Proving the Poisson differential is a coboundary map

Let $$(M, \pi)$$ be a Poisson manifold, that is, $$M$$ is a smooth real manifold and $$\pi \in \mathfrak{X}^2(M)$$ is a (possibly degenerate) skew-symmetric bivector field satisfying $$[\pi, \pi] = 0$$. Here and in the following, square brackets are the Schouten-Nijenhuis bracket.

To define Poisson cohomology, we basically notice $$[\pi,-]: \mathfrak{X}^k(M) \to \mathfrak{X}^{k+1}(M)$$ squares to zero: $$[\pi, [\pi, A]] = 0, \quad \forall A \in \mathfrak{X}^k(M).$$ Many texts (all those I could find) do not prove this, and simply states it follows from the graded Jacobi identity satisfied by the SN bracket plus the property of $$\pi$$ stated above. However, the Jacobi identity for $$B=C=\pi$$ becomes: $$(-1)^{k-1}[\pi,[\pi,A]] - [\pi,[A,\pi]] + (-1)^{k-1}[A,\underbrace{[\pi,\pi]}_{=0}]=0\\ (-1)^{k-1}[\pi,[\pi,A]] + [\pi,[\pi,A]] = 0$$ which gives the wanted result only if $$k$$ is odd, otherwise the LHS terms cancel out!

Am I missing something trivial? How do we fill the gap for $$k$$ even?

What form of the graded Jacobi identity for the Schouten bracket do you prefer? I prefer the one of the form "(graded) commutator of adjoint actions is adjoint action of commutator" (viewed as acting on $$C$$):

$$[A,[B,C]] - (-1)^{(b-1)(a-1)}[B,[A,C]] = [[A,B],C],$$ where $$a = |A|, b=|B|$$. This is the one written on Wikipedia (in reverse).

Putting $$A=B=\pi$$ (so $$a=b=2$$) yields

$$2[\pi,[\pi,C]] = [[\pi,\pi],C] = [0,C] = 0.$$

as desired.

The fact that we view the equation as acting on $$C$$ explains why we choose to put $$A=B=\pi$$.

Let me also do it using the symmetric form:

$$(-1)^{(a-1)(c-1)}[A,[B,C]]+(-1)^{(b-1)(a-1)}[B,[C,A]]+(-1)^{(c-1)(b-1)}[C,[A,B]] = 0.$$

Putting $$B=C=\pi$$ like you did, I get (as you did)

$$(-1)^{a-1}[A,[\pi,\pi]] + (-1)^{a-1}[\pi,[\pi,A]] - [\pi,[A,\pi]] = 0.$$

Now the trick is to use $$[A,\pi] = -(-1)^{a-1}[\pi,A]$$ in the last term (this is where you missed signs):

$$(-1)^{a-1}[A,[\pi,\pi]] + 2(-1)^{a-1}[\pi,[\pi,A]] = 0;$$

divide by $$(-1)^{a-1}$$ and use $$[\pi,\pi]=0$$ to get the result.

• Thanks for answering, but I can't really follow your reasoning. The Jacobi identity I know of is the same as in the wiki page, yes. Assuming some trick to convert it to yours, how does it solve the problem I highlight in my proof? That puzzles me – mattecapu Apr 23 at 19:39
• Now I did it the other way also, pointing out where you missed signs. – Ricardo Buring Apr 23 at 20:04
• Alright!! Thank you so much. I used naive skew-symmetry without thinking. – mattecapu Apr 24 at 7:36