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Consider the heat kernel $K(t)$ for some operator $M$ which acts on the Hilbert space $L^2[0,1]$: $$ K(t) = Tr e^{-t M} = \int dx \langle{x}|e^{-tM}|x\rangle $$ Naively I would have expected a Taylor series expansion of the integrand and written it as: $$ K(t) = \int dx \langle{x}|1|x\rangle - \langle{x}|tM|x\rangle + \langle{x}|\frac{t^2M^2}{2}|x\rangle \ldots $$ However, this is not what happens. The asymptotic expansion for $K(t)$ is generally given by: $$K(t) = \frac{1}{\sqrt{4 \pi t}}\sum_{k} a_k t^{k/2} $$ This befuddles me because it is a power series in $\sqrt{t}$ instead of a power series in $t$, which would be my naive expectation. So is there any intuition for why these half integer powers show up?

Remark: What is also interesting is that for heat kernels on some general manifold $M$ (that can be much more complicated than the interval $[0,1]$ we considered. People have figured out that for manifold without boundaries, only the integral coefficients survive (in other words, k even). Is there a connection between the validity of Taylor expansion and the structure of the manifold? I would love to get some references addressing these questions.

Edit 1: A note that the expansion is indeed for small t, not large t. For reference, see this link below: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.218.814&rep=rep1&type=pdf

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  • $\begingroup$ you mean $t^{-1/2}\sum_{k \ge 0} a_k t^{-k}$ $\endgroup$ – reuns Aug 11 '17 at 1:22
  • $\begingroup$ A Taylor series is for small $t$. This asymptotic series is for large $t$. $\endgroup$ – Robert Israel Aug 11 '17 at 1:34
  • $\begingroup$ I am pretty sure this series is also for small t. See edits. $\endgroup$ – Zhengyan Shi Aug 11 '17 at 7:50
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Not really an answer but your question still makes sense in the simplest case.


Let the one-dimensional heat equation $$\partial_t u = \partial_x^2 u, \qquad {\scriptstyle\text{ with initial condition }} \quad u(.,0)\in H^2(\mathbb{R}) \tag{1}$$

The fundamental solution is $$f(x,t)=\frac{1}{\sqrt{t 2\pi}} e^{-x^2/4t}$$ (which means $f$ is solution to the heat equation with initial condition $\delta(x)$)

so the solution of $(1)$ is $$u(x,t) = \int_{-\infty}^\infty u_0(y)f(x-y,t)dy=\int_{-\infty}^\infty u_0(y)\frac{1}{\sqrt{t 2\pi}} e^{-(y-x)^2/4t}dy$$ making it clear where the expansion in $t^{-k-1/2}$ comes from (expansion at $t= \infty$).

With the Fourier transform $(1)$ can be expressed as

$$\partial_t \hat{u} = -4\pi \xi^2 \hat{u}$$ whose solution is $$\hat{u}(\xi,t)= e^{-4\pi \xi^2 t} \hat{u}(\xi,0)$$ which is smooth in $t$, but where $\partial_t^k\hat{u}(.,t)$ makes sense only as a distribution.

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