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?



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


