2 The heat kernel of a generalized Laplacian

Let $\mathcal {E}$ be a Hermitian vector bundle over a compact Riemannian manifold M of dimension n. Consider $\operatorname {\Delta }$ to be a generalized Laplacian, i.e., a second-order differential operator which satisfies

$$\begin{align*}\sigma_2(\operatorname{\Delta})(x,\xi)=|\xi|^2 \cdot \operatorname{id}_{\mathcal{E}}. \end{align*}$$

For example, if $\nabla $ is a connection on $\mathcal {E}$ and $F \in \Gamma (\operatorname {End} \mathcal {E})$ , $F^*=F$ , then $\nabla ^*\nabla +F$ is a symmetric generalized Laplacian on $\mathcal {E}$ .

Suppose that $\operatorname {\Delta }$ is self-adjoint. Since M is compact, the spectrum of $\operatorname {\Delta }$ is discrete and $L^2(M,\mathcal {E})$ splits as an orthogonal Hilbert direct sum

$$\begin{align*}L^2(M,\mathcal{E})=\bigoplus_{\lambda \in \operatorname{Spec} \operatorname{\Delta}}^{\perp} E_{\lambda}, \end{align*}$$

where $E_{\lambda }$ is the eigenspace corresponding to the eigenvalue $\lambda $ of $\operatorname {\Delta }$ . Moreover, ${\dim E_{\lambda } < \infty }$ and by elliptic regularity, the eigensections are smooth (see, e.g., [Reference Bourguignon, Hijazi, Milhorat, Moroianu and Moroianu8]). Let $e^{-t\operatorname {\Delta }}$ be the heat operator defined as

$$\begin{align*}e^{-t\operatorname{\Delta}}\Phi=e^{-t\lambda} \Phi, \end{align*}$$

for any $\Phi \in E_{\lambda }$ , $\lambda \in \operatorname {Spec} \operatorname {\Delta }$ .

If we denote by $\lbrace \Phi _j \rbrace $ an orthonormal Hilbert basis of $\operatorname {\Delta }$ -eigensections, then the heat kernel $p_t(x,y)$ satisfies

$$\begin{align*}p_t(x,y)=\sum_{j}e^{-t\lambda_j} \Phi_j(x) \otimes \Phi_j^*(y) \end{align*}$$

in ${\mathcal C}^{\infty } \left ( (0,\infty ) \times M \times M \right )$ .

Recall that the $L^2$ -product of two sections $s_1, s_2 \in \Gamma (\mathcal {E})$ is given by

$$\begin{align*}\langle s_1,s_2 \rangle_{L^2(\mathcal{E})}=\int_M h_{\mathcal{E}}(s_1,s_2) \operatorname{dvol}_g ,\end{align*}$$

where g is the metric on M and $h_{\mathcal {E}}$ is the Hermitian product on $\mathcal {E}$ .

Let $y \in M$ be a fixed point. We work in geodesic normal coordinates defined by the exponential map

$$\begin{align*}\exp_y: T_yM\longrightarrow M. \end{align*}$$

Since M is compact, there exists a global injectivity radius $\epsilon $ . For x close enough to y ( $d(x,y) \leq \epsilon $ ), take $\operatorname {x} \in T_yM$ the unique tangent vector of length smaller than $\epsilon $ such that $x=\exp _y\operatorname {x}$ . Let

$$\begin{align*}\operatorname{j}(\operatorname{x})=\frac{\exp_y^* dx}{d\operatorname{x}}, \end{align*}$$

namely the pull-back of the volume form $dx$ on M through the exponential map $\exp _{y}$ is equal with $\operatorname {j}(\operatorname {x})d\operatorname {x}$ . More precisely,

$$\begin{align*}\operatorname{j}(\operatorname{x})=\vert \det \left( d_{\operatorname{x}}\exp_{x_0} \right) \vert={\det }^{1/2} \left( g_{ij}(\operatorname{x}) \right) .\end{align*}$$

Denote by $\tau _{x}^{y}: \mathcal {E}_x \longrightarrow \mathcal {E}_y$ the parallel transport along the unique minimal geodesic $x_s=\exp _{y} (s\operatorname {x})$ , where $s \in [0,1]$ , which connects the points x and y. The heat kernel $p_t(x,y)$ belongs to the space $ {\mathcal C}^{\infty } \left ( (0,\infty ) \times M \times M, \mathcal {E}_x \otimes \mathcal {E}_y^* \right )$ and $p_t(x,y)$ satisfies the heat equation

$$\begin{align*}\left( \partial_t+{\operatorname{\Delta}}_x \right) p_t(x,y)=0.\end{align*}$$

Furthermore, $ \lim _{t\rightarrow 0} P_t s=s,$ in $\Vert \cdot \Vert _0$ , for any smooth section $s \in \Gamma (M, \mathcal {E})$ , where

$$\begin{align*}(P_t s)(x)=\int_M p_t(x,y)s(y)dg(y),\end{align*}$$

where $dg(y)$ is the Riemannian density of the metric g. The next theorem is due to Minakshisundaram and Pleijel (see, for instance, [Reference Berger, Gauduchon and Mazet4, Reference Minakshisundaram and Pleijel21]).

Theorem 2.1 The heat kernel $p_t$ has the following asymptotic expansion near the diagonal:

$$ \begin{align*} p_t(x,y) \stackrel{t \searrow 0}{\sim} (4 \pi t)^{-n/2} e^{-\frac{d(x,y)^2}{4t}} \sum_{i=0}^{\infty} t^i \Psi_i (x,y), \end{align*} $$

where $\Psi _i: \mathcal {E}_y \longrightarrow \mathcal {E}_x $ are ${\mathcal C}^{\infty }$ sections defined near the diagonal. Moreover, the $\Psi _i$ ’s are given by the following explicit formulæ:

$$ \begin{align*} {}&\Psi_0(x,y)=\operatorname{j}^{-1/2}(\operatorname{x})\tau_{y}^{x}, \\ {}&\tau_{x}^y \Psi_i(x,y)=-\operatorname{j}^{-1/2}(\operatorname{x}) \int_{0}^{1}s^{i-1} \operatorname{j}^{-1/2}(x_s)\tau_{x_s}^{y} \operatorname{\Delta}_x \Psi_{i-1}(x_s,y)ds. \end{align*} $$

The asymptotic sum in Theorem 2.1 can be understood using truncation and bounds of derivatives as in [Reference Berline, Getzler and Vergne5]. We prefer the interpretation given in [Reference Melrose19], where the heat kernel $p_t$ is shown to belong to $\rho ^{-n} {\mathcal C}^{\infty }(M_H^2)$ on the parabolic blow-up space $M_H^2$ and to vanish rapidly at the temporal boundary face $\{ \omega _0 =0 \}$ (see Section 10).

Example 2.2 Let $\mathbb {T}^n=\left ( S^1\right )^n={\mathbb R}^n/(2 \pi {\mathbb Z})^n$ be the n-dimensional torus with the standard product metric $g=d\theta _1^2 \otimes \cdots \otimes d\theta _n^2$ . Consider the trivial bundle $\mathcal {E}= \underline {{\mathbb C}}$ over $\mathbb {T}^n$ with the standard metric $h_{\mathcal {E}}$ , the trivial connection $\nabla =d$ , and the zero endomorphism F. Let $\operatorname {\Delta }_1$ be the Laplacian on $\mathbb {T}^n$ given by the metric g. The eigenvalues of $\operatorname {\Delta }_1$ are $\{k_1^2+\cdots +k_n^2: k_1,\ldots ,k_n \in {\mathbb Z} \}$ . Let $\varphi _l(\xi )=\frac {1}{\sqrt {2\pi }} e^{il\xi }$ be the standard orthonormal basis of eigenfunctions of each $\operatorname {\Delta }_{S^1}$ . Then, for $\theta =(\theta _1,\ldots ,\theta _n) \in \mathbb {T}^n$ , the heat kernel $p_t$ of $\operatorname {\Delta }_1$ is the following:

$$\begin{align*}p_t(\theta,\theta)= \sum_{(k_1,\ldots,k_n) \in {\mathbb Z}^n} e^{-t(k_1^2+\cdots+k_n^2)} \varphi_{k_1}(\theta_1) \overline{\varphi_{k_1}(\theta_1)}\ldots \varphi_{k_n}(\theta_n) \overline{\varphi_{k_n}(\theta_n)}. \end{align*}$$

Since $\varphi _l(\xi )\overline {\varphi _l(\xi )}=\frac {1}{2\pi }$ , for any $\xi \in S^1$ , we get

$$\begin{align*}p_t(\theta,\theta)= \tfrac{1}{(2\pi)^n} \sum_{(k_1,\ldots,k_n) \in {\mathbb Z}^n} e^{-t(k_1^2+\cdots+k_n^2)}. \end{align*}$$

Remark that the Fourier transform of the function $f_t: {\mathbb R}^n \longrightarrow {\mathbb R}$ , $f_t(x)=e^{-t \vert x \vert ^2}$ is given by

$$\begin{align*}\hat{f_t}(\xi)=\tfrac{\pi^{n/2}}{t^{n/2}} e^{-\frac{\vert \xi \vert^2}{4t}}. \end{align*}$$

Using the multidimensional Poisson formula (see, for instance, [Reference Bellman3]), we obtain that

$$ \begin{align*} p_t(\theta,\theta)=\tfrac{1}{(2 \pi)^n} \sum_{k \in {\mathbb Z}^n} f_t(k)=\sum_{k \in {\mathbb Z}^n} \hat{f_t}(2 \pi k)=\tfrac{\pi^{n/2}}{(2 \pi )^n} t^{-n/2} + \tfrac{\pi^{n/2}}{(2 \pi )^n} t^{-n/2} \sum_{k \in {\mathbb Z}^n \setminus \{ 0 \}} e^{-\frac{\pi^2 \vert k \vert^2}{t}}. \end{align*} $$

Since the last sum is of order ${\mathcal O} \left ( e^{-\frac {1}{t}} \right )$ as $t \rightarrow 0$ , it follows that the first coefficient in the asymptotic expansion at small-time t of $p_t$ is $\tfrac {\pi ^{n/2}}{(2 \pi )^n}$ and all the others vanish.

From now on, suppose that $\operatorname {\Delta }$ is non-negative (i.e., $h_{\mathcal {E}} \left ( \operatorname {\Delta } f, f \right ) \geq 0$ , for any $f \in {\mathcal C}^{\infty } (M,\mathcal {E})$ ). For $s \in {\mathbb C}$ , we define the complex powers $\operatorname {\Delta }^{-s} \in \Psi ^{-2s} \left ( M, \mathcal {E} \right )$ of $\operatorname {\Delta }$ as

$$\begin{align*}\operatorname{\Delta}^{-s} \Phi= \left\{ \begin{array}{ll} \lambda^{-s}\Phi, & \mbox{ if } \Phi \in E_{\lambda}, \ \lambda \neq 0, \\ 0, & \mbox{ if } \Phi \in \operatorname{Ker} \operatorname{\Delta}. \end{array} \right. \end{align*}$$

Remark that $(\operatorname {\Delta }^s)_{s \in {\mathbb C}}$ is a holomorphic family of pseudodifferential operators. Let $r \in (0,1)$ . We denote by $h_t$ the heat kernel of $\operatorname {\Delta }^r$ , namely the Schwartz kernel of the operator $e^{-t\operatorname {\Delta }^r}$ . We have seen that

(2.1) $$ \begin{align} p_t(x,x) \stackrel{t \searrow 0}{\sim} t^{-n/2} \sum_{j=0}^{\infty} t^{j}a_{j}(x,x), \end{align} $$

with smooth sections $a_j(x,x) \in \mathcal {E}_x \otimes \mathcal {E}_x^*$ .

