Over a field of characteristic not two, the tensor square of a vector space decomposes as the direct sum of the symmetric and exterior tensor squares

Let F be a field of characteristic not 2, and let V be an n-dimensional vector space over F. Recall that in this case, we can realize \mathcal{S}^2(V) and \bigwedge^2(V) as subspaces of V \otimes_F V. Prove that V \otimes_F V = \mathcal{S}^2(V) \oplus \bigwedge^2(V).


Recall that \mathsf{dim}_F\ V \otimes_F V = n^2.

Suppose z \in \mathcal{S}^2(V) \cap \bigwedge^2(V). Then we have \mathsf{Alt}_2(z) = z = \mathsf{Sym}_2(z). Expanding these, we see that z + (1\ 2) z = z - (1\ 2) z, so that 2(1\ 2) z = 0, so that z = 0. That is, \mathcal{S}^2(V) and \mathsf{Alt}_2(V) intersect trivially.

Counting dimensions, recall that \mathsf{dim}_F\ \mathcal{S}^2(V) = \frac{n(n+1)}{2} and \mathsf{dim}_F\ \bigwedge^2(V) = \frac{n(n-1)}{2}.

Thus V \otimes_F V = \mathcal{S}^2(V) \oplus \bigwedge^2(V).

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