Table of Contents
Notes
Some prediction about mirror symmetry
Forget about 3d MS for a second. Toric HMS suppose to be easy.
More about CY completion
In this note, I want to record how to do the CY completion for a smooth and proper category $C$. Let $S: C \to C$ denote the Serre functor, which has the property that $$ Hom(x, y) = Hom(y, Sx)^\vee. $$
What can you say about toric CY variety?
Let $X$ be a toric CY variety, by which we mean $X$ has a toric fan $\Sigma \In N_\R$, with ray generators $v_\rho$ for each ray $\rho$, lying on a affine hyperplane of distance $1$. That is, there is $m \in M$, such that $\la v_\rho, m \ra = 1$ for all $\rho$. That $m$ gives me a distinguished function $W: X \to \C$ (modulo a constant factor).
However, it is wrong to say $Coh(X)$ is a d-CY category, for one thing, this category is not hom finite. If given $E, F$ two coherent sheaves, we don't have $Hom(E, F) \cong Hom(F, E[d])^\vee$. (Just take the example of $X = \C$). One cure is to fix some compact subset (maybe torus invariant) $Z \In X$, and require that $Coh_{Z}(X)$, just sheaves supported along $Z$, this hom-finite sub-category would be OK. Or we can just blindly say, look at the category of 'proper' module?
What does homological mirror symmetry say? Given a Lagrangian skeleton $\La$, we can ask for those 'infinitesimally wrapped Lagrangians' or sheaves, they are the sheaves with finite rank stalks (hence microlocal stalks) everywhere.
The question now is: why Fukaya category with such skeleton gives CY property? For example, for $Tot(O_{P^1}(n))$, only for $n=-2$ it is CY. But it is kinda hard to see from the fan.
$\gdef\sl{\mathfrak{sl}}$ $\gdef\hf{\mathfrak{h}}$ $\gdef\asl{\widehat{\mathfrak{sl}}}$ $\gdef\ahf{\widehat{\mathfrak{h}}}$
What is affine Lie algebra $\asl_2$
As a vector space we have $$\asl_2 = sl_2[t,t^{-1}] \oplus \C K \oplus \C d$$ As Lie algebra, we have: $$ [X t^m, Y t^n] = [X, Y] t^{m+n} + m \delta_{m+n=0} K \quad \forall X, Y \in sl_2 $$ $$ [d, X t^m] = m X t^m, \quad d= t \d_t $$
The affine Cartan (where coroot lives) is $$ \ahf = \hf \oplus \C K \oplus \C d$$ The dual affine Cartan (where root, weights lives) is $$ \widehat{\mathfrak{h}} = \mathfrak{h}^* \oplus \C K^* \oplus \C \delta, \quad \delta:= d^*$$
To talk about root system, we need to first list all roots $\hat \Phi$, which is the (non-zero) eigenvalues (living in $\ahf^*$) of how stuff in $\ahf$ acts on $\asl_2$.
For example, eigenvalue for $e$ inside $\hf^*$ is $\alpha$, and the coroot $\alpha^\vee \in \hf$ is defined to be $\alpha^\vee = [e,f]$, which is just usually denoted $h$.
We will see that $\hat \Phi \cup \{0\} = (\Phi \cup \{0\}) + \Z \delta \In \ahf^*$. Here are the positive roots (the usual choice for Borel) $$ \hat \Phi_+ = \{ \alpha + n \delta, n \geq 0 \} \sqcup \{ n \delta, n > 0\} \sqcup \{-\alpha + n \delta, n > 0\}. $$ Finally, we can talk about simple roots, which are primitive elements in positive roots, they are $$ \alpha_1 = \alpha, \quad \alpha_0 = -\alpha + \delta. $$ The corresponding elements $e_{\alpha_i} \in \mathfrak{g}_{\alpha_i}$ are $$ e_1 = e, \quad e_0 = f \otimes t, \quad f_1 = f, \quad f_0 = e \otimes t^{-1} $$ Now, there is no mistake or typo that I mistook an f with an e. One should think of, $t$ is SO positive that, it overcomes the negativity of $f$ or whatsoever.
Finally, we can get to the coroot, which I will $\alpha_0^\vee, \alpha_1^\vee$. $$ \alpha_0^\vee = h_0 = [e_0, f_0] = [f \otimes t, e \otimes t^{-1}] = [f,e] + 1 \delta_{1-1=0} K = K - h $$ and $$ \alpha_1^\vee = h_1 = [e_1, f_1] = [e,f] = h. $$
Sanity check, does $(e_0, h_0, f_0)$ satisfies an $sl_2$-triple condition? $$ [h_0, e_0] = [K - h, f\otimes t] = [-h, f \otimes t] = 2 f \otimes t = 2 e_0. $$
Fix an integral level $K = k > 0$. All integrable representation is given by highest weight $v$, we determine $v$ by its two weights $$ \alpha_0^\vee v = a_0 v, \quad \alpha_1^\vee v = a_1 v $$ where since $K = \alpha_0^\vee + \alpha_1^\vee$, and $K v = k v$, so $a_0 + a_1 = k$. By the definition of integrable representation, the finite $\sl_2$ acts locally finitely. In particular, we need to have $a_1 \geq 0$. Can we have $a_0 < 0$? Somehow