The Heat Kernel — Heat Equation on Manifold with Boundary

7 Heat Equation on Manifold with Boundary

We now use the previous calculations to establish local formulas for the index of a differential operator on a manifold with boundary.

Let MM be a Riemannian manifold with boundary M\partial M, and D:Γ(E)Γ(F)D\colon \Gamma (E) \to \Gamma (F) be a first-order elliptic differential operator. We assume there is a choice of collar neighbourhood of M\partial M with a diffeomorphism to M×[0,1]\partial M \times [0, 1] such that

D=σ(u+A), D = \sigma \left(\frac{\partial }{\partial u} + A\right),

where σ=σD(du)\sigma = \sigma _ D(\mathrm{d}u) is the bundle isomorphism EFE \to F given by the symbol of DD, and A:Γ(E)Γ(E)A\colon \Gamma (E) \to \Gamma (E) is a first-order self-adjoint elliptic operator, independent of uu. This is the scenario we had in Section 5.

There is no loss in generality if we assume E=FE = F and σ=1\sigma = 1, and we will do so to simplify notation.

Lemma 7.1

D:Γ(E;P)Γ(E)D: \Gamma (E; P) \to \Gamma (E) has a two-sided parametrix RR which sends HsHs+1H^ s \to H^{s + 1}.

The lemma implies everything we did for manifolds without boundary go through without modification, and the index of D:Γ(E;P)Γ(E)D: \Gamma (E; P) \to \Gamma (E) can be calculated by tretDDtretDD\operatorname{tr}e^{-t D^*D} - \operatorname{tr}e^{-t DD^*}, etc.

Proof

Let Q1Q_1 be the parametrix on M×[0,1]\partial M \times [0, 1] given by restricting that on M×R0\partial M \times \mathbb {R}_{\geq 0} we constructed previously. Let Q2Q_2 be a parametrix on the interior of MM. To be precise, we consider the double of MM obtained by gluing MMMM \cup _{\partial M} M. Everything such as EE and DD extend to the double, since we assumed everything looked like a product near the boundary. We then obtain a parametrix on the double using usual elliptic regularity theory, and then restrict to the interior of MM.

We pick bump functions ϕ1,ϕ2,ψ1,ψ2\phi _1, \phi _2, \psi _1, \psi _2 as follows:

\begin{tikzpicture} [xscale=5]
      \draw [thick] (0, 0) -- (1, 0);
      \draw (0, -0.1) node [below] {$0$} -- (0, 0.1);
      \draw (1, -0.1) node [below] {$1$} -- (1, 0.1);
      \draw (0, 1) node [left] {$\phi_1$} -- (0.666, 1) .. controls (0.777, 1) and (0.777, 0) .. (0.888, 0) -- (1, 0);

      \begin{scope}[shift={(0, -2)}]
        \draw [thick] (0, 0) -- (1, 0);
        \draw (0, -0.1) node [below] {$0$} -- (0, 0.1);
        \draw (1, -0.1) node [below] {$1$} -- (1, 0.1);
        \draw (0, 1) node [left] {$\psi_1$} -- (0.333, 1) .. controls (0.5, 1) and (0.5, 0) .. (0.666, 0) -- (1, 0);
      \end{scope}

      \begin{scope}[shift={(1.1, 0)}]
        \draw [thick] (0, 0) -- (1, 0);
        \draw (0, -0.1) node [below] {$0$} -- (0, 0.1);
        \draw (1, -0.1) node [below] {$1$} -- (1, 0.1);
        \draw (1, 1) node [right] {$\phi_2$} -- (0.333, 1) .. controls (0.222, 1) and (0.222, 0) .. (0.111, 0) -- (0, 0);
      \end{scope}

      \begin{scope}[shift={(1.1, -2)}]
        \draw [thick] (0, 0) -- (1, 0);
        \draw (0, -0.1) node [below] {$0$} -- (0, 0.1);
        \draw (1, -0.1) node [below] {$1$} -- (1, 0.1);
        \draw (1, 1) node [right] {$\psi_2$} -- (0.666, 1) .. controls (0.5, 1) and (0.5, 0) .. (0.333, 0) -- (0, 0);
      \end{scope}
    \end{tikzpicture}

Importantly, here ψ1+ψ2=1\psi _1 + \psi _2 = 1, and so is a partition of unity.

Thinking of these as multiplication operators, we set

R=ϕ1Q1ψ1+ϕ2Q2ψ2, R = \phi _1 Q_1 \psi _1 + \phi _2 Q_2 \psi _2,

and standard gluing techniques shows that this works.

