The Heat KernelSignature for Manifolds with Boundary

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 DD be a first-order differential operator on a manifold MM with boundary. To define it as an unbounded operator, we need to specify its domain. If M=\partial M = \emptyset , then we can simply take it to be H1(M)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 H01(M)H^1_0(M).

Once we have picked a domain for DD, we can define DD^*. 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 H1(M)H^1(M), and is determined by the requirement that

Mf(x)g(x)  dx=0 for all fdom(D),gdom(D). \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,Dg)(Df, g) = (f, D^*g) with the boundary terms vanishing when integrating by parts. For example, if dom(D)=H01(M)\operatorname{dom}(D) = H^1_0(M), then dom(D)=H1(M)\operatorname{dom}(D^*) = H^1(M), and vice versa.

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

indexD=dimkerDdimkerD=dimkerDDdimkerDD. \operatorname{index}D = \dim \ker D - \dim \ker D^* = \dim \ker D^*D - \dim \ker D D^*.

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

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 d+d:Ω+Ω\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

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

We then have a bilinear form

Hd/2(M,M)Hd/2(M,M)Hd(M,M)H0(M)R 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 Hd/2(M)Hd/2(M,M)H^{d/2}(M) \otimes H^{d/2}(M, \partial M), we know the kernel of this pairing is contained in ker(Hd/2(M,M)Hd/2(M))\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 Hk(M,M)H^k(M, \partial M) in Hk(M)H^k(M).

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

M^=MMM×R0, \hat{M} = M \cup _{\partial M} \partial M \times \mathbb {R}_{\leq 0},

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

      H^*_c(\hat{M}) \ar[r] \ar[d, "\sim"] & H^*(\hat{M}) \ar[d, "\sim"]\\
      H^*(M, \partial M) \ar[r] & H^*(M).

So we equivalently want to understand the image of Hc(M^)H(M^)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):

Lemma 5.1

The image of Hc(M^)H(M^)H^*_c(\hat{M}) \to H^*(\hat{M}) is naturally isomorphic to the space of L2L^2 harmonic forms on M^\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 L2L^2 harmonic forms on M^\hat{M} look like. Consider the subset M×(,1)\partial M \times (-\infty , 1), writing uu for the second coordinate. Then we can write the operator D=d+dD = \mathrm{d}+ \mathrm{d}^* as

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

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

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

α=α0+α1du, \alpha = \alpha _0 + \alpha _1 \wedge \mathrm{d}u,

and so we have

α=±(α1)±(α0)du. *\alpha = \pm (*\alpha _1) \pm (*\alpha _0) \wedge \mathrm{d}u.

So the map Ω+(M^)MΩ(M)\Omega _+(\hat{M})|_{\partial M} \to \Omega (\partial M) that sends α\alpha to α0\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, AA is given by

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

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

We now decompose Ω+M\Omega _+|_{\partial M} into the AA-eigenspaces with an eigenbasis {ψλ}\{ \psi _\lambda \} , and we can write an L2L^2 harmonic form in Ω+\Omega _+ as

α=λfλ(u)ψλ(y). \alpha = \sum _\lambda f_\lambda (u) \psi _\lambda (y).

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

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

Since we allow λ(,1)\lambda \in (-\infty , 1), this is in L2L^2 if and only if fλ(0)=0f_\lambda (0) = 0 for all λ0\lambda \geq 0. Thus, we conclude that

Theorem 5.2

The L2L^2 harmonic forms in Ω+(M^)\Omega _+(\hat{M}) are in canonical bijection with harmonic forms in Ω+(M)\Omega _+(M) such that in the collar neighbourhood of the boundary, if we decompose

α=λfλ(u)ψλ(y), \alpha = \sum _\lambda f_\lambda (u) \psi _\lambda (y),

then fλ(0)=0f_\lambda (0) = 0 for all λ0\lambda \geq 0.

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

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

The adjoint boundary condition then says fλ(0)=0f_\lambda (0) = 0 for all λ<0\lambda < 0. Since we allow f0(0)f_0(0) to be non-zero, the space of solutions has a slightly more complicated description.

Theorem 5.3

The harmonic forms in Ω(M)\Omega _-(M) satisfying fλ(0)=0f_\lambda (0) = 0 for all λ<0\lambda < 0 are in canonical bijection with the harmonic forms in Ω(M^)\Omega _-(\hat{M}) that can be written as a sum of an L2L^2 harmonic form plus a form that is constant in uu in the M×(,1)\partial M \times (-\infty , 1) part.

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

Theorem 5.4

Under the boundary conditions described, we have

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

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

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

An exercise in Poincaré duality gives dimimj=12dimH(M)\dim \operatorname{im}j = \frac{1}{2} \dim H^*(\partial M). This gives us a bound

h12dimH(M). h^-_\infty \leq \frac{1}{2}\dim H^*(\partial M).

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