# SVD-type decomposition for the tensor product of three Hilbert spaces?


In what follows: $\tp$ denotes the algebraic tensor product of complex vector spaces; $\ptp$ denotes the projective tensor product of Banach spaces; $\itp$ denotes the injective tensor product of Banach spaces; $\tp_2$ denotes the Hilbertian tensor product of two Hilbert spaces.

Let $H_1$, $H_2$ and $H_3$ be Hilbert spaces (you can assume finite--dimensional of arbitrary dimension, although ultimately I am after the separable infinite-dimensional case).

A consequence of the singular value decomposition (or Schmidt decomposition) is that each $w\in H_1\tp H_2$ can be written as $$w = \sum_{i=1}^N \lambda_i e_i \tp f_i$$ where $$\Vert w \Vert_{H_1\ptp H_2} = \sum_{i=1}^N |\lambda_i| \tag{1}$$ $$\Vert w \Vert_{H_1\tp_2 H_2} = \left(\sum_{i=1}^N |\lambda_i|^2\right)^{1/2} \tag{2}$$ $$\Vert w \Vert_{H_1\itp H_2} = \max_{1\leq i\leq N} |\lambda_i| \tag{\infty}$$

Question. Can we do the same for $H_1\tp H_2\tp H_3$?

That is, for given $w\in H_1\tp H_2\tp H_3$ we want vectors $(e_i)$, $(f_i)$, $(g_i)$ and scalars $(\lambda_i)$ such that $w=\sum_i \lambda_i e_i\tp f_i \tp g_i$ and the 3-variable analogues of $(1)$, $(2)$ and $(\infty)$ hold. I must admit this seems overly optimistic to me, so I wondered if there were standard counterexamples known, perhaps recorded in the quantum computing literature, or perhaps just folklore for specialists in Banach space theory.

(Remark: the SVD decomposition actually tells us that $(e_1,\dots, e_N)$ and $(f_1,\dots,f_N)$ are orthonormal. It isn't immediately clear to me if this is already forced by requiring $(1)$, $(2)$ and $(\infty)$, although I haven't given it any thought. In any case it doesn't seem to be directly needed in my intended application.)

• Have you looked at: en.wikipedia.org/wiki/Higher-order_singular_value_decomposition $$\phantom{a}$$ and also at the disappointing news in: arxiv.org/abs/0911.1393 – Suvrit Mar 21 '17 at 0:00
• @Suvrit I did briefly look at that wikipedia link but found it hard to follow (or rather, I found it hard to extract an answer to my question). Thanks for the arXiv link, but as with the MO questions I linked to, I got the impression it is not really the same question as mine – Yemon Choi Mar 21 '17 at 0:29