1
$\begingroup$

A couple of posts relate to this topic and indicate that when the spaces are finite dimensional there is a canonical isomorphism, while in the infinite dimensional case there is only an injection $\mathrm{End}(V) \otimes \mathrm{End}(W) \to \mathrm{End}(V \otimes W)$.

Do we have $End(V \otimes V) = End(V) \otimes End(V)$?

Operators on a Tensor Product Space

https://mathoverflow.net/q/72013

So, is there an example that demonstrates this ?

I considered infinite dimensional spaces over the reals with dimension $c$ and concluded that

  1. With $Dim(V) = Dim(W) = c$
  2. $Dim (\text{End} (V)) = c^c$; so $Dim (\text{End} (V) \otimes \text{End}(W)) = c^c.c^c = c^c$;
  3. $Dim (V \otimes W) = c.c = c$
  4. $Dim(\text {End}(V \otimes W) ) = c^c = Dim (\text{End} (V) \otimes \text{End}(W))$

So, if my (cardinal) arithmetic is correct they are isomorphic in this case.


In response to the first two comments, I see that the fact that the given canonical injection is not an isomorphism doesn't preclude the spaces being isomorphic. So my question is are they always isomorphic in the infinite dimensional case or is there an example to show they are not ?

$\endgroup$
  • 1
    $\begingroup$ Imprecise language, I guess. What is meant is that the obvious map $\operatorname{End}(V) \otimes \operatorname{End} (W) \to \operatorname{End} (V \otimes W)$ is only an isomorphism in the finite-dimensional case, not that there can't be an abstract isomorphism between these space in the infinite-dimensional case. $\endgroup$ – Daniel Fischer Dec 24 '16 at 11:59
  • 2
    $\begingroup$ What is your question exactly? the two are isomorphic, but not canonically so (in the sense that the canonical injection is not onto). $\endgroup$ – gniourf_gniourf Dec 24 '16 at 11:59
  • $\begingroup$ @gniourf_gniourf - Thanks, see the added last paragraph. $\endgroup$ – Tom Collinge Dec 24 '16 at 13:53
  • $\begingroup$ @DanielFischer - Thanks, see the added last paragraph. $\endgroup$ – Tom Collinge Dec 24 '16 at 13:53
0
$\begingroup$

I'll assume the axiom of choice. I don't know what weirdness may happen otherwise here.

Let's suppose that $\kappa = \dim V \leqslant \lambda = \dim W$, and $\lambda$ is an infinite cardinal.

If $\kappa = 0$, then $\operatorname{End} (V\otimes W) = \operatorname{End} \{0\} = \{0\} = (\operatorname{End} \{0\}) \otimes (\operatorname{End} W)$. And if $\kappa > 0$, then we have

$$\dim (V \otimes W) = \kappa\cdot \lambda = \lambda = \dim W,$$

so it follows that $V \otimes W \cong W$, and consequently

$$\operatorname{End} (V \otimes W) \cong \operatorname{End} W.$$

Thus, with the canonical injection $(\operatorname{End} V) \otimes (\operatorname{End} W) \to \operatorname{End} (V \otimes W)$ we deduce

$$\dim (\operatorname{End} W) \leqslant \dim \bigl((\operatorname{End} V) \otimes (\operatorname{End} W)\bigr) \leqslant \dim \operatorname{End} (V \otimes W) = \dim (\operatorname{End} W),$$

whence there always is an abstract isomorphism between $(\operatorname{End} V) \otimes (\operatorname{End} W)$ and $\operatorname{End} (V \otimes W)$, whatever the dimensions of the spaces. But when both spaces are infinite-dimensional, the canonical map is not an isomorphism, and that's what people are interested in.

$\endgroup$
  • $\begingroup$ Many thanks: I follow that. We are off to eat turkey - have a good Christmas. $\endgroup$ – Tom Collinge Dec 24 '16 at 15:26
  • $\begingroup$ Happy Hanukkah. $\endgroup$ – Daniel Fischer Dec 24 '16 at 15:26

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.