# Question re. tensor product of v. spaces

1. Jul 23, 2010

### Newtime

So I'm reading about tensor products and wanting to make sure I understand the notion completely.

I understand that $$V^* \otimes W$$ is the space of linear functions from $$V \text{to} W$$. And since $$V^{**} \backsimeq V$$, we have that $$V \otimes W$$ is the space of linear functions from $$V^* \text{to} W$$.

However, in a paper that I'm reading, it is stated that $$V \otimes W$$ can be thought of also as $$(V\otimes W)^*$$. But since $$V^* \otimes W^* \backsimeq (V\otimes W)^*, \text{we have that} V\otimes W \backsimeq V^* \otimes W^*$$ and this is where my understanding stops. Was there a type or is the previous statement true? Thanks.

Last edited: Jul 23, 2010
2. Jul 23, 2010

### element4

(Warning: I not an expert on this!)

I guess that by "thought of", you mean "isomorphic". Are there any conditions on these vector spaces? In some cases a vector space can be isomorphic to its dual space, for example $$\mathbb R^n$$. See for example this http://en.wikipedia.org/wiki/Riesz_representation_theorem" [Broken].
As for $$V^* \otimes W^* \backsimeq (V\otimes W)^*$$, I think you should be careful. I doubt this holds in infinite dimensions.

I think what you say is correct, under the assumption that $$V \otimes W \backsimeq (V\otimes W)^*$$ and everything is finite dimensional.

EDIT: See http://en.wikipedia.org/wiki/Tensor_product#Relation_with_the_dual_space !

Last edited by a moderator: May 4, 2017
3. Jul 23, 2010

### quasar987

Is there another structure involved? For instance are the spaces V and W endowed with a scalar product?

4. Jul 23, 2010

### Newtime

Thanks for the quick replies guys, and no we don't assume that the space is isomorphic to its dual space. Also, we are only working with finite dimensional spaces and yes there is a scalar product. Also I should have been more clear:

There was an exercise directly above the statement which asked one to verify that $$V^* \otimes W^* \backsimeq (V\otimes W)^*$$. However, below this it says "use the above exercise to show that $$V \otimes W$$ may be thought of as (isomorphic to) the space of linear maps from $$V^*$$ to $$W$$, the space of linear maps from $$W^*$$ to $$V$$, the space of bilinear maps $$V^* \times W^* \longrightarrow \mathbb{C}$$ or $$(V \otimes W)^*$$.

Now, I agree with the first two statements, but the last two (and in particular the last one) have be confused and because of these last two statements I thought the author might have meant to write "use the above exercise to show that $$V^* \otimes W^*$$ . . . but then I wouldn't agree with the first two statements. Clearly I'm misunderstanding something.

BIG TIME EDIT: I misread the statement in the paper...it actually says that the tensor product of V with W can be identified as the dual space of $$V^* \otimes W^*$$ which makes sense because by the exercise above it, we know this is isomorphic to $$V^{**} \otimes W^{**}$$ which is naturally isomorphic to $$V \otimes W$$. Thanks for the help though!

Last edited by a moderator: May 4, 2017
5. Jul 23, 2010

### Fredrik

Staff Emeritus
The isomorphism between V and V** is only "natural" in the sense that you can define it without using an inner product. If an inner product is defined on V, you can use it to define an isomorphism between V and V*. Just let x* be the map $y\mapsto \langle x,y\rangle$. Now the map $x\mapsto x^*$ is an isomorphism.

6. Jul 23, 2010

### Newtime

Good point, and thanks for that link. However, in the paper I don't think it is assumed we have an inner product on $$V$$ so although such an isomorphism may exist, it isn't really considered.

7. Jul 26, 2010

### Landau

More precisely, there is a natural isomorphism between the identity functor and the doube dual functor in the category of finite dimensional vector spaces.

8. Jul 26, 2010

### quasar987

I have a question for you. You know how we commonly say that given two general vector spaces V, W of the same finite dimension, they are isomorphic because you can construct an isomorphism between them by choosing bases for both spaces and defining a linear map that send one basis to the other.
But this isomorphism is not considered natural because of the choice involved which is not "natural": there is no reason to prefer one basis over another.

If we adopt as formal definition of "natural isomorphism" the category theory you just alluded to, then is there a way to express the non-naturality of the above isomorphism between V and W by cleverly choosing the categories and the functors?

At first sight, it seems to me that the notion of naturality in this case is not covered by the category theory definition.

On the other hand, when an isomorphism IS natural in the intuitive sense, I'm pretty sure that there is always a way to choose categories and functors such that the isomorphism is a natural one between these functors.

9. Aug 20, 2010

### Landau

Hi quasar, apologies for my late reply. (I have been busy with my bachelor's thesis the past month

A natural transformation is defined for two parallel functors, that is, functors with the same domain and codomain. The identity functor I and the single dual functor * are not parallel:

$$I:\sf{Vect_K}\to \sf{Vect_K}$$
$$*:\sf{Vect_K}^{op}\to \sf{Vect_K}$$

In other words: I is covariant and * is contravariant.

But let's ignore that. Suppose we have a collection of linear isomorphisms
$$\{a_V:V\to V^*\}_{V\in\sf{Vect_K}}.$$

$$\begin{array}{ccccc} & V & \longrightarrow^{a_V} & V^* & \\ L & \downarrow & & \uparrow & L^*\\ & W & \longrightarrow^{a_W} & W^* & \end{array}$$

where L* is the linear map defined by

$$L^*f := [v\mapsto f(Lv)],\ \ \ \ \ f\in W^*$$

Now this diagram commutes if

$$\alpha_V = L^*\alpha_W L$$

for all linear maps L:V->W. But this cannot be since $\alpha_V$ is an isomorphism, and L not necessarily.

10. Aug 20, 2010

### quasar987

You've shown that there is no natural isomorphism btw a vector space and its dual space. That's pretty cool but it's not what I asked. :tongue2: