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--- blog:2023-07-20 [2023/07/20 18:35] – pzhou
+++ blog:2023-07-20 [2023/07/20 20:09] (current) – pzhou
@@ Line 29: / Line 29: @@
 $V_{12}$ 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,
 $$ \frac{\C[ [t]]^3}{V_{12} + V_{23}} \cong \C[ [t]] / t \C[ [t]] \cong \C. $$
-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 $