3 The link between the heat kernel and complex powers of the Laplacian

Proposition 1 (Mellin Formula)

With the notations above, for $\Re s>0$ , we have

$$\begin{align*}\operatorname{\Delta}^{-s}=\frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1} \left( e^{-t\operatorname{\Delta}} - \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}} \right) dt , \end{align*}$$

where $\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}$ is the orthogonal projection onto the kernel of $\operatorname {\Delta }$ .

We will write $\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ for the Schwartz kernel $\sum _{k} \varphi _k(x) \otimes \varphi _k^*(y)$ , where $\{ \varphi _k \}$ is an orthonormal basis in $\operatorname {Ker} \operatorname {\Delta }$ . Denote by $q_{-s}$ the Schwartz kernel of the operator $\operatorname {\Delta }^{-s}$ . Let us first study the poles and the zeros of $q_{-s}$ away from the diagonal.

Proof For $\Re s>0$ , let $f_{x,y}(s)= \int _{0}^{\infty } t^{s-1}\left ( p_t(x,y)-\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y) \right ) dt $ . Remark that

$$ \begin{align*} f_{x,y}(s)&=\int_{0}^{\infty} t^{s-1}\left( p_t(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) \right) dt \\&={} \int_{1}^{\infty} t^{s-1} \left( p_t(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) \right) dt \\ &\quad+{} \int_{0}^{1} t^{s-1}p_t(x,y) dt - \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) \cdot \int_{0}^{1} t^{s-1}dt. \end{align*} $$

Since $p_t(x,y)-\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ decays exponentially fast as t goes to $\infty $ , the first integral is absolutely convergent in $ C^k$ norms. The heat kernel $p_t$ vanishes with all of its derivatives as $t \searrow 0$ in the compact K, thus the second integral is also absolutely convergent. The last integral term is well-defined for $\Re s>0$ , and it extends to a meromorphic function on ${\mathbb C}$ with a simple pole in $s=0$ . Therefore, $s \mapsto f_{x,y}(s)$ extends to a meromorphic function on ${\mathbb C}$ . By Proposition 1 and the identity theorem, the equality of meromorphic functions

$$\begin{align*}\Gamma(s)q_{-s}(x,y)= f_{x,y}(s) \end{align*}$$

holds for any $s \in {\mathbb C}$ . In particular, we obtain $q_0(x,y)=- \operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ . Furthermore, ${q_{-s}}_{\vert _{K}}$ is an entire function and vanishes in $s=-1,-2,\ldots $ .

Now we check the behavior of $q_{-s}$ along the diagonal. It is no longer holomorphic there, and the coefficients $a_j(x,x)$ from (2.1) appear as residues of $q_{-s}(x,x)$ .

Proof Consider the function $f_{x,x}(s)=\int _{0}^{\infty } t^{s-1}\left ( p_t(x,x)- \operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,x) \right ) dt$ for $\Re s> \frac {n}{2}$ . We have

$$ \begin{align*} f_{x,x}(s)&=\int_{0}^{\infty} t^{s-1}\left( p_t(x,x)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x) \right) dt \\&={} \int_{1}^{\infty} t^{s-1} \left( p_t(x,x)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x) \right) dt \\ &\quad+{} \int_{0}^{1} t^{s-1}p_t(x,x) dt - \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x) \cdot \int_{0}^{1} t^{s-1}dt. \end{align*} $$

The first integral is absolutely convergent, as seen in the proof of Proposition 2. The last integral term is meromorphic with a simple pole at $s=0$ with residue $-\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,x)$ . Let us analyze the behavior of the second term $A_x(s)=\int _{0}^{1} t^{s-1}p_t(x,x)dt$ .

Using (2.1), we have that for $N \geq 0$ ,

$$\begin{align*}t^{n/2}p_t(x,x)=\sum_{j=0}^N t^j a_j(x,x)+R_{N+1}(t,x), \end{align*}$$

where $R_{N+1}$ is of order ${\mathcal O} (t^{N+1})$ as $t \to 0$ . Furthermore, we obtain

$$ \begin{align*} A_x(s)=\!\int_{0}^{1}\! t^{s-\frac{n}{2}-1} t^{\frac{n}{2}}p_t(x,x)dt=\!&\sum_{j=0}^N \int_{0}^{1}\! t^{s-\frac{n}{2}-1} t^j a_j(x,x)dt + \!\int_{0}^{1} \!t^{s-\frac{n}{2}-1} R_{N+1}(t,x)dt \\ =\!&\sum_{j=0}^N a_j(x,x) \frac{1}{s-\frac{n}{2}+j} + \int_{0}^{1} t^{s-\frac{n}{2}-1} R_{N+1}(t,x)dt. \end{align*} $$

Thus $s \mapsto A_x(s)$ extends to a meromorphic function on ${\mathbb C}$ with simple poles in ${\{ \frac {n}{2}-j : \ j=\overline {0,N+1} \}}$ . Using again Proposition 1 and the identity theorem, we deduce the equality

$$\begin{align*}\Gamma(s)q_{-s}(x,x)=f_{x,x}(s),\end{align*}$$

for any $s \in {\mathbb C}$ . It follows that $\Gamma (s) q_{-s}(x,x)$ is meromorphic on ${\mathbb C}$ with simple poles in $s \in \lbrace 0 \rbrace \cup \lbrace \frac {n}{2}-j : j \in {\mathbb N} \rbrace $ . Moreover, the residue of $\Gamma (s)q_{-s}(x,x)$ in a pole $\frac {n}{2}-j$ is $a_j(x,x)$ , and the conclusion follows.

For $p\in {\mathbb C}$ and $\epsilon>0$ , let $B_{\epsilon }(p)$ be the open disk centered in p of radius $\epsilon $ . We need the following technical result.

4 The behavior of quotients of Gamma functions along vertical lines

A fundamental result used in [Reference Bär and Moroianu2] is the Legendre duplication formula

$$\begin{align*}\frac{\Gamma(s)}{\Gamma \left( \frac{s}{2} \right) } = \tfrac{1}{\sqrt{2\pi}} 2^{s- \frac{1}{2}} \Gamma\left( \frac{s+1}{2} \right), \end{align*}$$

together with the rapid decay of the Gamma function in vertical lines $\Re s = \tau $ (see, e.g., [Reference Paris and Kaminski22]). These results are replaced in our case by the following estimate.

Proof For $z \in {\mathbb C} \setminus {\mathbb R}_{-}$ , recall the Stirling formula (see, for instance, [Reference Whittaker and Watson23])

$$\begin{align*}\log \Gamma(z)=\left( z-\frac{1}{2} \right) \log z -z + \frac{1}{2} \log (2\pi) + \Omega(z), \end{align*}$$

where $\log $ is defined on its principal branch, and $\Omega $ is an analytic function of z. For $|\arg z|<\pi $ and $|z| \to \infty $ , $\Omega $ can be written as

$$\begin{align*}\Omega(z)=\sum_{j=1}^{N-1} \frac{B_{2j}}{2j(2j-1)z^{2j-1}}+R_N(z), \end{align*}$$

where $B_{2j}$ are the Bernoulli numbers $\left ( B_2=\frac {1}{6}, \ B_4=-\frac {1}{30}, \ B_6=\frac {1}{42}, \text {etc.} \right )$ . Moreover, the error term satisfies

$$\begin{align*}|R_N(z)| \leq \frac{|B_{2N}|}{2N(2N-1)} \cdot \frac{\sec^{2N}(\frac{\arg z}{2})}{|z|^{2N-1}}; \end{align*}$$

thus, $R_N(z)$ is of order ${\mathcal O} \left ( |z|^{-2N+1} \right )$ as $|z| \to \infty $ (see, for instance, [Reference Paris and Kaminski22, equation (2.1.6)]). For $s \notin (-\infty ,0)$ , it follows that

$$ \begin{align*} \frac{\Gamma(s)}{\Gamma(rs)}=s^{-s(r-1)}e^{s(r-1)}r^{\frac{1}{2}-rs}e^{\Omega(s)-\Omega(rs)}. \end{align*} $$

Let $s=a+ib$ , $a \in {\mathbb R}$ fixed. As $|b| \to \infty $ , the difference $\vert \Omega (s)-\Omega (rs) \vert \to 0$ ; thus, $\vert e^{\Omega (s)-\Omega (rs)} \vert \to 1$ . Note that $\vert r^{\frac {1}{2}-rs} \vert = \vert r^{\frac {1}{2}-ra} \vert $ and $\vert e^{(r-1)s} \vert = e^{(r-1)a}$ , so these terms are bounded. We show in Lemma 4.1 that for any $k \geq 0$ , $\vert s \vert ^{k} \vert s^s \vert $ goes to $0$ as $\Re s =a$ is fixed and $\vert \operatorname {Im} s \vert $ tends to $\infty $ . It follows that the quotient $\frac {\Gamma (s)}{\Gamma (rs)}$ indeed decreases in vertical lines faster than $\vert s \vert ^{-k}$ , for any $k \geq 0$ , uniformly in vertical strips.

Proof Let $s=a+ib \notin (-\infty ,0)$ and set $\log (a+ib)=x+iy$ . Then $x=\log \sqrt {a^2+b^2}$ , $y=\arg s \in (-\pi ,\pi )$ ; hence,

$$\begin{align*}\vert s^{s+k} \vert=\vert e^{(k+a+ib) \log (a+ib)} \vert= e^{(k+a)x-by} = e^{(k+a) \log \sqrt{a^2+b^2}-b \arg s}. \end{align*}$$

Since $b=\tan \arg s \cdot a$ , the exponent is equal to

(4.1) $$ \begin{align} &(k+a)\log \sqrt{a^2+b^2}-b \arg s \\&\quad= (k+a) \log a +\frac{k+a}{2} \log \left( 1+\tan^2 \arg s \right) -a \tan \arg s \cdot \arg s. \nonumber\end{align} $$

If $a>0$ , then $\arg s\nearrow \frac {\pi }{2}$ or $ \arg s \searrow -\frac {\pi }{2}$ , and in both cases $t:=\tan \arg s$ tends to $\infty $ . The exponent (4.1) behaves as the function $t \longmapsto \log (1+t^2)-t$ ; therefore, as $t \to \infty $ , the exponent goes to $-\infty $ and the statement of the claim follows.

If $a<0$ , then $\arg s \searrow \frac {\pi }{2}$ or $\arg s \nearrow -\frac {\pi }{2}$ . In the first case when $\arg s \searrow \frac {\pi }{2}$ , it follows that $ t = \tan \arg s \to -\infty $ . The exponent (4.1) behaves as $ \pm \log (1+t^2)+t $ ; hence, the conclusion follows. While if $\arg s \nearrow -\frac {\pi }{2}$ , then $t \to \infty $ , and the exponent (4.1) behaves as $ \pm \log (1+t^2)-t $ ; thus, the exponent tends again to $-\infty $ . Therefore, $\vert s^{k+s} \vert $ goes to zero, which ends the proof.

