# Correspondence between between tensor product and bilinear maps

1. Jan 15, 2009

### jojo12345

When considering finite dimensional vector spaces V and W over a field K, there exists a natural isomorphism between their tensor product and the space of bilinear maps from the cartesian product of the dual spaces to the underlying field. However, the text I'm reading asserts that if V and W are infinite dimensional, then the two spaces are not isomorphic.

In the finite dimensional case, one natural isomorphism between the two spaces is given by

$$T:V\otimes W\rightarrow (V^{*},W^{*})^{*}$$

where

$$T(A)=\bar{A},\forall A\in V\otimes W$$

and

$$\bar{A}(\rho,\psi)=(\bar{a^{r}v_{r}\otimes w_{r}})(\rho,\psi)=a^{r}\rho(v_{r})\psi(w_{r})$$ where I'm using the summation convention. There isn't some fixed range for the sums- they must only be finite.

What I am trying to show is that this mapping is not surjective when V and W are not finite dimensional. I have an argument to suggest this is the case, but I would appreciate being corrected or having my thoughts confirmed.

Let $$S_{1}=\{ \rho{}_{1},\rho{}_{2},...\}$$ be a countable subset of a basis for $$V^{*}$$ and $$S_{2}=\{ \psi{}_{1},\psi{}_{2},...\}$$ be a countable subset of a basis for $$W^{*}$$. Now let $$F\in(V^{*},W^{*})^{*}$$ be a bilinear map such that $$F(\rho{}_{i},\psi{}_{j})=k_{i,j}$$. There will not always be a tensor that maps to this function because every tensor is specified by a finite number of scalar components (they are vectors after all) and the system of equations one would have to solve in order to determine the components is $\displaystyle very$ overdetermined.

The things I'm most unsure about are my ability to restrict F as I have and how the fact that a tensor can be represented by an arbitrarily large number of components affects the last sentence in my argument.

As a side note, I am also wondering about how to go about proving no isomorphism exists in the case of infinite dimensional spaces for even if what I tried to prove above is true, it only rules out a single map from being an isomorphism. It seems to me that perhaps my argument applies to any linear map between the two spaces. However, if my argument above doesn't hold water, then I have a bit more thinking to do.

2. Jan 15, 2009

### Hurkyl

Staff Emeritus
Dimension (cardinality of a basis) gives a bijection from {isomorphism classes of vector spaces} to {cardinal numbers}.

What is (X, Y), for vector spaces X and Y?

3. Jan 15, 2009

### jojo12345

Sorry I didn't make it clear enough. I tried to suggest it in my first paragraph.$$(V^{*},W^{*})^{*}$$ is the space of all bilinear maps $$\psi : V^{*}\times W^{*}\rightarrow \mathbb{K}$$ where $$\mathbb{K}$$ is the underlying field.

4. Jan 16, 2009

### jojo12345

It's not obvious to me why the two bases would not have the same cardinality.