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 be a first-order differential operator on a manifold with boundary. To define it as an unbounded operator, we need to specify its domain. If , then we can simply take it to be . 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 .
Once we have picked a domain for , we can define . 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 , and is determined by the requirement that
This ensures we always have with the boundary terms vanishing when integrating by parts. For example, if , then , and vice versa.
Once we have done all these, we can define the index to be
A word has to be said about . Its domain consists of the functions such that . Elements in the kernel of definitely satisfy this boundary condition, and so we have . This index depends on the choice of domain of , and for our purposes, the right choice is the one whose index relates best to the signature of .
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 , 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
We then have a bilinear form
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 , we know the kernel of this pairing is contained in , and in fact is equal to it. Thus, to understand the signature, we have to understand the image of in .
Here we assume that there is a collar neighbourhood of the boundary that is isometric to . It is useful to consider the manifold
which then has a natural Riemannian structure. Topologically, deformation retracts to , and we have a commutative diagram
So we equivalently want to understand the image of . What we need is the following upgrade of the Hodge decomposition theorem (which we shall not prove):
The image of is naturally isomorphic to the space of harmonic forms on .□
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 harmonic forms on look like. Consider the subset , writing for the second coordinate. Then we can write the operator as
where is an isomorphism given by the symbol of . Up to some signs, it is given by .
The operator is some first-order self-adjoint elliptic operator on , and in particular is independent of . To understand it better, we observe that we can identify with . Indeed, a general differential form on can be written as
and so we have
So the map that sends to 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, is given by
We will work sufficiently formally that the only thing we need to know about is that its kernel consists of forms with , i.e. harmonic forms. The bored reader can figure out the signs carefully themselves.
We now decompose into the -eigenspaces with an eigenbasis , and we can write an harmonic form in as
We then see that a solution to must be given by
Since we allow , this is in if and only if for all . Thus, we conclude that
The harmonic forms in are in canonical bijection with harmonic forms in such that in the collar neighbourhood of the boundary, if we decompose
then for all .□
The adjoint boundary condition then says for all . Since we allow to be non-zero, the space of solutions has a slightly more complicated description.
The harmonic forms in satisfying for all are in canonical bijection with the harmonic forms in that can be written as a sum of an harmonic form plus a form that is constant in in the part.□
Let be the space of such harmonic forms constant in , and set . Then we have
Under the boundary conditions described, we have
We will say one final thing about . As before, we can identify with . This sends injectively into via the composition
An exercise in Poincaré duality gives . This gives us a bound
This does not seem very useful, but we will later bound from below via other means, and show that this is in fact an equality.