# Is $k_t(x,x)\sim(4\pi t)^{-n/2}\mathrm{e}^{-tF}$ if $R=0$? (Dirac operator, heat kernels)

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: $$$$\tag{1} \mathrm{ind}(D_+)=\int_M\frac{1}{k!}\mathrm{tr}\left[\left(\frac{\mathrm{i}F}{2\pi}\right)^k\right],\mathrm{dim}\;M=n=2k$$$$ Consider $$\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)$$ is based on the realisation that $$$$\mathrm{ind}(D_+)=\mathrm{Str}(K_t)=\int_M k_t(x,x)\,\mathrm{d}x\quad\text{for all }t>0.$$$$ I obtain the correct result by assuming that $$$$\tag{2} k_t(x,x)\sim(4\pi t)^{-n/2}\exp(-t\mathcal{F})\quad\text{if }R=0,$$$$ where $$$$\mathcal{F}=\frac{1}{2}\gamma^\mu\gamma^\nu F_{\mu\nu}$$$$ is the clifford curvature.

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

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?
• 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 $k_t^0(0)=(4\pi t)^{-n/2}\exp(tF)$. Apr 29, 2021 at 18:57
• Ah! I see that your $k$ is after you took the supertrace. So the answer to your original question is "yes" Apr 29, 2021 at 19:41
• 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 $\lim_{t\to 0}\mathrm{str}\;k_t=\lim_{t\to 0}(4\pi t)^{-n/2}\mathrm{str}\;\exp(t\mathcal{F})$... Apr 30, 2021 at 9:11