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When introducing the tensor product of vectors, Serge Winitzki in his work Linear Algebra via Exterior Products claims that

It turns out to be impossible to define a nontrivial product of vectors in a general vector space, such that the result is again a vector in the same space.

I can assume that he means a bilinear product here. In the footer on that page he says

The impossibility of this is proved in abstract algebra but I do not know the proof.

I forgot to look this up when I was reading through Dr. Winitzki's book, and perusing through the book, again, I realize that I've still never seen the proof.

Does someone know of a reference where I can see the proof of this claim?

  • It depends on what one means by "product". Brown and Gray classified the so-called cross products, in all (finite) dimensions (at least over $\Bbb R$). See http://math.stackexchange.com/questions/185991/is-the-vector-cross-product-only-defined-for-3d/462895#462895 If this is a satisfactory interpretation of "product", let me know and I'll write up a proper answer with a reference/link later tonight. – Travis Willse Oct 27 '15 at 15:13
  • Based on the language he used before the above quote I'm led to believe that Winitzki used the word product to mean a binary operation that's linear in each argument. On the other hand, that answer looks very interesting and I'm going to read through Brown and Gray's "Vector Cross Products" and couple of other papers I found referencing that paper today to learn more. So thanks! –  Oct 27 '15 at 15:21
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    You're welcome, I'm glad you found my comment useful! I'm not sure that what Winitzki means, as the map $(X, Y) \to X$ is a perfectly good nontrivial bilinear map $\Bbb V \times \Bbb V \to \Bbb V$. So, one really needs some stronger hypotheses if one wants to get nonexistence for most (say, f.d.) vector spaces. (Incidentally, there are nontrivial binary cross products in only two dimensions, namely, $3$ and $7$, and these are connected with the quaternions and octonions, resp.) – Travis Willse Oct 27 '15 at 15:50
  • I'm also dubious of the statement. Take your favorite field $K$ and let $V$ be an $n$-dimensional $K$-vector space. Identifying $V$ with $K[x]/(x^n)$ certainly defines a product on $V$. You can define a whole lot of other algebra structures by identifying $V\cong K[x]/(f)$ for a degree $n$ polynomial $f$. – David Hill Oct 27 '15 at 16:08

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I interpret the claim to be about natural maps

$$V \otimes V \to V$$

(where natural means at the very least $GL(V)$-equivariant). This follows from the fact that $V \otimes V$ decomposes as a $GL(V)$ representation into a direct sum $S^2(V) \oplus \Lambda^2(V)$; these are irreps and neither of them are isomorphic to $V$ as an irrep, so the only $GL(V)$-equivariant map $S^2(V) \oplus \Lambda(V) \to V$ is the zero map.

Qiaochu Yuan
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  • This seems like a valid interpretation. Accepted. –  Oct 27 '15 at 16:26
  • A stronger version of "natural" would mean a natural transformation between two functors $\text{Vect} \to \text{Vect}$ and that is even more impossible. By contrast, the exterior product is natural in this stronger sense. – Qiaochu Yuan Oct 27 '15 at 16:43
  • @QiaochuYuan Could you provide a hint on why no natural transformation $\otimes \Rightarrow \operatorname{Id}_{Vect}$ can exist? – lisyarus May 14 '16 at 17:01