5 Link between the complex powers of $\operatorname {\Delta }$ and the heat kernel of $\operatorname {\Delta }^r$

Proposition 6 (Inverse Mellin Formula)

For $\Re \tau>0$ , the operators $e^{-t\operatorname {\Delta }^r}$ and $\operatorname {\Delta }^{-s}$ are related by the following formula:

$$\begin{align*}e^{-t \operatorname{\Delta}^r} -\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}= \frac{1}{2\pi i}\int_{\Re s= \tau} t^{-s} \Gamma(s) \operatorname{\Delta}^{-rs} ds. \end{align*}$$

Set $\tau> \frac {n}{2r}$ . Then the Schwartz kernel $q_{-rs}$ of $\operatorname {\Delta }^{-rs}$ is continuous and by the inverse Mellin formula, we get an identity which relates the Schwartz kernels $h_t$ and $q_{-rs}$ :

$$ \begin{align*} h_t(x,y)- \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) ={}& \tfrac{1}{2 \pi i} \int_{\Re s=\tau} t^{-s}\Gamma(s) q_{-rs}(x,y) ds \\ ={}&\tfrac{1}{2 \pi i} \int_{\Re s = \tau} t^{-s}\frac{\Gamma(s)}{\Gamma(rs)} \cdot \Gamma(rs)q_{-rs}(x,y) ds. \end{align*} $$

Now let $k>0$ . By changing $\tau $ to $\tau + \epsilon $ (for a small $\epsilon>0$ ) if needed, we can assume that $\tau -k \notin \lbrace \frac {n}{2}-j :j \in {\mathbb N} \rbrace \cup \{ 0 \}$ . Using Propositions 4 and 5, we can apply the residue formula and move the line of integration to the left:

(5.1) $$ \begin{align} \begin{aligned} h_t(x,y) ={}&\tfrac{1}{2 \pi i} \int_{\Re s = \tau -k} t^{-s}\frac{\Gamma(s)}{\Gamma(rs)} \cdot \Gamma(rs)q_{-rs}(x,y) ds \\ {}& +\sum_{s \in -{\mathbb N} \cup \lbrace \frac{n-2j}{2r} : \ j \in {\mathbb N} \rbrace} \operatorname{Res}_{s} \left( t^{-s} \Gamma(s) q_{-rs}(x,y) \right) + \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y). \end{aligned} \end{align} $$

Notice that $-{\mathbb N} \cup \lbrace \frac {n-2j}{2r} : \ j \in {\mathbb N} \rbrace $ is the set of all possible poles of $s \mapsto \Gamma (s) q_{-rs}(x,y)$ , but some of them might actually be regular points. We will study the sum (5.1) in detail in Theorems 6.1 and 7.1.

Let K be a compact set in $M \times M \setminus \operatorname {Diag}$ and $l \in {\mathbb N}$ . Remark that the integral term in (5.1) is of order ${\mathcal O} \left ( t^{k-\tau } \right )$ in ${\mathcal C}^l (K, \mathcal {E} \boxtimes \mathcal {E}^*)$ . Indeed,

$$ \begin{align*} \left\Vert \int_{\Re s = \tau-k} t^{-s}\Gamma(s){q_{-rs}}_{\vert_K} ds \right\Vert{}_l \leq t^{-\tau+k} \cdot \int_{ s = \tau-k+i u} \left\Vert \frac{\Gamma(s)}{\Gamma(rs)} \cdot \Gamma(rs) {q_{-rs}}_{\vert_K} \right\Vert{}_l du, \end{align*} $$

and using again Propositions 4 and 5, the claim follows. Furthermore, when k goes to $\infty $ , we get

(5.2) $$ \begin{align} {h_t}_{\vert_K} \stackrel{t \searrow 0}{\sim} \sum_{\alpha=0}^{\infty} t^{\alpha} \cdot \operatorname{Res}_{s=-\alpha} \left( \Gamma(s) {q_{-rs}}_{\vert_K} \right) + t^0 \cdot {\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}}_{\vert_K}, \end{align} $$

The meaning of the asymptotic sign in (5.2) is that if we set $h_t^N$ to be the right-hand side in (5.2) restricted to $\alpha \leq N$ , then the difference $\vert \partial _t^j \left ( {h_t}_{\vert _K}- h_t^N \right ) \vert $ is of order ${\mathcal O} (t^{N+1-j})$ in ${\mathcal C}^l(K, \mathcal {E} \boxtimes \mathcal {E}^*)$ , for any $N,j \in {\mathbb N}$ .

Remark that using again Propositions 4 and 5, the integral term in (5.1) is of order ${\mathcal O} \left ( t^{k-\tau } \right )$ in ${\mathcal C}^l (\operatorname {Diag}, \mathcal {E} \otimes \mathcal {E}^*)$ . Therefore when k tends to $\infty $ , we obtain

(5.3) $$ \begin{align} {h_t}_{\vert_{\operatorname{Diag}}} \stackrel{t \searrow 0}{\sim} \sum_{\alpha \in \left( -{\mathbb N} \right) \cup \lbrace \frac{n-2j}{2r}: j \in {\mathbb N} \rbrace } t^{-\alpha} \cdot \operatorname{Res}_{s=\alpha} \left( \Gamma(s) {q_{-rs}}_{\vert_{\operatorname{Diag}}} \right) + t^0 \cdot {\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}}_{\vert_{\operatorname{Diag}}}, \end{align} $$

in the sense of the following:

Definition 5.1 Consider $l \in {\mathbb N}$ and let $A,B \subset {\mathbb R}$ . We say that $ {h_t}_{\vert _{\operatorname {Diag}}} \stackrel {t \searrow 0}{\sim } \sum _{\alpha \in A} t^{\alpha } {c_{\alpha }} + \sum _{\beta \in B} t^{\beta } \log t \cdot c_{\beta } $ if for any $k,N \in {\mathbb N}$ , the difference

$$\begin{align*}\partial_t^j \left( {h_t}_{\vert_{\operatorname{Diag}}}- \sum_{\alpha \leq N} t^{\alpha} {c_{\alpha}} - \sum_{\beta \leq N} t^{\beta} \log t \cdot c_{\beta} \right) \end{align*}$$

is of order ${\mathcal O} (t^{N+1-j} \log t)$ in ${\mathcal C}^l(\operatorname {Diag}, \mathcal {E} \otimes \mathcal {E}^*)$ .

6 The asymptotic expansion of $h_t$ away from the diagonal

Theorem 6.1 The Schwartz kernel $h_t$ of the operator $e^{-t\operatorname {\Delta }^r}$ is ${\mathcal C}^{\infty }$ on $[0,\infty )\times \left ( M \times M \setminus \operatorname {Diag} \right )$ . Furthermore, let $K \subset M \times M \setminus \operatorname {Diag}$ be a compact set. Then the Taylor series of ${h_t}_{\vert _{K}}$ as $t \searrow 0$ is the following:

$$\begin{align*}{h_t}_{\vert_K} \stackrel{t \searrow 0}{\sim} \sum_{j=1}^{\infty} t^{j} {q_{rj}}_{\vert_K} \frac{(-1)^j}{j!}. \end{align*}$$

Moreover, if $r=\frac {\alpha }{\beta }$ is rational with $\alpha ,\beta $ coprime, then the coefficient of $t^j$ vanishes for $j \in \beta {\mathbb N}^*$ .

Proof Let $j \in {\mathbb N}$ . Using Propositions 4 and 5, $(-s)(-s-1)\ldots (-s-j+1) t^{-s-j} \frac {\Gamma (s)}{\Gamma (rs) } \Gamma (rs){q_{-rs}}_{\vert _{K}} $ is $L^1$ integrable on $ \Re s = \tau -k$ in ${\mathcal C}^l (K, \mathcal {E} \boxtimes \mathcal {E}^*)$ , for sufficiently large k and for any $l \in {\mathbb N}$ . It follows that $h_t$ is ${\mathcal C}^{\infty }$ on $(0,\infty )\times \left ( M \times M \setminus \operatorname {Diag} \right )$ . By Proposition 2, the function $s \mapsto q_{-rs}(x,y)$ is entire for any $(x,y) \in K$ . Since $\operatorname {Res}_{s=-j} \Gamma (s)= \frac {(-1)^j}{j!} $ , using (5.2) we get

$$\begin{align*}{h_t}_{\vert_K} \stackrel{t \searrow 0}{\sim} \sum_{j=0}^{\infty} t^{j} {q_{rj}}_{\vert_K} \frac{(-1)^j}{j!} +{\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}}_{\vert_K}.\end{align*}$$

We obtained in the proof of Proposition 2 that ${q_0}_{\vert _K}=-{\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}}_{\vert _K}$ ; thus,

$$\begin{align*}{h_t}_{\vert_K} \stackrel{t \searrow 0}{\sim} \sum_{j=1}^{\infty} t^{j} {q_{rj}}_{\vert_K} \frac{(-1)^j}{j!}, \end{align*}$$

and therefore $h_{t_{\vert _K}}$ is ${\mathcal C}^{\infty }$ also at $t=0$ , and vanishes at order $1$ . Moreover, using again Proposition 2, if $r=\frac {\alpha }{\beta }$ is rational and j is a non-zero multiple of $\beta $ , then $q{_{rj}}_{\vert _K} \equiv 0$ and the conclusion follows.

7 The asymptotic expansion of $h_t$ along the diagonal

To obtain the coefficients in the asymptotic of $h_t$ along the diagonal as $t\searrow 0$ , we need to compute the residues from (5.3). Some of them are related to the heat coefficients $a_j$ ’s of $p_t$ due to Proposition 3. We will distinguish three cases. If n is even, $\Gamma (s)q_{-rs}(x)$ has simple poles in $\lbrace \frac {n}{2r},\frac {n-2}{2r},\ldots ,\frac {2}{2r} \rbrace \cup \lbrace 0,-1,\ldots \rbrace $ and the residues will give rise to real powers of t. If n is odd and either r is irrational or r is rational with odd denominator, $\Gamma (s)q_{-rs}(x)$ has simple poles in $\lbrace 0,-1,\ldots \rbrace \cup \lbrace \frac {n-2j}{2r} : j=0,1,\ldots \rbrace $ . Otherwise, if n is odd and r is rational with even denominator, then there exist some double poles which give rise to logarithmic terms in the asymptotic expansion of $h_t$ .

Theorem 7.1 Let $a_j(x,x)$ be the coefficients in ( 2.1 ) of the heat kernel $p_t$ of the non-negative self-adjoint generalized Laplacian $\operatorname {\Delta }$ . The asymptotic expansion of the Schwartz kernel $h_t$ of the operator $e^{-t\operatorname {\Delta }^r}$ , $r \in (0,1)$ along the diagonal when $t\searrow 0$ is the following:

1. (1) If n is even, then

   $$\begin{align*}{h_t}_{\vert_{\operatorname{Diag}}} \stackrel{t \searrow 0}{\sim} \sum_{j=0}^{n/2-1} t^{- \frac{n-2j}{2r}} \cdot A_{-\frac{n-2j}{2r}} + a_{n/2} + \sum_{j=1}^{\infty} t^j \cdot A_j. \end{align*}$$

   If $r=\frac {\alpha }{\beta }$ is rational, for $j=l \beta $ , $ l \in {\mathbb N}^*$ , we obtain that $q_{rj}(x,x)=(-1)^j \cdot j! \cdot a_{\frac {n}{2}+l \alpha }(x,x)$ , and the coefficient of $t^{l \beta }$ can be described more precisely as

   $$\begin{align*}A_{l \beta}=a_{\frac{n}{2}+l \alpha}. \end{align*}$$

2. (2) If n is odd and either $r \in {\mathbb R} \setminus {\mathbb Q}$ or the denominator of r is odd, then

   $$\begin{align*}{h_t}_{\vert_{\operatorname{Diag}}} \stackrel{t \searrow 0}{\sim} \sum_{j=0}^{(n-1)/2} t^{- \frac{n-2j}{2r}} \cdot A_{-\frac{n-2j}{2r}} + \sum_{j=1}^{\infty} t^j \cdot A_j + \sum_{j=1}^{\infty} t^{\frac{2j+1}{2r}} \cdot A_{\frac{2j+1}{2r}}. \end{align*}$$

   Moreover, if $r=\frac {\alpha }{\beta }$ is rational and $\beta $ is odd, then $ A_{l \beta } \equiv 0$ for any $l \in {\mathbb N}^*$ .

