# Exponential of operators not defined on a Hilbert space (Dirac operator)

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).

• 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. – DanielC Apr 9 at 12:17
• 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. – Filippo Apr 9 at 12:22
• 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 $L^2(M, S\otimes E))$ – mike stone Apr 9 at 13:44
• @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. – Filippo Apr 9 at 14:21
• 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. – 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