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Suppose $V$ and $W$ are vector spaces over the same field. Is $\text{End}(V \otimes W)$ the same as $\text{End}(V) \otimes \text{End}(W)$? Are there special names for these spaces?

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up vote 8 down vote accepted

Naive answer: in finite dimension, the vector space $\rm{End}(V)\otimes \rm{End}(W)$ has dimension $(\dim V)^2\cdot(\dim W)^2$, while $\rm{End}(V\otimes W)$ has dimension $(\dim V\cdot\dim W)^2$. So they are isomorphic.

The real question is: is there a canonical isomorphism? In the finite-dimensional case, yes. In the infinite-dimensional case, we only get a canonical injection from $\rm{End}(V)\otimes \rm{End}(W)$ into $\rm{End}(V\otimes W)$. That's the map constructed in the second answer below. As pointed out by Martin Brandenburg, it is still injective in infinite dimension, but it fails to be surjective. So from now on, I will assume that $V$ and $W$ are finite-dimensional.

With duals: each step below follows from canonical isomorphisms

$$ \rm{End}(V)\otimes \rm{End}(W)\simeq (V\otimes V^*)\otimes (W\otimes W^*)\simeq (V\otimes W)\otimes (V^*\otimes W^*) $$$$ \simeq (V\otimes W)\otimes (V\otimes W)^*\simeq \rm{End}(V\otimes W).$$

Let's verify these steps now. First, we will show that $\rm{End}(V)$ is canonically isomorphic to $V\otimes V^*$, where $V^*$ denotes the dual of $V$. Indeed, for every $v\in V$ and $v^*\in V^*$, we can define $\theta_{v,v^*}$ in $\rm{End}(V)$ by $\theta_{v,v^*}(x):=v^*(x)v$. This yields a bilinear map $(v,v^*)\longmapsto \theta_{v,v^*}$ on $V\times V^*$ which factors through $V\otimes V^*$ by the universal property of the tensor product. This gives us $$\overline{\theta}:V\otimes V^*\longrightarrow \rm{End}(V).$$ Now take $\{v_i\}$ a basis of $V$, and denote $\{v_j^*\}$ the basis of $V^*$ canonically associated with it: that is $v_j^*$ is the linear form which takes $v_j$ to $1$ and other $v_i$'s to $0$. Observe that the $\overline{\theta}(v_i\otimes v_j^*)=\theta_{v_i,v_j^*}=v_j^*(\cdot)v_i$ constitute the canonical basis of $\rm{End}(V)$ associated with $\{v_i\}$. So $\overline{\theta}$ is linear and takes a basis onto a basis: this is an isomorphism from $V\otimes V^*$ onto $\rm{End}(V)$. Since its definition does not depend on the choice of a basis, this is a canonical isomorphism. In particular, people usually invoke it implicitly to denote $v\otimes v^*$ what is actually $v^*(\cdot)v$.

So we have three canonical isomorphisms $\rm{End}(V)\simeq V\otimes V^*$, $\rm{End}(W)\simeq W\otimes W^*$, and $\rm{End}(V\otimes W)\simeq (V\otimes W)\otimes (V\otimes W)^*$.

Now we need to show that $ V^*\otimes W^*\simeq (V\otimes W)^*$ canonically. For every $v^*\in V^*$ and $w^*\in W^*$, the bilinear form $(v,w)\longmapsto v^*(v)w^*(w)$ factors through the tensor product to give a linear form $\psi_{v^*,w^*}\in (V\otimes W)^*$. It is easy to see that $(v^*,w^*)\longmapsto \psi_{v^*,w^*}$ is bilinear, as it is pointwise bilinear on the spanning set $v\otimes w$. This yields a canonical linear map $$\overline{\psi}:V^*\otimes W^*\longrightarrow (V\otimes W)^*.$$ It only remains to see that for our earlier choices of bases, $\{v_i^*\}$ and $\{w_j^*\}$ are the corresponding bases of $V^*$ and $W^*$, while $\{v_i\otimes w_j\}$ is the corresponding basis of $V\otimes W$, and $\{v_i^*\otimes w_j^*\}$ the one of $V^*\otimes W^*$. This yields also $\{(v_i\otimes w_j)^*\}$ the associated basis of $(V\otimes W)^*$. Now it is easy to see that $\overline{\psi}(v_i^*\otimes w_j^*)=(v_i\otimes w_j)^*$, as they coincide on the basis elements $v_k\otimes w_l$. So $\overline{\psi}$ is linear and takes a basis onto a basis: that's our canonical isomorphism between $V^*\otimes W^*$ and $(V\otimes W)^*$.