3. (3) If n is odd, $r=\frac {\alpha }{\beta }$ is rational and its denominator $\beta $ is even, then

   

In all these cases, the coefficients are

$$ \begin{align*} {}&A_{-\frac{n-2j}{2r}}(x)= \frac{\Gamma \left( \frac{n-2j}{2r} \right)} {\Gamma \left( \frac{n-2j}{2} \right)} \cdot \frac{1}{r} \cdot a_j(x,x), && A_{j}(x)=\frac{(-1)^j}{j!} \cdot q_{rj}(x,x), \\ {}&A_{\frac{2j+1}{2r}}(x)=\frac{\Gamma \left( - \frac{2j+1}{2r} \right)} {\Gamma \left( - \frac{2j+1}{2} \right)} \cdot \frac{1}{r} \cdot a_{\frac{n+2j+1}{2}}(x,x), && B_{l\frac{\beta}{2}}(x)=\frac{(-1)^{l\frac{\beta}{2}}}{r \left( l\frac{\beta}{2} \right) ! \Gamma \left( - l\frac{\beta}{2} \cdot r \right)} \cdot a_{\frac{n+l \alpha}{2}}(x,x), \end{align*} $$

$$\begin{align*}A_{l \frac{\beta}{2}}(x)= \frac{(-1)^{l \frac{\beta}{2}}}{(l \frac{\beta}{2})! \Gamma(-rl \frac{\beta}{2})} \cdot \operatorname{FP}_{s=-l \frac{\beta}{2}} \left( \Gamma(rs)q_{-rs}(x,x) \kern-1pt\right) + \operatorname{FP}_{s=-l \frac{\beta}{2}} \!\left(\kern-1pt \frac{\Gamma(s)}{\Gamma(rs)} \kern-1pt\right) \cdot \frac{a_{\frac{n+l\alpha}{2}(x,x)}}{r}. \end{align*}$$

7.1 The case when n is even

For $j \in \{0,1,\ldots ,n/2-1 \}$ , we have

(7.1) $$ \begin{align} \operatorname{Res}_{s=\frac{n-2j}{2r}} \left( t^{-s} \frac{\Gamma(s)}{\Gamma(rs)} \Gamma(rs)q_{-rs}(x,x) \right) =t^{-\frac{n-2j}{2r}} \cdot \frac{\Gamma(\frac{n-2j}{2r})}{\Gamma(\frac{n-2j}{2})} \cdot \frac{a_j(x,x)}{r}. \end{align} $$

The residue in $s=0$ is given by

$$ \begin{align*} \operatorname{Res}_{s=0} \left( t^{-s} \Gamma(s) q_{-rs}(x,x) \right) ={}& \operatorname{Res}_{s=0} \left( t^{-s} \frac{\Gamma(s)}{\Gamma(rs)} \Gamma(rs)q_{-rs}(x,x) \right) \\ ={}& r \cdot \frac{1}{r} \kern-1.2pt\left(\kern-1pt a_{\frac{n}{2}}(x,x) \kern1pt{-}\kern1pt \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x) \kern-1pt\right)\kern1.2pt{=}\kern1.2pt a_{\frac{n}{2}}(x,x)\kern1pt{-}\kern1pt\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x), \end{align*} $$

thus the coefficient of $t^0$ in the asymptotic expansion (5.3) is $a_{\frac {n}{2}}(x,x)$ .

7.1.1 The case when n is even and r is irrational

Let $j \in {\mathbb N}^*$ . Then

(7.2) $$ \begin{align} \operatorname{Res}_{s=-j} \left( t^{-s} \Gamma(s) q_{-rs}(x,x) \right) = t^j \frac{(-1)^j}{j!} \cdot q_{rj}(x,x). \end{align} $$

Therefore, in this case, the asymptotic expansion of $h_t$ is the following:

$$\begin{align*}h_t(x,x) \stackrel{t \searrow 0}{\sim} \sum_{j=0}^{n/2-1} t^{-\frac{n-2j}{2r}} \frac{\Gamma \left( \frac{n-2j}{2r} \right)}{\Gamma \left( \frac{n-2j}{2} \right)} \frac{a_j(x,x)}{r} + a_{\frac{n}{2}}(x,x) + \sum_{j=1}^{\infty} t^j \frac{(-1)^j}{j!} q_{rj}(x,x). \end{align*}$$

7.1.2 The case when n is even and $r=\frac {\alpha }{\beta }$ is rational with $(\alpha ,\beta )=1$

Some of the coefficients $q_{rj}(x,x)$ from (7.2) can be expressed in terms of the $a_{k}$ ’s from (2.1). Remark that $\frac {\Gamma (s)}{\Gamma (rs)}$ has simple poles in $\{-1,-2,\ldots \} \setminus \{ \frac {-1}{r}, \frac {-2}{r},\ldots \}$ . For $j \in {\mathbb N}^*$ , $s:=-\frac {j}{r} \in \{-1,-2,\ldots \}$ if and only if j is a multiple of $\alpha $ , which is equivalent to $s=\frac {-l\alpha }{r}=-l\beta $ for some $l \in {\mathbb N}^*$ . In this case, we obtain

$$ \begin{align*} \operatorname{Res}_{s=-l\beta} \left( t^{-s} \Gamma(s) q_{-rs}(x,x) \right) ={}& \operatorname{Res}_{s=-l\beta} \left( t^{-s} \frac{\Gamma(s)}{\Gamma(rs)} \Gamma(rs)q_{-rs}(x,x) \right) \\ ={}& t^{l\beta} r \cdot \frac{1}{r} a_{\frac{n}{2}+l\alpha}(x,x)=t^{l\beta} a_{\frac{n}{2}+l \alpha}(x,x). \end{align*} $$

Hence, for rational $r=\frac {\alpha }{\beta }$ , if $j=l\beta $ , $l \in {\mathbb N}^*$ , we conclude that

(7.3) $$ \begin{align} q_{rj}(x,x)=(-1)^j \cdot j! \cdot a_{\frac{n}{2}+l\alpha}(x,x), \end{align} $$

and $h_t(x,x)$ has the following asymptotic expansion as $t \searrow 0$ :

7.2 The case when n is odd

For $j \in \lbrace 0,1,\ldots ,(n-1)/2 \rbrace $ , the coefficient of $t^{-\frac {n-2j}{2r}}$ is computed as in (7.1). Furthermore, in $s=0$ ,

$$ \begin{align*} \operatorname{Res}_{s=0} \left( t^{-s} \Gamma(s) q_{-rs}(x,x) \right) ={}& \operatorname{Res}_{s=0} \left( t^{-s} \frac{\Gamma(s)}{\Gamma(rs)} \cdot \Gamma(rs)q_{-rs}(x,x) \right) \\ ={}& r \cdot \frac{-1}{r} \cdot \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x) =-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,x); \end{align*} $$

hence, there is no free term in the asymptotic expansion of $h_t$ as t goes to zero.

Now we have to compute the residues of the function $t^{-s} \Gamma (s) q_{-rs}(x,x)$ in ${s \in \{ -1,-2,\ldots \}}$ and $ s \in \{ \frac {-1}{2r}, \frac {-3}{2r},\ldots \}$ .

7.2.1 The case when n is odd and r is irrational

Then these sets are disjoint; thus, all poles of the function $\Gamma (s) q_{-rs}(x)$ are simple. For $j \in {\mathbb N}^*$ , the coefficient of $t^j$ is obtained as in (7.2). Furthermore, for $j \in {\mathbb N}$ , we get

(7.4) $$ \begin{align} \operatorname{Res}_{s=-\frac{2j+1}{2r}} \left( t^{-s} \frac{\Gamma(s)}{\Gamma(rs)} \cdot \Gamma(rs)q_{-rs}(x,x) \right) =t^{\frac{2j+1}{2r}} \cdot \frac{\Gamma(-\frac{2j+1}{2r})}{\Gamma(-\frac{2j+1}{r})} \cdot \frac{a_{\frac{n+2j+1}{2}}(x,x)}{r}. \end{align} $$

Therefore, the small-time asymptotic expansion of $h_t$ is the following:

$$ \begin{align*} h_t(x,x) \stackrel{t \searrow 0}{\sim}{}& \sum_{j=0}^{n/2-1} t^{-\frac{n-2j}{2r}} \cdot \frac{\Gamma \left( \frac{n-2j}{2r} \right)}{\Gamma \left( \frac{n-2j}{2} \right)} \cdot \frac{a_j(x,x)}{r} + \sum_{j=1}^{\infty} t^j \cdot \frac{(-1)^j}{j!} q_{rj}(x,x) \\ +{}& \sum_{j=0}^{\infty} t^{\frac{2j+1}{2r}} \cdot \frac{\Gamma \left( -\frac{2j+1}{2r} \right)}{\Gamma \left( - \frac{2j+1}{2} \right)} \cdot \frac{a_{\frac{n+2j+1}{2}}(x,x)}{r}. \end{align*} $$

7.2.2 The case when n is odd and $r=\frac {\alpha }{\beta }$ is rational

Consider the sets

$$ \begin{align*} A:=\{ -1,-2,\ldots \}, && B:=\{ \tfrac{-1}{2r},\tfrac{-3}{2r},\ldots \}, && C:=\{ \tfrac{-1}{r},\tfrac{-2}{r},\ldots \}. \end{align*} $$

Remark that A is the set of negative poles of $s \longmapsto t^{-s}\Gamma (s)q_{-rs}(x,x)$ , and $A \setminus C$ is the set of poles of the function $s \longmapsto \frac {\Gamma (s)}{\Gamma (rs)}$ . Clearly B and C are disjoint. Moreover, $A \cap C=\{ -l\beta : \ l \in {\mathbb N}^* \}$ . Furthermore, if $\beta $ is odd, then $A\cap B=\emptyset $ , and otherwise if $\beta $ is even, then $A\cap B=\{-l\frac {\beta }{2}: \ l \in 2 {\mathbb N}+1 \}$ . Such an $s=-\frac {2j+1}{2r}=l\frac {\beta }{2} \in A \cap B$ is a double pole for $\Gamma (s)q_{rs}(x)$ .

7.2.3 Suppose that $\beta $ is odd

Then A and B are disjoint. Thus, for $s=-\frac {2j+1}{2r} \in B$ , $j \in {\mathbb N}$ , the residue of $t^{-s}\Gamma (s)q_{rs}(x,x)$ is the one computed in (7.4).

For $s=-j \in A\setminus C$ (which means that $j \in {\mathbb N}^*$ , $\beta \nmid j$ ), the residue of $t^{-s}\Gamma (s)q_{-rs}(x,x)$ in s is the one computed in (7.2).

