How do differential 2-forms give us bilinear forms on tangent spaces?

Question

Suppose $\omega$ is a 2-form on a manifold $M$. So, by definition, $\omega$ is a section $M \to \bigwedge^2 T^*M$ where $\bigwedge^2 T^* M :=\sum_{p \in M} \bigwedge^2 (T_pM)^*$.

It seems to be a commonly used fact that $\omega$ can be interpreted as a bilinear form $\omega : T_pM \times T_pM \to \mathbb R$ on each tangent space. I know this is because there is an isomorphism $\bigwedge^2 V^* \to (\bigwedge^2 V)^*$ for any vector space $V$. I've seen several sources refer to this isomorphism as "natural," which usually means we don't need to chose a basis to construct it.

So how do we construct this isomorphism? Any other explantions/intuition are welcome.

Answer 1

1. The wedge-product $\wedge$ is the anti-symmetric subspace of the tensor-product $\otimes$. I.e. for any two vector spaces $V,W$, the wedge product $V\wedge W$ is a subspace of $V\otimes W$. and for vectors $v\in V, w\in W$ one can define $$v\wedge w = v\otimes w - w\otimes v$$ (though there might be a factor of $1/2$ depending on your literature. Thanks @Ted-Shifrin)
2. There is an isomorphism between $(V\otimes W)^*$ and $V^*\otimes W^*$, which you can simply write as $$(\alpha\otimes\beta)(v\otimes w)=\alpha(v)\cdot \beta(w)$$ and extending by linearity, where "$\cdot$" is just multiplication of real/complex numbers.
3. writing this down for the wedge-product case becomes $$(\alpha\wedge\beta)(v\otimes w)=\alpha(v)\beta(w)-\beta(v)\alpha(w)$$ which you can also write as $(\alpha\otimes\beta)(v,w)$