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).

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.