If $s=-l\beta =-\frac {l\alpha }{r} \in A\cap C$ for some $l \in {\mathbb N}^*$ , then $\Gamma (s)$ has a simple pole in s and by Proposition 3, (the meromorphic extension of) $q_{-rs}(x,x)$ vanishes at $s=-l\beta $ . Hence, the product $t^{-s}\Gamma (s)q_{-rs}(x,x)$ is holomorphic in $s=-l\beta $ and $t^{l\beta }$ , $l \in {\mathbb N}^*$ , does not appear in the asymptotic expansion.

Therefore, if $r=\frac {\alpha }{\beta }$ is rational and $\beta $ is odd, we obtain

7.2.4 Assume now that $\beta $ is even

For $s=-\frac {2j+1}{2r}\in B \setminus A$ ( $j \in {\mathbb N} $ with $\alpha \nmid 2j+1$ ), the residue is computed as in (7.4). For $s=-j \in A\setminus \left ( B\cup C \right )$ (namely $j \in {\mathbb N}^*$ , $\frac {\beta }{2} \nmid j$ ), the residue is computed as in (7.2).

For $s \in C \cap A$ (namely $s=-l\beta $ , $l \in {\mathbb N}^*$ ), the residue is again $0$ . Indeed, $\Gamma (s)$ has a simple pole in $-l\beta $ and by Proposition 3, (the meromorphic extension of) $q_{-rs}(x,x)$ vanishes in $-l\beta $ , thus $t^{l\beta }$ does not appear in the asymptotic expansion of $h_t$ .

Finally, if $s=-\frac {l\alpha }{2r}=-l\frac {\beta }{2} \in A \cap B$ , $l \in 2 {\mathbb N} +1$ , then s is a double pole for $\Gamma (s)q_{-rs}(x,x)$ . We write the Laurent expansions of the functions $t^{-s}$ , $\frac {\Gamma (s)}{\Gamma (rs)}$ , and $\Gamma (rs)q_{-rs}(x,x),$ respectively, in $s=-\frac {l\alpha }{2r}=-l\frac {\beta }{2}=:-k$ :

$$ \begin{align*} {}&t^{-s}=t^k-t^k \log t + {\mathcal O} (s+k)^2, \\ {}&\frac{\Gamma(s)}{\Gamma(rs)}=\frac{(-1)^k}{k! \cdot \Gamma(-kr)} (s+k)^{-1}+\cdots, \\ {}&\Gamma(rs)(q_{-rs}(x,x))=\frac{1}{r}a_{\frac{n+l\alpha}{2}}(x,x) (s+k)^{-1}+\cdots. \end{align*} $$

Thus, we finally obtain that

$$ \begin{align*} \operatorname{Res}_{s=-k}\left( t^{-s} \cdot \frac{\Gamma(s)}{\Gamma(rs)} \cdot \Gamma(rs)q_{-rs}(x,x)\right)={}& t^k \cdot \frac{(-1)^k}{k! \Gamma(-kr)} \cdot \operatorname{FP}_{s=-k} \left( \Gamma(rs)q_{-rs}(x,x) \right) \\ {}&+ t^k \operatorname{FP}_{s=-k} \left(\frac{\Gamma(s)}{\Gamma(rs)} \right) \cdot\frac{a_{\frac{n+l\alpha}{2}(x,x)}}{r} \\ {}&+t^k \log t \cdot \frac{(-1)^k}{k! \Gamma(-kr)} \frac{a_{\frac{n+l\alpha}{2}(x,x)}}{r}. \end{align*} $$

8 Non-triviality of the coefficients

Let us prove Theorem 1.1. Recall the definition of the zeta function of a non-negative self-adjoint generalized Laplacian $\Delta $ :

$$\begin{align*}\zeta_{\Delta}(s):=\sum_{\lambda \in \operatorname{Spec} \Delta \setminus \{ 0 \}} \lambda^{-s}=\int_{M} q_{-s}(x,x)dg(x). \end{align*}$$

This series is absolutely convergent for $\Re s> \frac {n}{2}$ and extends meromorphically to ${\mathbb C} $ with possible simple poles in the set

$$\begin{align*}\left\{ \frac{n}{2}-j : j \in {\mathbb N} \setminus \left\{ \frac{n}{2} \right\} \right\}\end{align*}$$

(see, for instance, [Reference Gilkey13]).

Consider the trivial bundle ${\mathbb C}$ over a compact Riemannian manifold M. As in [Reference Loya, Moroianu and Ponge17], let $\left ( \operatorname {\Delta } + \xi \right )_{\xi> 0}$ be a family of generalized Laplacians indexed by $\xi>0$ , and denote by $ q_{-s}^{\xi } $ the Schwartz kernels of the operators $(\operatorname {\Delta }+\xi )^{-s}$ . Note that for $\Re s> \frac {n}{2}$ ,

(8.1) $$ \begin{align} \int_M q_{-s}^{\xi} (x,x) dx = \operatorname{Tr} \left( \operatorname{\Delta} +\xi \right)^{-s}=\zeta_{\operatorname{\Delta}+\xi}(s) = \sum_{\lambda_j \in \operatorname{Spec} \operatorname{\Delta}} \left( \lambda_j +\xi \right)^{-s}. \end{align} $$

Since for $\Re s> \frac {n}{2}$ the sum is absolutely convergent, we obtain

$$\begin{align*}\frac{d}{d\xi} \zeta_{\operatorname{\Delta}+\xi} (s) =-s \cdot \sum_{\lambda_j \in \operatorname{Spec} \operatorname{\Delta}} \left( \lambda_j +\xi \right)^{-s-1}= -s\cdot \zeta_{ \operatorname{\Delta} +\xi}( s+1). \end{align*}$$

By induction, it follows that for $\Re s> \frac {n}{2}$ ,

(8.2) $$ \begin{align} \frac{d}{d\xi^k} \zeta_{\operatorname{\Delta}+\xi}(s)=(-1)^k s(s+1)\ldots(s+k-1) \cdot \zeta_{\operatorname{\Delta} +\xi}( s+k). \end{align} $$

Using the identity theorem, (8.2) holds true on ${\mathbb C}$ as an equality of meromorphic functions. Consider $s \in {\mathbb R} \setminus (-{\mathbb N})$ and $k \in {\mathbb N}$ large enough such that $s+k> \frac {n}{2}$ . Since $\zeta _{\operatorname {\Delta }+\xi }(s+k)$ is a convergent sum of strictly positive numbers, the right-hand side is non-zero. Thus, for any fixed $s \in {\mathbb R} \setminus (- {\mathbb N}) $ , on any open set $U \subset (0, \infty )$ , the function $\xi \longmapsto \zeta _{\operatorname {\Delta }+\xi }(s)$ is not identically zero on U, and by (8.1), $q_{-s}^{\xi }(x,x)$ cannot be constant zero on M. Hence, for $s=-rj \notin - {\mathbb N}$ , there exist $\xi _0 \in (0, \infty )$ and $x_0 \in M$ such that the coefficient $q_{rj}^{\xi _0}(x_0,x_0)$ of the asymptotic expansion of the Schwartz kernel $h_t$ of $e^{-t (\operatorname {\Delta }+\xi _0)^r}$ is non-zero.

Now suppose that $rj \in {\mathbb N}$ . Then $r=\frac {\alpha }{\beta }$ is rational and j is a multiple of $\beta $ , $j:=l\beta $ . If n is odd, we already proved in Theorem 7.1 that $t^{l\beta }$ does not appear in the asymptotic expansion of $h_t$ as $t \searrow 0$ . Furthermore, if n is even, by (7.3), $q_{rj}(x,x)$ is a non-zero multiple of the coefficient $a_{\frac {n}{2}+l\alpha }(x,x)$ in the asymptotic expansion (2.1) of the heat kernel $p_t$ . It is well known that the heat coefficients in (2.1) are non-trivial (see, for instance, [Reference Gilkey13]). It follows that all coefficients obtained in Theorem 7.1 indeed appear in the asymptotic expansion, proving Theorem 1.1.

9 Non-locality of the coefficients $A_j(x)$ in the asymptotic expansions

Let us prove Theorem 1.3. We give an example of an n-dimensional manifold and a Laplacian for which the coefficients $A_{j}(x)=\frac {(-1)^j}{j!} q_{rj(x,x)}$ , $j \in {\mathbb N}^*$ , $rj \notin {\mathbb N}$ appearing in Theorem 7.1 are not locally computable in the sense of Definition 1.1 (i). Let ${\mathbb {T}^n={\mathbb R}^n / \left ( 2\pi {\mathbb Z} \right )^n}$ be the n-dimensional torus from Example 2.2. Let $\operatorname {\Delta }_g$ be the Laplacian on $\mathbb {T}^n$ given by the metric $g=d\theta _1^2 +\cdots + d\theta _n^2$ .

Remark that the eigenvalues of $\operatorname {\Delta }_g$ are $\{ k_1^2 +\cdots +k_n^2 : k_1,\ldots ,k_n \in {\mathbb Z} \}$ . Let $ {\varphi _l(t)= \frac {1}{\sqrt {2\pi }} e^{il t} }$ be the standard orthonormal basis of eigenfunctions of each $\operatorname {\Delta }_{S^1}$ . Then, for $\Re s> \frac {n}{2}$ and $\theta =(\theta _1,\ldots ,\theta _n) \in \mathbb {T}^n$ , the Schwartz kernel of $\operatorname {\Delta }_g^{-s}$ is given by

$$\begin{align*}q_{-s}^{\operatorname{\Delta}_g} \left( \theta , \theta \right) = \sum_{ (k_1,\ldots,k_n) \in {\mathbb Z}^n \setminus \{ 0 \} } \left( k_1^2+\cdots+k_n^2 \right)^{-s} \varphi_{k_1}(\theta_1) \overline{\varphi_{k_1}(\theta_1)}\ldots \varphi_{k_n}(\theta_n) \overline{\varphi_{k_n}(\theta_n)}. \end{align*}$$

Consider the n-dimensional zeta function

$$\begin{align*}\zeta_n(s):= \sum_{(k_1,\ldots,k_n) \in {\mathbb Z}^n \setminus \{ 0 \}} \left( k_1^2 +\cdots+k_n^2 \right)^{-s}= \sum_{k \in {\mathbb N}^*} k^{-s} R_n(k), \end{align*}$$

where $R_n(k)$ is the number of representations of k as a sum of n squares. Since $\varphi _{l}(t) \overline {\varphi _{l}(t)} =\frac {1}{2\pi }$ for any $t \in S^1$ , it follows that

(9.1) $$ \begin{align} q_{-s}^{\operatorname{\Delta}_g} \left( \theta,\theta \right) = \frac{1}{(2 \pi)^n} \zeta_n(s), \end{align} $$

for any $\Re s> \frac {n}{2}$ , and clearly $q_{-s}^{\operatorname {\Delta }_g}$ is independent of $\theta $ .

