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Let $D$ be the Dirac operator. In his proof of the Atiyah Singer index theorem, Fujikawa considers the operator $\exp{D^2}$.

However, as far as I know, $D$ is not defined on a Hilbert space (the domain is equipped with a bilinear form, though), so how is the exponential defined?

More details Let $G\to P\to M$ be a principal fiber bundle, $\rho\colon G\to\mathrm{GL}(V)$ a representation, $E=P\times_\rho V$ the associated vector bundle, $\mathrm{Spin}(s,t)\to S(M)\to M$ the spin structure, $\Delta$ the vector space of Dirac spinors, $\kappa\colon\mathrm{Spin}(s,t)\to\mathrm{GL}(\Delta)$ the spinor representation and $S=S(M)\times_\kappa\Delta$ the spinor bundle. We consider the twisted Dirac operator $D\colon\Gamma(M,S\otimes E)\to\Gamma(M,S\otimes E)$.

If $\langle\,\cdot\,,\,\cdot\,\rangle$ is a bundle metric on $S\otimes E$ and $\omega$ is the volume form,

\begin{align} \Gamma(M,S\otimes E)\times\Gamma(M,S\otimes E)&\to\mathbf{C}\\ (\phi,\psi)&\mapsto\int_M\langle\phi,\psi\rangle\cdot\omega \end{align} is bilinear (to be precise, we only consider sections with compact support), but not an inner product (and $D$ becomes symmetric operator).

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  • $\begingroup$ Thanks for providing the complicated diff. geom. context, but it's still not clear for what you call $D$ a) its domain, b) its action upon elements (vectors?) of its domain. $\endgroup$ – DanielC Apr 9 at 12:17
  • $\begingroup$ I provided the diff. geom. context in order to specify the domain. After mentioning what $E$ and $S$ are, I wrote that $\Gamma(M,S\otimes E)$ is the domain. $D$ is the standard Dirac operator for twsited spinor bundles. $\endgroup$ – Filippo Apr 9 at 12:22
  • $\begingroup$ The Dirac opertor is not bounded, so its domain is strictly smaller than the bundle $\Gamma(M,S\otimes E)$. It is defined as a differential operator on a dense subset of $L^2(M, S\otimes E))$ $\endgroup$ – mike stone Apr 9 at 13:44
  • $\begingroup$ @mikestone Thank you for the clarification! The author of the book I'm reading mentions that the Dirac operator $D\colon\Gamma(M,S\otimes E)\to\Gamma(M,S\otimes E)$ maps sections with compact support to sections with compact support, so I thought we are considering the restriction of the Dirac Operator to the subset of sections with compact support. $\endgroup$ – Filippo Apr 9 at 14:21
  • $\begingroup$ But if I understand your comment correctly, we do want a hilbert space (so the question is not how to define the exponential of an operator without assuming a hilbert space), and to do so, we do basically the same we do in introductory QM - that is, we consider equivalence classes (of functions/sections) instead of functions. $\endgroup$ – Filippo Apr 9 at 14:21
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The Dirac operator $D$ is self adoint on a dense subset of the Hilbert space, it has a complete set of $V$-valued eigenfunctions $\varphi_n(x)$ with real eigenvalues $\lambda_n$. As a consequence one can define $\exp\{-tD^\dagger D\}$ on the entire Hilbert space by the integral kernel $$ K_t(x,x')=\exp\{-tD^\dagger D\}_{xx'}=\sum_n \varphi_n(x) e^{-t\lambda_n^2}\varphi_n^*(x'). $$
The rapid exponential suppression of the large $|\lambda_n|$ terms ensures convergence of the sum on all of $L^2(M)\otimes GL(V)$. Thus the domain of the heat kernel can be larger than the domain of $D$ itself. This is one of the features that makes the heat kernel such a useful tool

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