Rescaled heat kernels for positive matrix principal symbols

Work in one precompact bundle chart and write $p(y,\xi )=\sum A^{ij}(y){\xi }_i{\xi }_j$. The matrices need not commute. For a parameter derivative, the matrix exponential satisfies $\mathrm{\partial }{e}^{-p}=-{\int }_0^1{e}^{-(1-s)p}(\mathrm{\partial }p)e^{-sp}ds$, obtained by differentiating the matrix ODE defining the exponential. Iterating this identity and differentiating in $\xi$ expresses every derivative in $y,\xi$ as finitely many simplex integrals of products of polynomial derivatives of $p$ and exponential factors whose time lengths sum to one. Since $p\ge c^{\mathrm{\prime }}\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$ uniformly on the chart closure, the product norm is at most a polynomial in $\mathrm{\mid }\xi \mathrm{\mid }$ times $e^{-c^{\mathrm{\prime }}\mathrm{\mid }\xi {\mathrm{\mid }}^2}$. Thus $e^{-p(y,\cdot )}$ is a uniformly bounded smooth Schwartz family. By [F1] its inverse transform $k_y(w)$ has all uniformly bounded Schwartz seminorms, and $\int k_y(w)dw=e^{-p(y,0)}=I$. This uses Fourier inversion on Schwartz functions, not an exchange of an undamped double Fourier integral.

Let ${\kappa }_i$ be finitely many charts with smooth unitary frames, and let ${\varphi }_i$ be a partition subordinate to compact subsets of those charts. Choose smooth cutoffs ${\chi }_i(y,z)$ supported where both points lie in chart $i$, equal to one near $y=z$ whenever $z\in \mathrm{supp}{\varphi }_i$. Write volume as $w_i(z)d{\kappa }_i(z)$. Define $Z_t(y,z)$ in these frames by the finite sum of ${\varphi }_i(z){\chi }_i(y,z)w_i(z{)}^{-1}{t}^{-n\mathrm{/}2}{k}_{i,z}(({\kappa }_i(y)-{\kappa }_i(z))\mathrm{/}\sqrt{t})$, interpreting each summand as a bundle map. In each summand the principal coefficients are frozen at the source $z$. Step 1.1, local equivalence of chart distance and $d$, and cutoff support give $\mathrm{\parallel }{Z}_t(y,z)\mathrm{\parallel }\le C_L{H}_{t,L}(y,z)$, where $H_{t,L}=t^{-n\mathrm{/}2}(1+d(y,z)\mathrm{/}\sqrt{t}{)}^{-L}$. Moreover $Z_{t}f\to f$ in $L^2$ for smooth $f$: put ${\kappa }_i(y)-{\kappa }_i(z)=\sqrt{t}v$, use the uniform integrable Schwartz bound, and use ${\sum }_i{\varphi }_i=1$ and $\int k_{i,y}=I$. For $R_t=({\mathrm{\partial }}_t+P_y)Z_t$, the frozen principal part cancels ${\mathrm{\partial }}_t$. The remaining principal term contains $A(y)-A(z)=O(d(y,z))$ and two derivatives of the frozen kernel; lower terms contain at most one derivative. The bound $d(y,z)t^{-1}{H}_{t,L+1}\le t^{-1\mathrm{/}2}{H}_{t,L}$ therefore gives $\mathrm{\parallel }{R}_t(y,z)\mathrm{\parallel }\le C_L{t}^{-1\mathrm{/}2}{H}_{t,L}(y,z)$. Derivatives of cutoffs are supported away from the diagonal and obey the same bound by choosing additional Schwartz decay.

Fix $L>n$. Uniform volume bounds in the finite atlas give ${\sup }_y\int H_{t,L}(y,z)dz+{\sup }_z\int H_{t,L}(y,z)dy\le C_L$. Also $\int H_{a,L}(y,u)H_{b,L}(u,z)du\le C_L{H}_{a+b,L}(y,z)$ for $a,b>0$, $a+b\le 1$. To check the second estimate, suppose $a\le b$ and put $r=d(y,z)$. If $r\le \sqrt{b}$, bound the second factor by $b^{-n\mathrm{/}2}$ and integrate the first. If $r>\sqrt{b}$, split into $d(y,u)\ge r\mathrm{/}2$ and its complement. On the first set, the first factor is bounded by $C{a}^{(L-n)\mathrm{/}2}{r}^{-L}\le C{b}^{(L-n)\mathrm{/}2}{r}^{-L}$; integrate the second. On the complement, $d(u,z)\ge r\mathrm{/}2$, so bound the second factor by $C{b}^{(L-n)\mathrm{/}2}{r}^{-L}$ and integrate the first. These bounds are comparable to $H_{a+b,L}(y,z)$; exchanging the factors treats $b\le a$.

