6 hours.

• skeleton for hypertoric base mfd.
• implement the idea of path conormal.
• exact WKB, what's going on.

Reading Gammage-Mcbreen-Webster again, I notice that they indeed would require unimodular. But their phrasing of unimodularity is a bit strange. they require the projections to these coordinates.

It is understandable, because the basic model, $\C^n$ has a prefered factorization, so we only have $S_n$ symmetry. Instead of having the full $GL(n,\Z)$ symmetry. No, that is not a valid argument. Even the usual mirror symmetry for $\C^n$, has a favorite cone.

Suppose I want to have $\C^*$ act on $\C^3$ by weight $(2,3,5)$. This is an example where it is not unimodular. Let's see what would be wrong.

Consider the classical mirror symmetry for toric variety, here we would get $\P^2_{2,3,5}$, a not so bad smooth DM stack. It's mirror is just fine. We can use GIT to know its skeleton.

And, from the partition information on the 2-torus, we should be able to reconstruct the moment polytope, hence build the corresponding toric variety.

Consider a simpler example, $\C^*$ acting on $\C^2$ with weight $(1,2)$. We know what's the mirror is. Now, the question is, what the story on the multiplicative side? OK, we run Gale duality, we have $$ L \xto{(1,2)} \Z^2 \xto{(2,-1)} N $$ $L= \Z$, and the quotient is $N=\Z$ again, no problem.

The Gale dual is somehow isomorphis, up to a sign change. The other example is similar, any primitive vector can be completed to a $\Z$-basis. The correct condition should be saturated. Anyway.

In the weight $(1,2)$ case, we need to quotient out $(T^*\C^2)^o$, with variables $(x_1, y_1, x_2, y_2)$. We have complex moment map condition being $$ (x_1 y_1 + 1) (x_2 y_2 + 1)^2 = \beta $$ for some generic $\beta \in \C^*$. And then, we just take the usual GIT quotient, quotient by $\C^*$. Now, this subspace is smooth, and I believe that the action is also fine. For example, we can use coordinate, $$ z = x_2 y_2 + 1, \quad u = x_1^2 y_2, \quad v = y_1^2 x_2 $$ subject to $z \in \C^*$ $$ u v = (z-1) (\beta / z^2 - 1). $$ so we have three special fibers, $z = \pm \sqrt{\beta}$ and $z=1$, and the fiber become singular.

This one dimensional example is OK.

Try another one. Say the weight is $(1,1,2)$, we can complete that to a basis by $(0,1,0), (0,0,1)$, then take the dual basis, that defines a map $$ \Z \xto{(1,1,2)} \Z^3 \xto{(1,-1,0), (2,0,-1)} \Z^2 $$ OK, so what is the space? We should have $$ (x_1 y_1 + 1) (x_2 y_2 + 1) (x_3 y_3+1)^2 = \beta $$ then quotient by $\C^*$. We have base coordinates like $$ z_3 = x_3 y_3+1, \quad z_2 = x_2 y_2 + 1. $$ We have singularity over $z_2 z_3^2 = \beta$. So, we know the singularities downstairs. It shouldn't matter what is the value of $\beta$ as long as it is generic.

Now the question is: which circle get contracted?

Of course, there are many invariant functions. You can do the affine quotient, then use the complex moment map to cut-out a smooth piece, no problem. (interesting question how does the skeleton change).

Let's see, what is the generic fiber? Well, we are just quotienting $(\C^*)^3_{x}$ by $\C^*$. So far, this is very algebraic. And, as you move to a bad divisors, labelled by $1,2,3$, one of the factors $\C^*$ become $xy=0$. So, what happens here? In terms of the quotient.