= Solution
Let $M$ be a <closed manifold> of even <dimension> $n=2m$, with a <Riemannian metric> and a <spin structure>. Let $S=S^+\oplus S^-$ be its <spinor bundle>, and let $W$ be a <Hermitian vector bundle> with a <unitary connection> of <vector-bundle curvature> $F^W=(\nabla^W)^2$. We use <Clifford multiplication> satisfying $c(v)c(w)+c(w)c(v)=-2\langle v,w\rangle$, so the <Twisted Dirac operator>
$$
D_W=\sum_i c(e^i)\nabla^{S\otimes W}_{e_i}
=\begin{pmatrix}0&D_W^-\\D_W^+&0\end{pmatrix}
$$
is <self-adjoint> and $D_W^-=(D_W^+)^*$. Here $(e_i)$ is an oriented local <orthonormal basis> and $(e^i)$ its dual. The <Atiyah-Singer index theorem> in this setting is
$$
\boxed{\operatorname{ind}D_W^+=\int_M\left[\widehat A(R^{TM})\wedge\operatorname{tr}\exp\!\left(\frac{iF^W}{2\pi}\right)\right]_{2m}.}
$$
The second factor is the <Chern character>; the first is the <A-hat form>
$$
\widehat A(R)=\det{}^{1/2}\!\left(\frac{R/(4\pi i)}{\sinh(R/(4\pi i))}\right).
$$
The square root has constant term one. All functions of the <vector-bundle curvature> matrices mean <power series> truncated above the <dimension> of $M$. The brackets select the top-degree <differential form>. The <Bianchi identity> shows that these forms are closed, and <Chern-Weil connection transgression> shows that their <de Rham cohomology> classes do not depend on the chosen <connections on a vector bundle>.
The global step is <supersymmetric quantum mechanics>. On the <Hilbert space> $L^2(S\otimes W)$, take the grading operator $\Gamma$, the odd supercharge $D_W$, and the nonnegative Hamiltonian $H=D_W^2$. They obey $\Gamma D_W=-D_W\Gamma$ and $D_W^2=H$. The <principal symbol> of $D_W$ is invertible away from the zero covector, so <elliptic regularity> and <Rellich-Kondrachov compactness theorem> give a smooth discrete <orthonormal eigenbasis> for $H$, finite-dimensional <eigenspaces>, and a <trace-class operator> $e^{-tH}$ for $t>0$. Define its <supertrace> by $\operatorname{Str}A=\operatorname{Tr}(\Gamma A)$.
For every positive <eigenvalue> $\lambda$ of $H$, the map $\lambda^{-1/2}D_W$ is an isomorphism between its positive and negative grading <eigenspaces>, with inverse itself. Their contributions to the <supertrace> cancel. The zero <eigenspaces> are $\ker D_W^+$ and $\ker D_W^-$. Consequently
$$
\operatorname{Str}e^{-tD_W^2}=\dim\ker D_W^+-\dim\ker D_W^-
=\operatorname{ind}D_W^+\qquad(t>0).
$$
This also proves independence of $t$ without differentiating an unbounded operator inside a <trace>. If $K_t$ is the <heat kernel>, the <heat kernel trace formula> gives
$$
\operatorname{ind}D_W^+=\int_M\operatorname{str}_{S\otimes W}K_t(x,x)\,d\mathrm{vol}_g(x).
$$
The remaining task is to compute this local integrand as $t\downarrow0$.
Fix $p\in M$, use <geodesic normal coordinates> $x$, and trivialize $S\otimes W$ by radial <parallel transport>. Expanding $D_W^2$ at $p$ gives the <Lichnerowicz formula for a twisted Dirac operator>:
$$
D_W^2=\nabla^*\nabla+\frac14\operatorname{Scal}+\sum_{i<j}c(e^i)c(e^j)F^W_{ij}.
$$
Indeed the terms with identical indices give the <rough Laplacian>. The other terms are $\sum_{i<j}c(e^i)c(e^j)[\nabla_i,\nabla_j]$. Decompose this <vector-bundle curvature> into spin curvature and $F^W$; substituting the spin curvature into the <Clifford algebra> relations and the algebraic <Bianchi identity> leaves $\operatorname{Scal}/4$. Thus the calculation retains the curvature of the twisting bundle explicitly.
There is an especially strong cancellation in the spinor <supertrace>. With $\Gamma=i^mc(e^1)\cdots c(e^{2m})$, every ordered product of distinct Clifford generators of degree less than $2m$ has zero <supertrace>, while
$$
\operatorname{str}_S(c(e^1)\cdots c(e^{2m}))=(-2i)^m.
$$
For an even product of lower degree, conjugate $\Gamma$ times that product by a missing Clifford generator: it changes sign, so its <matrix trace> is zero. An odd product exchanges the grading summands and also has zero <supertrace>. For the full product, $(c(e^1)\cdots c(e^{2m}))^2=(-1)^m$, and $\dim S=2^m$, giving the displayed constant.
To exploit this, identify the <Clifford algebra> as a filtered <vector space> with the <exterior algebra> by sending an ordered Clifford product to the corresponding <wedge product of differential forms>. Implement <Getzler rescaling>: put $x=\varepsilon y$, $t=\varepsilon^2s$, multiply the degree-$k$ exterior component of the kernel by $\varepsilon^{-k}$, and multiply the whole kernel by $\varepsilon^{2m}$. A derivative and a Clifford generator have rescaling degree one; a coordinate has degree minus one. Clifford contractions lower the degree by two relative to the <wedge product of differential forms>, so the leading product becomes <exterior multiplication>.
