# Non-degeneracy of the Poisson structure.

I am reading a book on Poisson manifold. There I found the notion of non-degenerate Poisson structure.

Definition $$:$$ Let $$M$$ be a Poisson manifold with Poisson bivector field $$\pi.$$ Then $$\pi$$ defines a non-degenerate Poisson structure on $$M$$ if the induced map $$\pi^{\sharp} : T^{\ast} M \longrightarrow TM$$ given by $$\alpha \mapsto \pi (\alpha, \cdot)$$ $$(\alpha \in \Omega^1 (M))$$ is a vector bundle isomorphism.

I don't understand the definition of $$\pi^{\sharp}.$$ How can $$\pi^{\sharp}$$ act on a $$1$$-form rather than acting on an element of $$T^{\ast} M\$$? Also $$\pi (\alpha, \cdot) : \Omega^{1} (M) \longrightarrow C^{\infty} (M)$$ is a $$C^{\infty} (M)$$-linear map. So $$\pi (\alpha, \cdot) \in \mathfrak {X} (M).$$ So it seems to me that $$\pi^{\sharp} : \Omega^1 (M) \longrightarrow \mathfrak {X} (M).$$ What goes wrong in my argument? Could anyone please clear it to me?

• A one form $\alpha$ is a section of the vector bundle $T^*M$, i.e. for every $p\in M: \ \alpha_p \in T_p^*M$. Then $\pi^{\sharp}(\alpha)_p:=\pi_p (\alpha_p,\cdot )\in T_pM$. Indeed $\pi^{\sharp}$ induces a map $\Omega^1(M)\to \mathfrak{X}^1(M)$. Jan 12 at 16:48
• @Florian$:$ But the problem of defining $\pi^{\sharp}$ in this way is that there might be several liftings of $\alpha_p$ to an $1$-form in $\Omega^1 (M).$ So there is a question of well-definedness of $\pi^{\sharp}$ which needs to be guaranteed at first. Am I right? Jan 12 at 18:02

Maybe let's formulate everything a bit more simple and only introduce notation gradually, then it should become clearer. First, $$\pi$$ as a bi-vector field is a smooth map from $$M$$ to $$TM\times TM$$, i.e. there exist two vector fields $$X,Y\in C^\infty(M,TM) =: \Gamma TM =: \mathfrak{X}^1$$, so that $$\pi = (X,Y)$$.
More formally (but this is not so important for your questions, just to clarify notation), vector fields $$X,Y$$ are smooth sections in the tangent bundle $$TM$$ over $$M$$, which is denoted by $$X,Y\in \Gamma TM (=\mathfrak{X}^1)$$. Similarly, $$\pi$$ is a smooth section in $$TM\otimes TM$$, i.e. one writes $$\pi\in \Gamma (TM\otimes TM)$$ and $$\pi = X\otimes Y$$.
A $$1$$-form is a smooth map $$\alpha \in C^\infty(M,T^*M) =: \Gamma T^*M =:\Omega^1M$$, or formally a smooth section in the cotangent bundle $$T^*M$$, which is why one writes $$\alpha\in\Gamma T^*M (= \Omega^1M)$$.
Now, alternatively, such a bi-vector $$\pi$$ can be seen as a bilinear form on $$\Gamma T^*M$$: $$\pi(\alpha_1,\alpha_2) = (X\otimes Y)(\alpha_1,\alpha_2) = \alpha_1(X)\cdot\alpha_2(Y)$$ for any $$1$$-forms $$\alpha_1,\alpha_2\in \Gamma T^*M$$.
If one plugs in only $$\alpha_1$$, the second component of $$\pi$$ is still "free", and one gets (essentially) the vector field $$Y$$: $$\pi(\alpha_1,\cdot) = \alpha(X)\cdot Y\in\Gamma TM$$
This map from $$C^\infty(M,T^*M) = \Gamma T^*M = \Omega^1M$$ to $$C^\infty(M,TM)=\Gamma TM = \mathfrak{X}^1$$ induced by $$\pi$$ is then called $$\pi^\sharp$$.