# Asymptotic Expansion of the Schrödinger kernel?

My stackexchange post [http://math.stackexchange.com/questions/275830/schrodinger-kernels-on-manifolds] was somewhat unsatisfactory (also because I may not have stated clear enough what my interest was). So here it goes!

Let $M$ be a compact Riemannian manifold and $\Delta$ be the Laplace-Beltrami operator. It is well-known that the solution operator to the heat equation $e^{t \Delta}$ is smoothing for $t>0$ and has a smooth integral kernel $k_t(x, y) \in C^\infty(M \times M)$. Furthermore, $k_t$ has an asymptotic expansion $$k_t(x, y) \sim \underbrace{(4 \pi t)^{-n/2} \exp \left( -\frac{1}{4t} \mathrm{dist}(x, y)^2 \right)}_{:= e_t(x, y)} \sum_{j=0}^\infty t^j \Phi_j(x, y)$$ meaning that $$\left| k_t(x, y) - e_t(x, y) \sum_{j=0}^N t^j \Phi_j(x, y) \right| \leq C t^{N+1}$$ uniformly in $x$ and $y$ in a neighborhood of the diagonal.

Now by by formally substituting $t \rightarrow it$, one gets the formal asymptotic series $$e_{it}(x, y) \sum_{j=0}^\infty (it)^j \Phi_j(x, y),$$ which has the property that it formally (i.e. termwise, as asymptotic series in $t$) solves the Schrödinger equation $\left(i \frac{\partial}{\partial t} + \Delta\right)k_t = 0.$

Now my question is the following: Does this asymptotic series have any relation to the solution operator $e^{it\Delta}$ of the Schrödinger equation, or to its distribution kernel?

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No, in general, the expansion in small times of the heat kernel $k_t(x,y)$ does not tell much about the Schrodinger semigroup. You may think about the circle case for which the kernel of $e^{t \Delta}$ has an expansion at any order
$k_t(x,y)=\frac{1}{\sqrt{4\pi t}} e^{-\frac{(y-x)^2}{4t}} (1 +O(t^m))$
but the Schrodinger semigroup $e^{it \Delta}$ has a rather complicated behavior. The Schrodinger kernel in that case is actually a distribution. If $t$ is a rational multiple of $2\pi$, it is a finite linear combination of delta functions which is interestingly connected to Gauss sums.