Skip to main content
2 of 4
added 20 characters in body; edited tags
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

When the exterior derivation $d$ is a Lie algebra morphism?

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 mirphism 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 $\chi^{\infty}(M)$. 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 a 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 Lie algebra morphism for all domension $i$. 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 $\forall i>0 \;\;d:\Omega^i(M)\to \Omega^{i+1}(m)$ preserve the corresponding Lie brackets?

Ali Taghavi
  • 356
  • 8
  • 31
  • 123