We now know that if we can find a fundamental solution HH for etDDetDDe^{-t D^*D} - e^{-t DD^*}, then we can calculate

indexD=MHt(x,x)  dx \operatorname{index}D = \int _ M H_ t(x, x)\; \mathrm{d}x

for any tt. So our job will be to construct HH and understand its asymptotic behaviour as t0t \to 0.

As in the previous lemma, we can construct (approximate) fundamental solutions H1H_1 and H2H_2 near the boundary and in the interior respectively, and set

Kt(x,y)=H1,t(x,y)ϕ1(x)ψ1(y)+H2,tϕ1(x)ψ2(y). K_ t(x, y) = H_{1, t}(x, y) \phi _1(x) \psi _1(y) + H_{2, t} \phi _1(x) \psi _2(y).

We see that this satisfies the properties required to run the proof of Theorem 2.3. So we know there is a true heat kernel HH such that HtKt0H_ t - K_ t \to 0 exponentially as t0t \to 0. So we have

indexDM×[0,1]H1,t(x,x)ψ1(x)  dx+MH2,t(x,x)ψ2(x)  dx. \operatorname{index}D \sim \int _{\partial M \times [0, 1]} H_{1, t}(x, x) \psi _1(x) \; \mathrm{d}x + \int _ M H_{2, t}(x, x)\psi _2(x)\; \mathrm{d}x.

In the first integral, as t0t \to 0, the contributions of any positive uu is exponentially suppressed. So we can replace the integral with one over M×R0\partial M \times \mathbb {R}_{\geq 0} and get rid of ψ1\psi _1.

In the second integral, we know that

H2,t(x,x)k0tkd/2ak(x), H_{2, t}(x, x) \sim \sum _{k \geq 0} t^{k - d/2} a_ k(x),

where ak(x)a_ k(x) are given by local formulas, which are the same as the ones in the without boundary case. Moreover, on the collar neighbourhood, DDD^*D and DDDD^* are literally equal, both being 2u2+A2-\frac{\partial ^2}{\partial u^2} + A^2 (near the boundary, they are equal as differential operators but have different domains. Here we do not have boundary conditions). So the upshot is H2,t(x,x)0H_{2, t}(x, x) \sim 0 on the collar neighbourhood, and so we can drop the ψ2(x)\psi _2(x) in the integral.

Rearranging these, we know that

K(t)indexDk0tkd/2Mak(x)  dx. K(t) \sim \operatorname{index}D - \sum _{k \geq 0} t^{k - d/2} \int _ M a_ k(x)\; \mathrm{d}x.

We are now in the situation of the end of the previous section. After rearranging, we are allowed to conclude

indexD=Mad/2(x)  dxh+η(0)2. \operatorname{index}D = \int _ M a_{d/2}(x) \; \mathrm{d}x - \frac{h + \eta (0)}{2}.

We now apply this to the case where DD is the signature operator d+d:Ω+Ω\mathrm{d}+ \mathrm{d}^*\colon \Omega _+ \to \Omega _-. We have already found that

ad/2(x)=L(p1,,pd/4). a_{d/2}(x) = L(p_1, \ldots , p_{d/4}).

Recall that kerA\ker A consists of harmonic forms, so

h=dimH(M), h = \dim H^*(\partial M),

Combining these with Theorem 5.4, we deduce that

signM=ML(p)h+η(0)2+h. \operatorname{sign}M = \int _ M L(p) - \frac{h + \eta (0)}{2} + h^-_\infty .

What we have to do now is to do something slightly sneaky. The reason hh^-_\infty shows up is that our boundary condition for DD required fλ(0)=0f_\lambda (0) = 0 for λ0\lambda \geq 0, while the adjoint boundary condition requires it for λ<0\lambda < 0, and the λ=0\lambda = 0 case is not treated “symmetrically”.

We can run the whole calculation all over again where the boundary condition for DD is now fλ(0)=0f_\lambda (0) = 0 for λ>0\lambda > 0. The main difference is that in the formula for K(t)K(t), we now declare sign0=1\operatorname{sign}0 = -1 instead of +1+1. Then the result is that

signM=ML(p)h+η(0)2h+. \operatorname{sign}M = \int _ M L(p) - \frac{-h + \eta (0)}{2} - h_\infty ^+.

Subtracting these two equations give

h=h+h+. h = h_\infty ^- + h_\infty ^+.

Knowing also that h±h2h_\infty ^{\pm } \leq \frac{h}{2}, we know that they must in fact be equal. So we get

Theorem 7.2
signM=ML(p)η(0)2. \operatorname{sign}M = \int _ M L(p) - \frac{\eta (0)}{2}.\fakeqed