Showing that the dual space of bilinear maps $V \times W \to \mathbb{R}$ satisfies the tensor product property, for finite dimensional vector spaces.

Question

Let $U,V$ and $W$ be finite dimensional vector spaces, and define $B$ to be the vector space of all bilinear maps $V \times W \to \mathbb{R}$. Given a bilinear map $\alpha : V \times W \rightarrow U$, define $\tilde{\alpha}: B^* \rightarrow U^{**}$ by $\alpha(\psi)(\sigma) = \psi (\sigma \circ \alpha)$. Define a map $\pi : V \times W \rightarrow B^*$ by $\pi(v,w) (f:V \times W \rightarrow \mathbb{R}) = f(v,w).$

$\mathbf{CORRECTION:}$ $B$ should be the space of bilinear maps $V \times W \to \mathbb{R}$, not $V \times W \to U$ as previously stated.

In order to show that $B^*$ satifies the universal property of the tensor product, I have to show that given a map $\alpha : V \times W \rightarrow U$, then there is a unique $\tilde{\alpha} : B^* \rightarrow U^{**}$ such that $\Theta \circ \tilde{\alpha} \circ \pi = \alpha$, where $\Theta:U^{**} \to U$ is the canonical isomorphism.

It is quite clear that $\tilde{\alpha}$ defined above satisfies this property, but I am having trouble proving uniqueness. I would like to show that given $f:B^* \to U^{**}$ such that $\Theta\circ f \circ \pi = \alpha$, then $f= \tilde{\alpha}$, however I am getting nowhere. Any help would be appreciated, thank you.

Answer 1

If you already know that the tensor product of $V \times W$ exists (via the usual business of quotiening out the free module on the set $V \times W$ by the submodule generated by those usual relations) we can do the following: Given finite dimensional vector spaces $U,V,W$ over a field $F$ we have an isomorphism

$$\operatorname{Bil}( V,W;U) \cong \operatorname{Hom}( V\otimes W, U).$$

The notation $\operatorname{Bil}( V,W;U)$ is exactly your $B$ above. Now if we put $U =F$, we have that

$$\operatorname{Bil}( V,W;F) \cong \operatorname{Hom}( V\otimes W, F) \cong (V \otimes W)^{\ast},$$

in particular that $\operatorname{Bil}( V,W;F) \cong (V \otimes W)^{\ast}$. Since we are in the finite dimensional setting, taking the dual of both sides gives that

$$\operatorname{Bil}( V,W;F)^{\ast} \cong (V \otimes W)^{\ast \ast} \cong (V \otimes W).$$

Does this help?

Answer 2

After a great deal of soul searching and chats with fellow mathematicians, here is the solution.

Suppose that $V$ has basis $\{e_1, \ldots e_n \}$ and $W$ has basis $\{f_1, \ldots f_m \}$. Then it is quite easy to show that the vector space $B$ has basis $$\{ (e_if_j)^* \mid i=1,\ldots, n \ ; \ j = 1,\ldots m \} $$ where $(e_if_j)^* (e_{k},f_{l}):=0 \iff (k,l) = (i,j)$, and $1$ otherwise. Hence $\dim B = \dim V \dim W$. It follows that the dual space $B^*$ has $\dim B^* = \dim V \dim W$. I claim that $$\{ \pi(e_i,f_j) \mid i = 1, \ldots, n \ ; \ j=1,\ldots, m \}$$ is a basis for $B^*$. This is easy to see since we need only check that this set is linearly independent, which is clearly is. Hence $B^*$ is generated by the image of $\pi$.

Suppose for $g: B^* \to U$, we have that $g\pi = 0$. Suppose that $g \not= 0$. Then there is some basis element $ \pi(e_i, f_j)$ such that $g\pi(e_i,f_j) \not= 0$. This is a contradiction, so $g=0$. This of course implies that the map above is unique, so $B^*$ is the tensor product of $V$ and $W$.