde{y})+R_{N}(r,\tilde{x},\tilde{y})$

where $w_{\nu}\in C^{\infty}({\mathbb{R}}^{n-1}\times{\mathbb{R}}^{n-1})$,

(4.22)
$W_{0}(r,\tilde{x},\tilde{y})=(2\pi)^{-(n-1)}\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}\cos
r|\xi|\,d\xi,$

while for $\nu=1,2,\dots$, $W_{\nu}(t,\tilde{x},\tilde{y})$ is a finite linear
combination of Fourier integrals of the form

(4.23)
$\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}e^{\pm
ir|\xi|}\alpha_{\nu}(|\xi|)\,d\xi,\,\,\text{with
}\,\,\alpha_{\nu}(\tau)=0,\,\text{for }\,\tau\leq 1\,\,\text{and
}\,\,\partial^{j}_{\tau}\alpha_{\nu}(\tau)\lesssim\tau^{-\nu-j},$

and, if $N_{0}$ is given, then if $N$ is large enough,

(4.24) $|\partial_{r}^{j}R_{N}(r,\tilde{x},\tilde{y})|\leq C\exp(Cr),\quad
0\leq j\leq N_{0}$

for a fixed constant $C$. Furthermore, the leading coefficient
$w_{0}(\tilde{x},\tilde{y})$ reflects the geometry of
$({\mathbb{R}}^{n-1},\tilde{g})$. Specifically, in geodesic normal coordinates
about $\tilde{x}$

$w_{0}(\tilde{x},\tilde{y})=\bigl{(}\text{det
}\tilde{g}_{ij}(\tilde{y})\bigr{)}^{-1/4}.$

Thus, if in geodesic polar coordinates the volume element is given by

$dV_{\tilde{g}}(\tilde{y})=\bigl{(}{\mathcal{A}}(r,\omega)\bigr{)}^{n-2}drd\omega,\,\,r=d_{\tilde{g}}(\tilde{x},\tilde{y}),$

then

$w_{0}(\tilde{x},\tilde{y})=\bigl{(}r/{\mathcal{A}}(r,\omega)\bigr{)}^{\frac{n-2}{2}}.$

By the classical Günther comparison theorem from Riemannian geometry (see [12,
§III.4])

(4.25) $w_{0}(\tilde{x},\tilde{y})\leq 1,$

and, moreover, for later use,
${\mathcal{A}}(r,\omega)\geq\tfrac{1}{K}\sinh(Kr)$ if all the sectional
curvatures are $\leq-K^{2}<0$, and so

(4.26) $w_{0}(\tilde{x},\tilde{y})\leq C_{K,N}\mu^{-N}\\\ \text{if
}\,\,d_{\tilde{g}}(\tilde{x},\tilde{y})\approx\mu\,\,\text{and }\,\,\text{all
the sectional curvatures of }\,M^{n-1}\,\,\text{are}\,\,\leq-K^{2}<0.$

The other coefficients in (4.21) are not as well behaved; however, Bérard [2]
showed that if $N_{0}$ is fixed

(4.27) $|\partial^{\beta}_{x}w_{\nu}(\tilde{x},\tilde{y})|\leq
C\exp(Cr),\,\,|\beta|,\nu\leq N_{0},\quad
r=d_{\tilde{g}}(\tilde{x},\tilde{y}),$

for some uniform constant $C$ (depending on $\tilde{g}$ and $N_{0}$).

The facts that we have just recited are well known. One can see, for instance,
[2] or [28, §1.1, §3.6] for background regarding the Hadamard parametrix, and
[31] for a discussion of properties of $w_{0}$.

Let us next use the Hadamard parametrix to prove (4.19). By (4.21), it
suffices to see that if we replace $(\cos
r\sqrt{-\Delta_{\tilde{g}}})(\tilde{x},\tilde{y})$ in (4.14) by each of the
terms in the right side of (4.21) then each such expression will satisfy the
bounds in (4.19).

Let us start with the contribution of the main term in the Hadamard parametrix
which is the $\nu=0$ term in (4.21). In view of (4.22) and (4.25) it would
give rise to these bounds if

(4.28)
$(2\pi)^{-n}\int_{-\infty}^{\infty}\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}\cos(r|\xi|)\,a(r)\,\hat{m}(\lambda,t-s;r)\,drd\xi\\\
=O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}})\quad\text{when
}\,\,\,|t-s|\leq 1.$

However, by (4.2) and (4.5) and the support properties of $a$, if $|t-s|\leq
1$

(4.29) $\displaystyle(2\pi)^{-1}$
$\displaystyle\int_{-\infty}^{\infty}\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}\cos(r|\xi|)\,a(r)\,\hat{m}(\lambda,t-s;r)\,drd\xi$
$\displaystyle=(2\pi)^{-1}\int_{-\infty}^{\infty}\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}\cos(r|\xi|)\hat{m}(\lambda,t-s;r)\,drd\xi+O(\lambda^{-N})$
$\displaystyle=\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}\beta^{2}(|\xi|/\lambda)e^{i(t-s)\lambda^{-1}|\xi|^{2}}\,d\xi+O(\lambda^{-N}).$

A simple stationary phase argument shows that the last integral is
$O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}})$, and so we conclude that
the main term in the Hadamard parametrix leads to the desired bounds.

To estimate the contributions of the higher order terms $\nu=1,2,\dots$, we
note that by the first part of (4.18) we may assume that
$d_{\tilde{g}}(\tilde{x},\tilde{y})$ is bounded. So, by (4.23) the higher
order terms would lead to the desired bounds since

(4.30) $(2\pi)^{-1}\iint e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}e^{\pm
ir|\xi|}\alpha_{\nu}(|\xi|)\,a(r)\hat{m}(\lambda,t-s;r)\,drd\xi\\\
=\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}\beta^{2}(|\xi|/\lambda)e^{i(t-s)\lambda^{-1}|\xi|^{2}}\,\alpha_{\nu}(|\xi|)\,d\xi+O(\lambda^{-N}),\,\,\text{if
}\,\,|t-s|\leq 1,$

and, by (4.23), together with a stationary phase argument, the last integral
is $O(\lambda^{\frac{n-1}{2}-\nu}|t-s|^{-\frac{n-1}{2}})$.

We also need to see that the remainder term in (4.21) leads to the bounds

(4.31)
$\int^{\infty}_{-\infty}a(r)\hat{m}(\lambda,t-s;r)R(r,\tilde{x},\tilde{y})\,dr\\\
=\int^{\infty}_{-\infty}\beta^{2}(|\tau|/\lambda)e^{i(t-s)\lambda^{-1}\tau^{2}}\,\bigl{[}a(\,\cdot\,)R(\,\cdot\,,\tilde{x},\tilde{y})\bigr{]}\,\widehat{}\,\,(\tau)\,d\tau=O(\lambda^{\frac{n-1}{2}}),\,\,\text{if
}\,\,d_{\tilde{g}}(\tilde{x},\tilde{y})=O(1).$

By (4.24), the last factor in the integral in the right, which is the Fourier
transform of $r\to a(r)R(r,\tilde{x},\tilde{y})$, is $O(1)$ if
$d_{\tilde{g}}(\tilde{x},\tilde{y})=O(1)$. So, by the support properties of
$\beta$, the last integral in (4.31) is
$O(\lambda)=O(\lambda^{\frac{n-1}{2}})$, as desired, since $n\geq 3$.

Since each term in the Hadamard parametrix has the desired contribution, the
proof of (4.19) is complete.

Similar arguments will yield (4.20). We need to see that if we replace $(\cos
r\sqrt{-\Delta_{\tilde{g}}})(\tilde{x},\tilde{y})$ in (4.16) by each of the
terms in the right side of (4.21), then each will satisfy the bounds in (4.20)
if $d_{\tilde{g}}(\tilde{x},\tilde{y})\leq C2^{j}$ and
$|t-s|\in[2^{j-1},2^{j}]$ with $j=1,2,\dots$ and $2^{j}\leq c_{0}\log\lambda$
as above.

