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

$D = \sigma \left(\frac{\partial }{\partial u} + A\right),$ where $\sigma = \sigma _D(\mathrm{d}u)$ is the bundle isomorphism $E \to F$ given by the symbol of $D$, and $A\colon \Gamma (E) \to \Gamma (E)$ is a first-order self-adjoint elliptic operator, *independent of $u$*. This is the scenario we had in Section 5.

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

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

*double*of $M$ obtained by gluing $M \cup _{\partial M} M$. Everything such as $E$ and $D$ 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 $M$.

We pick bump functions $\phi _1, \phi _2, \psi _1, \psi _2$ as follows:

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

Thinking of these as multiplication operators, we set

$R = \phi _1 Q_1 \psi _1 + \phi _2 Q_2 \psi _2,$and standard gluing techniques shows that this works.

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

$\operatorname{index}D = \int _M H_t(x, x)\; \mathrm{d}x$for any $t$. So our job will be to construct $H$ and understand its asymptotic behaviour as $t \to 0$.

As in the previous lemma, we can construct (approximate) fundamental solutions $H_1$ and $H_2$ near the boundary and in the interior respectively, and set

$K_t(x, y) = H_{1, t}(x, y) \phi _1(x) \psi _1(y) + H_{2, t} \phi _1(x) \psi _2(y).$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 $H$ such that $H_t - K_t \to 0$ exponentially as $t \to 0$. So we have

$\operatorname{index}D \sim \int _{\partial M \times [0, 1]} H_{1, t}(x, x) \psi _1(x) \; \mathrm{d}x + \int _M H_{2, t}(x, x)\psi _2(x)\; \mathrm{d}x.$In the first integral, as $t \to 0$, the contributions of any positive $u$ is exponentially suppressed. So we can replace the integral with one over $\partial M \times \mathbb {R}_{\geq 0}$ and get rid of $\psi _1$.

In the second integral, we know that

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

Rearranging these, we know that

$K(t) \sim \operatorname{index}D - \sum _{k \geq 0} t^{k - d/2} \int _M a_k(x)\; \mathrm{d}x.$We are now in the situation of the end of the previous section. After rearranging, we are allowed to conclude

$\operatorname{index}D = \int _M a_{d/2}(x) \; \mathrm{d}x - \frac{h + \eta (0)}{2}.$We now apply this to the case where $D$ is the signature operator $\mathrm{d}+ \mathrm{d}^*\colon \Omega _+ \to \Omega _-$. We have already found that

$a_{d/2}(x) = L(p_1, \ldots , p_{d/4}).$Recall that $\ker A$ consists of harmonic forms, so

$h = \dim H^*(\partial M),$Combining these with Theorem 5.4, we deduce that

$\operatorname{sign}M = \int _M L(p) - \frac{h + \eta (0)}{2} + h^-_\infty .$What we have to do now is to do something slightly sneaky. The reason $h^-_\infty$ shows up is that our boundary condition for $D$ required $f_\lambda (0) = 0$ for $\lambda \geq 0$, while the adjoint boundary condition requires it for $\lambda < 0$, and the $\lambda = 0$ case is not treated “symmetrically”.

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

$\operatorname{sign}M = \int _M L(p) - \frac{-h + \eta (0)}{2} - h_\infty ^+.$Subtracting these two equations give

$h = h_\infty ^- + h_\infty ^+.$Knowing also that $h_\infty ^{\pm } \leq \frac{h}{2}$, we know that they must in fact be equal. So we get