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=

  • $\begingroup$ you mean $t^{-1/2}\sum_{k \ge 0} a_k t^{-k}$ $\endgroup$
    – reuns
    Aug 11, 2017 at 1:22
  • $\begingroup$ A Taylor series is for small $t$. This asymptotic series is for large $t$. $\endgroup$ Aug 11, 2017 at 1:34
  • $\begingroup$ I am pretty sure this series is also for small t. See edits. $\endgroup$ Aug 11, 2017 at 7:50

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.