Define Volterra convolution by $(A\ast B{)}_t(y,z)={\int }_0^t{\int }_M{A}_{t-s}(y,u)B_s(u,z)duds$. The residual bound and step 3.1 show $\mathrm{\parallel }R\ast R\mathrm{\parallel }\le C{H}_{t,L}$, since ${\int }_0^t((t-s)s{)}^{-1\mathrm{/}2}ds=\pi$ (substitute $s=t{\sin }^2\theta$). Iterating this bounded time kernel gives $\mathrm{\parallel }{R}_t^{\ast (2m)}\mathrm{\parallel }\le C^{2m}{t}^{m-1}{H}_{t,L}\mathrm{/}(m-1)!$ after enlarging $C$. Convolving once more with $R$ gives $\mathrm{\parallel }{R}_t^{\ast (2m+1)}\mathrm{\parallel }\le 2C^{2m+1}{t}^{m-1\mathrm{/}2}{H}_{t,L}\mathrm{/}(m-1)!$ for $m\ge 1$, by bounding $s^{m-1}\le t^{m-1}$ in the time integral; $R$ itself has its original bound. Consequently $Q={\sum }_{m\ge 1}(-1{)}^m{R}^{\ast m}$ converges absolutely on every positive-time compact set, has $\mathrm{\parallel }{Q}_t\mathrm{\parallel }\le C{t}^{-1\mathrm{/}2}{H}_{t,L}$ for $t\le 1$, and satisfies $Q+R+R\ast Q=0$. Fubini and associativity hold because every displayed absolute convolution bound is finite. Hence $S=Z+Z\ast Q$ satisfies $\mathrm{\parallel }{S}_t-Z_t\mathrm{\parallel }\le C\sqrt{t}{H}_{t,L}$ and $S_{t}f\to f$ in $L^2$ for smooth $f$.

Extend kernels causally by zero for negative time. The initial limit of $Z$ gives $({\mathrm{\partial }}_t+P_y)Z={\delta }_{0}I+R$ in distributions. Applying this identity to the absolutely convergent convolutions of step 4.1 yields $({\mathrm{\partial }}_t+P_y)S={\delta }_{0}I+R+Q+R\ast Q={\delta }_{0}I$. These distribution identities can equivalently be verified first on $\epsilon <s<t-\epsilon$, tested against a compactly supported smooth function of time and space, and then passed to the endpoints using the integrable bounds from step 4.1. Thus no differentiation of a singular convolution is assumed pointwise. For smooth initial $f$, let $\psi$ be a smooth eigensection of $P$ with eigenvalue $\lambda$. Formal symmetry and the distribution equation give $a^{\mathrm{\prime }}(t)+\lambda a(t)=0$ for $a(t)=⟨S_{t}f,\psi ⟩$, with $a(0)=⟨f,\psi ⟩$. Therefore $a(t)=e^{-t\lambda }⟨f,\psi ⟩$. The two-sided integral bounds in step 3.1 make $S_t$ bounded on $L^2$ (apply Cauchy–Schwarz with weight $\mathrm{\parallel }{S}_t(y,z)\mathrm{\parallel }$). Completeness of the given eigenbasis proves $S_t=e^{-tP}$ on $L^2$. The series and convolutions define continuous kernels for $t>0$: on time intervals away from the endpoints use uniform convergence, and bound the omitted endpoint intervals by the integrable convolution estimates. Equality as distributions with the smooth given $K_t$ therefore gives $S_t(y,z)=K_t(y,z)$ everywhere. In particular the bound for $S$ is the stated bound for $K$.

Fix $x$ and put $y={\exp }_x(\sqrt{t}a)$, $z={\exp }_x(\sqrt{t}b)$. In a chart, Taylor expansion gives $({\kappa }_i(y)-{\kappa }_i(z))\mathrm{/}\sqrt{t}\to d{\kappa }_i{\mathrm{\mid }}_x(a-b)$, the frozen symbol tends to $p_i(x,\xi )$, and all frame maps, partition weights and density factors tend to their values at $x$. Linear change of cotangent variables in the absolutely convergent Fourier integral transforms each limiting chart expression into ${\varphi }_i(x)k_x(a-b)$: the coordinate Jacobian cancels $w_i(x)$ because normal-coordinate volume density at $x$ is one. Summing gives $k_x(a-b)$. All convergences are uniform on bounded $a,b$ by step 1.1. By step 4.1, $t^{n\mathrm{/}2}\mathrm{\parallel }{K}_t-Z_t\mathrm{\parallel }\le C\sqrt{t}$, so it has zero limit. This proves the rescaled limit and all assertions for $n>0$. If $M$ is empty the claims are vacuous; for $n=0$, compact $M$ is finite and differential operators act separately on its fibers, so ordinary matrix exponentiation gives exactly the asserted finite-dimensional limit.