• Reading about symmetric product, by Dykerhoff-Jasso-Lekili and Auroux's ICM.
• Idea about generation

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.

Consider the simple case. $Hilb^2_{hor}(C^* x C)$, where $W = y_1^2 + y_2^2 + (x_1- x_2)/(y_1-y_2)$. We reduce it. Let $A = x_1 + x_2$, $B = (y_1-y_2)^2$, $u = (x_1- x_2)/(y_1-y_2)$, and $u^2 B - A^2 \neq 0$. Superpotential is $W = B + u$.

Or $Y = {x^2 - y^2 z \neq 0}$ and $W = y + z$.

We need to consider what is the divisor $x^2 - y^2 z = 0 $.

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?

Fiber better be R^2