Relative tangent-groupoid heat-trace limit at simple fixed points

At a simple fixed point $x$, the inverse function theorem makes $x$ isolated in $M^g$; since $M^g$ is closed in compact $M$, it is finite. Choose disjoint sufficiently small normal neighborhoods of these points. In normal coordinates the isometry satisfies $g^{-1}{\exp }_x(\sqrt{t}v)={\exp }_x(\sqrt{t}d{g}_x^{-1}v)$. For the trace-integral identity, factor $U_g{e}^{-t\mathrm{\Delta }}=(U_g{e}^{-t\mathrm{\Delta }\mathrm{/}2})e^{-t\mathrm{\Delta }\mathrm{/}2}$. Both factors have smooth kernels on compact $M$ and finite Hilbert–Schmidt norm. Expansion in an orthonormal basis and Cauchy–Schwarz make the integrated product absolutely summable, giving its trace as the integral of the composed diagonal kernel. The section-action convention in [F1] therefore identifies the trace integrand with ${\sum }_j(-1{)}^j\mathrm{tr}({\rho }_g^j(g^{-1}y)K_t^j(g^{-1}y,y))$.

Use the positive-frequency convention ${\sigma }_1(d_j)(x,\xi )={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=1}{a}_{j,\alpha }(x)(i\xi {)}^{\alpha }$. Integration by parts gives ${\sigma }_1(d_j^{\ast })={\sigma }_1(d_j{)}^{\ast }$ in this convention. Put $B_j={\sigma }_1(d_j)(x,\xi )$, with $B_{-1}=B_N=0$. The principal symbol on $E^j$ of $\mathrm{\Delta }$ is $p_j=B_j^{\ast }{B}_j+B_{j-1}{B}_{j-1}^{\ast }$. If its quadratic form vanishes on $v$, then $B_{j}v=0$ and $B_{j-1}^{\ast }v=0$; exactness gives $v\in \mathrm{im}{B}_{j-1}$ and also orthogonality to that image, hence $v=0$. Compactness of the unit cotangent sphere bundle gives $p_j(x,\xi )\ge c\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$ uniformly. Thus [L1] applies to $\mathrm{\Delta }$, using exactly the given spectral heat data. Its principal symbol may be matrix-valued.

Write $n=\dim M$ and $J_x$ for the normal-coordinate volume density. For $y={\exp }_x(\sqrt{t}v)$ in the chosen neighborhood, [L1] and step 1.1 imply that $t^{n\mathrm{/}2}{J}_x(\sqrt{t}v){\sum }_j(-1{)}^j\mathrm{tr}({\rho }_g^j(g^{-1}y)K_t^j(g^{-1}y,y))$ tends pointwise to ${\sum }_j(-1{)}^j\mathrm{tr}({\rho }_g^j{\mathrm{\mid }}_{E_x^j}{k}_x^j(d{g}_x^{-1}v-v))$. Simplicity makes $d{g}_x^{-1}-I$ invertible. Local distance comparison in the normal chart therefore gives $d(g^{-1}y,y)\ge c_x\sqrt{t}\mathrm{\mid }v\mathrm{\mid }$. Choose $L>n$; bounded density, unitary ${\rho }_g$ and [L1] bound the extended-by-zero rescaled density by $C(1+c_x\mathrm{\mid }v\mathrm{\mid }{)}^{-L}$, integrable on $T_{x}M$. Outside these neighborhoods compactness gives $d(g^{-1}y,y)\ge \delta >0$, so [L1] bounds the full complementary trace integral by $C{t}^{(L-n)\mathrm{/}2}\to 0$. By [F2], the heat-supertrace limit is consequently the sum of these tangent integrals. No uniform-integral-bound-to-convergence implication is used.

Since $d{g}_x$ is orthogonal, $\mathrm{\mid }\det (d{g}_x^{-1}-I)\mathrm{\mid }=\mathrm{\mid }\det (I-d{g}_x)\mathrm{\mid }$. The change of variables $w=(d{g}_x^{-1}-I)v$ therefore contributes $\mathrm{\mid }{\det }_{\mathbb{R}}(I-d{g}_x){\mathrm{\mid }}^{-1}$. By [L1], the constant-coefficient tangent heat kernel has integral $I_{E_x^{\bullet }}$. Its graded fiber trace after composition with ${\rho }_g$ is ${\sum }_j(-1{)}^j\mathrm{tr}({\rho }_g{\mathrm{\mid }}_{E_x^j})$. For $n=0$, the tangent integral and determinant are both the empty-dimensional values $1$, so the same expression holds. Summing the finitely many contributions proves the stated limit.