5 Signature for Manifolds with Boundary

We will move on to discuss the case of a manifold with boundary. We start with some generalities, and then move on to discuss the signature specifically. Here the boundary conditions play a crucial role, and the best way to think about them is via the theory of unbounded operators.

For simplicity, we will work with first-order differential operators only. Let $D$ be a first-order differential operator on a manifold $M$ with boundary. To define it as an unbounded operator, we need to specify its domain. If $\partial M = \emptyset$, then we can simply take it to be $H^1(M)$. If not, we have to be a bit more careful. For example, if we want to impose Dirichlet boundary conditions, then we can take the domain to be $H^1_0(M)$.

Once we have picked a domain for $D$, we can define $D^*$. As an operator, this will still be given by the usual formulas coming from integration by parts. The domain will again be a subspace of $H^1(M)$, and is determined by the requirement that

$$\int _{\partial M} f(x)g(x) \; \mathrm{d}x = 0\text{ for all }f \in \operatorname{dom}(D), g \in \operatorname{dom}(D^*).$$

This ensures we always have $(Df, g) = (f, D^*g)$ with the boundary terms vanishing when integrating by parts. For example, if $\operatorname{dom}(D) = H^1_0(M)$, then $\operatorname{dom}(D^*) = H^1(M)$, and vice versa.

Once we have done all these, we can define the index to be

$$\operatorname{index}D = \dim \ker D - \dim \ker D^* = \dim \ker D^*D - \dim \ker D D^*.$$

A word has to be said about $D^*D$. Its domain consists of the functions $f \in \operatorname{dom}(D)$ such that $Df \in \operatorname{dom}(D^*)$. Elements in the kernel of $D$ definitely satisfy this boundary condition, and so we have $\ker D = \ker D^*D$. This index depends on the choice of domain of $D$, and for our purposes, the right choice is the one whose index relates best to the signature of $M$.

The point of saying all this is that the boundary condition is pretty important. To understand the signature, we can still compute the index of $\mathrm{d}+ \mathrm{d}^*: \Omega _+ \to \Omega _-$, as long as we pick the right boundary conditions.

To figure out the right boundary condition, we need to recall some facts about the signature. Recall that Poincaré duality gives us an isomorphism

$$H^{d - \ell }(M, \partial M) \cong H_\ell (M).$$

We then have a bilinear form

$$H^{d/2}(M, \partial M) \otimes H^{d/2}(M, \partial M) \to H^{d}(M, \partial M) \to H_0(M) \cong \mathbb {R}$$

given by the cup product, and the signature is defined to be the difference between the dimensions of the positive and negative eigenspaces. This is a degenerate bilinear form. Since the map factors through $H^{d/2}(M) \otimes H^{d/2}(M, \partial M)$, we know the kernel of this pairing is contained in $\ker (H^{d/2}(M, \partial M) \to H^{d/2}(M))$, and in fact is equal to it. Thus, to understand the signature, we have to understand the image of $H^k(M, \partial M)$ in $H^k(M)$.

Here we assume that there is a collar neighbourhood of the boundary that is isometric to $\partial M \times [0, 1]$. It is useful to consider the manifold

$$\hat{M} = M \cup _{\partial M} \partial M \times \mathbb {R}_{\leq 0},$$

which then has a natural Riemannian structure. Topologically, $\hat{M}$ deformation retracts to $M$, and we have a commutative diagram

So we equivalently want to understand the image of $H^*_c(\hat{M}) \to H^*(\hat{M})$. What we need is the following upgrade of the Hodge decomposition theorem (which we shall not prove):

Note that this is peculiar to manifolds of this type. It is not in general true for all open manifolds.

Let us analyze what the $L^2$ harmonic forms on $\hat{M}$ look like. Consider the subset $\partial M \times (-\infty , 1)$, writing $u$ for the second coordinate. Then we can write the operator $D = \mathrm{d}+ \mathrm{d}^*$ as

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

where $\sigma = \sigma _D(\mathrm{d}u)$ is an isomorphism given by the symbol of $D$. Up to some signs, it is given by $\sigma = \mathrm{d}u \wedge +\, \iota _{\frac{\partial }{\partial u}}$.

The operator $A$ is some first-order self-adjoint elliptic operator on $\Omega _+(\hat{M})|_{\partial M}$, and in particular is independent of $u$. To understand it better, we observe that we can identify $\Omega _+(\hat{M})|_{\partial M}$ with $\Omega (\partial M)$. Indeed, a general differential form on $\Omega _+(\hat{M})$ can be written as

$$\alpha = \alpha _0 + \alpha _1 \wedge \mathrm{d}u,$$

and so we have

$$*\alpha = \pm (*\alpha _1) \pm (*\alpha _0) \wedge \mathrm{d}u.$$

So the map $\Omega _+(\hat{M})|_{\partial M} \to \Omega (\partial M)$ that sends $\alpha$ to $\alpha _0$ is in fact an isomorphism. Note that this map is simply the pullback of differential forms.

Under this identification, it is not hard to see that, up to some signs, $A$ is given by

$$A = \pm * \mathrm{d}\pm \mathrm{d}*.$$

We will work sufficiently formally that the only thing we need to know about $A$ is that its kernel consists of forms $\beta$ with $\mathrm{d}\beta = \mathrm{d}^*\beta = 0$, i.e. harmonic forms. The bored reader can figure out the signs carefully themselves.

We now decompose $\Omega _+|_{\partial M}$ into the $A$-eigenspaces with an eigenbasis $\{ \psi _\lambda \}$, and we can write an $L^2$ harmonic form in $\Omega _+$ as

$$\alpha = \sum _\lambda f_\lambda (u) \psi _\lambda (y).$$

We then see that a solution to $D\alpha = 0$ must be given by

$$\alpha = \sum _\lambda e^{\lambda u} f_\lambda (0) \psi _\lambda (y).$$

Since we allow $\lambda \in (-\infty , 1)$, this is in $L^2$ if and only if $f_\lambda (0) = 0$ for all $\lambda \geq 0$. Thus, we conclude that

This is the boundary condition we will be dealing with. To talk about the adjoint, it is easier to identify $\Omega _+$ with $\Omega _-$ via $\sigma$, and drop $\sigma$ from the definition of $D$. Then by the self-adjointness of $A$, we have

$$D^* = -\frac{\partial }{\partial u} + A$$

The adjoint boundary condition then says $f_\lambda (0) = 0$ for all $\lambda < 0$. Since we allow $f_0(0)$ to be non-zero, the space of solutions has a slightly more complicated description.

Let $\mathcal{H}_\infty ^-$ be the space of such harmonic forms constant in $u$, and set $h_\infty ^- = \dim \mathcal{H}_\infty ^-$. Then we have

We will say one final thing about $\mathcal{H}^-_\infty$. As before, we can identify $\Omega _-(\hat{M})|_{\partial M}$ with $\Omega (\partial M)$. This sends $\mathcal{H}_\infty ^-$ injectively into $H^*(\partial M)$ via the composition

$$\mathcal{H}^-_\infty \to H^*(M) \overset {j}{\to } H^*(\partial M).$$

An exercise in Poincaré duality gives $\dim \operatorname{im}j = \frac{1}{2} \dim H^*(\partial M)$. This gives us a bound

$$h^-_\infty \leq \frac{1}{2}\dim H^*(\partial M).$$

This does not seem very useful, but we will later bound $h^-_\infty$ from below via other means, and show that this is in fact an equality.