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 be a Riemannian manifold with boundary , and be a first-order elliptic differential operator. We assume there is a choice of collar neighbourhood of with a diffeomorphism to such that
where is the bundle isomorphism given by the symbol of , and is a first-order self-adjoint elliptic operator, independent of . This is the scenario we had in Section 5.
There is no loss in generality if we assume and , and we will do so to simplify notation.
has a two-sided parametrix which sends .
The lemma implies everything we did for manifolds without boundary go through without modification, and the index of
can be calculated by
be the parametrix on
given by restricting that on
we constructed previously. Let
be a parametrix on the interior of
. To be precise, we consider the double
obtained by gluing
. Everything such as
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
We pick bump functions as follows:
Importantly, here , and so is a partition of unity.
Thinking of these as multiplication operators, we set
and standard gluing techniques shows that this works.
We now know that if we can find a fundamental solution for , then we can calculate
for any . So our job will be to construct and understand its asymptotic behaviour as .
As in the previous lemma, we can construct (approximate) fundamental solutions and near the boundary and in the interior respectively, and set
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 such that exponentially as . So we have
In the first integral, as , the contributions of any positive is exponentially suppressed. So we can replace the integral with one over and get rid of .
In the second integral, we know that
where are given by local formulas, which are the same as the ones in the without boundary case. Moreover, on the collar neighbourhood, and are literally equal, both being (near the boundary, they are equal as differential operators but have different domains. Here we do not have boundary conditions). So the upshot is on the collar neighbourhood, and so we can drop the in the integral.
Rearranging these, we know that
We are now in the situation of the end of the previous section. After rearranging, we are allowed to conclude
We now apply this to the case where is the signature operator . We have already found that
Recall that consists of harmonic forms, so
Combining these with Theorem 5.4, we deduce that
What we have to do now is to do something slightly sneaky. The reason shows up is that our boundary condition for required for , while the adjoint boundary condition requires it for , and the case is not treated “symmetrically”.
We can run the whole calculation all over again where the boundary condition for is now for . The main difference is that in the formula for , we now declare instead of . Then the result is that
Subtracting these two equations give
Knowing also that , we know that they must in fact be equal. So we get