If the Riemannian curvature is zero and $\mathrm{dim}(M)=n=2k$, the Atiyah-Singer index theorem for the twisted Dirac operator reduces to the following equation: \begin{equation}\tag{1} \mathrm{ind}(D_+)=\int_M\frac{1}{k!}\mathrm{tr}\left[\left(\frac{\mathrm{i}F}{2\pi}\right)^k\right] \end{equation} Let \begin{equation} \nabla_i=e^\mu_i(\partial_\mu+\omega_\mu+A_\mu) \end{equation} be the twisted covariant derivative and \begin{equation} F'=\frac{1}{2}\gamma^\mu\gamma^\nu F_{\mu\nu} \end{equation} the clifford curvature.

Equation (2.15) in Fujikawa's paper from 1980 implies that we can prove $(1)$ by showing that \begin{equation}\tag{2} \lim_{t\to 0}\langle x|\gamma_{n+1}\exp(t\nabla{}^*\nabla+tF')|x\rangle=\lim_{t\to 0}\frac{1}{(4\pi t)^{n/2}}\mathrm{Tr}(\gamma_{n+1}\exp[tF'(x)]) \end{equation} for all $x\in M$.$^1$ (If you're interested in more details, please let me know in the comments).

Q: How can $(2)$ be proved? Has anyone seen this equation before?

The equation \begin{equation} \frac{1}{(4\pi t)^k}=\int\frac{\mathrm{d}^nk}{(2\pi)^n}\exp(-tk^2)=\int\frac{\mathrm{d}^nk}{(2\pi)^n}\langle k|\exp\left(-t\sum_{i=1}^n\partial_i\partial_i\right)|k\rangle=\langle x|\exp\left(-t\sum_{i=1}^n\partial_i\partial_i\right)|x\rangle \end{equation} might be useful...

$^1$If $O\colon\Gamma(M,E)\to\Gamma(M,E)$ is a trace class operator, \begin{equation} \langle x|O|x\rangle:=\sum_i\langle\psi_i(x)|(O\psi_i)(x)\rangle \end{equation} is independent of the basis.


1 Answer 1


If i understand correctly, you are asking why the flat space result continues to hold on more general backgrounds. In a hand-waving way, one can say that by taking the limit $t\rightarrow 0$, and on the diagonal, one considers very small-scale structures, and therefore the metric becomes less important.

If you are interested in a more rigourous perspective: if you search for "asymptotics of the heat kernel" you'll find some references, for example this here:


There are also a couple of references in that paper which might be helpful.

Sadly, i cannot reproduce that argument here, but i can point out some of its feats which might indicate whether the above sources contain the kind of information you are looking for.

It is possible to show that the Heat kernel

$$ K(t,x,y) := \langle x | \exp(t\Delta) | y \rangle $$

has an asymptotic expansion for $t\rightarrow 0$,

$$K(t,x,x) \sim t^{-n/2}\sum_i a_i(x) t^i \ .$$

The $a_i(x)$ are smooth functions and depened on the metric and its derivatives. I will just shortly steal from the paper i cited above the explanation of how this works. Basically one defines a $K_1(t,x,y)$, which is just the Euclidean heat kernel, for $x$ very close to $y$:

$$ K_1^{g}(t,x,y) = (4\pi t)^{-n/2} \exp(-\|x -y\|_g^2/4t) \ .$$

Here the distance is measured with the help of $g = g(x)$, the metric at a given point $x$. Then one shows that this differs from the real heat kernel by corrections of order $t^{-n/2+1}$, then repeat for the next order.

  • $\begingroup$ What exactly do you mean by the "flat space result"? I am only interested in the special case where the index theorem reduces to $(1)$. In other words, I'm assuming $\hat{A}(M)=1+0+0+\ldots$ or $R=0$, where $R$ is the Riemannian curvature tensor. $\endgroup$
    – Filippo
    Commented Apr 26, 2021 at 11:55
  • $\begingroup$ If that's what you mean by "flat space", then the "flat space result" might be exactly what I'm looking for. $\endgroup$
    – Filippo
    Commented Apr 26, 2021 at 12:04
  • $\begingroup$ But in flat space you already proved it, $\endgroup$ Commented Apr 26, 2021 at 14:09
  • $\begingroup$ I'm afraid I have created confusion by using the symbol $\Delta$ for the connection Laplacian. I replaced $\Delta$ by $\nabla{}^*\nabla$ in equation $(2)$ to make clear that $\Delta\neq\sum_i\partial_i\partial_i$. $\endgroup$
    – Filippo
    Commented Apr 27, 2021 at 5:43

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