$(ST)_{[13]}= S_{[13]}T_{[13]}$ for $S,T \in B(\mathcal{H}\otimes \mathcal{H}).$

Question

Let $T\in B(\mathcal{H} \otimes \mathcal{H})$ where $\mathcal{H}$ is a Hilbert space. We can define operators $$T_{[12]}= T \otimes 1;\quad T_{[23]}= 1 \otimes T$$ and if $\Sigma: \mathcal{H} \otimes \mathcal{H} \to \mathcal{H} \otimes \mathcal{H}$ is the "flip" map, then we can define $$T_{[13]}= \Sigma_{[23]}T_{[12]}\Sigma_{[23]}= \Sigma_{[12]}T_{[23]}\Sigma_{[12]}$$

Question: Given $S,T \in B(\mathcal{H} \otimes \mathcal{H})$, is it true that $$(ST)_{[13]}= S_{[13]}T_{[13]}?$$

I attempted this as follows:

We know that the algebraic tensor product $B(\mathcal{H}) \odot B(\mathcal{H})$ is weak$^*$-dense (= $\sigma$-weakly dense) in $B(\mathcal{H} \otimes \mathcal{H})$. It is easy to see that the identity holds for $S,T \in B(\mathcal{H}) \odot B(\mathcal{H})$.

Can I conclude from this that the equality holds for all $S,T \in B(\mathcal{H}) \overline{\otimes} B(\mathcal{H})= B(\mathcal{H}\otimes \mathcal{H})$ (here, the first tensorproduct is the von-Neumann algebraic tensor product).

It is natural to try to use results involving weak$^*$-continuity and Kaplansky-density-like results, but I'm having trouble finishing the proof. Any ideas?

Answer 1

You can probably do it the way that you suggested, but my instinct is to try and exploit the idea that the leg notation is really just about relabelling of the factors in the triple tensor product of Hilbert spaces, and "anything you do spatially" will be weak-star continuous at the level of B(triple tensor product).

Note that if $H$ and $K$ are Hilbert spaces then there is a unital weak-star continuous HM $B(K) \to B(H\otimes K)$ which sends $S$ to $I\otimes S$.

Then $T_{[13]}= (\Sigma\otimes I)(I \otimes T)(\Sigma\otimes I)$ as you observe above. So using the homomorphism property of $T \mapsto I\otimes T$ I think you obtain the result you're after.

(I am writing off the top of my head so if I have overlooked some tacit assumptions/steps please let me know.)

Answer 2

There is no need to get topology involved at all.

• We have that $T_{[12]} S_{[12]} = (TS)_{[12]}$ on $(H\otimes H)\otimes H$ just from basic properties of the Hilbert space tensor product.
• $\Sigma$ is involutive, so \begin{align*} (TS)_{[13]} &= \Sigma_{[23]} (TS)_{[12]}\Sigma_{[23]} = \Sigma_{[23]} T_{[12]} S_{[12]} \Sigma_{[23]} \\ &= \Sigma_{[23]} T_{[12]} \Sigma_{[23]}\Sigma_{[23]}S_{[12]} \Sigma_{[23]} = T_{[13]} S_{[13]}. \end{align*}

Actually, there is a slightly subtle point, which both the original question, and most treatments of this in the literature, gloss over. Namely, $T_{[12]}$ acts on $(H\otimes H)\otimes H$ while $\Sigma_{[23]}$ acts on $H\otimes (H\otimes H)$. These are different Hilbert spaces, so how can we multiply these operators? In fact, of course, $(H\otimes H)\otimes H$ and $H\otimes (H\otimes H)$ are unitarily equivalent for the linear and continuous extension of the obvious map which sends $$ (\xi\otimes\eta)\otimes\alpha \mapsto \xi\otimes(\eta\otimes\alpha). $$ This math is the associator, see the theory of Monoidal categories. Once we identify these two Hilbert spaces, the above proof works fine.