Let $S\to M$ be the spinor bundle and consider a vector bundle $E\to M$ with a covariant derivative $\nabla$ and associated curvature $F=F^\nabla$.

If $R=0$, the Atiyah-Singer index theorem 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],\mathrm{dim}\;M=n=2k \end{equation} Consider $\mathcal{E}=S\otimes E$ and let $k_t\in\Gamma(M\times M,\mathcal{E}\boxtimes\mathcal{E}^*)$ be the heat kernel associated to $K_t:=\exp(-tDD)$. The heat equation proof of $(1)$ is based on the realisation that \begin{equation} \mathrm{ind}(D_+)=\mathrm{Str}(K_t)=\int_M k_t(x,x)\,\mathrm{d}x\quad\text{for all }t>0. \end{equation} I obtain the correct result by assuming that \begin{equation}\tag{2} k_t(x,x)\sim(4\pi t)^{-n/2}\exp(-t\mathcal{F})\quad\text{if }R=0, \end{equation} where \begin{equation} \mathcal{F}=\frac{1}{2}\gamma^\mu\gamma^\nu F_{\mu\nu} \end{equation} is the clifford curvature.

Q: Is $(2)$ correct? Is it a special case of a more general formula? Where can I find a proof?


1 Answer 1


No. There are other terms. The $F$ ones are the only ones that survive the trace with the generalization of $\gamma^5$ to higher dimensions. As to the proof, have you not yet read the Getzler paper I recommended?

  • $\begingroup$ Thank you! I found it difficult to understand Getzler's paper. However, I think the main result is the equation just above the Appendix: $\lim_{t\to 0}\mathrm{Str}\;k_t(0,0)=\lim_{t\to 0}\mathrm{Str}\;k^0_t(0)=(2\pi i)^{n/2}(\hat{A}(\Omega)\mathrm{ch}(F))_n$. The RHS is defined on page 3: $k_t^0(x):=(4\pi t)^{-n/2}\hat{A}(t\Omega)\exp\left[tF-\frac{1}{4t}\left(\frac{t\Omega/2}{\mathrm{tanh}\;t\Omega/2}\right)_{ij}x^ix^j\right]$. Thus, $k_t^0(0)=(4\pi t)^{-n/2}\hat{A}(t\Omega)\exp(tF)$ and if $\Omega=0$, $\hat{A}(M)=1$ and therefore $k_t^0(0)=(4\pi t)^{-n/2}\exp(tF)$. $\endgroup$
    – Filippo
    Apr 29, 2021 at 18:57
  • $\begingroup$ But I wouldn't be surprised if I have misunderstood something. $\endgroup$
    – Filippo
    Apr 29, 2021 at 18:58
  • $\begingroup$ Ah! I see that your $k$ is after you took the supertrace. So the answer to your original question is "yes" $\endgroup$
    – mike stone
    Apr 29, 2021 at 19:41
  • 1
    $\begingroup$ I think that one of Getler's scaling equations has a typo: math.stackexchange.com/questions/3317426/… $\endgroup$
    – mike stone
    Apr 29, 2021 at 19:43
  • $\begingroup$ I think there's a problem: Either Getzler denotes two different curvatures by $F$ (my $F$ as well as my $\mathcal{F}$) or the equation $k_t^0(0)=(4\pi t)^{-n/2}\exp(tF)$ is not what I thought it is. However, I get the right result by assuming $\lim_{t\to 0}\mathrm{str}\;k_t=\lim_{t\to 0}(4\pi t)^{-n/2}\mathrm{str}\;\exp(t\mathcal{F})$... $\endgroup$
    – Filippo
    Apr 30, 2021 at 9:11

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