1
$\begingroup$

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).

$\endgroup$
  • 2
    $\begingroup$ In the special case of the trivial bundle, applying the Hodge star shows this is equivalent to $b_{n-1} =0$. $\endgroup$ – user98602 Mar 20 '17 at 14:51
1
$\begingroup$

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 $H^2(M) = 0$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Asaf Shachar Mar 21 '17 at 13:54
  • 1
    $\begingroup$ @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 $\endgroup$ – levap Mar 21 '17 at 18:21
  • 1
    $\begingroup$ 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 $\endgroup$ – levap Mar 21 '17 at 18:24
  • 1
    $\begingroup$ math.harvard.edu/~brantner/hodge.pdf and I expect this carries verbatim for Euclidean vector bundles. $\endgroup$ – levap Mar 21 '17 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.