Let $H$ be a $\mathbb R$-Hilbert space, $A\in\mathfrak L(H)$ be compact and $$|A|:=\sqrt{A^\ast A}$$ denote the square-root of $A$. By definition, the $k$th largest singular value $\sigma_k(A)$ of $A$ is equal to the $k$th largest eigenvalue $\lambda_k(|A|)$ of $A$.

Now, if $S,T:H\to H$ are linear, let $S\otimes T$ denote the linearization of $$H^2\ni(x,y)\mapsto Sx\otimes Ty\tag1$$ and $S^{\otimes 2}:=S\otimes S$. I would like to endow $H^{\otimes2}$ with an inner product so that $S\otimes T$ is bounded, whenever $S,T$ are bounded, and for its completion $H^{\hat\otimes2}$ the operator $S\otimes T$ has a unique bounded linear extension $S\hat\otimes T$.

Moreover, it should hold that the largest singular value $\sigma_1(A^{\hat\otimes k})$ of $A^{\hat\otimes k}$ is given by $\prod_{i=1}^k\sigma_i(A)$. Note that this is a property which is fulfilled by the exterior product $A^{\wedge k}$.

It's trivial to see that if $S:H\to H$ is linear, $\lambda_i\in\mathbb R$ and $e_i\in\mathcal N(\lambda_i-S)$, then $$S^{\otimes k}\underbrace{\left(\bigotimes_{i=1}^ke_i\right)}_{=:\:e}=\underbrace{\prod_{i=1}^k\lambda_i}_{=:\:\lambda}\bigotimes_{i=1}^ke_i\tag2$$ and hence $e\in\mathcal N(\lambda-S^{\otimes k})$.

The problem with this is that the largest eigenvalue of $|A|^{\otimes k}$ is clearly $\lambda_1(|A|)^k$; not $\prod_{i=1}^k\lambda_i(|A|)$ as I would like. The crucial difference to what I remarked about the exterior product before is that the exterior product consists of alternating multilinear forms (and hence $e_1\wedge\cdots\wedge e_1$ cannot be an eigenvector of $A^{\wedge k}$).

So, we would need some kind of alternating tensor product. Is such a construction possible?

Remark: Note that the natural choice for the inner product on $H^{\otimes2}$ is given by the Hilbert-Schmidt tensor product; maybe we can consider some kind of closed subspace of alternating tensors of it.

EDIT: What I'm trying to find is an analogue of Proposition 3.2.7 in Random Dynamical Systems: enter image description here

  • $\begingroup$ The proposition you quote is only stated in finite dimensions. Is it important to you to have a generalization that works in infinite dimensions, or is your question more about the algebraic construction (full tensor powers of a vector spaces versus its exterior powers)? $\endgroup$ – Yemon Choi Jun 4 at 13:22

If $\sigma_1(B) = 0$ then $B = 0$. Now pick your favorite $2\times 2$ matrix $A$ with $\sigma_1(A) = 1$, $\sigma_2(A) = 0$ and get a counterexample ($A\otimes A \neq 0$, but you want $\sigma_1(A\otimes A) = 1*0 = 0$).

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  • $\begingroup$ Thank you for your answer. Please consider the exterior/wedge product of your example and see my edit of the question for clarification. $\endgroup$ – 0xbadf00d May 5 at 4:18
  • $\begingroup$ @0xbadf00d perhaps I do not understand what you are asking for, sorry. If you want the norm on $H\otimes H$ so that your conditions on the singular values are satisfied, then I gave an example showing that is not possible. If on the other hand you are OK with using some other space, then why $\bigwedge^2 H$ is not suitable for you? $\endgroup$ – Aleksei Kulikov May 5 at 12:43
  • $\begingroup$ $\wedge^2H$ is suitable, but I wondered whether the same construction is possible using tensor products. It should be possible to construct an "alternating" tensor product as the subspace of $H^{\otimes2}$ spanned by the thensor $(x\otimes y-x\otimes y\rangle/2$. $\endgroup$ – 0xbadf00d May 5 at 13:16
  • $\begingroup$ I've found something in the book Ryan, but cannot manage to understand his notation: math.stackexchange.com/q/3660077/47771. $\endgroup$ – 0xbadf00d May 5 at 14:53
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    $\begingroup$ @0xbadf00d I am having trouble understanding your comments. If one actually wants to construct the wedge product then you do it precisely by taking a tensor power and applying antisymmetrisation. But the question in the body of your post asks for a tensor power that behaves like the wedge product. Clearly Aleksei's answer addresses the question of the tensor power, showing it does not behave like the wedge product. If you want to antisymmetrise then you just get back the wedge product, which seems to be covered by the proposition you quote $\endgroup$ – Yemon Choi Jun 4 at 13:20

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