Peng Zhou

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====== 2023-09-07 ======

Now, what is the expectation? We should use the KLRW algebra as the benchmark to tell me what grading on the endo of the T-brane I should get. Basically, dot has positive grading 2, crossing with a puncture has grading 1.

First question, you claim that, you have an $S^1$-family of symplectic form, show me. Consider $\C^2 / \mu_n$, with weight $(1,-1)$, $n=2$. We consider the coordinate ring
$$ \C[x,y]^{\mu_2} = \C[xy, x^2, y^2] $$
when you blow-up, you put in the ratio coordinate $u = x/y$ and $v = y/x$, you find that this happens to be invariant since $n=2$. So, you have two patches, one with coordinate $(x/y, y^2)$ call it $U_y$ (it does not mean $y \neq 0$, it just means when $x,y$ both goes to 0, $x$ goes to zero faster, the other with coord $(y/x, x^2)$ call it $U_x$. Is this the total space of $O(-2)$? yes (tangent bundle to fano has section, since we have many automorphism)

So, now, what is the hol'c symplectic form? We can use the old one, like
$$ \Omega = dx \wedge dy = (1/2) d(x/y) \wedge d(y^2) = (1/2) d x^2 \wedge d(y/x). $$
It is the one that makes sense in local coordinate. 

Suppose we do $\C^2 / \mu_3$, weight $(1,-1)$, then the invariants are $x^3, y^3, xy$. How do we blow-up? Well, we can consider more invariant functions, like $x^2/y, x/y^2, y^2/x, y/x^2$. So, we have 3 patche, with coordinate
  * $(y/x^2, x^3)$. 
  * $(y^2/x, x^2/y)$
  * $(y^3,  x/y^2)$
I don't know how I get it, I guessed it, and it worked. 

So, we have local coordinates, and we have holomorphic symplectic form. (why it is non-degenerate? well, just check it locally. )

We define $\omega_\theta = Re(e^{i\theta} \Omega)$. 

What does extra grading mean? Well, you can have if you have a Kahler manifold $M$, you can have $H^*(M)$ equipped with an extra grading $p-q$, Hodge grading. Is it possible to get this from the $S^1$-family? In the case of $\P^n$, all the weight grading are zero, because you only have $(i,i)$ class. That's not the case for elliptic curve, where you have $(1,0)$ and $(0,1)$.