= Solution
The positive sign printed in the exponential uses the <skew-adjoint> convention for the <Dirac operator>: $D^*=-D$, so $D_-=-D_+^*$. Put $A=D_+$ and define the nonnegative <self-adjoint operators>
$$
P=A^*A=-D_-D_+,\qquad Q=AA^*=-D_+D_-.
$$
For $t>0$, the assumed <compact elliptic spectral theorem> for these <elliptic differential operators> gives discrete <eigenvalues>, finite multiplicities and <trace-class operators> $e^{-tP}$, $e^{-tQ}$.
If $Pu=\lambda u$ with $\lambda>0$, then $Q(Au)=A(Pu)=\lambda Au$ and $Au\ne0$. Conversely $A^*$ sends the $\lambda$ <eigenspace> of $Q$ to that of $P$. On these <eigenspaces> the inverse of $A$ is $\lambda^{-1}A^*$. Thus the positive spectra agree with multiplicity, and absolute convergence of the <operator traces> permits cancellation term by term:
$$
\operatorname{Tr}e^{-tP}-\operatorname{Tr}e^{-tQ}
=\dim\ker P-\dim\ker Q
=\dim\ker A-\dim\ker A^*.
$$
The last expression is the <Fredholm index> of $A$, since its closed range has <orthogonal complement> $\ker A^*$. Consequently
$$
\boxed{\operatorname{ind}D_+=\operatorname{Tr}(e^{tD_-D_+})-\operatorname{Tr}(e^{tD_+D_-}),\qquad t>0.}
$$
This is the <supertrace> cancellation used in Question 1. If the <Dirac operator> is instead defined to be <self-adjoint>, $D_-=(D_+)^*$ and the heat exponentials must have a minus sign. With that convention the positive exponentials in the printed formula are generally unbounded and have no finite <operator trace>.
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