Finally, for every vector spaces $X,Y, Z$, we have canonical isomorphisms $X\otimes Y\simeq Y\otimes X$ and $(X\otimes Y)\otimes Z\simeq X\otimes (Y\otimes Z)$. That is: the tensor product is commutative and associative.

Without duals: Consider the bilinear map $$ \phi:\rm{End}(V)\times \rm{End}(W)\longrightarrow \rm{End}(V\otimes W) $$ defined by $$ \phi(S,T)(u\otimes v):=Su\otimes Tv. $$ The fact that this is yields a well-defined endomorphism $\phi(S,T)$ on the tensor product follows from the universal property of the tensor product. Indeed, the map $(u,v)\longmapsto Su\otimes Tv$ is bilinear, so it factors through the tensor product and yields $\phi(S,T)$ in $End(V\otimes W)$.

Since $(S,T)\longmapsto\phi(S,T)(u\otimes v)$ is bilinear for each $u\otimes v$ and since the latter span $\rm{End}(V)\otimes \rm{End}(W)$, the map $(S,T)\longmapsto \phi(S,T)$ is bilinear. So another application of the universal property yields a linear factorization $$ \overline{\phi}:\rm{End}(V)\otimes \rm{End}(W)\longrightarrow \rm{End}(V\otimes W). $$ This finishes the construction of our canonical isomorphism. It remains to check that it is indeed bijective. To see that, we will fix two bases $\{v_i\}$ for $V$ and $\{w_j\}$ for $W$. Then $\{v_i\otimes w_j\}$ is a basis of $V\otimes W$. Now denote $v_i\otimes v_k^*$ the endomorphism of $V$ which sends $v_k$ to $v_i$ and is null elsewhere on the basis. The family $\{v_i\otimes v_k^* \}$ is a basis of $\rm{End} (V)$. Likewise, $\{w_j\otimes w_l^*\}$ is a basis of $\rm{End} (W)$. Therefore $\{ (v_i\otimes v_k^*)\otimes(w_j\otimes w_l^*)\}$ is a basis of $\rm{End} (V)\otimes\rm{End} (W)$. On the other side, $\{ (v_i\otimes w_j)\otimes (v_k\otimes w_l)^*\}$ is a basis of $ \rm{End}(V\otimes W)$. Now $$ \overline{\phi}((v_i\otimes v_k^*)\otimes(w_j\otimes w_l^*))=\phi(v_i\otimes v_k^*,w_j\otimes w_l^*)=(v_i\otimes w_j)\otimes (v_k\otimes w_l)^*. $$ The first equality is by definition of $\overline{\phi}$. The second one is true on the basis of $V\otimes W$, so the operators coincide. Hence $\overline{\phi}$ is linear and takes a basis onto a basis: that's an isomorphism.

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Should your bilinear map actually be $ u \otimes v \mapsto Su \otimes Tv$ ? – Ragib Zaman Mar 16 '13 at 10:35
@RagibZaman Yes, of course. Thanks for catching the typo! – 1015 Mar 16 '13 at 13:26
@ChickenGod There were a few awful typos in my answer. Apparently, you had figured them out. But just in case, note the edit. – 1015 Mar 16 '13 at 13:31
@julien. There's still a type, "blilinear" :) but thanks! This was very helpful! – ChickenGod Mar 30 '13 at 1:37
@ChickenGod I have keyboard dyslexia... – 1015 Mar 30 '13 at 1:39

Julian's proof only works in the finite-dimensional case (dual bases etc.).

More generally, for $R$-modules $M,N,M',N'$ there is a canonical $R$-linear map $\hom(M,M') \otimes \hom(N,N') \to \hom(M \otimes N,M' \otimes N')$, which is an isomorphism when $M$ and $N$ are finitely generated projective (since it holds for $M=R$ and the $M$s satisfying this are closed under finite direct sums and direct summands; see also here). In particular, $\mathrm{End}(M) \otimes \mathrm{End}(N) \to \mathrm{End}(M \otimes N)$ is an isomorphism.

If $V,W$ are infinite-dimesional vector spaces, then the canonical homomorphism $\mathrm{End}(V) \otimes \mathrm{End}(W) \to \mathrm{End}(V \otimes W)$ is still injective, but not surjective. See my answer here.

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A short explanation of why it isn't surjective would be nice. – Peter Shor Mar 30 '13 at 13:47
That's a pretty good point...I have worked all the way as if I had finite dimensional spaces, but I had never mentioned it...+1. – 1015 Mar 30 '13 at 14:20

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