https://ems.press/content/serial-article-files/12429
He really starts from the very very beginning. The abelian theory. But, how does $G(O)$ acts on $G(K)/G(O)$, when $G = \C^*$. I guess it must be trivial. Similarly, how does the compact group $G$ acts on the based loop space $\Omega G$? I guess it must be by conjugation. But, how do I reconcile with the two descriptions?
Then, we have $C_3(U_1, 0) = T^*\C^* = \C_\tau \times \C_z^*$.
Here comes the cool thing. Say $U_1$ has a standard representation, call it $L = \C$. This is the case for the quiver$[1] - (1)$.
I know what the answer should be, it is going to be $$ \{(u,v, y) \in (\C^*)^2 \times \C \mid u v = y - a\} $$ No, that was wrong, it should be $$ \{(u,v, y) \in (\C)^2 \times \C \mid u v = y - a\}. $$ The parameter $a$ is like, turning on the mass parameter in Teleman's Language (here we could have set $a=0$). And the two copies $(u,y)$ and $(v,y)$ should be the two copies of pure Coulomb branch.
The first surprising thing is that, Teleman tells me, this is related to my well-known and loved, toric mirror symmetry for $\C$ (but with $\C^*$ acting on it). The mirror is a space with superpotential, $\varphi: \C^* \to \C$, $\varphi_V(z) = z$. I am going to look for critical point, but I am not going to find it.
If we take the universal cover of $\C^*_z$, we get $\varphi: \C_u \to \C$, $\varphi(u) = e^u$, and we can run the game of the Legendre transformation. $\psi(w) = crit-val(-e^u + w u)$, something like that, and we get, $w = e^{u}$, so $\psi(w) = w (\log w - 1)$, which is not very much well-defined! However, $d \psi(w) = \log w dw $, is somehow well-defined. Indeed, the graph Lagrangian $\Gamma_{d e^u}$ is multivalued as one projects to the $w$ direction. So, what am I doing here?
So, forget about Legendre transformation, what do I want to do with this Lagrangian? It is a meromorphic section over $\C_\tau$, with $z = \tau$ (meromorphic because when $\tau=0$, we get stuck, since $z$ cannot be $0$).
So, you say. Let's glue in another copy of pure Coulomb branch, after identify fiber by fiber, this Lagrangian become $1$ in the second copy. (but why should I do this? Why not glue in more copies? Why identify this particular Lagrangian? This toric mirror symmetry seems to come from NOWHERE.)