Clifford Bundles, Dirac Operators, and Index Interfaces

Define $f:M\to \mathrm{\mid }N\mathrm{\mid }$ by $f(x)={\sum }_i{\varphi }_i(x)i$. Its finite support consists of cover members containing $x$, so it spans a nerve simplex. The map is continuous for the weak realization topology: near a point only finitely many supports occur, and after shrinking the neighborhood every remaining assigned $U_i$ contains it. Thus the local map lands continuously in one finite simplex. Empty cover members contribute no vertex and may be omitted.

Define the barycentric subdivision $N^{\mathrm{\prime }}$ whose vertices are nonempty simplices $\sigma$ of $N$ and whose simplices are strict finite chains ${\sigma }_0\subset \cdots \subset {\sigma }_k$. The map $h:\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }\to \mathrm{\mid }N\mathrm{\mid }$ sends the vertex $\sigma$ to its barycenter. It is a homeomorphism by an explicit simplexwise inverse: if a point has distinct positive coordinate levels ${\lambda }_1>\cdots >{\lambda }_r>0$, put ${\sigma }_j=\{i:t_i\ge {\lambda }_j\}$ and ${\lambda }_{r+1}=0$. Its expression is ${\sum }_j\mathrm{\mid }{\sigma }_j\mathrm{\mid }({\lambda }_j-{\lambda }_{j+1})b_{{\sigma }_j}$. These nonnegative weights sum to one and recover each coordinate. When levels coincide the intervening weight becomes zero, and on a face zero coordinates disappear; hence the inverse is continuous on each closed simplex and for the weak topologies in [F3].

Using [A1], choose $x_{\sigma }\in U_{\sigma }$ for every nonempty simplex. Construct $g:\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }\to M$ on a chain simplex recursively. Its vertex $\sigma$ maps to $x_{\sigma }$. For ${\sigma }_0\subset \cdots \subset {\sigma }_k$, cone the already defined opposite face to $x_{{\sigma }_0}$ using the unique minimizing geodesics in $U_{{\sigma }_0}$. The opposite face lies in $U_{{\sigma }_1}\subset U_{{\sigma }_0}$ by induction. The cone agrees with the constructions on all other faces because deleting a vertex leaves the same ordered coning formula. It is continuous at the cone point by compactness of each face image and smooth dependence of the convex geodesics in [F1]. The resulting map has image in $U_{{\sigma }_0}$ on this chain simplex and is globally continuous by the weak topology.

For $x\in M$, each simplex in the chain representing $h^{-1}f(x)$ is a face of the support simplex of $f(x)$. Thus both $x$ and $g{h}^{-1}f(x)$ lie in the same $U_{{\sigma }_0}$. Join them by its unique minimizing geodesic. This gives a homotopy $g{h}^{-1}f\simeq {\mathrm{id}}_M$. It is continuous even when the chain changes: on every convex-pair neighborhood geodesics depend continuously on both endpoints, and the local formulas agree on overlaps by uniqueness of the minimizing geodesic. No CW topology on a thick nerve is assumed.

We construct the other inverse homotopy by carriers. For the chain simplex with smallest face ${\sigma }_0$, let $C_{{\sigma }_0}$ be the subcomplex of nerve simplices $\tau$ for which $\tau \cup {\sigma }_0$ is a simplex. It contracts to $b_{{\sigma }_0}$ by straight barycentric interpolation. The map $h$ lies in this carrier, since its support is contained in the largest chain face; $fg$ lies there because $g$ lands in $U_{{\sigma }_0}$ and all members in the support of $f(g(-))$ contain that same point. A face of a chain either has the same smallest simplex or a larger one; its carrier is therefore contained in $C_{{\sigma }_0}$. Induct over chain-simplex dimension to homotope $fg$ to $h$ within these carriers. On a closed $k$-simplex, the two endpoint maps and the already fixed boundary homotopy define a map from the boundary of its prism into the contractible carrier. Extend across the prism by coning that boundary map with the displayed contraction. This extends the homotopy relative to all previously defined faces. For continuity, transpose each characteristic-prism map into the compact-open interval mapping space by [F6]. The maps agree on faces, so the weak topology of [F3] gives a continuous global transpose; untransposing by [F6] gives the required homotopy on the ordinary product with the interval. Consequently $fg{h}^{-1}\simeq {\mathrm{id}}_{\mathrm{\mid }N\mathrm{\mid }}$. Together with step 3.1 this proves $M\simeq \mathrm{\mid }N\mathrm{\mid }$ by actual inverse maps.

