3
$\begingroup$

In this question we search for some conditions under which the exterior derivation $d:\Omega^i(M)\to \Omega^{i+1}(M)$ on a differentiable manifold $M$ is a Lie algebra morphism in a certain sense. We consider $2$ different cases:

For our first question we consider $(M,\omega)$ a symplectic manifold. Then $\Omega^0(M)$ has a natural Lie algebra structure via Poisson bracket. On the other hand for every Riemannian metric on $M$ we get a Lie algebra structure on $\Omega^1(M)$ since the metric gives us a linear isomorphism between $\Omega^1(M)$ and $\chi^{\infty}(M)$, the Lie algebra of smooth vector fields on $M$. In the simplest case, $M=\mathbb{R}^2$ with its standard symplectic and Riemannian structure, we observe that the differential operator $d:\Omega^0(M) \to \Omega^1(M)$ does not preserve the corresponding Lie brackets. This motivates us to ask the following question:

Question 1: Let $(M,\omega)$ be a symplectic manifold. Does there exist a Riemannian metric on $M$ such that $d:\Omega^0(M) \to \Omega^1(M)$ is a Lie algebra morphism?

In our next question we search for possible Lie algebra structures on higher order differential forms $\Omega^i(M)$, $i>1$, of a Riemannian manifold such that the exterior derivation $d$ would be a Lie algebra morphism for all dimensions $i$. More precisely:

Question 2:

Let $(M,g)$ be a Riemannian manifold. Can we equip each $\Omega^i(M)$ with a Lie algebra structure such that $\forall i>0$, $d:\Omega^i(M)\to \Omega^{i+1}(M)$ preserves the corresponding Lie brackets?

$\endgroup$
  • 6
    $\begingroup$ On Question 1: The Poisson bracket (or its negative, depending on the sign convention) has the property that $d:\Omega^0(M)\to \Omega^1(M)$ becomes a Lie algebra morphism if $\Omega^1(M)$ is identified with the space of smooth vector fields $\chi^\infty(M)$ using the symplectic form. No need for a Riemannian metric here. I even doubt that there can exist one as in the question, as such a metric would probably have to induce the same isomorphism from $\Omega^1(M)$ to $\chi^\infty(M)$ as the symplectic form, which would make it an antisymmetric tensor, a contradiction. $\endgroup$ – B K Jun 25 at 21:38
  • 1
    $\begingroup$ I feel like it's more natural to use the Schouten–Nijenhuis bracket on higher degree forms (along with the symplectic form vector-covector identification), rather than a plain Lie bracket. $\endgroup$ – Paul Reynolds Jun 25 at 21:59
  • $\begingroup$ @BK Thank you very much for your interesting comment. Could you please more explain about its last part?(About the contradiction you pointed out) $\endgroup$ – Ali Taghavi Jun 28 at 7:46
  • $\begingroup$ @PaulReynolds Thank you very much for informing me of this graded bracket. $\endgroup$ – Ali Taghavi Jun 28 at 13:04
3
+50
$\begingroup$

On question 1, to expand on what @BK said: If you have a symplectic structure $\omega$ on a manifold $M$, you get a natural Lie bracket on $\Omega^1(M)$ by the following rule: $$ [\alpha, \beta ] = \omega^\flat([\omega^\sharp (\alpha), \omega^\sharp(\beta)]) $$ Where: $$ \omega^\sharp \colon \Omega^1(M) \to \mathfrak{X}(M) \quad \omega^\sharp(\alpha) = X \ \Leftrightarrow \iota_X \omega = \alpha $$ $$ \omega^\flat \colon \mathfrak{X}(M) \to \Omega^1(M) \quad \omega^\flat(X) = \alpha \ \Leftrightarrow \iota_X \omega = \alpha $$ If you have a metric $ g $ on $M$ then you can define similar sharp and flat maps by pairing with the metric tensor instead. The problem is that the sharp and flat maps completely characterize the tensor. Therefore if $g$ induces the same isomorphism between $\Omega^1(M)$ and $\mathfrak{X}(M)$ as $\omega$ it follows that they are equal. The problem with this is that a metric cannot ever be equal to $\omega$.

In fact, symplectic structures are much more closely related to Lie algebra structures than metrics. So I think if you're trying to construct some Lie theoretic object, my recommendation is that you look more into the symplectic universe.

That said, regarding Question 2: There is no canonical way to make the $\Omega^i(M)$ into Lie algebras by using a symplectic structure. Is it possible? Sure, why not? By picking a basis for each vector space and splitting them along the images, kernels and cokernels of the differentials you can construct a variety of split Lie algebra structures on the resulting infinite dimensional vector spaces. But there is nothing geometrically interesting about this.

If you want a more natural Lie-theoretic structure that has some real geometric meaning, you can continue the suggestion of @PaulReynolds to look at graded brackets. I'll refer you to wikipedia for the definition of the Schouten-Nijenhuis bracket:

https://en.wikipedia.org/wiki/Schouten%E2%80%93Nijenhuis_bracket

Since the symplectic structure on $M $ produces a bunch of isomorphisms $ \omega^\flat \colon \Omega^i(M) \to \wedge^i \mathfrak{X}(M) $, you can transport the Schouten-Nijenhuis bracket to the complex of differential forms.

A closely related, but different option is to weaken the symplectic structure to a Poisson structure. Basically, this is just a Lie bracket: $$ \{ \cdot , \cdot \} \colon C^\infty(M) \times C^\infty(M) \to C^\infty(M) $$ which satisfies: $$ \{ f, gh \} = g \{ f, h \} + h \{ f, g \} $$

By using this bracket, you can actually construct a Lie bracket on $\Omega^1(M)$. By using the exact same formulas for the Schouten-Nijehuis bracket, except using 1-forms instead, you can get a graded bracket on the whole complex of forms. Depending on the Poisson bracket you started with, these carry a lot of geometric meaning regarding symplectic foliations and other cool stuff.

To be clear, these structures I just described do not make the differential into a Lie algebra homomorphism. Rather, I believe you get that the differential is a derivation of the graded bracket. That is: $$ \forall \alpha \in \Omega^i(M) , \ \beta \in \Omega^j(M) \quad d[\alpha,\beta] = [d \alpha , \beta] + (-1)^{i}[\alpha , d \beta] $$

Maybe try looking at Gerstenhaber algebras and BV-algebras in Poisson geometry by Ping Xu for some more advanced reading on the topic.

| cite | improve this answer | |
$\endgroup$

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.