• Discussion with all

If you point a gun at me, and ask me the gluing formula, I would say:

• introduce two identical copies of the pure Coulomb branch algebra (I don't know why two?)
• ok, you have two sets monopole operators, and the equivariant coefficient ring.
• then what?

Let $G$ be a compact Lie group. all nice stuff. $U(n), SU(n), SO(n)$, whatever you want.

ok, the basic space, pure gauge theory Coulomb branch $M_C(G,0)$ is spec of $H^G_*(\Omega G)$. The geometry is the algebra. It is like, taking the group algebra of the loop group. So, the resulting space is the dual group, like, the dual torus.

Here is another thing: if you consider $O(G /_{ad} G)$, that gives a subring of $O(G)$, which has a nice basis with the irreducible representations of $G$, we go like $$ O(\frac{G}{G}) = \oplus_{R \in irrep(G)} \chi_R $$ We have indeed orthogonality condition.

And even better, we have $$ O(G) \cong \oplus_{R \in irr(G)} End(R) $$ as $G$ representation. (? is this also as Hilbert space isom?)

Now, what are the eigenspace of $O(G)$? For a matrix element in $End(R)$, it would have eigenvalue proportional to $c_2(R)$.

Why do we like the $O(\frac{G}{G})$? It is the K-theory of $Rep(G)$. That's why it is called the character theory. direct sum goes to sum, product goes to product, very beautiful.

Back to that spherical BFM convolution algebra. It is roughly the coordinate ring of $T^* T^\vee_\C / W$. Some symplectic affine resolution.

what does root mean? root, $\alpha$, is an element in $\mathfrak h^\vee$, it is used to label the weights of the adjoint representation $\mathfrak g$. co-root, hmmmm, what is co-root, it is not an actual element (raising/lowering operator) in $\mathfrak g_\alpha$.

Root is a function of the Lie algebra $\mathfrak h$. $e^{\alpha^\vee} -1$ is a function on $T^\vee_\C$ on the dual torus. It vanishes on the corresponding loci. So, suffice to say, the Weyl group is generated by these reflections. Why do we want to treat all roots on the same footing? not just the simple roots? the positive roots, but all roots? Well, maybe all positive roots is enough. But for the sake of Weyl invariants, all the roots are in conjugation of one another, if you add one, you better add all.

We have a $\C^*$-action on the cotangent fiber (but I would rather say, the Lie algebra direction). The symplectic form comes easily.

consider $T_\C \times T_\C^\vee / W$. Now, both the numerator and denominators should be written using function on a torus. $\frac{e^{\alpha^\vee} - 1}{e^{\alpha} - 1}$.

Warning: the space has singularity when $\pi_1(G)$ has torsion. So, for $G=PSU(2)$, the torus is $T_\C = \C^* / \Z_2$. It is like, you can take the universal cover, to kill torsion. But then, you have to quotient back by $\pi_1(G)$. Is there such a trouble previously?

For example, the previous case, if $G = PSU(2)$, then we have $\C^*$ coordiantes $x_1, x_2$ for $G^\vee_\C$'s torus, $SL(2, \C)$, so that $x_1 x_2 = 1$; and we have torus $y_1, y_2$, with condition that $y_1/y_2$ is good coordinate. So, we have coordinate $x =x_1, y=y_1/y_2$. and the quotient is by $$ (x^2 - 1) / (y-1) $$ and $(x^{-2} - 1) / (y^{-1} -1)$ (unnecessary), and the Weyl group action goes like $x \to 1/x$ and $y \to 1/y$. So, what's the trouble? And why if we flip the denominator and numerator, we have no singularity?

So, the definition of $C_4$, should be as a stack? smooth symplectic stack.

We follow BFN, and build the linear bundle $L_V$ over the algebraic model of loop to $G$, which is like the free loop quotient by equivalence $Gr_G = G((z) )/ G[[z ] ]$. Given a Laurent loop $\gamma \in G((z) )$, we can look for those $\gamma$-robust pairs, namely those nice global section of $V$ over the disk $D$, such that, even if we do a singular gauge transformation / multiply by the section to the group, we are still holomorphic. This $\gamma$ really looks like a transition function of a principal $G$ bundle on a raviolo $B$, and we take the associated $V$-bundle, and then $L_V|_\gamma$ by definition is the global section, $\Gamma(B, V_\gamma)$.

aha, the massive version: why do we call that the massive version? Well, we have extra $S^1$ action, which rescales the fiber of $L_V$. So, we have additional equivariance parameter.