By (4.6) and (4.25) and the above argument, the $\nu=0$ term in the Hadamard
parametrix will lead to a contribution of

$(2\pi)^{-(n-1)}\int_{{\mathbb{R}}^{n-1}}e^{id_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}}e^{i(t-s)\lambda^{-1}|\xi|^{2}}\beta^{2}(|\xi|/\lambda)\,d\xi+O(\lambda^{-N})=O(\lambda^{\frac{n-1}{2}}),$

by stationary phase and the fact that we are assuming $|t-s|\geq 1$. By (4.27)
and the above arguments each of the $\nu=1,2,\dots$ terms will have
contributions which are
$O(\lambda^{\frac{n-1}{2}-\nu}\cdot\exp(C2^{j}))=O(\lambda^{\frac{n-1}{2}})$
if $2^{j}\leq c_{0}\log\lambda$ with $c_{0}>0$ small enough. If we repeat the
argument above for the contribution of the remainder term, we see that the
contributions here will be of the form

$\int^{\infty}_{-\infty}\beta^{2}(|\tau|/\lambda)e^{i(t-s)\lambda^{-1}\tau^{2}}\bigl{[}a(2^{-j}\,\cdot\,)R(\,\cdot\,,\tilde{x},\tilde{y})\bigr{]}\,\widehat{}\,\,(\tau)\,d\tau,$

which, by (4.24) and the support properties of $a$, is
$O(\lambda\exp(C2^{j}))=O(\lambda^{\frac{n-1}{2}}\exp(C2^{j}))$, as desired.

Since each term in the Hadamard parametrix has the desired contribution, the
proof of (4.20) is complete, which finishes the proof of Proposition 4.1.

We should point out that the small $|t-s|$ estimates are universally true as
in [11].

4.2. Estimates for kernels of microlocalized operators

Let us prove the kernel estimates, (2.69) and (2.70), that we used in the
proof of $L^{q_{c}}(A_{-})$-estimates.

######  Proposition 4.2.

For each $m\in{\mathbb{Z}}$ pick $\nu(m)\in{\mathbb{Z}}^{2n-3}$ as in (2.46)
and let

(4.32) $K_{\lambda}(x,t,m;y,s,m^{\prime})=\\\
\alpha_{m}(t)\alpha_{m^{\prime}}(s)\Bigl{(}A_{\nu(m)}^{\theta_{0}}\circ\bigl{(}\,\beta^{2}(P/\lambda)e^{-i\lambda^{-1}(t-s)\Delta_{g}}\,\bigr{)}\circ(A^{\theta_{0}}_{\nu(m^{\prime})})^{*}\Bigr{)}(x,y).$

Then if $M=M^{n-1}$ has nonpositive curvature

(4.33) $|K_{\lambda}(x,t,m;y,s,m^{\prime})|\leq
C\lambda^{\frac{n-1}{2}}\,|t-s|^{-\frac{n-1}{2}},$

provided that $|t-s|\leq c_{0}\log\lambda$ with $c_{0}=c_{0}(M)>0$
sufficiently small. Moreover, for such $|t-s|$ we have

(4.34) $|K_{\lambda}(x,t,m;y,s,m^{\prime})|\leq
C\lambda^{\frac{n-1}{2}}\,|t-s|^{-N}\,\,\forall\,N,\\\ \text{if }\,|t-s|\geq
1\,\,\text{and all the sectional curvatures of }\,M^{n-1}\,\text{are
negative}.$

The uniform constants $C=C(M^{n-1})$ do not depend on the particular choice of
the $\nu(m)$.

###### Proof.

Since, as we mentioned before, the kernels of the $A_{\nu}^{\theta_{0}}$
operators satisfy the uniform bounds

(4.35)
$\int|A_{\nu}^{\theta_{0}}(x,y)|\,dx,\,\,\,\int|A_{\nu}^{\theta_{0}}(x,y)|\,dy\,\leq
C,$

by Proposition 4.1, we obtain the above bounds when $|t-s|\leq 1$.

Also, by (4.35), if we let

(4.36) $\tilde{K}_{\lambda}(x,t,m;y,s,m^{\prime})=\\\
\alpha_{m}(t)\alpha_{m^{\prime}}(s)\Bigl{(}A_{\nu}^{\theta_{0}}\circ\bigl{(}\,\beta^{2}(P/\lambda)e^{-i\lambda^{-1}(t-s)\Delta_{g}}\,\bigr{)}\Bigr{)}(x,y),\quad\nu=\nu(m),$

it suffices to see that this kernel, which does not include the microlocal
cutoffs in the right satisfies the bounds in (4.33) and (4.34) when $|t-s|\geq
1$.

Let us start by proving that (4.36) satisfies the bounds in (4.34) for
$|t-s|\geq 1$ if $M^{n-1}$ has nonpositive curvatures.

To do this recall that, by (2.42) and (2.46) with
$\theta=\theta_{0}=\lambda^{-1/8}$ , if $\nu(m)=(\theta_{0}k,\theta_{0}\ell)$
then

(4.37) $A_{\nu}^{\theta_{0}}(x,D)=\tilde{A}_{k}^{\theta_{0}}(x,D)\circ
b(\lambda^{-7/8}(P-\lambda\kappa_{\ell}^{\theta_{0}})),\quad\kappa^{\theta_{0}}_{\ell}=1+\theta_{0}\ell,\,\,|\ell|\lesssim\theta_{0}^{-1},\\\
\text{if
}\,\,\tilde{A}_{k}^{\theta_{0}}(x,D)=A_{k}^{\theta_{0}}(x,D)\circ\tilde{\Upsilon}(P/\lambda),$

with $b\in C^{\infty}_{0}((-1,1))$ and $\Upsilon$ as in (2.43). Here the
$A_{k}^{\theta_{0}}$ operators localize at scale $\theta_{0}=\lambda^{-1/8}$
about a geodesic $\overline{\gamma}_{k}$ in $\Omega$ due to (2.37)–(2.41).

By (4.37) we now have the following variant of (4.3)

(4.38) $\displaystyle
A_{\nu}^{\theta_{0}}\circ\bigl{(}\beta^{2}(P/\lambda)e^{-i(t-s)\lambda^{-1}\Delta_{g}}\bigr{)}$
$\displaystyle=\tilde{A}_{k}^{\theta_{0}}\circ\bigl{(}\beta^{2}(P/\lambda)\,b(\lambda^{-7/8}(P-\lambda\kappa^{\theta_{0}}_{\ell}))\,e^{-i(t-s)\lambda^{-1}\Delta_{g}}\bigr{)}$
$\displaystyle=\Bigl{(}\,(2\pi)^{-1}\int\hat{m}_{\ell}(\lambda,t-s;r)\,\bigl{(}\tilde{A}_{k}^{\theta_{0}}\circ\cos
r\sqrt{-\Delta_{g}}\bigr{)}\,dr\Bigr{)},$

if now (4.4) is replaced by

(4.39) $\hat{m}_{\ell}(\lambda,t-s;r)=\int_{-\infty}^{\infty}e^{-i\tau
r}\beta^{2}(|\tau|/\lambda)\,b(\lambda^{-7/8}(|\tau|-\lambda\kappa_{\ell}^{\theta_{0}}))e^{i(t-s)\lambda^{-1}\tau^{2}}\,d\tau.$

By a simple integration by parts argument we have the following analog of
(4.6)

(4.40)
$\partial_{r}^{k}\hat{m}_{\ell}(\lambda,t-s;r)=O(\lambda^{-N}(1+|r|)^{-N})\,\,\forall\,N\\\
\text{if }\,\,|t-s|\leq c_{0}\log\lambda\,\,\,\text{and }\,\,|r|\geq
100c_{0}\log\lambda.$

Thus, if $a$ is as in the proof of Proposition 4.1, since the dyadic operators
$\tilde{A}_{k}^{\theta_{0}}$ have kernels as in (4.35), if we insert a factor
of $(1-a(r/c_{0}\log\lambda))$ into the integral in the last term in (4.38)
the resulting kernels will be $O(\lambda^{-N})$ for all $N$. So, we have
reduced the proof of (4.33) to showing that we have the kernel estimates

