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

It's well known that $U^*\otimes V\cong L(U,V)$ for finite dimensional spaces. However, why people say that $U^*\otimes V$ is not isomorphic to $L(U,V)$ for infinite dimensional spaces and many books don't say anything related to this?

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.

• The condition that $|\Bbbk|^\kappa \lambda \ne |\Bbbk|^\kappa \lambda^\kappa$ seems to be more complex than $\lambda > |\Bbbk|$. See math.stackexchange.com/questions/1365632/… which I asked after reading this answer. Jul 18, 2015 at 15:53
• Yes, you're right @TomCollinge, it's more complicated than what I thought at the time... Jul 20, 2015 at 7:59
• The example with $Dim(U) = \aleph_0$; $Dim(V) = 2^{\aleph_0}$; $|k| = \aleph_0$ gives a result that the two are isomorphic, but so far I haven't been able to construct an example where they aren't - can you ? Jul 20, 2015 at 8:08
• I'm working on it @Tom. One at least needs $\lambda < \lambda^\kappa$ (which can be done with König's theorem for example), but I'm not able to deal with the $|\Bbbk|^\kappa$. Bear in mind that I'm far from being a set theory expert... Jul 20, 2015 at 8:09
• @Tom I've found an example, see my edit. Jul 20, 2015 at 8:16

If $$U$$ or $$V$$ has finite dimension, then the map $$\varphi\otimes v\mapsto (u\mapsto \varphi(u)v)$$ is an isomorphism. To visualize why this map fails to be surjective when both $$U$$, $$V$$ are infinite-dimensional, let's think as follows:

Pick a basis $$\{u_i\}_{i\in I}$$ of $$U$$ and $$\{v_j\}_{j\in J}$$ of $$V$$, then what does an element in $$L(U,V)$$ look like? It has the form $$u_i\mapsto \sum_{j\in J} a_{ij} v_j$$, where for any $$i\in I$$ there are only finitely many $$j$$ such that $$a_{ij}\neq 0$$. What does an element in $$U^*\otimes V$$ look like? It has the form $$\sum_{j\in J} (u_i\mapsto a_{ij})\otimes v_j$$, where for only finitely many $$j$$ the map $$u_i\mapsto a_{ij}$$ is not the zero map in $$U^*$$, namely there are only finitely many $$j$$ such that there exists some $$i$$ such that $$a_{ij}\neq 0$$. Now, the map $$\varphi\otimes v\mapsto (u\mapsto \varphi(u)v)$$ sends the element $$\sum_{j\in J} (v_i\mapsto a_{ij})\otimes w_j$$ to the element $$u_i\mapsto \sum_{j\in J} a_{ij} v_j$$.

If you think $$\{a_{ij}\}_{i\in I,j\in J}$$ as an infinite matrix, then the element $$v_i\mapsto \sum_{j\in J} a_{ij} w_j$$ is well-defined if and only if each row has only finitely many nonzero entries, while the element $$\sum_{j\in J} (v_i\mapsto a_{ij})\otimes w_j$$ is well-defined if and only if all but finitely many columns consist of entirely zeros! If $$U$$ or $$V$$ has finite dimension (namely the matrix has only finitely many rows of columns), then these two conditions are the same; but the latter condition is much stronger when $$U$$, $$V$$ are both infinite-dimensional.