# How to prove that $V \otimes W \cong V^{\ast \ast} \otimes W$?

Suppose $$V$$ is a vector space of finite dimension. We know that $$V$$ and $$V^{\ast \ast}$$ are canonically isomorphic via the isomorphism $$v \mapsto E_v$$, with $$E_v: V^\ast \to \mathbb{C}: f \mapsto f(v)$$.

Now I want to prove that $$V \otimes W \cong V^{\ast \ast} \otimes W$$, what is the easiest way to do this?

My initial idea: define a bilinear map $$\phi: V \times W \to V^{\ast \ast} \otimes W: (v,w) \mapsto E_v \otimes w$$. This is bilinear since $$\otimes$$ is bilinear and $$v \mapsto E_v$$ is linear. Thus, by the universality property this gives us a unique linear map $$\overline{\phi}: V \otimes W \to V^{\ast \ast} \otimes W$$ with $$\overline{\phi}(v \otimes w) = E_v \otimes w$$.

But how can we show that this is an isomorphism? We only know it is linear, what about bijectivity? I don't know how to go about showing that, since the tensor product is defined indirectly.

• $V$ and $V^{**}$ are in general not canonically isomorphic. Do we assume finite dimensions here? – Hagen von Eitzen Oct 21 '18 at 20:13
• Yes indeed finite dimension, good point. – Sigurd Oct 21 '18 at 20:13

This works more generally. Let $$V\cong U$$ with isomorphisms $$\varphi\colon V\to U$$ and $$\psi\colon U\to V$$ being inverse to each other.

Similar to what you've done, you can now induce maps $$\varphi\otimes \operatorname{id}_W\colon V\otimes W\to U\otimes W,\\ v\otimes w\mapsto \varphi(v)\otimes w,$$ and $$\psi\otimes \operatorname{id}_W\colon U\otimes W\to V\otimes W,\\ u\otimes w\mapsto \psi(u)\otimes w.$$

So, the question is, what are the compositions $$(\varphi\otimes \operatorname{id}_W)\circ(\psi\otimes \operatorname{id}_W)$$ and $$(\psi\otimes \operatorname{id}_W)\circ(\varphi\otimes \operatorname{id}_W)$$?

• Nice higher level solution, thanks. This is indeed the kind of construction I was looking for. – Sigurd Oct 21 '18 at 21:02
• @Sigurd, you are welcome. Notice how it avoids issues about basis/generators. Everything can be concluded just from the universal property. – Ennar Oct 21 '18 at 21:03
• Indeed! That is the simplicity that I particularly like about this solution. It avoids my confusion of going from generators to arbitrary elements. – Sigurd Oct 21 '18 at 21:34
• @Sigurd, generators are quite natural concept. The universal property of tensor product basically tells you that simple tensors $v\otimes w$ generate tensor product $V\otimes W$. Think about it, it says for bilinear map $B$, there is unique linear map $\bar B$ such that $\bar B(v\otimes w) = B(v,w)$. But, there is only one way that such uniqueness can hold: any element of $V\otimes W$ is of the form $\sum \alpha_{v,w}v\otimes w$. And then you can see the uniqueness: $\bar B(\sum \alpha_{v,w}v\otimes w) = \sum \alpha_{v,w}\bar B(v\otimes w) = \sum \alpha_{v,w}B(v,w)$. – Ennar Oct 21 '18 at 21:41
• You will quickly get used to just checking if something holds on generators and automatically conclude that it holds on the whole space. – Ennar Oct 21 '18 at 21:42

Hint: What does it mean to be in the kernel of this map? What element maps to $$E_v \otimes w$$ (which is an arbitrary generating element)?

# Elaboration

Suppose $$\bar{\phi}(v \otimes w) = 0_{V^** \otimes W}$$. What can we say about $$v \otimes w$$? You could also consider two simple tensors $$v\ otimes w$$ and $$v' \otimes w'$$ which map to the same place.

$$\bar{\phi}(v \otimes w)=\bar{\phi}(v' \otimes w')$$

Let $$E_v \otimes w \in V^** \otimes W$$. Can you find an element that maps to it? (like $$v \otimes w$$).

• Could you elaborate on this a bit more? – Sigurd Oct 21 '18 at 20:03
• The problem is that I'm trying to consider arbitrary elements, and not just pure tensors. This is in an exercise in which I'm trying to prove that $e_i \otimes f_j$ forms a basis for $V \otimes W$, so I'm not suppose to use that fact yet. – Sigurd Oct 21 '18 at 20:09
• What I've done is considering elements which generate the tensor product. You can use this in your problem since this is how the tensor product is defined. It is generated by the elements of the form $v \otimes w$. Since any arbitrary element is sums of these elements we need only consider the simple tensors. – Aaron Zolotor Oct 21 '18 at 20:11
• It seems I'm working with a different definition of tensor product then. What do you mean by `generated by'? – Sigurd Oct 21 '18 at 20:13
• Like a basis but not necessarily linearly independent. That is, any element can be written as a sum of simple tensors. They generate any element in the tensor product. – Aaron Zolotor Oct 21 '18 at 20:14