A multiplicative unitary on a Hilbert space $H$ is a unitary $V: H \otimes H \to H\otimes H$ such that the pentagon identity $V_{[12]} V_{[13]}V_{[23]}= V_{[23]} V_{[12]}$ holds. Here the leg-numbering notation is employed. For example, $$V_{[12]}= V \otimes 1 \in B(H \otimes H \otimes H).$$
Is there any intuition behind this pentagon identity? I can't memorise the identity, but I need it often (in the context of quantum groups), and I think if I would have some intuition I would be able to memorise it better.
If $G$ is a locally compact group and $H=L^2(G)$ (for the right invariant Haar measure), then $H\otimes H$ identifies naturally with $L^2(G\times G)$ and one may consider the so called Kac-Takesaki operator $$ V:L^2(G\times G)\to L^2(G\times G) $$ defined by $$ (V\xi )(x, y)=\xi (xy, y),\quad \forall \xi \in L^2(G\times G) ,\quad \forall x, y\in G. $$ One may easily prove that $V$ is unitary.
Notice that $V$ codifies the multiplication operation of $G$, so it may be seen as an incarnation of the very group structure of $G$, so much so that $G$ itself can be recovered from $V$.
It is easy to show that both $V_{[12]} V_{[13]}V_{[23]}$ and $V_{[23]} V_{[12]}$ send the square integrable function $\xi $ on $G\times G\times G$ to the function $\eta $ given by $$ \eta (x, y, z)=\xi (xyz, yz, z), $$ so $V$ satisfies the pentagon identity.
When $H$ is an abstract Hilbert space, and $V$ is an operator on $H\otimes H$ satisfying the pentagon identity, one may therefore suspect that something resembling a group must be in the background!
[1] Baaj, Saad; Skandalis, Georges, Multiplicative unitaries and duality for crossed products of (C^*)-algebras, Ann. Sci. Éc. Norm. Supér. (4) 26, No. 4, 425-488 (1993). ZBL0804.46078.
[2] Sołtan, Piotr M.; Woronowicz, Stanisław L., From multiplicative unitaries to quantum groups. II, J. Funct. Anal. 252, No. 1, 42-67 (2007). ZBL1134.46044.