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

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.

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Ideally I'd like to show that $\widetilde{\Theta \circ f \circ \pi} = f$ but I can't seem to derive that from the definition. – Paul Slevin Apr 30 '12 at 3:09
May I know how do you send a pair in $V \times W$ to an element in $B^{\ast}$? – BenjaLim Apr 30 '12 at 12:09
It's the map $\pi: V \times W \to B^*$. Given an element $f \in B$, $\pi(v,w)$ sends $f$ to $f(v,w) \in \mathbb{R}$. I defined this map as part of the construction, so I'm not sure it's correct. However it seems like the only reasonable choice. Apologies if this wasn't clear in my original post. – Paul Slevin Apr 30 '12 at 12:12
$f(v,w)$ is an element of $U$ no? Maybe I'm misunderstanding something. – BenjaLim Apr 30 '12 at 12:15
Ah ok I get it now, because on elementary tensors the map is surjective we can define our linear map $L$ out of the tensor product in terms of $f$. However I got confused because $\pi : V \times W$ to $V \otimes W$ is actually a bilinear map and not a linear one. – BenjaLim Apr 30 '12 at 22:57

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?

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That is the way I would do it, but for this question I am supposed to verify the tensor product directly... I know it is annoying because we can definitely construct $V \otimes W$ in terms of free vector spaces and then by definition $B \cong (V \otimes W)^*$ as you say. I guess in the question we are trying to prove existence of $V \otimes W$. – Paul Slevin Apr 30 '12 at 11:38
@PaulSlevin I am in the process of trying to do it directly. – BenjaLim Apr 30 '12 at 11:40
Ben, I made a mistake. $B$ should be the set of bilinear maps $V \times W \to \mathbb{R}$, which allows my original definition of $\pi$ to work. I've corrected that in the post, perhaps this will make things easier. – Paul Slevin Apr 30 '12 at 13:13

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 $W$.

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