(4.41)
$\bigl{(}W_{\lambda,\ell,t-s}\bigr{)}(x,y)=O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}}),\quad\text{if
}\,\,1\leq|t-s|\leq c_{0}\log\lambda,$

for small enough $c_{0}>0$ if

(4.42) $W_{\lambda,\ell,t-s}=(2\pi)^{-1}\int
a(r/c_{0}\log\lambda)\,\hat{m}_{\ell}(\lambda,t-s;r)\,\bigl{(}\tilde{A}_{k}^{\theta_{0}}\cos
r\sqrt{-\Delta_{g}}\bigr{)}\,dr.$

To estimate (4.41), we shall argue as in the last subsection. We first lift
the calculation up to the universal cover exactly as before by rewriting

(4.43)
$W_{\lambda,\ell,t-s}(x,y)=\sum_{\alpha\in\Gamma}W^{\alpha}_{\lambda,\ell,t-s}(\tilde{x},\tilde{y}),$

where

(4.44) $W^{\alpha}_{\lambda,\ell,t-s}(\tilde{x},\tilde{y})=\\\
(2\pi)^{-1}\int_{-\infty}^{\infty}a(r/c_{0}\log\lambda)\,\hat{m}_{\ell}(\lambda,t-s;r)\,\bigl{(}\tilde{A}_{k}^{\theta_{0}}\cos
r\sqrt{-\Delta_{\tilde{g}}}\bigr{)}(\tilde{x},\alpha(\tilde{y}))\,dr,$

and $\tilde{x}$, $\tilde{y}$ denote the lift of $x,y$, respectively, to the
universal cover $({\mathbb{R}}^{n-1},\tilde{g})$. By the support properties of
$a$ and Huygens principle

(4.45)
$W^{\alpha}_{\lambda,\ell,t-s}(\tilde{x},\alpha(\tilde{y}))=0\quad\text{if
}\,\,d_{\tilde{g}}(\tilde{x},\tilde{\alpha}(\tilde{y}))\geq
Cc_{0}\log\lambda,$

with $C$ being a fixed constant.

In the last subsection we had to deal with the fact that the sums that arose
after lifting the calculations up to the universal cover involved
$O(\exp(Cc_{0}\log\lambda))$ terms. Here, because of the $\nu=(k,\ell)$
localizations, it will turn out that, given $|t-s|\geq 1$, there are only
$O(1)$ summands above which are nontrivial.

Let us start by exploiting the localization coming from the
$\tilde{A}_{k}^{\theta_{0}}$ operators which localize about the geodesic
$\overline{\gamma}_{k}$ in $\Omega$. If we argue exactly in [6], just by using
this operator and elementary arguments involving the calculus of Fourier
integral operators and Toponogov’s triangle comparison theorem, we shall be
able to see that the overwhelming majority of the terms in (4.43) are
$O(\lambda^{-1/2})$, which is much better than the bounds posited above.

To do this, just as in earlier works we start by modifying the coordinates in
$\Omega$ so that the $0\in\overline{\gamma}_{k}$. Then, as in [6], we let
$\tilde{\gamma}(t)$, $t\in{\mathbb{R}}$ denote the lift of the geodesic
$\overline{\gamma}_{k}$ to the universal cover and

${\mathcal{T}}_{R}(\tilde{\gamma})=\\{x:\,d_{\tilde{g}}(\tilde{\gamma},\tilde{z})\leq
R\\}.$

Then, just as in [6], if $R$ is fixed large enough and
$\alpha(D)\cap{\mathcal{T}}_{R}(\tilde{\gamma})=\emptyset$, with, as before,
$D\simeq M^{n-1}$ being our fundamental domain, then the summand in (4.43)
involving $\alpha$ must by $O(\lambda^{-1})$ by Toponogov’s theorem and
microlocal arguments. This is exactly how (3.9) in [6] was proved, and one can
simply repeat the arguments there to obtain this bound.

Since there are $O(\lambda^{1/2})$ non-zero terms in (4.43) if $c_{0}$ in
(4.44) is fixed small enough, we obtain in this case

$\sum_{\\{\alpha:\,\alpha(D)\cap{\mathcal{T}}_{R}(\tilde{\gamma})=\emptyset\\}}W_{\lambda,\ell,t-s}^{\alpha}(\tilde{x},\tilde{y})=O(\lambda^{-1/2}),\quad\text{if
}\,1\leq|t-s|\leq c_{0}\log\lambda,$

which is much better than the bounds in (4.41).

In order to obtain (4.41) we still have to deal with the terms for which
$\alpha(D)\cap{\mathcal{T}}_{R}(\tilde{\gamma})\neq\emptyset$; however,
fortunately for us, by (4.45) there are only $O(\log|t-s|)$ such non-zero
terms in (4.43). Having reduced out task to only considering such summands we
no longer need to use the microlocal cutoff $\tilde{A}_{k}^{\theta_{0}}$.
Since it satisfies the bounds in (4.35) we would have (4.41) if

(4.46)
$\sum_{\\{\alpha:\,\alpha(D)\cap{\mathcal{T}}_{R}(\tilde{\gamma})\neq\emptyset\\}}(2\pi)^{-1}\int_{-\infty}^{\infty}a(r/c_{0}\log\lambda)\,\hat{m}_{\ell}(\lambda,t-s;r)\,\bigl{(}\cos
r\sqrt{-\Delta_{\tilde{g}}}\bigr{)}(\tilde{x},\alpha(\tilde{y}))\,dr\\\
=O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}}),\quad\text{if
}\,\,1\leq|t-s|\leq c_{0}\log\lambda.$

To do this, just like before, we shall use the Hadamard parametrix (4.21). We
need to see that the contribution of each term gives a contribution satisfying
the these bounds.

If we argue as before, and use (4.23) the contribution of the higher order
terms to (4.44) will be a linear combination of terms of the form

(4.47)
$(2\pi)^{-1}w_{\nu}(\tilde{x},\alpha(\tilde{y}))\int_{-\infty}^{\infty}\int
e^{id_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\xi_{1}}e^{\pm
ir|\xi|}\alpha_{\nu}(|\xi|)a(r/c_{0}\log\lambda)\\\
\times\hat{m}_{\ell}(\lambda,t-s;r)\,d\xi dr.$

Assuming as we are that $|t-s|\leq c_{0}\log\lambda$, modulo a
$O(\lambda^{-N})$ term, just as in (4.30), this equals

(4.48)
$w_{\nu}(\tilde{x},\alpha(\tilde{y}))\int_{{\mathbb{R}}^{n-1}}\beta^{2}(|\xi|/\lambda)b(\lambda^{-7/8}(|\xi|-\lambda\kappa^{\theta_{0}}_{\ell}))\,e^{id_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\xi_{1}}\,\alpha_{\nu}(|\xi|)\,e^{i(t-s)\lambda^{-1}|\xi|^{2}}\,d\xi.$

By an easy stationary phase calculation if $|t-s|\geq 1$ the last integral is
$O(\lambda^{\frac{n-1}{2}-\nu})$ in view of the last part of (4.23). Since, as
we noted before there are only $O(\log\lambda)$ summands in (4.46), we
conclude that the contribution of the higher order terms, $\nu=1,2,\dots$, in
Hadamard parametrix will be $O(\lambda^{\frac{n-1}{2}-\frac{1}{2}})$, which is
much better than we need for (4.46).

We next notice that, similar to (4.31), the contribution of the remainder term
in (4.21) will be

$\int_{-\infty}^{\infty}\beta^{2}(|\tau|/\lambda)e^{i(t-s)\lambda^{-1}\tau^{2}}b(\lambda^{-7/8}(|\tau|-\lambda\kappa^{\theta_{0}}_{\ell}))\bigl{[}a((c_{0}\log\lambda)^{-1}\,\cdot\,)R(\,\cdot\,,\tilde{x},\tilde{y})\bigr{]}\,\widehat{}\,\,(\tau)\,d\tau.$

By (4.24), the last factor in the integral is
$O(\exp(Cc_{0}\log\lambda))\leq\lambda^{1/16}$ if $c_{0}$ is small enough.
Since the rest of the integrand is bounded and supported on a set of size
$\lambda^{7/8}$, we conclude that the contribution of the remainder term in
the Hadamard parametrix to (4.41) also not only satisfies the bounds in
(4.46), but, moreover, like the above terms for $\nu=1,2,\dots$ in (4.21),
satisfies the improved ones in (4.34). Indeed, its contribution will be
$O(\lambda^{15/16}\log\lambda)$ for such $c_{0}$.

