# Operators and Tensor Products

For clarity I've re-written this question.

Let $V, \text{End} (V), V \otimes V, \text{End} (V\otimes V), \text{and } \text{End} (V) \otimes \text{End} (V)$ be a vector space, the space of linear operators on it, the tensor product of $V$ with itself, the space of linear operators on $V \otimes V$, and the tensor product of the space of $\text{End}(V)$ with itself. There is no topology involved in this question.

1. Define a map $\phi: \text{End}(V) \times \text{End}(V) \to \text{End}(V \otimes V)$ by $\phi(S, T)(u \otimes v) = S(u) \otimes T(v)$. Then $\phi$ is bilinear: e.g. $\phi(S_1 + S_2, T)(u \otimes v) = (S_1(u) + S_2(u)) \otimes T(v) = S_1(u) \otimes T(v) + S_2(u) \otimes T(v) = \phi(S_1 , T)(u \otimes v) + \phi(S_2, T)(u \otimes v)$
2. So by the universal property of the tensor product $\text{End}(V) \otimes \text{End}(V)$ there is a linear map $\overline \phi: \text{End}(V) \otimes \text{End}(V) \to \text{End}(V \otimes V)$ where for all $S, T \in \text{End}(V)$we have $\overline \phi (S \otimes T) (u \otimes v) = S(u) \otimes T(v)$
3. according to Operators on a Tensor Product Space and its answers, if $V$ is finite dimensional then $\overline \phi$ is a canonical isomorphism. Note that if $V$ has finite dimension $n$ then $Dim(\text{End}(V)) = n^2$; $Dim (\text{End}(V) \otimes \text{End}(V)) = n^ 4 = Dim(\text{End}(V \otimes V))$. So the spaces are isomorphic (but is $\overline \phi$ an isomorphism ?)
4. If $\overline \phi$ is an isomorphism then it has an inverse, $\overline \phi^{-1}: \text{End}(V \otimes V) \to \text{End}(V) \otimes \text{End}(V)$
5. Consider the map $f: V \times V \to V \otimes V$ where $f(u, v) = v \otimes u$. This is also bilinear, e.g. $f(u, v_1 + v_2) = (v_1 + v_2) \otimes u = v_1 \otimes u + v_2 \otimes u = f(u, v_1) + f(u, v_2)$
6. By the universal property of $V \otimes V$ there is a linear map $\overline f \in \text{End}(V\otimes V)$ where for all $u, v \in V$ we have $\overline f(u \otimes v) = v \otimes u$.
7. So, my question is if (V is finite dimensional and) $\overline \phi$ is an isomorphism, then what is $\overline \phi ^{-1}(\overline f)$ ?

The answer has been provided below by @martini

The original question wording is ....

I may be mistaken in my understanding of what follows.....

The answers to this question Operators on a Tensor Product Space say that for finite dimensional spaces there is a canonical isomporphism between $\text{End}(V \otimes W)$ and $\text{End}(V) \otimes \text{End}(W)$ based on extending $\phi(S,T)(u\otimes v):=Su\otimes Tv.$

On the other hand I am reading http://www.physik.uni-leipzig.de/~schmidtm/qm/tepr.pdf which includes the following statement which appears to conflict with this Remark: There are many operators on V ⊗W which are not of the form A ⊗ B with A and B being operators on V and W, respectively. For example, if W = V , the mapping V ×V → V ⊗V defined by (v,w) 7→ (w, v) induces an operator on V ⊗V (check this) which is not of that form.

This suggests (to me) that a mapping $\text{End}(V) \otimes \text{End}(W) \to \text{End}(V \otimes W)$ can't be surjective and so there can't be an isomorphism ?

I would appreciate help with this.

• You should be able to define the \L macro for this post by inserting $\newcommand{\L}{\mathcal{L}}$ at the beginning of the post (or whatever you want \L to show up as instead of $\mathcal{L}$). But, ouch, I see I was too late... – pjs36 Dec 21 '16 at 7:23
• @pjs36 Thanks, I gave up and just put in text - any ideas on an answer? – Tom Collinge Dec 21 '16 at 7:25
• None at all, this is above my pay grade, unfortunately. – pjs36 Dec 21 '16 at 7:28

No, the map $\phi \colon \def\L{\mathop{\rm End}}\L V \times \L W \to \L(V \otimes W)$, $(S,T) \mapsto S \otimes T$ is not surjective. But the map you get by using the universal property of the tensor product on $\phi$, the map $\L V \otimes\L W \to \L(V \otimes W)$ is surjective (but of course not every element in $\L V \otimes \L W$ is decomposable, that is of the form $S \otimes T$, the general element is of the form $\sum_{i=1}^k S_i \otimes T_i$).
To answer your question regarding the "swap map": Let $v_1, \ldots, v_n$ be a basis of $V$, $S_{ij} \colon V \to V$ the linear maps given by $S_{ij}(v_k) =\delta_{ik}v_j$ on the basis, that is $S_{ij}$ maps $v_i$ to $v_j$ and all other basis elements to $0$. Now consider $\sum_{i,j} S_{ij}\otimes S_{ji}$. For $u, w \in V$, written as $u = \sum_k u_k v_k$ and $w = \sum_\ell w_\ell v_\ell$, we have \begin{align*} \left(\sum_{i,j} S_{ij}\otimes S_{ji}\right)(u \otimes w) &= \left(\sum_{i,j} S_{ij}\otimes S_{ji} \right)\left(\sum_k u_kv_k \otimes \sum_\ell w_\ell v_\ell\right)\\ &= \sum_{i,j,k,\ell} u_kw_\ell (S_{ij} \otimes S_{ji})(v_k \otimes v_\ell)\\ &= \sum_{i,j,k,\ell} u_k w_\ell (\delta_{ik}v_j \otimes \delta_{j\ell} v_i)\\ &= \sum_{i,j} u_i w_j (v_j \otimes v_i)\\ &= \sum_j w_j v_j \otimes \sum_i u_iv_i\\ &= w \otimes u \end{align*}
• Thanks, but what about the map referred to which takes $v \otimes w \to w \otimes v$ ? – Tom Collinge Dec 20 '16 at 14:02
• This map is not decomposable for $n := \dim V > 1$ (for $n=1$ its the identity) that is it is an element of $\L V \otimes \L V$ which cannot be written as $S \otimes T$. – martini Dec 20 '16 at 14:07
• I think you mean an element of $\L(V \otimes V)$. That is my problem. If it can't be written as $\sum_{i=1}^k (S_i \otimes T_i)$ how is the map from $\L V \otimes\L W \to \L(V \otimes W)$ surjective ? – Tom Collinge Dec 20 '16 at 18:12