Now let us change the metric locally on each component $S^1$ . Let U be an open interval in $S^1$ , and $\psi :S^1 \longrightarrow [0,\infty )$ a smooth function with $\operatorname {supp} \psi \subset U$ . Consider the new metric $\left ( 1+\psi (\theta ) \right ) d\theta ^2$ on each $S^1$ . Then there exist $p>0$ and an isometry $\Phi : \left ( S^1, \left ( 1+\psi (\theta ) \right ) d\theta ^2 \right ) \longrightarrow \left ( S^1, p^2 d\theta ^2 \right )$ . Remark that the Laplacian on $S^1$ given by the metric $p^2 d \theta ^2$ corresponds under this isometry to $p^{-2}$ times the Laplacian for the metric $d\theta ^2$ . Let

$$ \begin{align*} \tilde{g}= \sum_{j=1}^n \left( 1+ \psi(\theta_j) \right) d\theta_j^2 && g_p=\sum_{j=1}^n p^2 d\theta_j^2=p^2g. \end{align*} $$

Then clearly $\Phi \times \cdots \times \Phi : (\mathbb {T}^n, \tilde {g}) \longmapsto (\mathbb {T}^n, g_p) $ is an isometry, and let $\tilde {\Delta }$ , $\Delta _{p }$ be the corresponding Laplacians on $\mathbb {T}^n$ . Denote by $q_{-s}^{\tilde {\Delta }}$ and $q_{-s}^{\Delta _{p }}$ the Schwartz kernels of the complex powers $ \tilde {\Delta } ^{-s}$ and $\Delta _{p }^{-s}$ . We have for $\Re s>\frac {n}{2}$ ,

(9.2) $$ \begin{align} q_{-s}^{\operatorname{\Delta}_{p }} \left( \theta, \theta \right) = \frac{1}{(2 \pi p)^n} \sum_{k=(k_1,\ldots,k_n) \in {\mathbb Z}^n \setminus \{ 0 \}} \left( p^{-2}k_1^2 +\cdots+p^{-2}k_n^2 \right)^{-s} = \frac{p^{2s}}{(2 \pi p)^n} \zeta_n(s). \end{align} $$

Remark that

$$\begin{align*}q_{-s}^{\operatorname{\Delta}_{p}} \left( \theta,\theta \right) =q_{-s}^{\tilde{\operatorname{\Delta}}} \left( \Phi(\theta),\Phi(\theta) \right), \end{align*}$$

and both of them are independent of $\theta $ . By (9.2), for $\Re s>\frac {n}{2}$ , we obtain

(9.3) $$ \begin{align} q_{-s}^{\tilde{\operatorname{\Delta}}} \left( \theta,\theta \right) = \frac{p^{2s-n}}{(2 \pi )^n} \zeta_n(s). \end{align} $$

Now we prove that $\zeta _n(s)$ has a meromorphic extension on ${\mathbb C}$ with so-called trivial zeros at $s=-1,-2,\ldots $ . By Proposition 1, for $\Re s> \frac {n}{2}$ , we have

$$\begin{align*}\zeta_n(s) \Gamma(s)=\int_{0}^{\infty} t^{s-1} \sum_{k=(k_1,\ldots,k_n) \in {\mathbb Z}^n \setminus \{ 0 \}} e^{-t(k_1^2+\cdots+k_n^2)} dt = \int_0^{\infty} t^{s-1} F(t) dt, \end{align*}$$

where $F(t):= \sum _{k=(k_1,\ldots ,k_n) \in {\mathbb Z}^n \setminus \{ 0 \}} e^{-t(k_1^2+\cdots +k_n^2)}$ . Using the multidimensional Poisson formula (see, for instance, [Reference Bellman3]), it follows that

$$\begin{align*}1+F(t)= \sum_{k \in {\mathbb Z}^n} f_t(k)=\sum_{k \in {\mathbb Z}^n} \hat{f_t}(2 \pi k)=\pi^{n/2} t^{-n/2} \left( 1 + F \left( \frac{\pi^2}{t} \right) \right), \end{align*}$$

and therefore

$$\begin{align*}F(t)=-1+\pi^{n/2}t^{-n/2} + \pi^{n/2} t^{-n/2} F \left( \frac{\pi^2}{t} \right). \end{align*}$$

Since $F(t)$ goes to $0$ rapidly as $t \to \infty $ , the function $A(s)= \int _{1}^{\infty } t^{s-1} F(\pi t)dt $ is entire. Remark that

$$ \begin{align*} \zeta_n(s) \Gamma(s)={}& \int_{0}^{\pi} t^{-s} F(t)dt + \int_{\pi}^{\infty} t^{s-1}F(t) dt \\ ={}& \pi^s \left( -\frac{1}{s} + \frac{1}{s-\frac{n}{2}} + A\left( \frac{n}{2}-s \right) + A(s)\right), \end{align*} $$

so

(9.4) $$ \begin{align} \pi^{-s} \zeta_n(s) \Gamma(s)= -\frac{1}{s} + \frac{1}{s-\frac{n}{2}} + A\left( \frac{n}{2}-s \right) + A(s). \end{align} $$

Therefore, $\zeta _n$ extends meromorphically to ${\mathbb C}$ with a simple pole in $s=\frac {n}{2}$ and zeros at $s=-1,-2,\ldots $ . Furthermore, since the RHS is invariant through the involution $s \mapsto \frac {n}{2}-s$ , it follows that $\zeta _n(s)$ does not have any other zeros for $s \in (-\infty ,0)$ . We obtain the well-known functional equation of the Epstein zeta function

$$\begin{align*}\pi^{-s} \zeta_n(s) \Gamma(s)= \pi^{s-n/2} \zeta_n \left( \frac{n}{2}-s \right) \Gamma \left( \frac{n}{2}-s \right) \end{align*}$$

(see, for instance, [Reference Chandrasekharan and Narasimhan9, equation (63)]). Remark that for $r \in (0,1)$ and $j \in {\mathbb N}^*$ with $rj \notin {\mathbb N}$ , $\zeta _n (-rj )$ is not zero.

Using the identity theorem, it follows that (9.1) and (9.3) hold true as an equality of meromorphic functions on ${\mathbb C}$ , and furthermore, we get

$$\begin{align*}q_{rj}^{\operatorname{\Delta}_g}(\theta, \theta) \neq q_{rj}^{\tilde{\operatorname{\Delta}}}(\theta,\theta), \end{align*}$$

for $rj \notin {\mathbb N}$ . Since we modified the metric locally in $U^n \subset \mathbb {T}^n$ and the corresponding kernel $q_{rj}^{\tilde {\operatorname {\Delta }}}$ changed its behavior globally, it follows that it is not locally computable in the sense of Definition 1.1 (i).

Furthermore, let us see that the heat coefficients $A_j(x)=\frac {(-1)^j}{j!} q_{rj}(x,x)$ for $j={\mathbb N}^*$ , $rj \notin {\mathbb N}$ are not cohomologically local in the sense of Definition 1.1 (iii). We argue by contradiction. Let j be fixed. Suppose that there exists a function C, locally computable in the sense of Definition 1.1 (i), such that

(9.5) $$ \begin{align} \int_{\mathbb{T}^n} q_{rj}^{\operatorname{\Delta}_g} \operatorname{dvol}_g = \int_{\mathbb{T}^n} C(g) \operatorname{dvol}_g, && \int_{\mathbb{T}^n} q_{rj}^{\tilde{\operatorname{\Delta}}} \operatorname{dvol}_{\tilde{g}} = \int_{\mathbb{T}^n} C(\tilde{g}) \operatorname{dvol}_{\tilde{g}}. \end{align} $$

Using (9.1) and (9.3), it follows that

$$ \begin{align*} (2 \pi)^n \zeta_n(-rj) = \int_{\mathbb{T}^n} C(g) \operatorname{dvol}_g , && &{} (2 \pi p)^n p^{-2rj} \zeta_{n}(-rj)= \int_{\mathbb{T}^n} C(\tilde{g}) \operatorname{dvol}_{\tilde{g}}. \end{align*} $$

Remark that in the case of the trivial bundle with the trivial connection over a locally homogeneous Riemannian manifold $(M,h)$ (i.e., such that every two points have isometric neighborhoods), the function $C(M,h) \in {\mathcal C}^{\infty }(M)$ is constant on M. This follows directly from Definition 1.1 (i). Therefore, $C(g)$ , $C(\tilde {g}),$ and $C(g_p)$ are constant functions.

Since $(\mathbb {T}^n , \tilde {g})$ is (globally) isometric to $(\mathbb {T}^n , g_p)$ , it follows that $C(\tilde {g})=C(g_p)$ . Furthermore, since $(\mathbb {T}^n , g_p)$ is locally isometric to $(\mathbb {T}^n , g)$ and $C(g_p)$ , $C(g)$ are constant functions, it also follows that they are equal: $C(g_p)=C(g)$ . Hence we conclude that $C(\tilde {g})=C(g_p)=C(g)=:C$ , for some $C \in {\mathbb C}$ , and thus we have

(9.6) $$ \begin{align} \int_{\mathbb{T}^n} C \operatorname{dvol}_{\tilde{g}} = \int_{\mathbb{T}^n} C \operatorname{dvol}_{g_p}. \end{align} $$

Since $g_p=p^2 g$ , we obtain that

(9.7) $$ \begin{align} \int_{\mathbb{T}^n} C \operatorname{dvol}_{g_p} = p^n \int_{\mathbb{T}^n} C \operatorname{dvol}_g, \end{align} $$

and then using (9.5)–(9.7), we get

$$\begin{align*}(2 \pi p)^n p^{-2rj} \zeta_n(-rj) = p^n \cdot (2 \pi)^n \zeta_n(-rj). \end{align*}$$

But, we proved above that $\zeta _n (-rj)$ does not vanish for $rj \notin {\mathbb N}$ . We obtain a contradiction because $p^{-2rj} \neq 1$ for $r \in (0,1)$ , $j=1,2,\ldots $ .

10 Interpretation of $h_t$ on the heat space for $r=1/2$

In Theorems 6.1 and 7.1, we studied the asymptotic behavior of the heat kernel $h_t$ of ${ \operatorname {\Delta }^r}$ , $r \in (0,1)$ for small-time t in two distinct cases: when we approach $t=0$ along the diagonal in $M \times M$ , and when we approach a compact set away from the diagonal. We now give a simultaneous asymptotic expansion formula for both cases when $r=\frac {1}{2}$ . Furthermore, in order to understand the asymptotic behavior as t goes to zero in any direction (not just the case when t goes to $0$ in the vertical one),we will pull-back the formula on a certain linear heat space $\operatorname {M_{heat}}$ .

In [Reference Melrose19], Melrose used his blow-up techniques to give a conceptual interpretation for the asymptotic of the heat kernel $p_t$ . Recall that the heat space $M_H^2$ is obtained by performing a parabolic blow-up of $ \{ t=0 \} \times \operatorname {Diag} $ in $[0,\infty ) \times M \times M$ . The heat space $M_H^2$ is a manifold with corners with boundary hypersurfaces given by the boundary defining functions $\rho $ and $\omega _0$ . The heat kernel $p_t$ belongs to $\rho ^{-n} {\mathcal C}^{\infty }(M_H^2)$ , and vanishes rapidly at the boundary hypersurface $\{ \omega _0=0 \}$ (see [Reference Melrose19, Theorem 7.12]).

