General "Hodge theorem"

Question

I know basically zero Hodge theory, so this question might be weird.

Let $$A \stackrel{S}{\longrightarrow} B \stackrel{T}{\longrightarrow} C$$

be a sequence of closed, densely defined maps of Hilbert spaces, with $T \circ S = 0$. Note that the conditions imply that the adjoints of both maps are also densely defined.

Let $\Delta = SS^*+T^*T$ be the "Laplacian" of this complex.

Call an element $b$ of $B$ "harmonic" if $\Delta b = 0$.

Question 1 Is it true that every cohomology class $[b] = \{ b+ Sa : a \in \textrm{Dom}(S)\}$ has a harmonic representative?

I suspect the answer is "no", since otherwise proofs of the Hodge theorem would be presented in this generality.

Question 2 Can someone give me a natural example of where this fails?

Question 3 At this level of generality, are there any additional hypotheses which would yield a Hodge theorem?

Some motivation:

I frequently use the lemma that, in this situation, all closed forms are exact precisely when $\langle \Delta b,b \rangle \geq C\lVert b \rVert ^2$ for some constant $C$.

The proof of this is a slightly tricky application of Hahn-Banach.

If the "Hodge theorem" was true, the lemma would be a corollary of it, since the inequality implies 0 is the only harmonic form, and thus that the cohomology must be trivial.

Answer 1

Response to questions 1 and 2:

Take $A=B = L^2(\mathbb{Z})$ and $C=0$. Let $S((a_i)) = (a_i - a_{i-1})$ and $T$ be the zero map. Since $\sum_i (a_i- a_{i-1})^2 \leq \sum_i (2 a_i^2 + 2 a_{i-1}^2) = 4 \sum_i a_i^2$, the operator $S$ is bounded and hence closed. Note that any sequence $(b_i)$ in the image of $S$ obeys $\sum_{i=-\infty}^{\infty} b_i =0$, so $S$ is not surjective.

We have $S^{\ast}(b_i) = (b_i - b_{i+1})$, so $\Delta(b_i) = (-b_{i-1} + 2 b_i - b_{i+1})$. The kernel of $\Delta$ is sequences of the form $b_i = u i + v$. However, such a sequence is in $L^2(\mathbb{Z})$ if and only if $u=v=0$. So there are no nonzero harmonic functions in $B$.

I think you should be able to make a similar example with $L^2(\mathbb{R})$ and differentiation, but my lack of confidence in functional analysis terminology lead me to use the discrete version.

Answer 2

Notation: I'll denote the cohomology ($\ker T/ \operatorname{im} S$) by $H$ and the harmonic elements of $B$ by $\mathscr{H}$. Note that the harmonics $\mathscr{H}$ consist precisely of the intersection of the kernel of $T$ with the kernel of $S^\ast$, since $\Delta u = 0 \implies 0=\langle \Delta u, u \rangle = \langle S^\ast u, S^\ast u \rangle + \langle T u, T u \rangle \implies S^\ast u = Tu =0$.

For convenience, let $\tilde{A} = A \oplus C$, and view $S$ and $T$ as operators $S \oplus 0 : \tilde{A} \to B$ and $0 \oplus T: B \to \tilde{A}$, so that our complex is $$\tilde{A} \overset{S}{\to} B \overset{T}{\to} \tilde{A} $$ Let $D = S^\ast + T$, thought of as a "Dirac operator" $B \to \tilde{A}$. Its adjoint is $D^\ast = S + T^\ast: \tilde{A} \to B$. Then $D^\ast D = \Delta$, and $\mathscr{H} = \ker D$.

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Now, since $\mathscr{H} \subset \ker T$, there is a natural map $\mathscr{H} \to H$ mapping an element to its cohomology class. This map can be seen to be injective, under no additional assumptions. I will focus on the question of its surjectivity. David's answer shows that it is not surjective in general; I will try to answer Question 3, i.e., come up with sufficient conditions for surjectivity.

My source for the following is the proof of the Hodge theorem for elliptic complexes in Liviu Nicolaescu's notes on the index theorem (link). The usual assumption in the case of an elliptic complex is that $D$ is an elliptic differential operator over a compact manifold, from which several nice properties follow. I think I have identified the following two hypotheses that hold for elliptic complexes that are necessary to make Nicolaescu's proof work:

1. $\mathscr{H}$ is a closed subspace of $B$. (I'm not sure, but I think this holds if $T$ and $S^\ast$ are both closed unbounded operators. In the case $D$ is an elliptic differential operator, $\mathscr{H}$ is in fact finite-dimensional.)
2. A "Fredholm alternative" property for $D^\ast$, which I will state as follows: $(\ker D)^\perp$ is contained in the range of $D^\ast$ (i.e, for every $v \in \mathscr{H}^\perp = (\ker D)^\perp$, there exists $u \in \text{Dom}(D^\ast) \subset \tilde{A}$ such that $D^\ast u = v$).

(I'm a little rusty on unbounded operator theory, so it's possible that I'm being sloppy here somehow.)

Here's an outline, following Nicolaescu and using assumptions 1 and 2 above, of a proof of surjectivity of the map $\mathscr{H} \to H$, i.e., that every cohomology class contains a harmonic element:

For $h \in \ker T$, let $\tilde{h} \in \mathscr{H}$ be the orthogonal projection of $h$ onto $\mathscr{H}$ (which exists by the closedness of $\mathscr{H}$). We want to show that $h$ and $\tilde{h}$ are cohomologous, i.e., that $h = \tilde{h} + Su$ for some $u$. The Fredholm alternative property gives us that $h - \tilde{h} = D^\ast u$ for some $u$, since $h - \tilde{h} \in \mathscr{H}^\perp$ by the definition of orthogonal projection. Applying $T$ gives that $ 0 = T(h - \tilde{h}) = TD^\ast u = TSu + TT^\ast u = TT^\ast u$, proving $T^\ast u= 0$. Thus in fact $D^\ast u = S u$, and $h$ and $\tilde{h}$ are cohomologous.

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Comments: Perhaps assumption 2 (the Fredholm alternative property) could be rephrased, or a more useful sufficient condition for it could found. Nicolaescu's notes (p. 60) give a proof of the Fredholm property (assuming $D^\ast$ is an elliptic operator) that seems to use elliptic regularity and the fact that the range of $D^\ast$ is closed (which he proves using a Poincare inequality and the compact embedding from Rellich's theorem).

Perhaps weaker sufficient conditions could be found, but I'm not sure.

As may be evident, I feel a bit uncomfortable working in this abstract setting. I would rather work with elliptic operators and Sobolev spaces. Comments from readers would be appreciated.