We still have to deal with the main term in the Hadamard parametrix, i.e., the
contribution of the $\nu=0$ term in (4.21) to (4.47). Arguing as before, the
proof of (4.46) would be complete if we could show that

(4.49)
$\sum_{\\{\alpha:\,\alpha(D)\cap{\mathcal{T}}_{R}(\tilde{\gamma})\neq\emptyset\\}}w_{0}(\tilde{x},\alpha(\tilde{y}))\int_{{\mathbb{R}}^{n-1}}\beta^{2}(|\xi|/\lambda)\,b(\lambda^{-7/8}(|\xi|-\lambda\kappa^{\theta_{0}}_{\ell}))\\\
\times\cos(r|\xi|)e^{id_{\tilde{g}}(\tilde{x},\alpha(\tilde{y})\xi_{1}}e^{i(t-s)\lambda^{-1}|\xi|^{2}}\,dr\\\
=O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}}),\quad\text{if
}\,\,1\leq|t-s|\leq c_{0}\log\lambda.$

To obtain (4.49) we shall use the fact that for $|t-s|\geq 1$ we have:

(4.50)
$\int_{{\mathbb{R}}^{n-1}}\beta^{2}(|\xi|/\lambda)b(\lambda^{-7/8}(|\xi|-\lambda\kappa))e^{id_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\xi_{1}}e^{i(t-s)\lambda^{-1}|\xi|^{2}}\,d\xi=O(\lambda^{-N})\,\forall
N\\\ \text{if }\,\,d_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\notin
I_{t,s,\kappa}=\bigl{[}2|t-s|(\kappa-C\lambda^{-1/8}),\,2|t-s|(\kappa+C\lambda^{-1/8})\big{]},\,\,\text{if
}\,\kappa=\kappa^{\theta_{0}}_{\ell}$

with $C$ large enough, and

(4.51)
$\int_{{\mathbb{R}}^{n-1}}\beta^{2}(|\xi|/\lambda)b(\lambda^{-7/8}(|\xi|-\lambda\kappa))e^{id_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\xi_{1}}e^{i(t-s)\lambda^{-1}|\xi|^{2}}\,d\xi=O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}}),\\\
\text{if }\,\,d_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\in
I_{t,s,\kappa},\,\,\,\kappa=\kappa^{\theta_{0}}_{\ell}.$

The first estimate just follows from a simple integration by parts argument.
For if
$\phi=d_{\tilde{g}}(\tilde{x},\tilde{y})\xi_{1}+(t-s)\lambda^{-1}|\xi|^{2}$,
then, if $C$ in the definition of $I_{t,s,\kappa}$ is fixed large enough,
$|\nabla_{\xi}\phi|\geq\lambda^{-1/8}$ if
$d_{\tilde{g}}(\tilde{x},\tilde{y})\notin I_{\kappa,t,s}$, and, also,
derivatives of the amplitude of the integral are $O(\lambda^{-7/8})$. Thus, in
this case, every integration by parts gains a power of $\lambda^{-3/4}$,
resulting in (4.50). The other estimate, (4.51) just follows from stationary
phase.

If we note that the interval $I_{t,s,\kappa}$ has length
$O(|t-s|\,\lambda^{-1/8})$ which is much smaller than 1, if as above we assume
that $|t-s|\leq c_{0}\log\lambda$, we conclude that there can only be $O(1)$
terms in (4.49) which are not $O(\lambda^{-N})$, which leads to (4.49) since,
by (4.25) $w_{0}$, is bounded.

This completes the proof of (4.33).

To prove the much stronger bounds (4.34) which require that $M^{n-1}$ have
negative sectional curvatures, we note that the contribution of all of the
terms in the Hadamard parametrix other than the main one, corresponding to
$\nu=0$, involved a $\lambda$-power improvement of what was needed for (4.33)
and thus lead to bounds of the form (4.34) since we are assuming that
$|t-s|=O(\log\lambda)$. Thus, to prove (4.34), it is enough to show that under
these curvature assumptions we have the analog of (4.49) with
$O(\lambda^{\frac{n-1}{2}}|t-s|^{-\frac{n-1}{2}})$ replaced by
$O(\lambda^{\frac{n-1}{2}}|t-s|^{-N})$ for every $N$. To do this, we also use
the simple fact, which follows from an integration by parts argument, that we
have the $O(\lambda^{-N})$ bounds in (4.50) if
$d_{\tilde{g}}(\tilde{x},\alpha(\tilde{y}))\notin[C_{1}^{-1}|t-s|,C_{1}|t-s|]$
with $C_{1}$ fixed sufficiently large. In view of (4.26) each of the $O(1)$
nontrivial terms in the sum in the left side of (4.49) must be
$O(\lambda^{\frac{n-1}{2}}|t-s|^{-N})$ for every $N$, as desired. This
completes the proof of Proposition 4.2. ∎

4.3. Estimates for kernels involving local auxiliary operators.

Let us prove the kernel estimates we used in §3.

Proof of Lemma 3.5. Let us now prove Lemma 3.5, which allowed us to use
parabolic scaling and results from [23] to obtain the bilinear estimates
(3.63). This lemma follows from a straightforward variation on the stationary
phase arguments used to prove [29, Lemma 5.1.2]. Moreover, Lemma 3.5 is
essentially Lemma 3.2 in [5] or Lemma 4.3 in [7], and in fact the latter
result almost immediately gives our results given how we have constructed the
local operators in (2.5).

We first note that the kernel of our local operators are given by

(4.52)
$\tilde{\sigma}_{\lambda}(x,t;y,s)=\bigl{(}B\circ\sigma_{\lambda}\bigr{)}(x,t;y,s)\\\
=(2\pi)^{-2}\iint
e^{i(t-s)\tau}e^{ir\lambda^{1/2}\tau^{1/2}}\tilde{\beta}(\tau/\lambda)\,\hat{\sigma}(r)\,\bigl{(}B\circ
e^{-irP}\bigr{)}(x,y)\,drd\tau.$

Since $(r,x,y)\to(B\circ e^{-irP})(x,y)$ is smooth when $d_{g}(x,y)\neq|r|$
and $\hat{\sigma}$ is as in (2.6), by a simple integration by parts argument,

(4.53)
$\tilde{\sigma}_{\lambda}(x,t;y,s)=O(\lambda^{-N})\,\forall\,N\quad\text{if
}\,\,|d_{g}(x,y)-\delta|\geq\tfrac{3}{2}\delta_{0}\delta.$

This leads to the first part of (3.35) if $\theta$ is small since the kernels
of our microlocal cutoffs, $A_{\nu}^{c_{0}\theta}$, satisfy

(4.54) $A^{c_{0}\theta}_{\nu}(x,y)=O(\lambda^{-N})\,\forall\,N\quad\text{if
}\,\,d_{g}(x,y)\geq C_{1}\theta,$

for a uniform constant $C_{1}$ since we are assuming that
$\lambda^{-1/2}\ll\lambda^{-1/8}\leq\theta$. Also, since the symbols
$A_{\nu}^{c_{0}\theta}(x,\xi)=0$ if $\xi$ is not in a small conic neighborhood
of $(0,\dots,0,1)\in{\mathbb{R}}^{n-1}$, it follows that
$(r,x,y)\to\bigl{(}B\circ e^{-irP}\circ A_{\nu}^{c_{0}\theta}\bigr{)}(x,y)$ is
smooth when $x_{n-1}-y_{n-1}<0$ and $\hat{\sigma}\neq 0$, which yields the
other half of (3.35) via another simple integration by parts argument.