Simplicial and singular integral chains are free. Their natural comparison is a homology isomorphism by [F5]. Applying the natural cohomological universal-coefficient short exact sequences gives isomorphisms on the Ext and Hom terms, hence on cohomology by the elementary exact-sequence diagram argument. Therefore $H_{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}^{\ast }(N;{\mathbb{F}}_2)\cong H_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}^{\ast }(\mathrm{\mid }N\mathrm{\mid };{\mathbb{F}}_2)$. The homotopy equivalence of step 4.1 identifies this with $H_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}^{\ast }(M;{\mathbb{F}}_2)$. Two subordinate partitions give homotopic maps $f$ by convex interpolation: their combined support at $x$ consists entirely of sets containing $x$. For a refinement, the composed nerve map and the original partition map have the same property and are homotopic by the same interpolation. Thus the cohomology comparison is independent of those choices and natural under refinement; the choices of $g$ merely supply its inverse.

If $M$ is empty, the nerve has no nonempty simplex and both realizations/maps are empty. A singleton simplex is handled by its chosen point and the same geodesic contraction. Infinite covers cause no compact-image assumption on a thick realization: continuity follows simplexwise and local finiteness of the partition, and the homotopy is built on actual CW prisms. Full AC supplies the choices in [A1] and [F2]; no extra choice strength is introduced.

On an overlap let $e_i^{\mathrm{\prime }}={\sum }_j{a}_{ij}{e}_j$ be a change of oriented orthonormal frame, so $A=(a_{ij})$ is orthogonal. The dual coframe changes by the same orthogonal coefficient matrix with the corresponding inverse convention, while ${\mathrm{\nabla }}_{e_i^{\mathrm{\prime }}}={\sum }_k{a}_{ik}{\mathrm{\nabla }}_{e_k}$. Therefore ${\sum }_{i}c(e^{\mathrm{\prime }i}){\mathrm{\nabla }}_{e_i^{\mathrm{\prime }}}={\sum }_{i,j,k}{a}_{ij}{a}_{ik}c(e^j){\mathrm{\nabla }}_{e_k}={\sum }_{j}c(e^j){\mathrm{\nabla }}_{e_j}$ by ${\sum }_i{a}_{ij}{a}_{ik}={\delta }_{jk}$. Each chart expression has smooth coefficients, and these overlap equalities glue them to one global first-order operator.

If $\dim M=0$, the defining sum for $D_V$ is empty, so $D_V=0$ is formally symmetric. Now suppose $\dim M\ge 1$. Define the complex vector field $Y={\sum }_i⟨c(e^i)u,v⟩e_i$ for smooth sections $u,v$. At any chosen point take a local orthonormal frame with ${\mathrm{\nabla }}^{LC}{e}_i=0$ there. Using [F2], metric compatibility and $\mathrm{\nabla }{e}_i=0$ at that point gives $\mathrm{div}Y=⟨D_{V}u,v⟩-⟨u,D_{V}v⟩$: the second term has this sign because $c(e^i{)}^{\ast }=-c(e^i)$ and the Hermitian pairing is linear in its first variable. Since this is a pointwise identity independent of the chosen normal frame, it holds globally. Apply [F3] to real and imaginary parts of $Y$ and integrate to get ${\int }_M⟨D_{V}u,v⟩={\int }_M⟨u,D_{V}v⟩$. Thus $D_V$ is formally symmetric. Oddness follows termwise because each $c(e^i)$ is odd and $\mathrm{\nabla }$ preserves grading.

At $p$, ${\mathrm{\nabla }}^{LC}{e}_i=0$ and $[e_i,e_j]=0$. Compatibility [F2] gives ${\mathrm{\nabla }}_i^V(c(e^j))=0$ there, so expanding $D_V^2={\sum }_{i,j}{c}_i{c}_j{\mathrm{\nabla }}_i{\mathrm{\nabla }}_j$ yields diagonal terms $-{\sum }_i{\mathrm{\nabla }}_i^2$ and off-diagonal terms ${\sum }_{i<j}{c}_i{c}_j[{\mathrm{\nabla }}_i,{\mathrm{\nabla }}_j]$. The first term is $({\mathrm{\nabla }}^V{)}^{\ast }{\mathrm{\nabla }}^V$ at $p$; the commutators are $F_{ij}^V$. Since $p$ was arbitrary, $D_V^2=({\mathrm{\nabla }}^V{)}^{\ast }{\mathrm{\nabla }}^V+{\sum }_{i<j}{c}_i{c}_j{F}_{ij}^V$ globally.

