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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?

Thanks for your time.

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  • $\begingroup$ 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 $\Omega^1(M)\to \mathfrak{X}^1(M)$. $\endgroup$
    – Florian
    Jan 12 at 16:48
  • $\begingroup$ @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? $\endgroup$
    – Anil Bagchi.
    Jan 12 at 18:02

1 Answer 1

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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 $\pi^\sharp$.

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