Let $V$ be a (possibly infinite dimensional) vector space, and $W$ a finite dimensional one. Then, one can define the tensor product $V\otimes W$ as the free vector space on $V\times W$ modulo some relations.

Now, if $V$ is finite dimensional, and if $\{e_1,\ldots,e_m\}$ is a basis for $V$ and $\{e'_1,\ldots,e'_n\}$ for $W$, we get a basis $$\{e_i\otimes e'_j:1\le i\le m,q\le j\le n\}$$ for $V\otimes W$.

In particular, every element $u\in V\otimes W$ can be expressed as $$u=v_1\otimes e'_1+\cdots+v_n\otimes e'_n$$ for unique $v_1,\ldots,v_n\in V$.

Question: Does this statement also holds when $V$ is infinite dimensional?

I can easily show that the elements $v_i\in V$ exist, but I am not sure why they would be unique.

Existence: An element $u\in V\times W$ is the projection of a formal sum of finitely many $(v_i,w_i)\in V\times W$, i.e. $u=v_1\otimes w_1+\cdots v_n\otimes w_n$. Then, we can express each $w_i$ in terms of the basis.

  • $\begingroup$ Do you understand how to prove that $\{e_i\otimes e_j'\}$ is a basis for $V\otimes W$ in the finite-dimensional case? If so, what difficulty do you have generalizing that proof to the infinite-dimensional case? $\endgroup$ – Eric Wofsey Oct 26 '15 at 20:19

The infinite-dimensional case follows formally from the finite-dimensional case. Specifically, $V\otimes W$ is defined as the quotient of $V\times W$ by certain relations. So whenever two tensors are equal in $V\otimes W$, it is because the difference between them is equal to a certain finite linear combination of the relations. All of the vectors from $V$ and $W$ involved in expressing the two tensors and the finitely many relations form a finite set. From this, we conclude the following: if $u=\sum c_i v_i\otimes w_i=\sum d_j v_j'\otimes w_j'$ is an element of $V\otimes W$ which can be expressed as a sum of elementary tensors in two different ways, then there exist finite-dimensional subspaces $V_0\subseteq V$ and $W_0\subseteq W$ such that the equation $\sum c_i v_i\otimes w_i=\sum d_j v_j'\otimes w_j'$ is also valid in the tensor product $V_0\otimes W_0$.

So in particular, if you have a (possibly infinite) basis $\{e_i'\}$ for $W$ and you can write $u=\sum v_i\otimes e_i'=\sum v_i'\otimes e_i'$ in two different ways, then there is actually a finite-dimensional subspace $W_0$ (which we can take to be spanned by finitely many of the $e_i'$) for which the equation $\sum v_i\otimes e_i'=\sum v_i'\otimes e_i'$ is valid in $V\otimes W_0$. By the finite-dimensional version of the result, this then implies $v_i=v_i'$ for all $i$.

Alternatively, most methods that I can think of to prove that $\{e_i\otimes e_j'\}$ is linearly independent do not actually use finiteness anywhere, so they work equally well if $V$ and/or $W$ is infinite-dimensional.


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