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.

  • $\begingroup$ 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... $\endgroup$ – pjs36 Dec 21 '16 at 7:23
  • $\begingroup$ @pjs36 Thanks, I gave up and just put in text - any ideas on an answer? $\endgroup$ – Tom Collinge Dec 21 '16 at 7:25
  • $\begingroup$ None at all, this is above my pay grade, unfortunately. $\endgroup$ – 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*}

  • $\begingroup$ Thanks, but what about the map referred to which takes $v \otimes w \to w \otimes v$ ? $\endgroup$ – Tom Collinge Dec 20 '16 at 14:02
  • $\begingroup$ 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 $S \otimes T$. $\endgroup$ – martini Dec 20 '16 at 14:07
  • $\begingroup$ 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 ? $\endgroup$ – Tom Collinge Dec 20 '16 at 18:12
  • $\begingroup$ I rewrote the question for (hopefully) clarity: can you possibly take a look ? $\endgroup$ – Tom Collinge Dec 21 '16 at 7:33
  • $\begingroup$ @TomCollinge Thanks, now I understood your problem, added something. $\endgroup$ – martini Dec 21 '16 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.