Next, we recall that, by (2.46)
$A_{\nu}^{c_{0}\theta}=A^{c_{0}\theta}_{j}(x,D)\circ
A_{\ell}^{c_{0}\theta}(P)$, where $A_{j}$ localizes to a $c_{0}\theta$
neighborhood of a geodesic $\overline{\gamma}_{j}\in\Omega$ about which we
have chosen Fermi normal coordinates and $A_{\ell}^{c_{0}\theta}(P)$ is the
“height operator” given by (2.42). The other operator
$A_{\nu^{\prime}}^{c_{0}\theta}$ localizes at scale $c_{0}\theta$ to a
geodesic $\overline{\gamma}_{j^{\prime}}$ and height operator
$A_{\ell^{\prime}}^{c_{0}\theta}$ which are $\theta$-close to the above.

Next, let us use the fact that, by Lemma 3.2 in [4] or Lemma 4.3 in [7],222In
[7] different notation was used to denote the pseudodifferential cutoff $B$
due to the semiclassical notation there. if we just consider the localizations
coming from the ones arising from geodesics, we have, for
$\omega\approx\lambda$, that the following kernels on $M^{n-1}$ satisfy

(4.55) $\bigl{(}\tilde{\sigma}(\omega-P)\circ
A^{c_{0}\theta}_{\iota}\bigr{)}(x,y)=\omega^{\frac{n-2}{2}}e^{i\omega
d_{g}(x,y)}a_{\iota,\theta}(\omega;x,y)+O(\lambda^{-N}),\,\,\iota=j,j^{\prime},$

where the amplitude satisfies $a_{\iota,\theta}=0$ if (3.32) is valid, and,
additionally, since we are working in Fermi normal coordiates about
$\overline{\gamma}_{j}$

(4.56)
$\bigl{|}\partial^{i}_{\omega}\partial^{k}_{x_{n-1}}\partial^{k^{\prime}}_{y_{n-1}}D^{\beta}_{x,y}a_{\iota,\theta}(\omega;x,y)|\leq
C_{i,k,k^{\prime},\beta}\,\omega^{-i}\theta^{-|\beta|},\quad\iota=j,j^{\prime}.$

If $i=0$, this just follows [7, Lemma 4.3] and our choice of coordinates. In
order to get the $c_{0}\theta$-scale concentration as in (3.32) that we used
in the last section, we apply [7, Lemma 4.3] with $\theta$ there replaced by
$c_{0}\theta$. The fact that we also have a $\omega^{-1}$ improvement for each
$\omega$-derivative just comes from the fact that if we use parametrices for
$e^{-irP}$ to represent $e^{-i\omega d_{g}(x,y)}$ times the right side of
(4.55) as an oscillatory integral in the standard way, such as in [7], each
$\omega$-derivative brings down a factor of the phase function (normalized to
vanish at the stationary points) and so results in a $\omega^{-1}$
improvement, just as in standard stationary phase with parameters results (see
e.g., [29, Corollary 1.1.8]).

To obtain (3.32) for our kernels, we first note that, by (4.52) and (2.46),

(4.57) $\bigl{(}\tilde{\sigma}_{\lambda}\circ
A_{\nu}^{c_{0}\theta}\bigr{)}(x,t;y,s)\\\ =\Bigl{(}\bigl{[}\,(2\pi)^{-1}\int
e^{i\tau(t-s)}\,\tilde{\sigma}(\lambda^{1/2}\tau^{1/2}-P)\circ
A_{j}^{c_{0}\theta}\circ\tilde{\beta}(\tau/\lambda)\,d\tau\,\bigr{]}\circ
A_{\ell}^{c_{0}\theta}\Bigr{)}.$

If we consider the kernel of the operator inside the square brackets, by
(4.55), we can write it as

(4.58) $\displaystyle(2\pi)^{-1}$
$\displaystyle\int^{\infty}_{-\infty}e^{i\tau(t-s)}(\lambda\tau)^{\frac{n-2}{4}}e^{i\lambda^{1/2}\tau^{1/2}d_{g}(x,y)}a_{j,\theta}(\lambda^{1/2}\tau^{1/2};x,y)\,\tilde{\beta}(\tau/\lambda)\,d\tau+O(\lambda^{-N})$
$\displaystyle=(2\pi)^{-1}\lambda^{n/2}\int^{\infty}_{-\infty}e^{i\lambda[\tau(t-s)+\tau^{1/2}d_{g}(x,y)]}a_{j,\theta}(\lambda\tau^{1/2};x,y)\,\tau^{\frac{n-2}{4}}\tilde{\beta}(\tau)d\tau+O(\lambda^{-N}),$
$\displaystyle=\pi^{-1}\lambda^{n/2}\int_{0}^{\infty}e^{i\lambda[\tau^{2}(t-s)+\tau
d_{g}(x,y)]}a_{j,\theta}(\lambda\tau;x,y)\,\tau^{n/2}\,\tilde{\beta}(\tau^{2})\,d\tau+O(\lambda^{-N}),$

where $a_{j,\theta}$ is as in (4.55) and so vanishes when $x$ or $y$ is
outside a $O(c_{0}\theta)$-tube about $\overline{\gamma}_{j}$. Since the
kernel of $A_{\ell}^{c_{0}\theta}$ also satisfies (4.54), by combining (4.57)
and (4.58), we obtain (3.32) for $\nu=(c_{0}\theta j,c_{0}\theta\ell)$. The
same argument gives us this for $\nu^{\prime}=(c_{0}\theta
j^{\prime},c_{0}\theta\ell^{\prime})$.

Next, let us use (4.56)–(4.58) to prove the remaining parts of Lemma 3.5
saying that the kernels are also $O(\lambda^{-N})$ in the regions described by
(3.33) and (3.34) and that outside of these and the ones in (3.32) and (3.35)
(where we already know this), they are as in (3.30) and (3.31).

In order to do this we argue as in Hörmander [18] or more specifically as in
[29, §4.3] to see that the we can write the kernel of the height operators,
$m=\ell,\ell^{\prime}$, as

(4.59) $\displaystyle A_{m}^{c_{0}\theta}(x,y)$
$\displaystyle=\int_{{\mathbb{R}}^{n-1}}e^{i\varphi(x,y;\xi)}b\bigl{(}(c_{0}\theta\lambda)^{-1}(p(x,\xi)-\lambda\kappa_{m}^{c_{0}\theta})\bigr{)}\Upsilon(p(x,\xi)/\lambda)\,q(x,y,\xi)\,d\xi+O(\lambda^{-N})$
$\displaystyle=\lambda^{n-1}\int_{{\mathbb{R}}^{n-1}}e^{i\lambda\varphi(x,y;\xi)}b\bigl{(}(c_{0}\theta)^{-1}(p(x,\xi)-\kappa_{m}^{c_{0}\theta})\bigr{)}\Upsilon(p(x,\xi))\,q(x,y,\lambda\xi)\,d\xi+O(\lambda^{-N}),$

where $b\in C_{0}^{\infty}((-1,1))$ is as in (2.42), $\Upsilon$ as in (2.43),
and $p(x,\xi)$ is the principal symbol of $P$, $q\in S^{0}_{1,0}$,
$(2\pi)^{-(n-1)}-q\in S^{-1}_{1,0}$ and $\varphi$ is homogeneous of degree one
in $\xi$ and satisfies

(4.60) $\varphi(x,y;\xi)=\langle
x-y,\xi\rangle+O(|x-y|^{2}|\xi|),\quad\text{on supp }\,q.$

So, in particular,

(4.61) $\nabla_{\xi}\varphi=0\,\iff\,x=y,\,\,\,\text{and
}\,\nabla_{x}\varphi=\xi\,\,\text{as well as
}\,\frac{\partial^{2}\varphi}{\partial x\partial\xi}=I_{n-1}\,\,\text{if
}\,x=y.$

Indeed, to see this, one recalls that the Lax parametrix allows to write for
small $|t|$

$\bigl{(}e^{itP}\bigr{)}(x,y)=\int
e^{i\varphi(x,y;\xi)+itp(x,\xi)}q(x,y,t;\xi)\,d\xi,$

for $q\in S^{0}_{1,0}$ solving a transport equation and so
$(2\pi)^{-(n-1)}-q(0,x,y;\xi)\in S^{-1}_{1,0}$. Using this, and the fact that
the Fourier transform of $\tau\to
b\bigl{(}(c_{0}\theta\lambda)^{-1}(\tau-\lambda\kappa_{m}^{c_{0}\theta})\bigr{)}\Upsilon(\tau/\lambda)$
is $O(\lambda^{-N})$ and rapidly decreasing outside of a fixed interval about
the origin, allows one to argue as in [29, §4.3] or the previous two
subsections here to obtain (4.59).

