What are the elements in $\Gamma(\Lambda^2 TM)$?

Question

In the lecture notes, Proposition 1.19 on page 9, it is said that on every Poisson manifold there is a unique bivector field $\Pi \in \Gamma(\Lambda^2 TM)$ such that $$ \{f, g\} = \langle \Pi, df \wedge dg \rangle. $$

I think that $f, g \in C^{\infty}(M, \mathbb{R})$ and $df, dg \in T^*M$. Therefore $df \wedge dg \in \Lambda^2 T^*M$. Why here $\Pi \in \Gamma(\Lambda^2 TM)$ but not in $\Gamma(\Lambda^2 T^*M)$? What are the form of elements in $\Gamma(\Lambda^2 TM)$ Thank you very much.

Edit: is the following true? $\Pi^*: \Lambda^2(T_x^*M) \to C^{\infty}(M)$, $\langle \Pi, df \wedge dg \rangle = \Pi^*(df \wedge dg) \in C^{\infty}(M)$.

Answer 1

The author is using $\langle,\rangle$ to denote the natural pairing between $\Lambda^2TM$ and its dual $\Lambda^2 T^*M$.

Answer 2

I'm going to be a bit more general and deal with general multivectors. We have an isomorphism of $C^\infty(M)$-modules: $$\Gamma(\Lambda^k TM)\longrightarrow \textrm{Alt}^k_{C^\infty(M)}(\Omega^1(M), C^\infty(M)),$$ where $\Omega^1(M):=\Gamma(T^*M)$. The isomorphism is done as follows:

$$\Pi\longmapsto ((\omega_1, \ldots, \omega_k)\longmapsto \langle \Pi, \omega_1\wedge\ldots \wedge \omega_k\rangle),$$ where $\langle \Pi, \omega_1\wedge \ldots \wedge \omega_k\rangle$ is the smooth function on $M$ defined by:

$$\langle \Pi, \omega_1\wedge \ldots \wedge \omega_k\rangle(p):=(\omega_1\wedge \ldots \wedge \omega_k)_p(\Pi_p).$$ To understand the above definition notice:

(i) $\Pi_p\in \Lambda^k T_pM$

(ii) $(\omega_1\wedge \ldots \omega_k)_p:=(\omega_1)_p\wedge \ldots (\omega_k)_p\in \Lambda^k T_p^*M\simeq \textrm{Hom}_{\mathbb R}(\Lambda^k T_pM, \mathbb R)$ so that it makes sense $(\omega_1\wedge \ldots \wedge \omega_k)_p(\Pi_p)$.

As to the form of $\Pi$ we must work locally: Take $(U, x_1, \ldots, x_n)$ a local coordinate system of $M$ around $p$. Then $\{(\partial_{x_1})_p, \ldots, (\partial_{x_n})_p\}$ is a basis of $T_pM$ hence:

$$\{(\partial_{x_{j_1}})_p\wedge \ldots \wedge (\partial_{x_{j_k}})_p: j_1<\ldots<j_k\}$$ is a basis of $\Lambda^k T_pM$, hence:

$$\Pi_p=\sum_{j_1<\ldots<j_k} f_{j_1\ldots j_k}(p) (\partial_{x_{j_1}})_p\wedge \ldots (\partial_{x_{j_k}})_p,$$ for suitable scalars $f_{j_1\ldots j_k}(p)$. Hence, on $U$:$$\Pi=\sum_{j_1<\ldots <j_k} f_{j_1\ldots j_k} \partial_{x_{j_1}}\wedge \ldots \wedge \partial_{x_{j_k}}$$ where $f_{j_1\cdots j_k}\in C^\infty(U)$.