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.

  • $\begingroup$ 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 $\endgroup$ – mattecapu Apr 23 at 19:39
  • $\begingroup$ Now I did it the other way also, pointing out where you missed signs. $\endgroup$ – Ricardo Buring Apr 23 at 20:04
  • $\begingroup$ Alright!! Thank you so much. I used naive skew-symmetry without thinking. $\endgroup$ – mattecapu Apr 24 at 7:36

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