$U^*\otimes V$ versus $L(U,V)$ for infinite dimensional spaces

Solution 1:

Suppose we are over an infinite field $\Bbbk$, and let $\dim(U) = \kappa$, $\dim(V) = \lambda$.

• This great answer of Arturo Magidin shows that $|U^*| = |\Bbbk|^\kappa$. Basically $U^*$ is in bijection with maps from a basis of $U$ to $\Bbbk$.
• When the spaces are infinite, the cardinality of the tensor product is the product of the cardinalities.
  • There is a surjection $\bigsqcup_{n \geq 0} (X \times Y)^n$ to $X \otimes Y$ given by linear combinations.
  • $|X \otimes Y|$ is obviously at least $|X| |Y|$: there is an injection given by $x \otimes y$.
• Therefore $|U^* \otimes V| = |U^*| |V| = |\Bbbk|^\kappa \lambda$
• On the other hand, $L(U,V)$ is in bijection with maps from a basis of $U$ to $V$. Therefore $|L(U,V)| = |V|^\kappa = |\Bbbk|^\kappa \lambda^\kappa$.

So if $|\Bbbk|^\kappa \lambda < |\Bbbk|^\kappa \lambda^\kappa$ (equality can happen), then the two spaces don't have the same dimension. So they can't be isomorphic. For example, assume the continuum hypothesis and the axiom of choice (sorry set theorists) and take $|\Bbbk| = \aleph_0$ (e.g. $\Bbbk = \mathbb{Q}$), $\lambda = \aleph_\omega$ and $\kappa = \operatorname{cof}(\lambda) = \aleph_0$. Then by König's theorem $\lambda^{\operatorname{cof}(\lambda)} > \lambda$, so: $$|\Bbbk|^\kappa \lambda = \aleph_0^{\aleph_0} \aleph_\omega = 2^{\aleph_0} \aleph_\omega \overset{\mathsf{CH}}{=} \aleph_1 \aleph_\omega = \aleph_\omega = \lambda < \lambda^\kappa = |\Bbbk|^\kappa \lambda^\kappa.$$

In general though, the spaces can be isomorphic. For example is both spaces have countably infinite dimension over $\mathbb{R}$, then both $U^* \otimes V$ and $L(U,V)$ have dimensions $2^{\aleph_0}$ and so they are isomorphic, though one would be hard-pressed to describe an explicit isomorphism.

• A trivial example: if $V = \Bbbk$, then you're wondering whether $U^*$ is isomorphic to $L(U, \Bbbk)$...
• If $U$ is finite dimensional, then the usual map is an isomorphism (see below).

But if both $U$, $V$ are infinite dimensional, they are not naturally isomorphic, at least through the only obvious natural map.

In the finite dimensional case, the natural isomorphism looks like this: $\varphi \otimes v \mapsto (u \mapsto \varphi(u)v)$. When the spaces are infinite dimensional, this is not an isomorphism! Indeed any element of $L(U,V)$ in the image has finite dimensional range. But there are elements in $L(U,V)$ with infinite dimensional range, obviously.