In the radially parallel frame, the linear Taylor term of the connection is determined by its <vector-bundle curvature>: if $A_i$ is its matrix, then $A_i(x)=\frac12x^jF_{ji}(p)+O(|x|^2)$. To check the coefficient, radial gauge means $x^iA_i=0$, and integration of $x^jF_{ji}$ along the radial segment gives $A_i(x)=\int_0^1s x^jF_{ji}(sx)\,ds$. Combining this formula with the spin representation, the symmetries of <Riemann curvature tensor>, and the <Lichnerowicz formula for a twisted Dirac operator>, the rescaled square has leading model
$$
\mathcal H_p=-\sum_i\left(\partial_{y_i}-\frac14\sum_j\mathcal R_{ij}y_j\right)^2+F^W_p.
$$
Here $\mathcal R$ is the skew matrix of tangent-bundle curvature two-forms at $p$; changing the curvature-matrix sign convention changes the sign inside the covariant derivatives but not the determinant below. The coefficients act in $\Lambda T_p^*M\otimes\operatorname{End}(W_p)$. The scalar-curvature term has degree zero and disappears after multiplication by $\varepsilon^2$; the linear spin-connection term and the Clifford action of $F^W$ have degree two and survive. Higher Taylor terms have smaller degree and disappear.
For clarity, the analytic meaning of this limiting calculation is the following. Construct the local <heat kernel expansion> to any fixed order and insert the Taylor expansion of the rescaled coefficients into its transport equations. At each exterior degree, the leading equation is the heat equation of $\mathcal H_p$; the remaining terms acquire positive powers of $\varepsilon$. Equivalently, successive substitutions in
$$
e^{-sP_\varepsilon}-e^{-s\mathcal H_p}
=-\int_0^s e^{-(s-u)P_\varepsilon}(P_\varepsilon-\mathcal H_p)e^{-u\mathcal H_p}\,du
$$
give the same expansion. Polynomial spatial factors are integrable against the local Gaussian kernels, and only finitely many exterior degrees exist. A cutoff outside a fixed normal neighbourhood contributes $O(e^{-c/t})$. Choosing the initial expansion order arbitrarily large bounds its differentiated remainder to any required order. This proves convergence of the rescaled diagonal kernel to the model diagonal kernel, uniformly in $p$ on the <compact manifold>. Its degree-$2m$ component has no net power of $\varepsilon$, precisely the component detected by the spinor <supertrace>.
We can compute the model kernel rather than invoke a local index formula. Set $F=F^W_p$ and seek its kernel from the origin in the form
$$
k_s(y,0)=a(s)\exp\!\left(-\tfrac14 y^TQ(s)y\right)e^{-sF}.
$$
The scalar even exterior coefficients of $\mathcal R$ commute. On this ansatz the rotational first-order term vanishes because $Q$ is a function of $\mathcal R^2$. Expanding the model operator gives $-\sum\partial_i^2+\frac12(\mathcal R y)\cdot\nabla+\frac1{16}y^T\mathcal R^2y+F$, so substitution into the <heat equation> gives
$$
Q'=\tfrac14\mathcal R^2-Q^2,\qquad\frac{a'}a=-\tfrac12\operatorname{tr}Q.
$$
The Euclidean initial condition $Q(s)\sim s^{-1}I$, $a(s)\sim(4\pi s)^{-m}$ selects
$$
Q(s)=\frac{\mathcal R}{2}\coth\!\left(\frac{s\mathcal R}{2}\right),\qquad
a(s)=(4\pi s)^{-m}\det{}^{1/2}\!\left(\frac{s\mathcal R/2}{\sinh(s\mathcal R/2)}\right).
$$
These expressions have well-defined <power series> at $\mathcal R=0$. This is the curvature version of the Gaussian calculation underlying the <Mehler formula for Hermite functions>.
At $s=1$, $y=0$, multiply the top exterior coefficient by the spinor <supertrace> constant. Since $(4\pi)^{-m}(-2i)^m=(2\pi i)^{-m}$, and every curvature factor is a two-form, the result is
$$
\lim_{t\downarrow0}\operatorname{str}K_t(p,p)\,d\mathrm{vol}_g(p)
=\left[\det{}^{1/2}\!\left(\frac{R^{TM}/(4\pi i)}{\sinh(R^{TM}/(4\pi i))}\right)
\operatorname{tr}\exp\!\left(-\frac{F^W}{2\pi i}\right)\right]_{2m}(p).
$$
Integration and the constant <supertrace> established above prove the boxed <Atiyah-Singer index theorem>. The mechanism is that odd symmetry pairs the nonzero spectrum globally, while the Clifford filtration makes the surviving local contribution an explicitly solvable curvature oscillator. Odd-dimensional spin <Dirac operators> have no canonical chiral index of this type; their self-adjoint <Fredholm index> is zero. For comparison of normalization conventions, https://arxiv.org/pdf/math/0211333[Raphaël Ponge's local index calculation] writes the same unnormalized curvature model and spinor supertrace constant.
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