Peng Zhou

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blog:2023-02-10

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2023-02-10

Upstairs Skeleton

How to understand it? Downstairs, we have $Sym^2(\C^*)$, which we can shrink to $Sym^2(S^1)$. Upstairs, I just don't know what to put over the diagonal. We know the fiber of the diagonal is $\C^* \times \C_u$, and the superpotential is basically the variable $u$, so if we just ask for the fiberwise skeleton, it is zero.

Same thing, if we ask for the multiplicative case. The fiberwise skeleton of a smooth fiber is $(S^1)^2$, then become subcritical for the special fiber.

blog/2023-02-10.1676103028.txt.gz · Last modified: 2023/06/25 15:53 (external edit)