In the regions where we do not already know that the kernel
$K^{c_{0}\theta}_{\lambda,\mu}$ in (3.30) is $O(\lambda^{-N})$, by (4.57),
(4.58) and (4.59), we can write

(4.62)
$K^{c_{0}\theta}_{\lambda,\mu}(x,y)=c\lambda^{\frac{n}{2}}\int_{0}^{\infty}\Bigl{[}\,\lambda^{n-1}\int_{{\mathbb{R}}^{2(n-1)}}e^{i\lambda[\tau
d_{g}(x,z)+\varphi(z,y;\xi)]}a_{\iota,\theta}(\lambda\tau;x,z)\,\tau^{\frac{n}{2}}\tilde{\beta}(\tau^{2})\\\
\qquad\qquad\qquad\times
b((c_{0}\theta)^{-1}(p(z,\xi)-\kappa_{m}^{c_{0}\theta}))\Upsilon(p(z,\xi)/\lambda)q(z,y;\lambda\xi)dzd\xi\,\Bigr{]}\,e^{i\lambda\tau^{2}(t-s)}d\tau,\\\
\mu=\nu,\nu^{\prime},\,\iota=j,j^{\prime},\,m=\ell,\ell^{\prime}.$

If we consider the oscillatory integral over ${\mathbb{R}}^{2(n-1)}$ in the
square brackets here, the phase function is

$\phi(z,\xi)=\phi(x,y,\tau;z,\xi)=\tau d_{g}(x,z)+\varphi(z,y;\xi).$

It has a unique stationary point when

$y=z\quad\text{and
}\,\tau\nabla_{z}d_{g}(x,z)=-\nabla_{z}\varphi(z,y;\xi)=-\xi,$

with the last inequality coming from the second part of (4.61). This
stationary point is non-degenerate by the last part of (4.61), and $\phi=\tau
d_{g}(x,y)$ there. Also, since $p(z,\nabla_{z}d_{g}(x,z))=1$ and
$p(z,\xi)=p(z,-\xi)$, we conclude that

(4.63) $\tau=p(z,\xi)\,\,\text{and }\,\,\varphi=0\quad\text{when
}\,\,\nabla_{z,\xi}\phi=0.$

Since $\theta\geq\lambda^{-1/8}\gg\lambda^{-1/2}$ we may use (4.61) and (4.63)
along with stationary phase to evaluate $\lambda^{n-1}$ times the oscillatory
integral inside the square brackets in (4.62). It must be of the form

(4.64) $\displaystyle e^{i\tau
d_{g}(x,y)}\tilde{a}_{\iota,\theta}(\lambda\tau;x,y)\,\tau^{\frac{n}{2}}\tilde{\beta}(\tau^{2})\,\tilde{b}\bigl{(}(c_{0}\theta)^{-1}(\tau-\kappa_{m}^{c_{0}\theta})\bigr{)}\tilde{q}(x,y;\lambda\tau)$
$\displaystyle=e^{i\tau
d_{g}(x,y)}a_{\iota,\theta}(\lambda\tau;x,y)\,\tau^{\frac{n}{2}}\tilde{\beta}(\tau^{2})\,b\bigl{(}(c_{0}\theta)^{-1}(\tau-\kappa_{m}^{c_{0}\theta})\bigr{)}\Upsilon(\tau/\lambda)q(y,y;-\lambda\tau\nabla_{y}d_{g}(x,y))$
$\displaystyle\quad+O(\lambda^{-3/4}).$

Here $\tilde{a}_{\iota,\theta}$ satisfies the bounds in (4.56), like $b$, the
smooth function $\tilde{b}$ vanishes outside of $[-1,1]$, and, finally,
$\tilde{q}\in S^{0}_{1,0}$.

If we combine (4.62) and (4.63), we conclude that

(4.65)
$K^{c_{0}\theta}_{\lambda,\mu}(x,y)=c\lambda^{\frac{n}{2}}\int_{0}^{\infty}e^{i\lambda[\tau
d_{g}(x,y)+\tau^{2}(t-s)]}\tilde{a}_{\iota,\theta}(\lambda\tau;x,y)\tau^{\frac{n}{2}}\tilde{\beta}(\tau^{2})\\\
\times\tilde{b}\bigl{(}(c_{0}\theta)^{-1}(\tau-\kappa_{m}^{c_{0}\theta})\bigr{)}\tilde{q}(x,y;\lambda\tau)\,d\tau,\,\,\mu=\nu,\nu^{\prime},\,\iota=j,j^{\prime},\,m=\ell,\ell^{\prime}.$

Now we shall prove (3.32)-(3.34), by a simple integration by parts argument,
we obtain (3.33) from (4.65), and, by using the properties of the amplitude
function $\tilde{a}_{\iota,\theta}(\lambda\tau;x,y)$, we have the assertion in
(3.34) that the amplitudes are $O(\lambda^{-N})$ when
$|(x_{1},\dots,x_{n-2}|+|(y_{1},\dots,y_{n-2})|$ is larger than a fixed
multiple of $\theta$ for both $\mu=\nu,\nu^{\prime}$ since
$|\nu-\nu^{\prime}|=O(\theta)$. For the last part of (3.34), saying that the
amplitudes are also trivial when
$|(x_{n-1}-y_{n-1})+2\kappa^{c_{0}\theta}_{\ell}(t-s)|$ is larger than a fixed
multiple of $\theta$, we use the fact that
$d_{g}(x,y)=x_{n-1}-y_{n-1}+O(\theta)$ in our Fermi normal coordinates if
(3.35) is valid and $x,y$ are in a $O(\theta)$-tube about
$\overline{\gamma_{j}}$ as in (3.32). By (3.34), along with the earlier steps,
we conclude that these kernels are $O(\lambda^{-N})$ in the regions described
by (3.32)–(3.35).

Also, since the phase function in (4.65) has a unique stationary point when
$\tau=-d_{g}(x,y)/2(t-s)$ which is non-degenerate, and since the phase equals
$-(d_{g}(x,y))^{2}/4(t-s)$ there, we conclude that the kernels in (4.64) must
be of the form (3.30). It is also straightforward that the amplitudes must
satisfy the estimates in (3.31) in the special cases where both $m_{1}$ and
$m_{2}$ are zero due to (4.56).

