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Aaron Maroja
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Context: I have been trying to understand this paper from Y. Cho and K. Kim. More precisely, a specific argument in Lemma 2.2 where they say the ABBV localization formula on an integral over a symplectic cut $N_c$ on the extremum of the interval $[c - \varepsilon, c + \varepsilon]$ gives the following $$0 = \int_{N_c} 1 = \sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}+ \int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} + \int_{N_{c+ \varepsilon}} \frac{1}{-\lambda - e_+}$$ is equivalent to $$0 =\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2} = \int_{N_{c-\varepsilon}} e_- - \int_{N_{c+ \varepsilon}} e_+$$ where $d_p$ is the order of the group acting on $p$, $p_1$ and $p_2$ are the isotropy weights of the $S^1$-representation on $T_pN$, $\lambda$ is the weight of the $S^1$-action on the fiber of the line bundle $L$, and $e_-$ (respectively e_+) is the Euler class of $\phi^{-1} (c- \varepsilon)$ (respectively $\phi^{-1} (c + \varepsilon)$).

Question: I don't understand the 'is equivalent' part. MostlyMore precisely, why is the integral $\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-}$ equivalent to $\int_{N_{c-\varepsilon}} e_-$. Remark: I believe there is a typo in the paper where they write $M_{c - \varepsilon}$ where it should be $N_{c- \varepsilon}$.

My thoughts: The first equation is actually written $ -\int_{N_c} (-1) = 0$ where the Euler class of the bundle $\phi^{-1}(c) \to N_c =\phi^{-1}(c)/S^1$ is $-1$, since $N_c \cong B^2$ and the fibers are homeomorphic to $S^3$. The second equation comes from applying ABBV localization formula for orbifolds which is quite straightforward. This gives us that $$0=\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} - \int_{N_{c+ \varepsilon}} \frac{1}{\lambda + e_+} = -\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}.$$ This is as far as I got. Any additional comments or thoughts?

Context: I have been trying to understand this paper from Y. Cho and K. Kim. More precisely, a specific argument in Lemma 2.2 where they say the ABBV localization formula on an integral over a symplectic cut $N_c$ on the extremum of the interval $[c - \varepsilon, c + \varepsilon]$ gives the following $$0 = \int_{N_c} 1 = \sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}+ \int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} + \int_{N_{c+ \varepsilon}} \frac{1}{-\lambda - e_+}$$ is equivalent to $$0 =\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2} = \int_{N_{c-\varepsilon}} e_- - \int_{N_{c+ \varepsilon}} e_+$$ where $d_p$ is the order of the group acting on $p$, $p_1$ and $p_2$ are the isotropy weights of the $S^1$-representation on $T_pN$, $\lambda$ is the weight of the $S^1$-action on the fiber of the line bundle $L$, and $e_-$ (respectively e_+) is the Euler class of $\phi^{-1} (c- \varepsilon)$ (respectively $\phi^{-1} (c + \varepsilon)$).

Question: I don't understand the 'is equivalent' part. Mostly why is the integral $\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-}$ equivalent to $\int_{N_{c-\varepsilon}} e_-$. Remark: I believe there is a typo in the paper where they write $M_{c - \varepsilon}$ where it should be $N_{c- \varepsilon}$.

My thoughts: The first equation is actually written $ -\int_{N_c} (-1) = 0$ where the Euler class of the bundle $\phi^{-1}(c) \to N_c =\phi^{-1}(c)/S^1$ is $-1$, since $N_c \cong B^2$ and the fibers are homeomorphic to $S^3$. The second equation comes from applying ABBV localization formula for orbifolds which is quite straightforward. This gives us that $$0=\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} - \int_{N_{c+ \varepsilon}} \frac{1}{\lambda + e_+} = -\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}.$$ This is as far as I got. Any additional comments or thoughts?

