## June 14, 2006

### Kontsevich Lectures on Mirror Symmetry, I

#### Posted by Urs Schreiber

Right at the moment Daniel Huybrechts is talking about branes on K3 surfaces. Since I already know this talk ($\to$) I have time to type some notes.

I’ll try to reproduce something from Kontsevich’s ESI lectures on various things related to (homological) mirror symmetry ($\to$). I have to warn you though, that the following, as far as my reproduction makes any sense at all, is either well known to the experts or possibly close to incomprehensibe to those who are not. The fact that I am closer to the second group than to the first does not help, either.

The second lecture (which I might talk about in a later entry), had a lot of Kevin Costello’s results in it, and for all of the following it pays to look at his papers ($\to$).

While I still feel like I should just try (in fact, I have been urged) to try to absorb some of this, I cannot recommend that you try to read the following.

The setup we are concerned with is encoded by the data

(1)$\left(X,\tau ,\omega ,B\right)\phantom{\rule{thinmathspace}{0ex}},$

of a Calabi-Yau manifold with $B$-field. So $X$ is a compact manifold, $\tau$ a complex structure on $X$, $\omega$ a Kähler form on $X$ and $\mathrm{iB}\in {H}^{2}\left(X,ℝ/2\pi ℤ\right)$. The Ricci curvature vanishes.

The well-known procedure is the following.

By looking at the standard 2D sigma model for this target (the type II string) we obtain a unitary $N=2$ SCFT of central charge $\stackrel{^}{c}={\mathrm{dim}}_{ℂ}\in ℕ$. There is an “abstract mirror symmetry” acting on the “geometric data” of this model.

Instead of looking at that directly, we pass to the A-model or the B-model twist of the SCFT and obtain 2D topological field theory.

But we don’t stop here. Using the chiral ring from the TQFT a CohFT is obtained, a cohomological field theory.

This comes with a finite dimensional super vector space

(2)$ℋ$

equipped wth a symmetric and non-degenerate bilinear form

(3)$\left(\cdot ,\cdot \right):{\mathrm{Sym}}^{2}ℋ\to ℂ\phantom{\rule{thinmathspace}{0ex}}.$

Moreover, the $n$-point correlator for this CohFT provides us with a map

(4)$\left(\cdot ,\cdot {\right)}_{n}:{ℋ}^{\otimes n}\to {H}^{0}\left({\overline{M}}_{g,n},ℂ\right)$

taking $n$ states of the theory to their correlator, which is a function on the moduli space of 2D surfaces of fixed genus $g$, where we assume

(5)$2-2g-n<0\phantom{\rule{thinmathspace}{0ex}}.$

There is an axiom in the game which says that restricted to the boundary divisors of moduli space these corelators are expressible in terms of simpler correlators.

The vector space of spates $ℋ$ is $ℤ$-graded. With respect to this grading the above $n$-ary maps have a certain grading, which, if I can trust my notes, is given by $2\stackrel{^}{c}\left(2g-2-n\right)$.

The way from CY data to cohomological field thery via twisted SCFT is long. There is a shortcut, namely a direct way to assign to the data $\left(X,\tau ,\omega ,B\right)$ an A-twisted CohFT with $H={H}^{•}\left(C,ℂ\right)$ and nondegenerate symmetric bilinear form $\left(\cdot ,\cdot \right)$.

The correlators of this are given by the formula

(6)$\left({\varphi }_{1},\cdots ,{\varphi }_{n}{\right)}_{g,n}=\sum _{d\in {H}_{2}\left(ℂ\right)}\mathrm{exp}\left(-{\int }_{d}\left(\omega +iB\right)\left({\mathrm{GW}}_{g,n,{\varphi }_{\cdot }}\right)\right)\phantom{\rule{thinmathspace}{0ex}},$

where

(7)${\mathrm{GW}}_{g,n,d}\in {H}_{•}\left({\overline{M}}_{g,n}×{X}^{n},ℚ\right)$

are the Gromov-Witten invariants ($\to$).

(I hope I got this formula right, but don’t trust me. Same for all following formulas.)

My notes next say:

“Why do we get the same answer?”.

Then comes a definition of TQFT:

So define TQFT to be something coming from the following data:

An infinite-dimensional nuclear Fréchet (instead of just Hilbert) superspace $ℋ$. An odd map

(8)$Q:ℋ\to ℋ$

which is nilpotent

(9)${Q}^{2}=0\phantom{\rule{thinmathspace}{0ex}}.$

For $g,n>0$ a moduli space

(10)${\mathrm{Met}}_{g,n}=\text{moduli space of pairs}\left(\Sigma ,g,\text{markings}\right)\phantom{\rule{thinmathspace}{0ex}},$

wher $\Sigma$ is a compact ${C}^{inft}$, oriented surface, possibly with boundary; $g$ is a metric on $\Sigma$ which is flat in some neighbourhood of the boundary $\partial \Sigma$, which is hence a disjoint union of $n$ copies of small cylinders. On each each of the boundary components we have one marked point.

