4
$\begingroup$

I am presuming the dimension of all the vectors spaces is finite. It is a known fact that $V$ and $V^{**}$ are naturally isomorphic in the sense that there is a natural transformation-isomorphism between the functors $id$ and $-^{**}$.

Is there a natural isomorphism in some similar sense, between $V \otimes V$ and $W$, where $W$ is the vector space of bilinear forms on $V^*$? By bilinear form on $V^*$, I mean a bilinear map from $V^* \times V^*$ to $F$, $V$ vector space above $F$.

-

I would expect that yes, since similarly to the argument with $V$ and $V^{**}$, I can define a 'basis-indendent' isomorphism between $V \otimes V$ and $W$:

We can define a bilinear map $\varphi : V \times V \rightarrow W$, by $\varphi(v_1,v_2)(f,g)=f(v_1) \cdot g(v_2)$. From the universal property of $V \otimes V$, there is thus a unique linear map $\psi : V \otimes V \rightarrow W$, such that composing $\psi$ with the associated bilinear map $\phi: V\times V \rightarrow V\otimes V$ gives us $\varphi$ - in other words, $\varphi = \psi \circ \phi$.

This could maybe suggest, that this $\psi$ will be the wanted map, that will give us the natural isomorphism. If we want, perhaps a more explicit definition of $\psi$ could be deduced - it seems like it is necessarily true that $\psi (v_1 \otimes v_2) = \varphi (v_1,v_2)$, so we know what $\psi$ does on $v_1 \otimes v_2$ elements in $V\otimes V$ - this should induce what $\psi$ does on the whole vector space $V\otimes V$.

$\endgroup$
4
$\begingroup$

We have that $$ V \otimes V \cong V^{**} \otimes V^{**} \cong (V^* \otimes V^*)^* \cong \operatorname{BF}(V^*). $$


  • The first isomorphism comes from the natural linear map $$ \iota \colon V \to V^{**}, \quad v \mapsto ( \varphi \mapsto \varphi(v) ). $$ This map is always injective, and since $V$ is finite-dimensional it is already an isomorphism.

  • The second isomorphism comes from the fact that for all vector spaces $V$, $W$ there exists a natural linear map $$ V^* \otimes W^* \to (V \otimes W)^*, \quad \varphi \otimes \psi \mapsto (v \otimes w \mapsto \varphi(v) \psi(w)). $$ This map is always injective, and if both $V$ and $W$ are finite-dimensional (or one of them is $0$) then it is already an isomorphism.

  • The third isomorphism comes from the fact that for all vector spaces $V$, $W$ the universal property of the tensor product $V \otimes W$ results in a natural isomorphism $$ \operatorname{BF}(V,W) \to (V \otimes W)^*, \quad \beta \mapsto (v \otimes w \mapsto \beta(v, w)), $$ with inverse given by $$ (V \otimes W)^* \to \operatorname{BF}(V,W), \quad \varphi \mapsto ( (v, w) \mapsto \varphi(v \otimes w) ). $$

One can combine these descriptions of the three isomorphisms to see that the constructed isomorphism $\psi \colon V \otimes V \to \operatorname{BF}(V^*)$ is indeed given by $$ \psi(v_1 \otimes v_2)( \varphi_1, \varphi_2 ) = \varphi_1(v_1) \varphi_2(v_2). $$ Since each of the three isomorphisms is natural it follows that $\psi$ is also natural.


Note that the third isomorphism also holds for infinite-dimesional vector spaces while the first two maps will only be injective. So for infinite-dimensional vector spaces we still get the natural linear map $\psi$, and it will always be injective, but it won’t be an isomorphism anymore.

Because of this, there will (probably) also be no good way to write down $\psi^{-1}$ even for finite-dimensional vector spaces in a basis-independent way.

$\endgroup$
1
$\begingroup$

Yes, they are naturally isomorphic. The map you wrote is a natural isomorphism.

But be careful. If you want to be really nitpicky, everything should be in the same category. If we were just in the category of vector spaces, all you would have to do would be to show that $V \otimes V$ and $W$ have the same dimension. You want to be in the category of vector spaces equipped with a bilinear map from $V$, with morphisms that commute with the bilinear maps. Okay, we know what the bilinear map is from $V$ to $V \otimes V$. What is the bilinear map from $V$ to $W$? Given $v \in V$, what is the corresponding bilinear form $e_v$ on $V^*$? Without this, $W$ is not even in the same category as $V \otimes V$.

Well, it is not a huge mystery. We had better choose $e_v(f,g) = f(v)g(v)$. You can check that your $\varphi$ makes everything work (the relevant diagram commutes).

Next, you could try to show "directly" that $\varphi$ is injective and surjective. But... I don't recommend that. (Surjectivity is not too bad.)

Instead, how about showing that $W$ (together with its structure map $V \times V \to W$) satisfies the universal property of the tensor product? Then $W$ must be isomorphic to $V \otimes V$. And there is a unique map between them, compatible with the structure maps. And your $\varphi$ is a map between them, compatible with the structure maps. So $\varphi$ will be an isomorphism. And $\varphi$ is at least in an intuitive sense "natural". (For more technical senses of "natural", you'll have to ask someone who knows more than me.)

So, to show $W$ satisfies the universal property of the tensor product: Let $V \times V \to X$ be a bilinear map. Let $F \in W$ be a bilinear form on $V^*$. We want to send $F$ to an element of $X$. Well, we'll send it to an element of $X^{**}$. We send $F$ to a linear form also denoted $F \in X^{**}$, defined as follows. If $F$ is in the image of $V \times V \to W$, say $F$ is the image of $(v_1,v_2) \in V \times V$, then for $\ell \in X^*$, we set $F(\ell)=\ell(v_1,v_2)$ where we also identified $(v_1,v_2)$ with its image in $X$.

Well... I'm sure that can be simplified and cleaned up. It's still left to show this map $W \to X^{**}$ extends linearly to all of $W$ and is unique. Maybe someone else will have a quicker answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.