In order to study the Schwartz kernel $h_t$ of $e^{-t \operatorname {\Delta }^{1/2}}$ , we introduce the linear heat space $\operatorname {M_{heat}}$ , which is just the standard blow-up of $\{ 0 \} \times \operatorname {Diag}$ in $[0,\infty ) \times M \times M$ (see [Reference Melrose and Mazzeo20] for details regarding the blow-up of a submanifold). Let $\operatorname {ff}$ be the front face, i.e., the newly added face, and denote by $\operatorname {lb}$ the lateral boundary which is the lift of the old boundary $\{ 0 \} \times M \times M$ . The blow down map is given locally by

$$ \begin{align*} \beta_H: \operatorname{M_{heat}} \longrightarrow [0,\infty) \times M \times M && \beta_H(\rho, \omega, x')=(\rho \omega_0, \rho \omega'+x', x'), \end{align*} $$

where

$$\begin{align*}\omega \in {\mathbb S}^n_H= \{ \omega= (\omega_0, \omega') \in {\mathbb R}^{n+1} : \ \omega_0 \geq 0, \ \omega_0^2+|\omega'|^2 =1 \}. \end{align*}$$

Proof of Theorem 1.4

We want to show that $h_t \in \rho ^{-n} \omega _0 \cdot {\mathcal C}^{\infty } (\operatorname {M_{heat}}) + \rho \log \rho \cdot \omega _0 \cdot {\mathcal C}^{\infty } (\operatorname {M_{heat}}) $ , and in fact, the second (logarithmic) term does not occur when n is even. First, we deduce the unified formula for $h_t$ as $t \searrow 0$ both on the diagonal and away from it. By Mellin formula 1 and inverse Mellin formula 6, for $\tau>n$ , we get

$$ \begin{align*} h_t(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y)={}& \tfrac{1}{2\pi i} \int_{\Re s=\tau} t^{-s} \frac{\Gamma(s)}{\Gamma \left( \frac{s}{2} \right)} \Gamma \left(\frac{s}{2} \right)q_{-s/2}(x,y)ds \\ ={}& \tfrac{1}{2\pi i} \int_{\Re s = \tau} t^{-s} \frac{\Gamma(s)}{\Gamma \left( \frac{s}{2} \right)} \int_{0}^{\infty} T^{\tfrac{s}{2}-1} \left( p_T(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) \right) dT ds. \end{align*} $$

We use the Legendre duplication formula as in [Reference Bär and Moroianu2] (see, for instance, [Reference Paris and Kaminski22]):

$$\begin{align*}\frac{\Gamma(s)}{\Gamma\left( \frac{s}{2} \right) } = \frac{1}{\sqrt{2\pi}} 2^{s-\tfrac{1}{2}} \Gamma \left( \frac{s+1}{2} \right), \end{align*}$$

obtaining that $h_t(x,y)-\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ is equal to

$$ \begin{align*} \tfrac{1}{\sqrt{4 \pi}} \tfrac{1}{2\pi i} \int_{\Re s = \tau} \int_{0}^{\infty} \left( \frac{2 \sqrt{T}}{t} \right)^s \Gamma \left( \frac{s+1}{2} \right) \left( p_T(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) \right) dT ds. \end{align*} $$

Set $X:=\tfrac {2\sqrt {T}}{t}$ . Using Propositions 4, 5, and Fubini, we first compute the integral in s. Changing the variable $S=\frac {s+1}{2}$ and applying the residue theorem, we get

$$ \begin{align*} \tfrac{1}{2\pi i} \int_{\Re s = \tau} X^s \Gamma \left( \frac{s+1}{2} \right) ds={}& \tfrac{2}{2\pi i} \int_{\Re S = \frac{\tau+1}{2}} X^{2S-1} \Gamma(S) dS = 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} X^{-2k-1} \\ ={}& 2 X^{-1}e^{-X^{-2}}=\frac{t}{\sqrt{T}}e^{-\frac{t^2}{4T}}. \end{align*} $$

Thus, we obtain

(10.1) $$ \begin{align} \begin{aligned} h_t(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y)= \tfrac{t}{2\sqrt{\pi}} \int_{0}^{\infty} T^{-3/2} e^{-\frac{t^2}{4T}} \left( p_T(x,y)-\operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) \right) dT. \end{aligned} \end{align} $$

Since $p_T(x,y)-\operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ decays exponentially as T goes to infinity, it follows that the integral from $1$ to $\infty $ in the right-hand side of equation (10.1) is of the form $t \cdot {\mathcal C}^{\infty }_{t,x,y} \left ( [0,\infty ) \times M^2 \right )$ . Furthermore, by the change of variable $u=\tfrac {t}{2\sqrt {T}}$ , we have

$$ \begin{align*} -\tfrac{t}{2\sqrt{\pi}}\int_{0}^{1} T^{-3/2}e^{-\frac{t^2}{4T}} dT \cdot \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y) ={}&-\tfrac{2}{\sqrt{\pi}}\int_{t/2}^{\infty} e^{-u^2} du \cdot \operatorname{P}_{\operatorname{Ker} \operatorname{\Delta}}(x,y). \end{align*} $$

Since $ \int _{t/2}^{\infty } e^{-u^2}du$ tends to $\frac {\sqrt {\pi }}{2}$ as $t \searrow 0$ , the term $-\tfrac {t}{2\sqrt {\pi }}\int _{0}^{1} T^{-3/2}e^{-\frac {t^2}{4T}} dT \operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ will cancel in the limit as $t \to 0$ with $- \operatorname {P}_{\operatorname {Ker} \operatorname {\Delta }}(x,y)$ from the left-hand side of (10.1).

Let us study the remaining integral term $\tfrac {t}{2\sqrt {\pi }} \int _{0}^{1} T^{-3/2}e^{-\frac {t^2}{4T}} p_T(x,y)dT$ . By Theorem 2.1,

$$ \begin{align*} p_T(x,y)=T^{-n/2} e^{-\frac{d(x,y)^2}{4T}} \sum_{j=0}^N T^j a_j(x,y) + R_{N+1}(T,x,y), \end{align*} $$

where the remainder $R_{N+1}(T,x,y)$ is of order ${\mathcal O} (T^{N+1})$ ; therefore,

$$ \begin{align*} \tfrac{t}{2\sqrt{\pi}} \int_{0}^{1} T^{-3/2}e^{-\frac{t^2}{4T}} p_T(x,y)dT={}& \tfrac{t}{2\sqrt{\pi}} \int_{0}^{1} T^{-3/2}e^{-\frac{t^2}{4T}} R_{N+1}(T,x,y) dT \\ +{}& \tfrac{t}{2\sqrt{\pi}} \!\int_{0}^{1} \!T^{-3/2}e^{-\frac{t^2}{4T}} T^{-n/2}e^{-\frac{d(x,y)^2}{4T}} \sum_{j=0}^{N} T^j a_{j}(x,y) dT. \end{align*} $$

Since $R_{N+1}(T,x,y)$ is of order ${\mathcal O} (T^{N+1})$ , the first integral is again of type $ t \cdot {\mathcal C}^{\infty }_{t,x,y} $ . By changing the variable $u=\tfrac {t^2+d(x,y)^2}{4T}$ in the second integral, we get

$$ \begin{align*} {}&\tfrac{t}{2\sqrt{\pi}} \sum_{j=0}^N a_{j}(x,y) \int_{0}^{1} T^{-\frac{n+3}{2} +j}e^{-\frac{t^2+d(x,y)^2}{4T}} dT \\ ={}& \tfrac{t}{2 \sqrt{\pi}} \sum_{j=0}^N a_j(x,y) \left( \frac{t^2+d(x,y)^2}{4} \right)^{-\frac{n+1}{2}+j} \int_{\frac{t^2+d(x,y)^2}{4}}^{\infty} u^{\frac{n+1}{2}-j-1} e^{-u}du \\ ={}& \tfrac{t}{2 \sqrt{\pi}} \sum_{j=0}^N a_j(x,y) \Gamma \left( \frac{n+1}{2}-j, \frac{t^2+d(x,y)^2}{4} \right) \left( \frac{t^2+d(x,y)^2}{4} \right)^{-\frac{n+1}{2}+j}, \end{align*} $$

where $\Gamma (z,\xi ):=\int _{\xi }^{\infty } u^{z-1}e^{-u}du$ is the upper incomplete Gamma function. We conclude that $h_t(x,y)$ is equal to

(10.2) $$ \begin{align} t \cdot {\mathcal C}^{\infty}_{t,x,y} + \tfrac{t}{2 \sqrt{\pi}} \sum_{j=0}^N a_j(x,y) \Gamma \left( \frac{n+1}{2}-j, \frac{t^2+d(x,y)^2}{4} \right) \left( \frac{t^2+d(x,y)^2}{4} \right)^{-\frac{n+1}{2}+j}. \end{align} $$

10.1 The case when n is even

If $z>0$ , then one can easily check that $\Gamma (z,\xi ) \in \xi ^z {\mathcal C}^{\infty }_{\xi }[0,\epsilon ) + \Gamma (z)$ , for some $\epsilon>0$ . Furthermore, for $z \in (-\infty ,0] \setminus \lbrace 0,-1,-2,\ldots \rbrace $ ,

$$ \begin{align*} \Gamma(z,\xi)={}&-\frac{1}{z}\xi^ze^{-\xi}+\frac{1}{z} \Gamma(z+1,\xi) \\ ={}&\xi^z e^{-\xi} \sum_{k=0}^{a-1} \frac{-1}{z(z+1)\ldots(z+k)} \xi^k + \frac{1}{z(z+1)\ldots(z+a)} \Gamma(z+a,\xi) \\ ={}& \xi^z {\mathcal C}^{\infty}_{\xi}[0, \epsilon) + \frac{1}{z(z+1)\ldots(z+a-1)} \Gamma(z+a,\xi), \end{align*} $$

where a is a positive integer such that $z+a>0$ . Thus, for a non-integer $z<0$ , we have

$$ \begin{align*} \Gamma(z,\xi)= \xi^z {\mathcal C}^{\infty}_{\xi} [0, \epsilon) + \frac{1}{z(z+1)\ldots(z+a-1)} \Gamma(z+a). \end{align*} $$

We want to interpret equation (10.2) on the heat space $\operatorname {M_{heat}}$ ; thus, we pull back (10.2) through $\beta _H$ :

$$ \begin{align*} \beta_H^*h={}& \rho \omega_0 \beta_H^* {\mathcal C}^{\infty}_{t,x,y} + \tfrac{1}{2 \sqrt \pi} \rho \omega_0 \sum_{j=0}^{N} \left( \tfrac{\rho^2}{4} \right)^{-\frac{n+1}{2}+j} \beta_H^*a_j(x,y) \Gamma\left( \frac{n+1}{2}-j , \frac{\rho^2}{4} \right) \\ ={}& \rho \omega_0 \beta_H^* {\mathcal C}^{\infty}_{t,x,y} + \tfrac{1}{2 \sqrt{\pi}} \rho^{-n}\omega_0 \sum_{j=0}^{n/2} \rho^{2j} 2^{n+1-2j} \beta_H^*a_j(x,y) \Gamma \left( \frac{n+1}{2}-j \right) \\ {}&+ \tfrac{1}{2 \sqrt{\pi}} \rho \omega_0 \sum_{j=0}^{n/2} \beta_H^* a_j(x,y) {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) + \tfrac{1}{2 \sqrt{\pi}} \rho\omega_0 \sum_{j=n/2 +1}^{N} \beta_H^* a_j(x,y) {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) \\ {}&+\tfrac{1}{2 \sqrt{\pi}} \rho^{-n}\omega_0 \sum_{j=n/2+1}^{N} \rho^{2j} 2^{n+1-2j} \beta_H^*a_j(x,y) \frac{2^{-n/2+j}}{\left( n+1-2j \right) \left( n+3-2j \right)\ldots(-1)} \Gamma \left( \frac{1}{2} \right). \end{align*} $$

Since $\Gamma \left ( \frac {n+1}{2} -j \right ) = \frac {\sqrt {\pi } (n-2j-1)!!}{2^{n/2-j}} $ for $j \in \{0,1,\ldots ,n/2 \}$ , it follows that

(10.3) $$ \begin{align} \begin{aligned} \beta_H^* h={}& \rho \omega_0 \beta_H^* {\mathcal C}^{\infty}_{t,x,y} + \omega_0 \rho {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) + \rho^{-n} \omega_0 \sum_{j=0}^{n/2} \rho^{2j} 2^{n/2-j} (n-2j-1)!! \beta_H^* a_j(x,y) \\ {}&+ \rho^{-n}\omega_0 \sum_{j=n/2+1}^{N} \rho^{2j} \frac{ (-1)^{j-n/2} 2^{n/2-j} }{(2j-n-1)!!} \beta_H^*a_j(x,y). \end{aligned} \end{align} $$

The case $\rho \neq 0$ and $\omega _0 \to 0$ corresponds to $x \neq y$ and $t \searrow 0$ before the pull-back. We obtain that $\beta _H^*h$ is in ${\mathcal C}^{\infty }(\operatorname {M_{heat}})$ and it vanishes at first order on $\operatorname {lb}$ , which is compatible with Theorem 6.1.

