# Is every co-closed form co-exact?

Let $M$ be a Riemannian manifold, $E$ a vector bundle over $M$, equipped with a flat connection connection $\nabla$ and a $\nabla$-compatible metric $\eta$.

Denote by $\delta$ the adjoint of the covariant exterior derivative $d^{\nabla}$.

$$\delta: \Omega^k(M,E) \to \Omega^{k-1}(M,E)$$

Let $\sigma \in \Gamma(T^*M \otimes E)$ be a one $\,E$-valued form, and suppose that $\delta(\sigma)=0$.

Does there exist an $\,E$-valued form $\alpha \in \Omega^2(M,E)$ such that $\sigma=\delta(\alpha)$?

Comment:

The flatness of $\nabla$ implies $d^{\nabla} \circ d^{\nabla}=0$, hence $\delta^2=0$. So, the condition $\delta(\sigma)=0$ is necessary for the existence of such an $\alpha$. The question is whether it is sufficient.

If this is not true, is there some other representation theorem? What about the special case of $E=M \times \mathbb{R}$? (real-valued forms).

• In the special case of the trivial bundle, applying the Hodge star shows this is equivalent to $b_{n-1} =0$. – user98602 Mar 20 '17 at 14:51

## 1 Answer

Pretty much by definition, an $E$-valued $k$-form $\alpha$ which satisfies $\delta \alpha = 0$ will be exact if $[\alpha] = 0$ in the homology $H_{*}(\Omega(M;E), \delta)$ of the chain complex $(\Omega^{*}(M;E), \delta)$. Let us change the grading and define $A^i(M;E) := \Omega^{n-i}(M;E)$ (where $n = \dim M$). Then $d_{\nabla} \colon \Omega^i(M;E) \rightarrow \Omega^{i+1}(M;E)$ is of degree one and satisfies $d_{\nabla} \circ d_{\nabla} = 0$ and $\delta \colon A^i(M;E) \rightarrow A^{i+1}(M;E)$ is also of degree one and satisfies $\delta \circ \delta = 0$. The Hodge star map can be fixed by introducing signs to make it a (co)chain complex map $\star \colon A^{*}(M;E) \rightarrow \Omega^{*}(M;E)$ which is an isomorphism and so

$$H^{*}(M;E) := H^{*}(\Omega^{*}(M;E), d_{\nabla}) \cong H^{*}(A^{*}(M;E), \delta) = H_{n-*}(\Omega^{*}(M;E), \delta).$$

That is, the cohomology of $E$-valued differential forms (with respect to $d_{\nabla}$) is the same up to grading to the homology of $E$-valued differential forms (with respect to $\delta$). In particular, if $H^{n-1}(M;E) = 0$ then any one-form which satisfies $\delta \alpha = 0$ will be exact.

For concrete examples, take $M$ to be an open subset of $\mathbb{R}^3$ and $E$ to be the trivial vector bundle. If you identify one-forms and two-forms with vector fields on $M$, the equation $\delta X = 0$ is identical (maybe up to a sign) to the equation $\operatorname{div} X = 0$. The equation $\delta Y = X$ is identical to the equation $\operatorname{curl} Y = X$. A necessary and sufficient condition such that for any $X$ which satisfies $\operatorname{div} X = 0$ we can find a "primitive" $Y$ with $\operatorname{curl} Y = X$ is $H^2(M) = 0$.

• Thanks. I guess the last equality is "Poincare Duality"? BTW, do you know if there is a vector-valued version for the Hodge decomposition theorem? – Asaf Shachar Mar 21 '17 at 13:54
• @AsafShachar: It is much weaker but related to Poincare duality. What I wrote is purely algebraic - if $\alpha$ is $d$-closed (or exact) then $\star \alpha$ is $\delta$-closed (or exact) but it won't be necessarily $d$-exact. Poincare duality relates $H^{*}(\Omega(M), d)$ with $H^{n-*}(\Omega(M), d)$, not with $H^{n-*}(\Omega(M),\delta)$. However, if one uses Hodge decomposition, one can get a unique representative in each $d$-cohomology class which is also $\delta$-closed. Hence, those are also representatives of $\delta$-cohomology and $\star$ then will map the $d,\delta$-closed $k$-forms to – levap Mar 21 '17 at 18:21
• the $d,\delta$ closed $n-k$ forms so you get Poincare duality for both $H^{*}(\Omega(M),d)$ and $H^{*}(\Omega(M),\delta)$. Regarding your second question, the same statement should hold for the vector bundle valued case. There is a general notion of an elliptic complex for which one can prove Hodge decomposition - you need a complex of differential operators such that the corresponding Dirac operator $d + \delta$ (the "square root of the Laplacian") is elliptic and then you get Hodge decomposition. You can see a proof along the lines for Hermitian vector bundles in – levap Mar 21 '17 at 18:24
• math.harvard.edu/~brantner/hodge.pdf and I expect this carries verbatim for Euclidean vector bundles. – levap Mar 21 '17 at 18:25