The correlator map

(11)$\left(\cdots {\right)}_{g,n}:{ℋ}^{\otimes n}\to {\Omega }^{•}\left({\mathrm{Met}}_{g,n}\right)$

is an even, ${S}_{n}$-equivariant map under which the differential $Q$ goes over to the deRham differental on ${\mathrm{Met}}_{g,n}$.

There is a splitting axiom (which I otherwise know as the “sewing constraints”) which says that cutting a surface along some circle, thus producing a surface with two more boundary components and genus smaller by one than before must reproduce the original correlators when traced over the two new boundary spaces of states.

Formally, this means the following. Denote by $\mathrm{pr}$ the projection which forgets that we have cut the surface (i.e. that which glues the two boundary components). Notice that cutting the surface along some circle provides a map

(12)${\mathrm{Met}}_{g,n}\to {\mathrm{Met}}_{g-1,n+2}\phantom{\rule{thinmathspace}{0ex}}.$

Then the axiom says that the following diagram has to commute

(13)$\begin{array}{ccccc}{ℋ}^{\otimes n}& \to & {\Omega }^{•}\left({\mathrm{Met}}_{g,n}\right)& \stackrel{\mathrm{pr}}{\to }& \Omega \left(\text{surface with cut}\right)\\ \otimes {\mathrm{Diag}}_{n}↓\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& & & & \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}↑{\mathrm{pr}}^{*}\\ {ℋ}^{\otimes n}\otimes {ℋ}^{\otimes 2}& & \stackrel{\text{correlator}}{\to }& & \Omega \left({\mathrm{Met}}_{g-1,n+2}\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

Here ${\mathrm{Diag}}_{n}$ is the inverse map of the the 2-point correlator on the sphere, in the limit where the length $t$ of the cylinder (i.e. the sphere with two disks removed) becomes very small.

In more detail, this works as follows.

Assume that the propagation along finite cylinders

(14)${T}_{t}:ℋ\to ℋ$

forms a semigroup acting on our topological vector space $ℋ$.

Somehow we use this semigroup structure to define precisely what the dual space ${ℋ}^{\vee }$ of $ℋ$ is like, but I cannot decode my notes at this point.

In any case, we assume that $ℋ$ with the semigroup given by the cylinder propagators ${T}_{t}$ is isomorphic to ${ℋ}^{\vee }$ with the semigroup of the dual cylinde propagators ${T}_{x}^{\vee }$.

Next, my notes say that using this we get the map

(15)${\mathrm{Diag}}_{ℋ,t}\in ℋ\stackrel{^}{\otimes }ℋ$

for all $t>0$, used in the above diagram.

It is now a corollary (apparently), that the cohomology of $Q$

(16)${H}_{Q}:=\mathrm{Ker}Q/\mathrm{Im}Q$

is a finite dimensional vector space, ad we have a map

(17)${H}_{Q}^{\otimes n}\to {H}^{•}\left({\mathrm{Met}}_{g,n}\right)\phantom{\rule{thinmathspace}{0ex}}.$

Now we assume that the semigroup action from above actually extends to cylinders of vanishing length $t=0$. Then we get a Lie group called $\Pi T{S}^{1}$ acting on $ℋ$.

Its Lie algebra is

(18)$\mathrm{Lie}\left(\Pi T{S}^{1}\right)={ℝ}^{1\mid 0}\oplus {ℝ}^{0\mid 1}\phantom{\rule{thinmathspace}{0ex}}.$

If I understand my notes correctly (but possibly I do not), the first part contains the Lie algebra $\mathrm{Spin}$ (something is fishy here, I’ll check this), while the second part contains special nilpotent elements called $B$.

The $Q$-closure of the $B$ is in $\mathrm{Spin}$

(19)$\left[Q,B\right]=\mathrm{Spin}\phantom{\rule{thinmathspace}{0ex}},$

and all other (?) brackets vanish (my notes say).

(Well, as far as I understood the $B$ guys are generators of circle actions that rotate the boundary components of our surface. Hence they are essentially the $L-\overline{L}$ generator in the Virasoro algebra.)

Now comes a crucial construction.

We introduce a “formal variable” $u$, wich, I think, is supposed to be thought of as a rotation angle. Then we regard the two operator

(20)$Q±uB$

which square to 0 when restricted to ${ℋ}^{\left(0\right)}$.

(By what I said above they should square to $\propto \left[Q,B\right]$ in general.)

One finds now that

(21)${H}^{1}\left({ℋ}^{n}\left[\left[u\right]\right],Q+\mathrm{uB}\right)$

is a free module for $ℂ\left[\left[u\right]\right]$.

In physics this supposedly follows from $N=2$ SUSY and unitarity.

With respect to the inner product $\left(\cdot ,\cdot \right)$ (the 2-point correlator), $Q$ is antihermitian while $B$ is hermitian

(22)$\begin{array}{ccc}{Q}^{t}& =& -Q\\ {B}^{t}& =& +B\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

It follows that ${H}_{-n}$ is canonically isomorphic to ${H}_{n}^{*}$.

Gotta run to the next talk now. More later.

Posted at June 14, 2006 1:19 PM UTC

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