The geometric Dirac heat parametrix has a graph-norm remainder

Fix $y$ and use geodesic normal coordinates $x$ centered at $y$. Parallel transport identifies fibers with $V_y$; write $g(x)=\det (g_{ij}(x))$. Set $u_0(x,y)=g(x{)}^{-1\mathrm{/}4}{\tau }_{y\to x}$, where $\tau$ is radial parallel transport. For $j\ge 1$, solve $(r{\mathrm{\nabla }}_{{\mathrm{\partial }}_r}+j+\frac{r}{4}{\mathrm{\partial }}_r\log g)u_j=-P{u}_{j-1}$. Multiplying by $g^{1\mathrm{/}4}$ and transporting back to $V_y$ turns this into $(r{\mathrm{\partial }}_r+j)v_j=-g^{1\mathrm{/}4}{\tau }^{-1}P{u}_{j-1}$. The unique smooth solution is $v_j(r,\theta )=-{\int }_0^1{s}^{j-1}(g^{1\mathrm{/}4}{\tau }^{-1}P{u}_{j-1})(sr,\theta )ds$; substitution and differentiation verify the equation, and smoothness at $r=0$ excludes the singular homogeneous term $C{r}^{-j}$. Hence each $u_j$ is smooth in both variables on a smaller normal neighborhood.

The identity $({\mathrm{\partial }}_t+P)(h_t{\sum }_{j=0}^J{t}^j{u}_j)=t^J{h}_t{b}_J(x,y)$ follows by differentiating the Gaussian and using the displayed recursion to cancel each power $t^{j-1}$; $b_J$ is smooth with bounded derivatives on the smaller neighborhood. Choose $\chi =1$ near the diagonal and supported inside the normal neighborhood. Derivatives of $\chi$ are supported where $d(x,y)\ge \delta >0$; for any integers $q,s$, $t^{-q}{e}^{-{\delta }^2\mathrm{/}(4t)}\le C_{q,s}{t}^s$, so all fixed derivatives of the cutoff error are $O(t^s)$. Near the diagonal each fixed derivative of $t^J{h}_t{b}_J$ loses only finitely many powers of $t$, so increasing $J$ makes its required Sobolev norms $O(t^N)$ to any prescribed order.

The initial operator $H_J(t)$ tends to the identity on smooth sections: in normal coordinates put $x=\sqrt{t}z$ in its integral; $u_0(y,y)=I$, $g(y)=1$, and the normalized Gaussian integrates to one, while the terms $t^j{u}_j$ for $j>0$ vanish. For $R_J=({\mathrm{\partial }}_t+P)H_J$, variation of constants gives $e^{-tP}-H_J(t)=-{\int }_0^t{e}^{-(t-s)P}{R}_J(s)ds$; differentiating $e^{-(t-s)P}{H}_J(s)$ and using the zero initial difference proves the identity and its sign.

For each integer $a$, the graph norm $\mathrm{\parallel }(1+P{)}^{a}u{\mathrm{\parallel }}_2$ is equivalent to $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^{2a}}$ by repeated [F1] elliptic estimates. The semigroup $e^{-tP}$ is a contraction in that graph norm because its spectral multiplier is $e^{-t\lambda }\in [0,1]$ and commutes with $(1+P{)}^a$. The bounds of step 2.1 therefore make the integral in step 3.1 $O(t^{N+1})$ in $H_x^{2a}{C}_y^b$, after choosing $J$ large enough. Choose $2a>r+n\mathrm{/}2$ and Sobolev embedding to obtain the claimed $C_x^r{C}_y^b$ estimate. Differentiating the Duhamel identity finitely many times costs only finitely many powers of $t$, which are absorbed by increasing $J$.

The distance function is used only inside the normal neighborhood, where squared distance is smooth; the cutoff vanishes before the cut locus. Empty $M$ is vacuous, and dimension zero is finite-dimensional heat evolution. This proves the differentiated remainder without invoking Nicolaescu's deferred expansion criterion or Wernli's unproved rescaling limit.