Applying two covariant derivatives to ${\mathrm{\nabla }}_X^V(c(\xi )v)=c({\mathrm{\nabla }}_X^{LC}\xi )v+c(\xi ){\mathrm{\nabla }}_X^{V}v$ and antisymmetrizing in $X,Y$ gives $[F_{ij}^V,c(\xi )]=c(R_{ij}^{T^{\ast }M}\xi )$. Direct Clifford expansion gives $[c_a{c}_b,c_k]=2({\delta }_{ak}{c}_b-{\delta }_{bk}{c}_a)$; substituting the displayed coefficient $F_{ij}^S$ shows $[F_{ij}^S,c(\xi )]=c(R_{ij}^{T^{\ast }M}\xi )$ with the cotangent action induced by the fixed curvature convention. Subtracting proves $[F_{ij}^{V\mathrm{/}S},c(\xi )]=0$ for every $\xi$.

Substituting $F_{ij}^S$ gives ${\sum }_{i<j}{c}_i{c}_j{F}_{ij}^S=\frac{1}{8}{\sum }_{i,j,a,b}{\mathrm{Rm}}_{ijab}{c}_i{c}_j{c}_a{c}_b$. The Clifford relation expands $c_i{c}_j{c}_a{c}_b$ into exterior degrees $4,2,0$ as $c_i\wedge c_j\wedge c_a\wedge c_b-{\delta }_{ij}{c}_a\wedge c_b+{\delta }_{ia}{c}_j\wedge c_b-{\delta }_{ib}{c}_j\wedge c_a-{\delta }_{ja}{c}_i\wedge c_b+{\delta }_{jb}{c}_i\wedge c_a+{\delta }_{ij}{\delta }_{ab}-{\delta }_{ia}{\delta }_{jb}+{\delta }_{ib}{\delta }_{ja}$. The degree-four coefficient is the total alternation of ${\mathrm{Rm}}_{ijab}$ and is zero by the cyclic first-Bianchi identity [F3]. For the degree-two part, the ${\delta }_{ij}$ term vanishes by skewness in the first curvature pair; the other four contractions give $4{\sum }_i{\mathrm{Rm}}_{ipiq}$ as the coefficient of $c_p\wedge c_q$, which is symmetric in $p,q$ by pair interchange and therefore contracts to zero. The scalar coefficient is ${\sum }_{i,j,a,b}{\mathrm{Rm}}_{ijab}({\delta }_{ij}{\delta }_{ab}-{\delta }_{ia}{\delta }_{jb}+{\delta }_{ib}{\delta }_{ja})=2{\sum }_{i,j}{\mathrm{Rm}}_{ijji}=2{\mathrm{scal}}_g$: the first contraction vanishes and each of the last two equals the scalar contraction. Multiplication by $1\mathrm{/}8$ gives ${\mathrm{scal}}_g\mathrm{/}4$.

Substituting $F_{ij}^V=F_{ij}^S+F_{ij}^{V\mathrm{/}S}$ in step 1.1 and using step 3.1 gives the displayed formula. The principal symbol of $D_V$ at covector $\xi$ is $ic(\xi )$, so ${\sigma }_2(D_V^2)(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$ by [F1]. If $M$ is compact and boundaryless with $n\ge 2$, The geometric Dirac operator is globally well defined and formally symmetric supplies formal symmetry of $D_V$; therefore $D_V^2$ is formally symmetric and [F4] applies. In dimension zero, $T^{\ast }M=0$, so $D_V=0$ and the formula is direct. The empty and zero-rank cases have the zero operator and the same formula with all endomorphisms zero.