If $\rho \to 0$ and $\omega _0=1$ , which corresponds to $x=y$ and $t \searrow 0$ , then $\beta _H^*h= \rho ^{-n}\omega _0 \sum _{j=0}^{N} \rho ^{2j} A_{j}(x)$ , where we denoted by $A_j(x)$ the coefficients appearing in (10.3). Again, this result is compatible with Theorem 7.1, and moreover, the coefficients are precisely the ones from [Reference Bär and Moroianu2, Theorem 3.1].

Remark that formula (10.3) is stronger than Theorems 6.1 and 7.1. If both $\rho $ and $\omega _0$ tend to $0$ (with different speeds), it describes the behavior of $h_t$ as t goes to zero from any positive direction (not only the vertical one).

10.2 The case when n is odd

Remark that for small $\xi $ , we have

$$ \begin{align*} \Gamma(0,\xi)={}&\int_{\xi}^{\infty} t^{-1}e^{-t} dt = \int_{\xi}^{1} \frac{e^{-t}-1}{t} dt +\int_{\xi}^{1} t^{-1} dt + \int_{1}^{\infty}t^{-1}e^{-t}dt \\ ={}&- \log \xi + {\mathcal C}^{\infty}_{\xi}[0,\epsilon). \end{align*} $$

Furthermore, if p is a negative integer, inductively we obtain

$$ \begin{align*} \Gamma(-p,\xi)={}& \frac{e^{-\xi}\xi^{-p}}{p!} \sum_{k=0}^{p-1} (-1)^k (p-k-1)! \xi^k + \frac{(-1)^p}{p!} \Gamma(0,\xi) \\ ={}& \xi^{-p} {\mathcal C}^{\infty}_{\xi}[0,\epsilon)-\frac{(-1)^p}{p!} \log \xi +{\mathcal C}^{\infty}_{\xi}[0,\epsilon). \end{align*} $$

We pull-back equation (10.2) on the heat space $\operatorname {M_{heat}}$ :

$$ \begin{align*} \beta_H^*h={}& \rho \omega_0 \beta_H^* {\mathcal C}^{\infty}_{t,x,y} + \tfrac{1}{2 \sqrt \pi} \rho \omega_0 \sum_{j=0}^{N} \left( \tfrac{\rho^2}{4} \right)^{-\frac{n+1}{2}+j} \beta_H^*a_j(x,y) \Gamma\left( \frac{n+1}{2}-j , \frac{\rho^2}{4} \right) \\ ={}& \rho \omega_0 \beta_H^*a_j(x,y) + \tfrac{1}{2\sqrt{\pi}} \rho \omega_0 \sum_{l=0}^{(n-1)/2} \beta_H^* a_j(x,y) {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) \\ {}&+\frac{1}{\sqrt{\pi}} \rho^{-n}\omega_0 \sum_{j=0}^{(n-1)/2} \rho^{2j} \beta_H^*a_j(x,y) 2^{n-2j} \Gamma \left( \frac{n+1}{2}-j \right) \\ {}&+ \frac{2}{\sqrt{\pi}} \rho^{-n} \omega_0 \sum_{j=(n+1)/2}^{N} \rho^{2j} \log \rho \beta_H^{*} a_j(x,y) 2^{n-2j} \frac{(-1)^{j-\frac{n+1}{2}+1}}{\left( j-\frac{n+1}{2} \right)!} \\ {}&+ \frac{2}{\sqrt{\pi}} \rho^{-n} \omega_0 \sum_{j=(n+1)/2}^{N} \rho^{2j} \beta_H^{*} a_j(x,y) 2^{n-2j} \frac{(-1)^{j-\frac{n+1}{2}}}{\left( j-\frac{n+1}{2} \right)!} \log 2 \\ {}&+\tfrac{1}{2\sqrt{\pi}} \rho \omega_0 \sum_{j=(n+1)/2}^{N} \beta_H^*a_j(x.y) {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) \\ {}&+ \tfrac{1}{ \sqrt{\pi}} \rho^{-n}\omega_0 \sum_{j=(n+1)/2}^{N} \rho^{2j} \beta_H^*a_j(x,y) 2^{n-2j} \frac{(-1)^{j-\frac{n+1}{2}}}{\left( j-\frac{n+1}{2} \right)!} {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon). \end{align*} $$

Therefore, we obtain

(10.4) $$ \begin{align} \begin{aligned} \beta_H^* h={}& \rho \omega_0 \beta_H^* {\mathcal C}^{\infty}_{t,x,y} + \omega_0 \rho {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) + \omega_0 \rho^{-n} {\mathcal C}^{\infty}_{\rho^2}[0,\epsilon) \\ {}&+ \frac{1}{\sqrt{\pi}} \rho^{-n} \omega_0 \sum_{j=0}^{(n-1)/2} \rho^{2j} \beta_H^*a_j(x,y) 2^{n-2j} \left( \frac{n+1}{2}-j \right) ! \\ {}&+ \frac{2}{\sqrt{\pi}} \rho^{-n} \omega_0 \sum_{j=(n+1)/2}^{N} \rho^{2j} \log \rho \beta_H^{*} a_j(x,y) 2^{n-2j} \frac{(-1)^{j-\frac{n+1}{2}+1}}{\left( j-\frac{n+1}{2} \right)!} \\ {}&+ \frac{2}{\sqrt{\pi}} \rho^{-n} \omega_0 \sum_{j=(n+1)/2}^{N} \rho^{2j} \beta_H^{*} a_j(x,y) 2^{n-2j} \frac{(-1)^{j-\frac{n+1}{2}}}{\left( j-\frac{n+1}{2} \right)!} \log 2. \end{aligned} \end{align} $$

If $\rho \neq 0$ and $\omega _0 \to 0$ (corresponding to $x \neq y$ and $t \searrow 0$ before the pull-back on $\operatorname {M_{heat}}$ ), we obtain that $\beta _H^* h \in {\mathcal C}^{\infty }(M_{heat})$ and it vanishes at order $1$ at $\operatorname {lb}$ , which is compatible with the result of Theorem 6.1.

In the case $\rho \to 0$ and $\omega _0=1$ which corresponds to $x=y$ and $t \searrow 0$ , we obtain $\beta _H^* h= \rho ^{-n}{\mathcal C}^{\infty }_{\rho ^2} + \rho ^{-n}\sum _{j=0}^N \rho ^{2j} A_j(x) + \rho ^{-n} \sum _{j=(n+1)/2}^N \rho ^{2j} \log \rho B_j(x)$ , where we denoted by $A_j$ and $B_j$ the coefficients appearing in (10.4). This result is compatible with Theorem 7.1 and again, we find some of the coefficients appearing in [Reference Bär and Moroianu2, Theorem 3.1].

11 The heat kernel as a polyhomogeneous conormal section

Let us recall the notions of index family and polyhomogeneous conormal functions on a manifold with corners with two boundary hypersurfaces. (For an accessible introduction, see [Reference Grieser, Gil, Grieser and Lesch15], and for full details of the theory, see [Reference Melrose18].) A discrete subset $F \in {\mathbb C} \times {\mathbb N} $ is called an index set if the following conditions are satisfied:

1. 1) For any $N \in {\mathbb R}$ , the set $F \cap \{ (z,p): \Re z < N \} $ is finite.

2. 2) If $p> p_0$ and $(z,p) \in F$ , then $(z, p_0) \in F$ .

If X is a manifold with corners with two boundary hypersurfaces $B_1$ and $B_2$ given by the boundary defining functions x and y, a smooth function f on is said to be polyhomogeneous conormal with index sets E and F, respectively, if in a small neighborhood $[0,\epsilon ) \times B_1$ , f has the asymptotic expansion

$$\begin{align*}f(x,y) \stackrel{x \searrow 0}{\sim} \sum_{(z,p) \in F} a_{z,p}(y)\cdot x^z \log^p x, \end{align*}$$

where $a_{z,p}$ are smooth coefficients on $B_2$ , and for each $a_{z,p}$ there exists a sequence of real numbers $b_{w,q}$ , such that

$$\begin{align*}a_{z,p}(y) \stackrel{y \searrow 0}{\sim} \sum_{(w,q) \in E}b_{w,q} \cdot y^w \log^q y.\end{align*}$$

One can prove that f is a polyhomogeneous conormal function on X with index sets $F_p= \{ (k,0) : k \in {\mathbb Z}, k \geq -p \}$ and $F_0=\{ (n,0) : n \in {\mathbb N} \}$ if and only if $f \in y^{-p} {\mathcal C}^{\infty }(X)$ . Furthermore, f is a polyhomogeneous conormal function on X with index sets $F'=\{ (n,1) : n \in {\mathbb N}^* \}$ and $F_0$ if and only if $f \in {\mathcal C}^{\infty }(X)+ \log y \cdot {\mathcal C}^{\infty }(X)$ . Therefore, we can restate Theorem 1.4 as follows:

Theorem 11.1 For $r=\frac {1}{2}$ , the heat kernel $h_t$ of the operator $e^{-t \operatorname {\Delta }^{1/2}}$ is a polyhomogeneous conormal section on the linear heat space $\operatorname {M_{heat}}$ with values in $\mathcal {E} \boxtimes \mathcal {E}^*$ . The index set for the lateral boundary is

$$\begin{align*}F_{\operatorname{lb}} =\{ (k,0): k \in {\mathbb N}^* \}. \end{align*}$$

If n is even, the index set of the front face is

$$\begin{align*}F_{\operatorname{ff}}=\{(-n+k,0): k \in {\mathbb N} \}, \end{align*}$$

whereas for n odd, the index set toward $\operatorname {ff}$ is given by

$$\begin{align*}F_{\operatorname{ff}}=\{ (-n+k,0): k \in {\mathbb N} \} \cup \{ (k,1) : k \in {\mathbb N}^* \}. \end{align*}$$

It seems reasonable to expect that the Schwartz kernel $h_t$ of the operator $e^{-t\operatorname {\Delta }^r}$ for $r \in (0,1)$ can be lifted to a polyhomogeneous conormal section in a certain “transcendental” heat space $M^r_{Heat}$ depending on r with values in $\mathcal {E} \boxtimes \mathcal {E}^*$ . However, already in the case $r=1/3,$ our method leads to complicated computations involving Bessel modified functions. We therefore leave this investigation open for a future project.