Differences

This shows you the differences between two versions of the page.

--- blog:2023-07-20 [2023/07/20 18:09] – created pzhou
+++ blog:2023-07-20 [2023/07/20 20:09] (current) – pzhou
@@ Line 1: / Line 1: @@
-====== 2023-07-20 ======
+ ====== 2023-07-20 ======
 ===== the example on abelian Coulomb branch =====
@@ Line 23: / Line 23: @@
 %$$ G \ (G/B \times G/B) \cong B\G/B, \quad [ g_1 B, g_2 B] \mapsto B g_1^{-1} g_2 B. $$
 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 $\C[[t]]^3$. 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
+(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 $\C[ [t]]^3$. 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
-$$ V_{12} = \{(s_1, s_2, s_3) \in \C[[t]]^3 \mid t^{n_1} s_1 = t^{n_2} s_2, \text{i.e.} s_1 = t^{-1} s_2 \} $$
+$$ V_{12} = \{(s_1, s_2, s_3) \in \C[ [t]]^3 \mid t^{n_1} s_1 = t^{n_2} s_2, \text{i.e.} s_1 = t^{-1} s_2 \} $$
-$$ V_{23} = \{(s_1, s_2, s_3) \in \C[[t]]^3 \mid t^{n_2} s_2 = t^{n_3} s_3 , \text{i.e.} s_2 = t s_3 \}. $$
+$$ V_{23} = \{(s_1, s_2, s_3) \in \C[ [t]]^3 \mid t^{n_2} s_2 = t^{n_3} s_3 , \text{i.e.} s_2 = t s_3 \}. $$
 $V_{12}$ 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 $\C[[t]]^3$ cannot be generated. Thus, $V_{12}$ and $V_{23}$ does not intersect transversely, but with a cokernel,
+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 $\C[ [t]]^3$ cannot be generated. Thus, $V_{12}$ and $V_{23}$ does not intersect transversely, but with a cokernel,
-$$ \frac{\C[[t]]^3}{V_{12} + V_{23}} \cong \C[[t]] / t \C[[t]] \cong \C. $$
+$$ \frac{\C[ [t]]^3}{V_{12} + V_{23}} \cong \C[ [t]] / t \C[ [t]] \cong \C. $$
-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 $\C[[t]]^3$, they only have $(g_{12} s_2, g_{13} s_3) \in t^{a} N(O) \oplus t^{a+b} N(O)$, where $g_{ij} = g_i^{-1} g_j$, $g_{12}=t^a, g_{23} = t^b$.
+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 $\C[ [t]]^3$, 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 $
@@ Line 38: / Line 38: @@
 (b) $r_{1} \star r_{-1}$. We repeat the above setup, and we find that
-$$ V_{12} = \{(s_1, s_2, s_3) \in \C[[t]]^3 \mid t^{n_1} s_1 = t^{n_2} s_2, \text{i.e.} s_1 = t^{1} s_2 \} $$
+$$ V_{12} = \{(s_1, s_2, s_3) \in \C[ [t]]^3 \mid t^{n_1} s_1 = t^{n_2} s_2, \text{i.e.} s_1 = t^{1} s_2 \} $$
-$$ V_{23} = \{(s_1, s_2, s_3) \in \C[[t]]^3 \mid t^{n_2} s_2 = t^{n_3} s_3 , \text{i.e.} s_2 = t^{-1} s_3 \}. $$
+$$ V_{23} = \{(s_1, s_2, s_3) \in \C[ [t]]^3 \mid t^{n_2} s_2 = t^{n_3} s_3 , \text{i.e.} s_2 = t^{-1} s_3 \}. $$
 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.
 $$ r_{1} \star r_{-1} = u r_0.$$