For smooth $u$, formal symmetry [F4] and integration of [F1] against $u$ give $\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2=\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2+⟨R_{0}u,u⟩$. Since $R_0$ is bounded and the connection Sobolev norm is equivalent to a finite-chart $H^1$ norm, this gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le C(\mathrm{\parallel }Du{\mathrm{\parallel }}_2+\mathrm{\parallel }u{\mathrm{\parallel }}_2)$. Smooth density extends the estimate to $H^1$, while the first-order local formula gives $\mathrm{\parallel }Du{\mathrm{\parallel }}_2\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$. Thus the graph norm is equivalent to $H^1$, and smooth sections are dense in that graph norm.

Let $B=(P+\lambda {)}^{-1}$ from [F2]. It maps $L^2\to H^2$, so $DB:L^2\to H^1$ is bounded. On smooth $h$, $D(Bh)=B(Dh)$: both solve $(P+\lambda )z=Dh$ because $D$ commutes with $D^2$ on smooth sections and the solution is unique. Density extends the symmetry identity $⟨DBf,h⟩=⟨f,DBh⟩$ to $f,h\in L^2$.

Suppose $u\in L^2$ and $Du=f\in L^2$ distributionally. Then $(P+\lambda )u=Df+\lambda u$ distributionally. Test against $Bh$ for smooth $h$: formal symmetry gives $⟨u,h⟩=⟨f,DBh⟩+\lambda ⟨u,Bh⟩=⟨DBf+\lambda Bu,h⟩$, using step 1.2 and selfadjointness of $B$. Hence $u=DBf+\lambda Bu$ as distributions. The first term is in $H^1$ by the bounded map $DB$, and the second is in $H^2$ by [F2]; therefore $u\in H^1$. Conversely every $H^1$ section has $Du\in L^2$ by the first-order local formula, so the maximal distributional domain equals $H^1$.

The closure of smooth $D$ has domain $H^1$ by step 1.1, so its closure is contained in the maximal distributional operator. If $u\in \mathrm{dom}(D^{\ast })$, then the distribution $Du$ equals the $L^2$ vector $D^{\ast }u$ by the definition of adjoint testing against smooth sections. Step 2.1 gives $u\in H^1$, and formal symmetry plus density shows $D^{\ast }u=Du$. Thus $D^{\ast }=\bar{D}$; the closure is selfadjoint. No compact-resolvent theorem for a general first-order operator is assumed. Full AC supplies the choice assumptions of [F2]–[F4] as recorded in [A1].

Since $D$ is odd and selfadjoint, its blocks satisfy $(D^{+}{)}^{\ast }=D^{-}$ on the $H^1$ domain. If $D^{+}u=0$, then $P_{+}u=D^{-}{D}^{+}u=0$ distributionally, so [F2] makes $u$ a smooth element of the finite-dimensional $\ker P_{+}$. Conversely $P_{+}u=0$ implies $\mathrm{\parallel }{D}^{+}u{\mathrm{\parallel }}_2^2=⟨P_{+}u,u⟩=0$, hence $D^{+}u=0$. The same argument applies to $D^{-}$.

Define $Q=D^{-}{G}_{-}:L^2(V^{-})\to H^1(V^{+})$. The Green bound gives $G_{-}:L^2\to H^2$ and the first-order local formula gives $D^{-}:H^2\to H^1$ bounded. Therefore $D^{+}Q=P_{-}{G}_{-}=I-{\mathrm{\Pi }}_{-}$. The range of $D^{+}$ contains $\ker P_{-}^{\perp }$ and is orthogonal to $\ker P_{-}$ because $⟨D^{+}u,h_{-}⟩=⟨u,D^{-}{h}_{-}⟩=0$. Hence its range is exactly $\ker P_{-}^{\perp }$, which is closed, and its cokernel is $\ker P_{-}$.

For $u\in H^1(V^{+})$, approximate by smooth sections. For smooth $u$, $D^{-}{G}_{-}{D}^{+}u-u$ lies in $\ker D^{+}$ because applying $D^{+}$ gives $D^{+}u-{\mathrm{\Pi }}_{-}{D}^{+}u-D^{+}u=0$; here ${\mathrm{\Pi }}_{-}{D}^{+}u=0$ by the orthogonality in step 1.2. The term $D^{-}{G}_{-}{D}^{+}u$ is orthogonal to $\ker D^{+}$, because $D^{-}$ is the adjoint of $D^{+}$ and $D^{+}{h}_{+}=0$. The difference lies in that kernel and its projection is $-{\mathrm{\Pi }}_{+}u$, so it equals $-{\mathrm{\Pi }}_{+}u$. Therefore $D^{-}{G}_{-}{D}^{+}u=u-{\mathrm{\Pi }}_{+}u$. Boundedness and density extend the identity to $H^1$. The two-sided inverses modulo finite-rank projections and the finite kernel/cokernel prove Fredholmness and the index formula. Full AC is inherited through [A1] and the Green-operator supplier.