Context: I have been trying to understand this paper from Y. Cho and K. Kim. More precisely, a specific argument in Lemma 2.2 where they say the ABBV localization formula on an integral over a symplectic cut $N_c$ on the extremum of the interval $[c - \varepsilon, c + \varepsilon]$ gives the following $$0 = \int_{N_c} 1 = \sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}+ \int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} + \int_{N_{c+ \varepsilon}} \frac{1}{-\lambda - e_+}$$ is equivalent to $$0 =\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2} = \int_{N_{c-\varepsilon}} e_- - \int_{N_{c+ \varepsilon}} e_+$$ where $d_p$ is the order of the group acting on $p$, $p_1$ and $p_2$ are the isotropy weights of the $S^1$-representation on $T_pN$, $\lambda$ is the weight of the $S^1$-action on the fiber of the line bundle $L$, and $e_-$ (respectively e_+) is the Euler class of $\phi^{-1} (c- \varepsilon)$ (respectively $\phi^{-1} (c + \varepsilon)$).

Question: I don't understand the 'is equivalent' part. More precisely, why is the integral $\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-}$ equivalent to $\int_{N_{c-\varepsilon}} e_-$. Remark: I believe there is a typo in the paper where they write $M_{c - \varepsilon}$ where it should be $N_{c- \varepsilon}$.

My thoughts: The first equation is actually written $ -\int_{N_c} (-1) = 0$ where the Euler class of the bundle $\phi^{-1}(c) \to N_c =\phi^{-1}(c)/S^1$ is $-1$, since $N_c \cong B^2$ and the fibers are homeomorphic to $S^3$. The second equation comes from applying ABBV localization formula for orbifolds which is quite straightforward. This gives us that $$0=\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} - \int_{N_{c+ \varepsilon}} \frac{1}{\lambda + e_+} = -\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}.$$ This is as far as I got. Any additional comments or thoughts?

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Aaron Maroja
  • 83
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  • 10

Equivalence between integrals over a reduced space

Context: I have been trying to understand this paper from Y. Cho and K. Kim. More precisely, a specific argument in Lemma 2.2 where they say the ABBV localization formula on an integral over a symplectic cut $N_c$ on the extremum of the interval $[c - \varepsilon, c + \varepsilon]$ gives the following $$0 = \int_{N_c} 1 = \sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}+ \int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} + \int_{N_{c+ \varepsilon}} \frac{1}{-\lambda - e_+}$$ is equivalent to $$0 =\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2} = \int_{N_{c-\varepsilon}} e_- - \int_{N_{c+ \varepsilon}} e_+$$ where $d_p$ is the order of the group acting on $p$, $p_1$ and $p_2$ are the isotropy weights of the $S^1$-representation on $T_pN$, $\lambda$ is the weight of the $S^1$-action on the fiber of the line bundle $L$, and $e_-$ (respectively e_+) is the Euler class of $\phi^{-1} (c- \varepsilon)$ (respectively $\phi^{-1} (c + \varepsilon)$).

Question: I don't understand the 'is equivalent' part. Mostly why is the integral $\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-}$ equivalent to $\int_{N_{c-\varepsilon}} e_-$. Remark: I believe there is a typo in the paper where they write $M_{c - \varepsilon}$ where it should be $N_{c- \varepsilon}$.

My thoughts: The first equation is actually written $ -\int_{N_c} (-1) = 0$ where the Euler class of the bundle $\phi^{-1}(c) \to N_c =\phi^{-1}(c)/S^1$ is $-1$, since $N_c \cong B^2$ and the fibers are homeomorphic to $S^3$. The second equation comes from applying ABBV localization formula for orbifolds which is quite straightforward. This gives us that $$0=\int_{N_{c-\varepsilon}}\frac{1}{\lambda + e_-} - \int_{N_{c+ \varepsilon}} \frac{1}{\lambda + e_+} = -\sum_{p \in N^{S^1}\cap \phi^{-1}(c)} \frac{1}{d_p}\frac{1}{p_1p_2 \lambda^2}.$$ This is as far as I got. Any additional comments or thoughts?