Is the $n/2$-th heat kernel coefficient topological?

Question

I have asked the same question on math.SE, without much success so I'm trying my luck here too.

Let $M$ be an $n$-dimensional manifold, with $n$ even and consider the heat kernel of the Laplacian on $M$: $$K_M(t) := \mathrm{Tr}\Big(\mathrm{e}^{-t \Delta_M}\Big) \equiv \sum_{\lambda\in\mathrm{spec}(\Delta_M)} \mathrm{e}^{-t \lambda}.$$ The heat kernel admits an asympotic, small-$t$ expansion as $$K_M(t) = \frac{1}{(4\pi t)^{n/2}}\sum_{k=0}^\infty b_k^{(M)} t^k,$$ where $b_k^{(M)}$ are known as the heat kernel coefficients and are given by integrals of local geometric data of $M$.

I am particularly interested in what I'll call middle-dimensional heat kernel coefficient, namely $$ b_{n/2}^{(M)} = b_{\mathrm{dim}M/2}^{(M)}.$$ This coefficient is interesting because it is very closely related to the value of the Minakshisundaram-Pleijel zeta function (i.e. the spectral zeta function for the Laplacian) at zero.

Is an explicit formula for the calculation of $b_{n/2}^{(M)}$ known? In particular, is it a topological invariant? It seems to me that it wants to be related to the number of connected components of $M$ or to some (alternating?) sum of Betti numbers of $M$, but I haven't managed to show it.

If anyone feels brave, I'd be curious about how it generalizes to $p$-forms; namely what's the middle-dimensional heat kernel coefficient associated to the Laplacian, $\Delta_p = \mathrm{d}\delta + \delta \mathrm{d}$, acting on $p$-forms on $M$?

Answer 1

For $n=2$, the answer is given by $\frac{E}{6}$, where $E$ is the Euler characteristic of $M$, see McKean Jr, H. P., & Singer, I. M. (1967). Curvature and the eigenvalues of the Laplacian. Journal of Differential Geometry, 1(1-2), 43-69.

For $n>2$, if it is topological, then it is none of what you conjectured. Consider the case of hyperbolic manifolds: since those are locally isometric to the hyperbolic space, the coefficient $b^{M}_{n/2}$ is proportional to the hyperbolic volume (Which, incidentally, is indeed topological by Mostow rigidity.)

Update: It is not topological for $n>2$. Let $M_1=\mathbb{T}^2\times\mathbb{S}^2$ and $M_2=(2\mathbb{T}^2)\times \mathbb{S}^2,$ where $2\mathbb{T}^2$ denotes the torus scaled by the factor of $2$. Then, $M_1$ and $M_2$ are homeomorphic and $b_2^{M_2}=4b_2^{M_1},$ because both spaces are homogeneous and locally isometric to each other. On the other hand, $$P^{M_1}_t(x,x)=P^{\mathbb{T}^2}_t(x,x)P^{\mathbb{S}^2}_t(x,x)=\left(\frac{1}{4\pi t}+o(t^{100})\right)P^{\mathbb{S}^2}_t(x,x).$$ Therefore, we only need to show that $b^{\mathbb{S}^2}_2$ does not vanish. According to the table on page 63 of the above reference, this is indeed the case.

Answer 2

Let $D$ be a Dirac operator on an even-dimensional manifold and C be the chirality endomorphism. Let $a_k$ be the coefficients of the heat kernel diagonal of the operator $D^2$ (called the local heat kernel coefficients) so that $b_k=\int_M tr a_k $. Then $Tr C a_{k}=0$ for all $k\ne n/2$, and for $k=n/2$ it is the topological invariant (the index of the Dirac operator). So, $b_{n/2}$ is not a topological invariant.