= Solution
A <symplectic manifold> is a smooth manifold $P$ with a closed, nondegenerate <two-form> $\omega$. Nondegeneracy forces $\dim P=2n$. Examples are $\mathbb R^{2n}$ with $\omega=\sum_i dq^i\wedge dp_i$, an oriented surface with a nowhere-zero area form, and every <cotangent bundle> with its canonical form.
A <Poisson manifold> has a bivector $\Pi$ such that $\{f,g\}=\Pi(df,dg)$ is an antisymmetric bilinear <Poisson bracket>, is a derivation in each argument, and satisfies the <Jacobi identity>. Its rank need not be full or constant. The zero bivector gives a <Poisson manifold> on any smooth manifold, including an odd-dimensional or nonorientable one. A nonzero example on $\mathbb R^3$ is the <Lie-Poisson bracket> $\{x_i,x_j\}=\epsilon_{ijk}x_k$: it has rank two off the origin and rank zero at the origin, so it does not arise from a <symplectic form> on the whole three-dimensional space. Even on $\mathbb R^2$, $\Pi=x\partial_x\wedge\partial_y$ is Poisson, since its Jacobi condition is a three-vector and hence vanishes, but its degeneracy on $x=0$ prevents it from being the inverse of a global <symplectic form>. A manifold carrying a degenerate Poisson structure might separately admit some other symplectic structure; these are distinct claims.
At each point of a <symplectic manifold>, linear nondegeneracy gives a basis with $\omega=\sum_i e^i\wedge f^i$. Its top <wedge product of differential forms> is
$$
\frac{\omega^n}{n!}=e^1\wedge f^1\wedge\cdots\wedge e^n\wedge f^n\ne0.
$$
Thus $\omega^n/n!$ is a global nowhere-zero <volume form> and selects a consistent <symplectic orientation>. This proves \b[every symplectic manifold is orientable] without choosing a global frame.
Let $\pi:T^*M\to M$ and let a point of the <cotangent bundle> be $(x,p)$ with $p\in T_x^*M$. Define the canonical <Liouville one-form> intrinsically by $\theta_{(x,p)}(V)=p(d\pi(V))$. In local coordinates it is $\theta=p_i dq^i$, so
$$
\omega=-d\theta=dq^i\wedge dp_i.
$$
This definition is global, $d\omega=0$ by $d^2=0$, and its coordinate matrix has invertible blocks. Hence \b[$T^*M$ is symplectic for every smooth $M$], even if $M$ itself is not orientable.
The <coordinate Jacobi condition for a Poisson bivector> is
$$
\boxed{\Pi^{i\ell}\partial_\ell\Pi^{jk}+\Pi^{j\ell}\partial_\ell\Pi^{ki}+\Pi^{k\ell}\partial_\ell\Pi^{ij}=0.}
$$
Necessity follows by applying the <Jacobi identity> to three coordinate functions. For arbitrary functions, the second-derivative terms in the Jacobiator cancel by antisymmetry of $\Pi$; the terms left are exactly this coefficient multiplying $f_i g_j h_k$, proving sufficiency.
For a <symplectic form>, take $\iota_{X_f}\omega=df$ and $\{f,g\}=\omega(X_f,X_g)$. With these signs the matrix of $\Pi$ is $-\omega^{-1}$. Differentiation gives $\partial_\ell\Pi=\Pi(\partial_\ell\omega)\Pi$. The cyclic expression above consequently equals
$$
-\Pi^{ia}\Pi^{jb}\Pi^{kc}(\partial_a\omega_{bc}+\partial_b\omega_{ca}+\partial_c\omega_{ab}),
$$
which vanishes because $d\omega=0$. This directly proves that a <symplectic manifold> defines a <Poisson manifold>.
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