The Mehler coefficient has A-hat times the relative Chern character

Put $B(t)=(t\mathcal{R}\mathrm{/}2)\coth (t\mathcal{R}\mathrm{/}2)$, $z=t\mathcal{R}\mathrm{/}2$, and $a(t)=(4\pi t{)}^{-m}{\det }^{1\mathrm{/}2}(z\mathrm{/}\sinh z)$. All positive-degree curvature coefficients are nilpotent, so the power series for $\sinh$, $\coth$, determinant and the square root truncate in each exterior degree. Since $\mathcal{R}$ is skew, $B$ is an even power series in $\mathcal{R}$, hence symmetric and commuting with $\mathcal{R}$; $\mathcal{R}B$ is skew. Direct differentiation gives $t{B}^{\mathrm{\prime }}=B-B^2+\frac{t^2}{4}{\mathcal{R}}^2,$ using $z\frac{d}{dz}(z\coth z)=z\coth z-z^2{\mathrm{csch}}^{2}z$, and $\frac{a^{\mathrm{\prime }}}{a}=-\frac{\mathrm{tr}B}{2t},$ since $\dim M=2m$.

Let $\varphi =a(t)\exp (-x^{T}B(t)x\mathrm{/}(4t))e^{-tF}$. For $L_i={\mathrm{\partial }}_i+(\mathcal{R}x{)}_i\mathrm{/}4$, one has $L_i\varphi =(-(Bx{)}_i\mathrm{/}(2t)+(\mathcal{R}x{)}_i\mathrm{/}4)\varphi$. Summing $L_i^2$ for $L_i={\mathrm{\partial }}_i+(\mathcal{R}x{)}_i\mathrm{/}4$ gives ${\sum }_i{L}_i^2\varphi =(-\frac{\mathrm{tr}B}{2t}+\frac{x^T{B}^{2}x}{4t^2}-\frac{x^T{\mathcal{R}}^{2}x}{16})\varphi ,$ because the mixed term is $x^T\mathcal{R}Bx=0$ and $\mathrm{tr}\mathcal{R}=0$. The equations in step 1.1 cancel the constant and quadratic coefficients of $({\mathrm{\partial }}_t-{\sum }_i{L}_i^2+F\wedge )\varphi$, so $\varphi$ solves the oscillator heat equation. As $t\downarrow 0$, $B(t)\to I$, $a(t)\sim (4\pi t{)}^{-m}$, and the normalized Euclidean Gaussian tends to the delta distribution; this fixes the initial value.

At $x=0$ the Gaussian factor is $1$. The spinor supertrace kills every Clifford degree below $2m$ and multiplies the top term by $(-2i{)}^m$ from [F1]; combining this with $(4\pi {)}^{-m}$ contributes $(-i\mathrm{/}(2\pi ){)}^m$ to each product of $m$ curvature two-forms. The determinant factor is even in $\mathcal{R}$, so replacing each tangent-curvature symbol by $-i{\mathrm{\Omega }}^{LC}\mathrm{/}(2\pi )$ gives exactly $\widehat{A}$ in [F2]. The twisting factor $e^{-F}$ becomes $e^{iF\mathrm{/}(2\pi )}=e^{-F\mathrm{/}(2\pi i)}$ and its trace is the relative supertrace in [F2]. Their degree-$2m$ part is the asserted density.

No matrix of forms is diagonalized: the calculation is in a finite-dimensional graded-commutative algebra, and all formal series are degree-truncated. Endomorphisms on $W$ need not commute with one another; they commute with the scalar tangent-curvature matrix, which is all the Gaussian calculation uses. The sign is fixed by the positive-frequency symbol $ic(\xi )$ and the curvature convention in [F2]. The local spinor factor is the actual interface in [A1].

If $m=0$, the determinant of the empty matrix and the empty Gaussian factor are $1$, and all curvature two-forms vanish. Thus the displayed kernel is $I_V$, the normalized heat solution on a point as in [F1]. Its diagonal supertrace with respect to the unsigned counting density is ${\mathrm{str}}_{V}I$. Multiplying by the oriented volume form in [F3] gives the degree-zero form ${\epsilon }_y{\mathrm{str}}_{V}I$. By [F2] and the reference grading, ${\mathrm{str}}_{W_y}I={\epsilon }_y{\mathrm{str}}_{V_y}I$, so this oriented form is exactly $[\widehat{A}\mathrm{ch}(V\mathrm{/}S){]}_0$. This distinguishes the heat density from the differential form at a negatively oriented point.