blog:2023-02-10
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| ====== 2023-02-10 ====== | ====== 2023-02-10 ====== | ||
| - | * Reading about symmetric product, by [[https:// | + | * Reading about symmetric product, by [[https:// |
| + | * Idea about generation | ||
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| + | ===== Upstairs Skeleton ===== | ||
| + | How to understand it? Downstairs, we have $Sym^2(\C^*)$, | ||
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| + | 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. | ||
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| + | ===== Fiber of W ===== | ||
| + | Consider the simple case. $Hilb^2_{hor}(C^* x C)$, where $W = y_1^2 + y_2^2 + (x_1- x_2)/ | ||
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| + | Or $Y = {x^2 - y^2 z \neq 0}$ and $W = y + z$. | ||
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| + | We need to consider what is the divisor $x^2 - y^2 z = 0 $. | ||
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| + | The fiber over $w$ is $C^2 \ \{x^2 - y^2 (w-y) = 0 \}$ | ||
| + | Is that deleted guy smooth? Namely | ||
| + | $$ xdx - (2w y - 3y^2) dy = 0 $$ | ||
| + | means $x=0, y=0$ or $x=0, y= 2w/3$. So, what happens here? when $w=0$, the two critical points in the deleted divisor merge. What do you want to delete to make the complement the vanishing cycle the same? | ||
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| + | Fiber better be R^2 | ||
blog/2023-02-10.1676060277.txt.gz · Last modified: 2023/06/25 15:53 (external edit)