To prove the estimates (3.31) involving $(m_{1},m_{2})\neq(0,0)$, we first
note that

$(d_{g}(x,y))^{2}/4(t-s)+\tau
d_{g}(x,y)+\tau^{2}(t-s)=(t-s)\cdot\bigl{(}\tau+d_{g}(x,y)/2(t-s)\bigr{)}^{2},$

and, also, by (3.34) and (3.35) $t-s\approx-\delta$ when the kernel is non-
trivial. Therefore, by (4.65), the amplitude in (3.30) is of the form

(4.66) $a_{\lambda,\mu}(x,t;y,s)=\\\
\lambda^{\frac{1}{2}}\int_{-\infty}^{\infty}e^{-i\lambda\tau^{2}}h_{\lambda,\iota,\theta}\bigl{(}x,t;y,s;(\tau-\tfrac{(s-t)^{1/2}d_{g}(x,y)}{2(t-s)});\,\tfrac{1}{(s-t)^{1/2}c_{0}\theta}(\tau-(s-t)^{1/2}[\kappa_{m}^{c_{0}\theta}+\tfrac{d_{g}(x,y)}{2(t-s)}])\bigr{)}\,d\tau,\\\
\mu=\nu=(j,k),\nu^{\prime}=(j^{\prime},\ell^{\prime}),\,\,\iota=j,j^{\prime},\,\,m=\ell,\ell^{\prime},$

with

(4.67) $h_{\lambda,\iota,\theta}(x,t;y,s;u;r)\\\
=\tilde{a}_{\iota,\theta}(\lambda
u/(s-t)^{1/2};x,y)\,(u_{+}/(s-t)^{1/2})^{n/2}\,\tilde{\beta}\bigl{(}u^{2}/(s-t)\bigr{)}\,\,\tilde{q}\bigl{(}x,y;\lambda
u/(s-t)^{1/2}\bigr{)}\,\tilde{b}(r).$

Here $u_{+}=u$ if $u\geq 0$ and $0$ otherwise. What is important for us and
follows from the fact that $\tilde{a}_{\iota,\theta}$ satisfies(4.56), the
support properties of $\tilde{b}$ and $\tilde{\beta}$, as well as the fact
that $s-t$ is bounded away from zero and $\tilde{q}\in S^{0}_{1,0}$, is that
we have

(4.68) $h_{\lambda,\iota,\theta}(x,t;y,s;u;r)=0\quad\text{if }\,\,|u|+|r|\geq
C\,\,\,\text{or }\,\,|u|\leq C^{-1},$

for some fixed $C=C_{\delta}$, and, moreover

(4.69) $D^{\beta_{1}}_{t,s,u,r,x_{n-1},y_{n-1}}(\theta
D_{x^{\prime},y^{\prime}})^{\beta_{2}}h_{\lambda,\iota,\theta}(x,t;y,s;u;r)=O_{\beta_{1},\beta_{2}}(1).$

Let us now use this to prove (3.31) for $(m_{1},m_{2})\neq 0$. We shall first
consider the special case where $\mu=\nu$ and on
$x,y\in\overline{\gamma}_{j}$, which is a portion of the $(n-1)$-axis in the
Fermi normal coordinate system in which we are working. We then have
$d_{g}(x,y)=x_{n-1}-y_{n-1}$ if the kernel is nontrivial by (3.35). Note that
(3.33) tells us that the amplitude $a_{\lambda,\nu}$ in (3.30) is also very
highly concentrated on the Schrödinger curve where we also have
$x_{n-1}-y_{n-1}=-2\kappa^{c_{0}\theta}_{\ell}(t-s)$. With this in mind, let
us prove (3.31) when $\beta=0$, $m_{2}=0$ and $m_{1}=1$ and we are on this
Schrödinger curve. We then have
$-d_{g}(x,y)/2(t-s)\equiv\kappa_{\ell}^{c_{0}\theta}$. In this case we take
$\kappa_{m}^{c_{0}\theta}=\kappa^{c_{0}\theta}_{\ell}$, $\iota=j$ and
$\mu=\nu$ in (4.66) and see that we would have (3.31) for this special case if

(4.70) $\int
e^{-i\lambda\tau^{2}}\bigl{(}\partial_{u}h_{\lambda,\iota,\theta}\bigr{)}(x,t;y,s;\tau+(s-t)^{1/2}\kappa^{c_{0}\theta}_{\ell};((s-t)^{1/2}c_{0}\theta)^{-1}\tau)\,d\tau=O(\lambda^{-1/2}),$

as well as

(4.71) $\theta^{-1}\int\tau
e^{-i\lambda\tau^{2}}\bigl{(}\partial_{r}h_{\lambda,\iota,\theta}\bigr{)}(x,t;y,s;\tau+(s-t)^{1/2}\kappa^{c_{0}\theta}_{\ell};((s-t)^{1/2}c_{0}\theta)^{-1}\tau)\,d\tau\\\
=O(\lambda^{-1/2}).$

The first estimate, (4.70) just follows from stationary phase and (4.69). We
obtain the second estimate by realizing that, after integrating by parts, we
can rewrite the left side as

(4.72) $(2i\lambda\theta)^{-1}\int
e^{-i\lambda\tau^{2}}\,\frac{\partial}{\partial\tau}\Bigl{[}\bigl{(}\partial_{r}h_{\lambda,\iota,\theta}\bigr{)}(x,t;y,s;\tau+(s-t)^{1/2}\kappa^{c_{0}\theta}_{\ell};((s-t)^{1/2}c_{0}\theta)^{-1}\tau)\Bigr{]}\,d\tau\\\
=O((\lambda\theta^{2})^{-1}\lambda^{-1/2})=O(\lambda^{-1/2}),$

with the bounds in the right holding by (4.69) along with stationary phase and
the fact that $\theta\gg\lambda^{-1/2}$. In view of (4.69), it is clear that
by induction this argument will give the rest of (3.31) in this special case
where both $(x,t)$ and $(y,s)$ lie on this special Schrödinger curve.

If $x,y$ are in a $O(\theta)$-tube about $\overline{\gamma}_{j}$ with
$d_{g}(x,y)\approx\delta$ and we let $\gamma(\,\cdot\,)$ be the unit speed
geodesic in $M^{n-1}$ with $\gamma(0)=0$ and $\gamma(r)=x$, $r=d_{g}(x,y)$,
then the argument also yields

$\bigl{(}2\kappa^{c_{0}\theta}_{\ell}\partial_{r}-\partial_{t}\bigr{)}a_{\lambda,\nu}(\gamma(r),t;y,s)=O(1)\quad\text{if
}\,\,r=d_{g}(x,y)\,\,\text{and }\,r=-2\kappa^{c_{0}\theta}_{\ell}(t-s),$

due to the fact that the Schrödinger curve connecting $(x,t)$ and $(y,s)$ is
as in (3.27). Since we are working in Fermi normal coordinates about
$\overline{\gamma}_{j}$ this equals

$(2\kappa^{c_{0}\theta}\partial_{x_{n-1}}-\partial_{t})a_{\lambda,\nu}(x,t;y,s)+O(\theta|\nabla_{x}a_{\lambda,\nu}|),$

and the error term here is $O(1)$ by our known estimate in (3.31) where
$|\beta|=1$ and $m_{1}=m_{2}=0$. Thus, if there is a
$\kappa^{c_{0}\theta}_{\ell}$-speed Schrödinger curve connecting $(x,t)$ and
$(y,s)$ and the kernel is not $O(\lambda^{-N})$ we have (3.31) with $m_{1}=1$,
$m_{2}=0$ and $\mu=\nu$. By an induction argument, it must be valid for all
$(m_{1},m_{2},\beta)$ in this case.

If the kernel is non-trivial at at $(x,t;y,s)$, then by (3.34) and (3.35)
there is a Schrödinger curve as in (3.27) with associated speed
$\kappa=\kappa^{c_{0}\theta}_{\ell}+O(\theta)$, passing through $(x,t)$ and
$(y,s)$. So, by the above argument, we would we have (3.31) for $\mu=\nu$,
$m_{1}=1$ and $m_{2}=0$ if we had the following variants of (4.70) and (4.71):

(4.73) $\int
e^{-i\lambda\tau^{2}}\bigl{(}\partial_{u}h_{\lambda,\iota,\theta}\bigr{)}(x,t;y,s;\tau+(s-t)^{1/2}\kappa;((s-t)^{1/2}c_{0}\theta)^{-1}(\tau-(s-t)^{1/2}(\kappa^{c_{0}\theta}_{\ell}-\kappa))\,d\tau\\\
=O(\lambda^{-1/2}),$

as well as

(4.74) $\theta^{-1}\int(\tau-(\kappa^{c_{0}\theta}_{\ell}-\kappa))\cdot
e^{-i\lambda\tau^{2}}\\\
\times\bigl{(}\partial_{r}h_{\lambda,\iota,\theta}\bigr{)}(x,t;y,s;\tau+(s-t)^{1/2}\kappa;((s-t)^{1/2}c_{0}\theta)^{-1}(\tau-(s-t)^{1/2}(\kappa^{c_{0}\theta}_{\ell}-\kappa))\,d\tau\\\
=O(\lambda^{-1/2}).$

Just as with (4.70), (4.73) follows immediately from stationary phase
arguments and (4.68)–(4.69). We also get (4.74) since, as we mentioned before,
we must have $\kappa^{c_{0}\theta}_{\ell}-\kappa=O(\theta)$, and so the left
side of (4.74) splits into two terms, one of which is of the form (4.70),
while the other is of the form (4.71). Thus, (4.73) and (4.74) imply (3.31)
for $\mu=\nu$ when $m_{1}=1$ and $m_{2}=0$. Also, just as before, one obtains
the remaining cases of (3.31) by an induction argument.

