# Can a commutative, associative “multiplication” on an infinite-dimensional vector space be an isomorphism?

Let $V$ be a vector space (over $\mathbb C$, but I don't think it matters), and $m: V\otimes V \to V$ a "multiplication" that is associative and commutative (but I do not demand that it is unital). Is it possible that $m$ is an isomorphism $V\otimes V \overset \sim \to V$? Yes: $V$ can be zero-dimensional, or $V$ can be one-dimensional and $m$ non-zero.

Since $0$ and $1$ are the only finite solutions to $v^2 = v$, any other example must have $\dim V = \infty$. But there are many $\infty$s and many possible maps, and although I am sure that there are some examples, I am having trouble writing one down. Hence:

What is an example of an infinite-dimensional vector space $V$ and an isomorphism $m: V\otimes V \overset\sim\to V$ that is associative and commutative? Or is this impossible?

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If $V$ has a unit and the multiplication is an isomorphism, then $1 \otimes a = a \otimes 1$, which forces $a$ to be a multiple of $1$. I am not sure about the non-unital case, but it seems rather hopeless. –  Andreas Thom Oct 16 '10 at 14:30
@Andreas Thom: Yes, I had gotten that far, and almost wrote "but necessarily not unital" where I wrote "do not demand that it is unital". I changed my mind because (1) the zero- and one-dimensional examples are unital, and (2) I hadn't come to the question yet. –  Theo Johnson-Freyd Oct 16 '10 at 23:29

It is impossible. Let $x$ and $y$ be linearly independent vectors in $V$. Then $x \otimes y \neq y \otimes x$ but $m(x \otimes y) = m(y \otimes x)$.

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Well, that was easy. –  Ben Webster Oct 16 '10 at 7:38
Yes. Clearly I had not thought about the word "commutative" much. I do want commutative in my application, but I had been eying possible constructions for associative but not commutative. –  Theo Johnson-Freyd Oct 16 '10 at 7:46
I can see how it could be difficult if you've convinced yourself that examples exist. –  S. Carnahan Oct 16 '10 at 7:48

Even just associativity looks problematic. Following up on Scott's observation, for any $x$ we would have to have $x^2 \otimes x = x \otimes x^2$, so there is a linear dependency $x^2 = \lambda_x x$ for some scalar $\lambda_x$. Also, if $x \neq 0$, the function $x \cdot -$ would have to be injective (left cancellation). In particular, $\lambda_x \neq 0$.

Next, we have $(x^2)y = x(xy)$, and so $x(\lambda_x y) = x(xy)$. By left cancellation, $xy = \lambda_x y$ for all $y$.

By the same token, we have cancellation on the right. So if $y \neq 0$, from $x(yy) = (xy)y$, we derive $(\lambda_y x)y = (xy)y$, hence $xy = \lambda_y x$.

We now conclude $\lambda_x y = \lambda_y x \neq 0$ for any two nonzero $x, y$. Hence the dimension of $V$ is 1 at most.

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Nice. If I could accept this answer as well, I would. –  Theo Johnson-Freyd Oct 16 '10 at 23:30