# 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):

The image of $H^*_ c(\hat{M}) \to H^*(\hat{M})$ is naturally isomorphic to the space of $L^2$ harmonic forms on $\hat{M}$.□

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

The $L^2$ harmonic forms in $\Omega _+(\hat{M})$ are in canonical bijection with harmonic forms in $\Omega _+(M)$ such that in the collar neighbourhood of the boundary, if we decompose

$\alpha = \sum _\lambda f_\lambda (u) \psi _\lambda (y),$then $f_\lambda (0) = 0$ for all $\lambda \geq 0$.□

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.

The harmonic forms in $\Omega _-(M)$ satisfying $f_\lambda (0) = 0$ for all $\lambda < 0$ are in canonical bijection with the harmonic forms in $\Omega _-(\hat{M})$ that can be written as a sum of an $L^2$ harmonic form plus a form that is constant in $u$ in the $\partial M \times (-\infty , 1)$ part.□

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

Under the boundary conditions described, we have

$\operatorname{index}(D: \Omega _+(M) \to \Omega _-(M)) = \operatorname{sign}M - h^-_\infty .\fakeqed$□

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

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.