Finally, since
$|\kappa^{c_{0}\theta}_{\ell}-\kappa^{c_{0}\theta}_{\ell^{\prime}}|,\,\theta|j-j^{\prime}|=O(\theta)$,
it is also clear that (3.31) also must hold when $\nu=(c_{0}\theta
j,c_{0}\theta\ell)$ is replaced by $\nu^{\prime}=(c_{0}\theta
j^{\prime},c_{0}\theta\ell^{\prime})$, which completes the proof of Lemma 3.5.

## References

  * [1] N. Anantharaman and S. Nonnenmacher.  Half-delocalization of eigenfunctions for the Laplacian on an Anosov manifold.  volume 57, pages 2465–2523. 2007.  Festival Yves Colin de Verdière.
  * [2] P. H. Bérard.  On the wave equation on a compact Riemannian manifold without conjugate points.  Math. Z., 155(3):249–276, 1977.
  * [3] M. D. Blair, X. Huang, and C. D. Sogge.  Improved spectral projection estimates.  preprint, arXiv:2211.17266.
  * [4] M. D. Blair and C. D. Sogge.  Refined and microlocal Kakeya-Nikodym bounds for eigenfunctions in two dimensions.  Anal. PDE, 8(3):747–764, 2014.
  * [5] M. D. Blair and C. D. Sogge.  Refined and microlocal Kakeya-Nikodym bounds of eigenfunctions in higher dimensions.  Comm. Math. Phys., 356(2):501–533, 2017.
  * [6] M. D. Blair and C. D. Sogge.  Concerning Toponogov’s theorem and logarithmic improvement of estimates of eigenfunctions.  J. Differential Geom., 109(2):189–221, 2018.
  * [7] M. D. Blair and C. D. Sogge.  Logarithmic improvements in $L^{p}$ bounds for eigenfunctions at the critical exponent in the presence of nonpositive curvature.  Invent. Math., 217(2):703–748, 2019.
  * [8] J. Bourgain.  Besicovitch type maximal operators and applications to Fourier analysis.  Geom. Funct. Anal., 1(2):147–187, 1991.
  * [9] J. Bourgain.  Fourier transform restriction phenomena for certain lattice subsets and applications to nonlinear evolution equations. I. Schrödinger equations.  Geom. Funct. Anal., 3(2):107–156, 1993.
  * [10] J. Bourgain and C. Demeter.  The proof of the $l^{2}$ decoupling conjecture.  Ann. of Math., 182:351–389, 2015.
  * [11] N. Burq, P. Gérard, and N. Tzvetkov.  Strichartz inequalities and the nonlinear Schrödinger equation on compact manifolds.  Amer. J. Math., 126(3):569–605, 2004.
  * [12] I. Chavel.  Riemannian geometry—a modern introduction, volume 108 of Cambridge Tracts in Mathematics.  Cambridge University Press, Cambridge, 1993.
  * [13] Y. Deng, P. Germain, and L. Guth.  Strichartz estimates for the Schrödinger equation on irrational tori.  J. Funct. Anal., 273(9):2846–2869, 2017.
  * [14] Y. Deng, P. Germain, L. Guth, and S. L. Rydin Myerson.  Strichartz estimates for the Schrödinger equation on non-rectangular two-dimensional tori.  Amer. J. Math., 144(3):701–745, 2022.
  * [15] J. Gell-Redman, S. Gomes, and A. Hassell.  Propagation of singularities and Fredholm analysis of the time-dependent Schrödinger equation.  preprint, arXiv:2201.03140.
  * [16] A. Gray.  Tubes, volume 221 of Progress in Mathematics.  Birkhäuser Verlag, Basel, second edition, 2004.
  * [17] A. Hassell and M. Tacy.  Improvement of eigenfunction estimates on manifolds of nonpositive curvature.  Forum Mathematicum, 27(3):1435–1451, 2015.
  * [18] L. Hörmander.  The spectral function of an elliptic operator.  Acta Math., 121:193–218, 1968.
  * [19] L. Hörmander.  Oscillatory integrals and multipliers on $FL^{p}$.  Ark. Mat., 11:1–11, 1973.
  * [20] X. Huang and C. D. Sogge.  Quasimode and Strichartz estimates for time-dependent Schrödinger equations with singular potentials.  Math Research Letters, 29:727–762, 2022.
  * [21] M. Keel and T. Tao.  Endpoint Strichartz estimates.  Amer. J. Math., 120(5):955–980, 1998.
  * [22] T. LaGatta and J. Wehr.  Geodesics of random Riemannian metrics.  Comm. Math. Phys., 327(1):181–241, 2014.
  * [23] S. Lee.  Linear and bilinear estimates for oscillatory integral operators related to restriction to hypersurfaces.  J. Funct. Anal., 241(1):56–98, 2006.
  * [24] F. K. Manasse and C. W. Misner.  Fermi normal coordinates and some basic concepts in differential geometry.  J. Mathematical Phys., 4:735–745, 1963.
  * [25] D. C. Sánchez and L. Esquivel.  Sharp Strichartz estimates for the Schrödinger equation on the sphere.  Journal of Pseudo-Differential Operators and Applications, 12:1–14, 2021.
  * [26] C. D. Sogge.  Oscillatory integrals and spherical harmonics.  Duke Math. J., 53(1):43–65, 1986.
  * [27] C. D. Sogge.  Concerning the $L^{p}$ norm of spectral clusters for second-order elliptic operators on compact manifolds.  J. Funct. Anal., 77(1):123–138, 1988.
  * [28] C. D. Sogge.  Hangzhou lectures on eigenfunctions of the Laplacian, volume 188 of Annals of Mathematics Studies.  Princeton University Press, Princeton, NJ, 2014.
  * [29] C. D. Sogge.  Fourier integrals in classical analysis, volume 210 of Cambridge Tracts in Mathematics.  Cambridge University Press, Cambridge, second edition, 2017.
  * [30] C. D. Sogge.  Improved critical eigenfunction estimates on manifolds of nonpositive curvature.  Math. Res. Lett., 24:549–570, 2017.
  * [31] C. D. Sogge and S. Zelditch.  On eigenfunction restriction estimates and $L^{4}$-bounds for compact surfaces with nonpositive curvature.  In Advances in analysis: the legacy of Elias M. Stein, volume 50 of Princeton Math. Ser., pages 447–461. Princeton Univ. Press, Princeton, NJ, 2014.
  * [32] G. Staffilani and D. Tataru.  Strichartz estimates for a Schrödinger operator with nonsmooth coefficients.  Comm. Partial Differential Equations, 27(7-8):1337–1372, 2002.
  * [33] E. M. Stein.  Oscillatory integrals in Fourier analysis.  In Beijing lectures in harmonic analysis (Beijing, 1984), volume 112 of Ann. of Math. Stud., pages 307–355. Princeton Univ. Press, Princeton, NJ, 1986.
  * [34] R. S. Strichartz.  Restrictions of Fourier transforms to quadratic surfaces and decay of solutions of wave equations.  Duke Math. J., 44(3):705–714, 1977.
  * [35] T. Tao, A. Vargas, and L. Vega.  A bilinear approach to the restriction and Kakeya conjectures.  J. Amer. Math. Soc., 11(4):967–1000, 1998.
  * [36] M. E. Taylor.  Pseudodifferential operators.  Princeton Mathematical Series, No. 34. Princeton University Press, Princeton, N.J., 1981.