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Timeline for Tensor product is complete?

Current License: CC BY-SA 4.0

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Aug 24, 2022 at 10:51 comment added Martin Geller Yes, the symmetry requirement only works for copies of the same $V$ tensored with each other. This is an oversight
Aug 23, 2022 at 21:55 comment added KConrad I have rereread the question multiple times and there is something I don't get: what could it mean to say $||v \otimes w||_{V \otimes W} = ||w \otimes v||$ when $V$ and $W$ are not the same space? The elementary tensor $w \otimes v$ doesn't live in $V \otimes W$, so the right side of that equation doesn't make sense to me.
Aug 23, 2022 at 19:29 comment added Nik Weaver @terceira it sure doesn't look like it could be in the algebraic tensor product, but can you prove this?
Aug 23, 2022 at 18:41 comment added terceira A suggestion to show that the result is false if both spaces are infinite dimensional. One can find norm one linearly independent sequences $(x_n)$ and $(y_n)$. The series $\Sigma \frac 1 {n^2}x_n \otimes y_n$ converges in the completion but the limit is not in the algebraic tensor product.
Aug 23, 2022 at 16:21 comment added Nik Weaver Good idea ... is the set of rank $\leq r$ tensors closed? I guess this would do it.
Aug 23, 2022 at 15:55 review Close votes
Aug 28, 2022 at 3:01
Aug 23, 2022 at 15:51 answer added Gerald Edgar timeline score: 1
Aug 23, 2022 at 14:38 comment added Nik Weaver In fact, it's hard to think of any examples where the algebraic tensor product $V\otimes W$ is complete. Conjecture: if $V$ and $W$ are both infinite dimensional, then $V\otimes W$ is not complete.
Aug 23, 2022 at 14:31 comment added Onur Oktay Let's assume both V and W are infinite dimensional. Injective tensor product and projective tensor product are the very first examples that comes to mind for which $V\otimes W$ is not complete.
Aug 23, 2022 at 14:15 history asked Martin Geller CC BY-SA 4.0