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*: