The Dirac heat trace is smooth and its supertrace equals the index

Let $E_L$ be the span of eigensections with $0\le \lambda \le L$. For $u\in E_L$, the Lichnerowicz identity and formal symmetry give $\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2=⟨Pu,u⟩\le L\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2=\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2+⟨R_{0}u,u⟩$, where $R_0$ is bounded on compact $M$. Hence $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le C(1+L{)}^{1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Fix finitely many precompact bundle charts and a smooth partition of unity ${\chi }_a$ subordinate to them. In each chart, write ${\chi }_{a}u$ in its bundle frame and extend its coefficients by zero to a fixed Euclidean box. Tile that box by cubes of side between $h\mathrm{/}2$ and $h$, and map $u$ to the averages of all coefficients on all cubes. This linear map has rank at most $C{h}^{-n}$. The fundamental theorem of calculus along coordinate segments gives the cube Poincaré bound $\mathrm{\parallel }f-f_Q{\mathrm{\parallel }}_{L^2(Q)}\le C_{n}h\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_{L^2(Q)}$, where $f_Q$ is the average on $Q$. If all averages vanish, summing this bound over cubes and charts gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le Ch\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$. Take $h=c(1+L{)}^{-1\mathrm{/}2}$ with $c>0$ small enough that $Cc<1$; the averaging map is injective on $E_L$. Therefore $\dim E_L\le C^{\mathrm{\prime }}(1+L{)}^{n\mathrm{/}2}$. For an orthonormal eigenbasis $u_j$, iteration of [F1] on $P^a{u}_j={\lambda }_j^a{u}_j$ gives $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{H^{2a}}\le C_a(1+{\lambda }_j{)}^a$; if $2a>r+n\mathrm{/}2$, Sobolev embedding gives the same polynomial bound for every $C^r$ norm.

The polynomial counting bound and the polynomial $C^r$ eigenfunction bounds imply ${\sum }_j{e}^{-t{\lambda }_j}(1+{\lambda }_j{)}^q<\mathrm{\infty }$ for every fixed $t>0$ and integer $q$: split $[0,\mathrm{\infty })$ into dyadic shells, bound the number and $C^r$ norm on each shell polynomially, and use $e^{-t{2}^k}$ to dominate every polynomial. Thus $K_t(x,y)={\sum }_j{e}^{-t{\lambda }_j}{u}_j(x)\otimes u_j(y{)}^{\ast }$ and every fixed space/time derivative converge absolutely and uniformly. The series solves $({\mathrm{\partial }}_t+P_x)K_t=0$, and integrating it against a smooth section gives $e^{-tP}$ by the orthonormal expansion. Termwise integration on the diagonal gives $\mathrm{Tr}(e^{-tP})={\sum }_j{e}^{-t{\lambda }_j}<\mathrm{\infty }$.

On smooth sections $P_{-}{D}^{+}=D^{+}{P}_{+}$ and $D^{-}{D}^{+}=P_{+}$, $D^{+}{D}^{-}=P_{-}$. Thus for $\lambda >0$, $D^{+}$ maps its $P_{+}$ eigenspace isomorphically to the $P_{-}$ eigenspace with inverse ${\lambda }^{-1}{D}^{-}$, so their contributions to the graded trace $\mathrm{Str}(e^{-tP})$ cancel. At $\lambda =0$, the eigenspaces are $\ker D_V^{+}$ and $\ker D_V^{-}$; their signed dimension difference is the Fredholm index from [F3]. The diagonal kernel series is uniformly convergent, so integrating its pointwise supertrace equals the sum of these eigenspace traces. This proves the stated identity for every $t>0$.

If $M$ is empty all Hilbert spaces and integrals are zero. In dimension zero the spectrum is the finite-dimensional zero operator and the same identity is direct; the analytic estimates above are only invoked in positive even dimension. The trace equality was obtained from the explicit smooth kernel series, not from an unproved diagonal trace theorem.