$\gdef\acal{\mathcal A}$ $\gdef\tcal{\mathcal T}$ $\gdef\K{\mathcal K}$ $\gdef\O{\mathcal O}$

The next simplest example is the abelian gauge theory with matter.

Let $G = GL_1$ and $V=\C$. We find that $$Spec \acal^+(G,V) = \{(u,v,z) \in \C^3 \mid u v = z\}, \quad Spec \acal^\times(G,V) = \{(u,v,z) \in \C^2 \times \C^* \mid u v = 1-z\}.$$ where one can see \cite[Section 4.1]{BFN} for details.

Here we compute this using directly $$\acal^+(G,V) \cong H_*\left( \frac{T_{G,V} \times_{V(\K)} T_{G,V}}{G(\K)}\right).$$ Note that $\pi: T_{G,V} \to Gr_G$, with base $Gr_G=\Z$ and fiber $V(\O) = \C[ [t]]$.

Write $\tcal = T_{G,V}$ for short. Concretely, we have $\tcal = \cup_{n \in \Z} \{t^n\} \times \C[ [t]]$. Our goal is to compute the convolution of two ($G(\K)$-equivariant) homology classes in $\tcal^2$ that is supported on the fiber product $$\tcal \times_{V(\K)} \tcal = \{ (t^{n_1}, s_1, t^{n_2}, s_2) \mid n_i \in \Z, s_i \in \C[ [t]], t^{n_1} s_1 = t^{n_2} s_2 \}$$ For any $d \in \Z$, let $r_d$ denote the homology cycle supported on $$r_d = \{ (g_1 = t^{n_1}, s_1, g_2=t^{n_2}, s_2) \in \tcal \times_{V(\K)} \tcal \mid -n_1 + n_2 = d \}.$$ %where we used the isomorphism that %$$G \ (G/B \times G/B) \cong B\G/B, \quad [ g_1 B, g_2 B] \mapsto B g_1^{-1} g_2 B.$$ Let $\pi_{12}, \pi_{23}, \pi_{13}$ denote the projection of $\tcal^3 \to \tcal^2$ to the corresponding factors. Then, to compute $r_a \star r_b$, we need to compute the intersection (in homology) of $\pi_{12}^{-1}(r_a)$ with $\pi_{23}^{-1}(r_b)$, in the ambient space $\tcal^3$. We compute $r_{1} \star r_{-1}$ and $r_{-1} \star r_1$ to illustrate.

(a) $r_{-1} \star r_{1}$. We need to intersect the vector bundles $\pi_{12}^{-1}(r_{-1})$ and $\pi_{23}^{-1}(r_{1})$ inside $\tcal^3$. $\tcal^3$ fibers over $Gr^3=\Z^3$ with fiber $\Ct^3$. The intersection falls over the points$(n_1, n_2, n_3)$with$n_1 - n_2 = -1, n_2-n_3=1$. Over such a point in$\Z^3$, we have two (infinite dimensional) subspaces$$ V_{12} = \{(s_1, s_2, s_3) \in \Ct^3 \mid t^{n_1} s_1 = t^{n_2} s_2, \text{i.e.} s_1 = t^{-1} s_2 \} $$$$ V_{23} = \{(s_1, s_2, s_3) \in \Ct^3 \mid t^{n_2} s_2 = t^{n_3} s_3 , \text{i.e.} s_2 = t s_3 \}. $$ $V_{12}$ has a basis given by $(\{ t^{n+1}, t^n, 0) | n \in \Z_{\geq 0}\} \cup \{(0,0,t^m) \mid m \in \Z_{\geq 0}\}$. $V_{23}$ has a basis given by $(\{ 0, t^{n+1}, t^n) | n \in \Z_{\geq 0}\} \cup \{(t^m, 0,0) \mid m \in \Z_{\geq 0}\}$. We may check that $V_{12} + V_{23}$ has a basis of $\{(t^a, 0, 0)\} \cup \{(0, t^{b+1}, 0)\} \cup \{(0,0,t^c)\}$ where $a,b,c \in \Z_{\geq 0}$, and the basis vector $(0, 1, 0)$ in $\Ct^3$cannot be generated. Thus,$V_{12}$and$V_{23}$does not intersect transversely, but with a cokernel,$$ \frac{\Ct^3}{V_{12} + V_{23}} \cong \Ct / t \Ct \cong \C. $$ This cokernel lives in the middle section $s_2$, and we need to use $g_1^{-1} g_2 s_2$ to translate it to the first (left most) slot, to have the correct $\C^*_{rot}$ action. What BFN did is to translate all the computation to the first slot, drop data on $s_1$. So, instead of having $\Ct^3$, they only have$(g_{12} s_2, g_{13} s_3) \in t^{a} N(O) \oplus t^{a+b} N(O)$, where$g_{ij} = g_i^{-1} g_j$,$g_{12}=t^a, g_{23} = t^b$.

Why BFN can forget about $s_1$ slot? This is because at the end of the day, after we enforce the compatibility conditions, $s_1$ is determined by the group element $g_{13} \in G(K)$, and section $s_3$. It is like how we label a graph $\Gamma_f$ of a bijection $f$ $\Gamma \In M \times M$, not by the pair $\{(x_1, x_2) \mid x_1 = f(x_2) \}$, but by $x_1$ alone.

(b) $r_{1} \star r_{-1}$. We repeat the above setup, and we find that $$ V_{12} = \{(s_1, s_2, s_3) \in \Ct^3 \mid t^{n_1} s_1 = t^{n_2} s_2, \text{i.e.} s_1 = t^{1} s_2 \} $$$$ V_{23} = \{(s_1, s_2, s_3) \in \Ct^3 \mid t^{n_2} s_2 = t^{n_3} s_3 , \text{i.e.} s_2 = t^{-1} s_3 \}. $$ And $V_{12}$ intersects $V_{23}$ transversely. However, $\pi_{13}(V_{12} \cap V_{23})$ is not the full $r_0$ but is a codimension $1$ subspace. $$ r_{1} \star r_{-1} = u r_0.$$