Let $E_L$ be the span of eigensections with $0\le \lambda \le L$. For $u\in E_L$, the Lichnerowicz identity and formal symmetry give $\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2=⟨Pu,u⟩\le L\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2=\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2+⟨R_{0}u,u⟩$, where $R_0$ is bounded on compact $M$. Hence $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le C(1+L{)}^{1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Fix finitely many precompact bundle charts and a smooth partition of unity ${\chi }_a$ subordinate to them. In each chart, write ${\chi }_{a}u$ in its bundle frame and extend its coefficients by zero to a fixed Euclidean box. Tile that box by cubes of side between $h\mathrm{/}2$ and $h$, and map $u$ to the averages of all coefficients on all cubes. This linear map has rank at most $C{h}^{-n}$. The fundamental theorem of calculus along coordinate segments gives the cube Poincaré bound $\mathrm{\parallel }f-f_Q{\mathrm{\parallel }}_{L^2(Q)}\le C_{n}h\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_{L^2(Q)}$, where $f_Q$ is the average on $Q$. If all averages vanish, summing this bound over cubes and charts gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le Ch\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$. Take $h=c(1+L{)}^{-1\mathrm{/}2}$ with $c>0$ small enough that $Cc<1$; the averaging map is injective on $E_L$. Therefore $\dim E_L\le C^{\mathrm{\prime }}(1+L{)}^{n\mathrm{/}2}$. For an orthonormal eigenbasis $u_j$, iteration of [F1] on $P^a{u}_j={\lambda }_j^a{u}_j$ gives $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{H^{2a}}\le C_a(1+{\lambda }_j{)}^a$; if $2a>r+n\mathrm{/}2$, Sobolev embedding gives the same polynomial bound for every $C^r$ norm.

The polynomial counting bound and the polynomial $C^r$ eigenfunction bounds imply ${\sum }_j{e}^{-t{\lambda }_j}(1+{\lambda }_j{)}^q<\mathrm{\infty }$ for every fixed $t>0$ and integer $q$: split $[0,\mathrm{\infty })$ into dyadic shells, bound the number and $C^r$ norm on each shell polynomially, and use $e^{-t{2}^k}$ to dominate every polynomial. Thus $K_t(x,y)={\sum }_j{e}^{-t{\lambda }_j}{u}_j(x)\otimes u_j(y{)}^{\ast }$ and every fixed space/time derivative converge absolutely and uniformly. The series solves $({\mathrm{\partial }}_t+P_x)K_t=0$, and integrating it against a smooth section gives $e^{-tP}$ by the orthonormal expansion. Termwise integration on the diagonal gives $\mathrm{Tr}(e^{-tP})={\sum }_j{e}^{-t{\lambda }_j}<\mathrm{\infty }$.

On smooth sections $P_{-}{D}^{+}=D^{+}{P}_{+}$ and $D^{-}{D}^{+}=P_{+}$, $D^{+}{D}^{-}=P_{-}$. Thus for $\lambda >0$, $D^{+}$ maps its $P_{+}$ eigenspace isomorphically to the $P_{-}$ eigenspace with inverse ${\lambda }^{-1}{D}^{-}$, so their contributions to the graded trace $\mathrm{Str}(e^{-tP})$ cancel. At $\lambda =0$, the eigenspaces are $\ker D_V^{+}$ and $\ker D_V^{-}$; their signed dimension difference is the Fredholm index from [F3]. The diagonal kernel series is uniformly convergent, so integrating its pointwise supertrace equals the sum of these eigenspace traces. This proves the stated identity for every $t>0$.

If $M$ is empty all Hilbert spaces and integrals are zero. In dimension zero the spectrum is the finite-dimensional zero operator and the same identity is direct; the analytic estimates above are only invoked in positive even dimension. The trace equality was obtained from the explicit smooth kernel series, not from an unproved diagonal trace theorem.

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 $J$.