I started writing this post this past weekend, but got stuck really quickly and then kept putting it off. I don’t want to leave anyone following this hanging with no idea what the A-model is. This is harder for me to describe than the A-model for some reason. Mostly I’m running into the problem of either just saying what the A-side is without explanation or I’m getting too bogged down in details. Both seem bad. In conclusion, I think I’ll err on the side of too few details, and then hopefully make sense of what is going on by completely describing mirror symmetry in the easiest case possible: the one dimensional case, i.e. for an elliptic curve.

I’m going to semi-cheat right off and refer to posts over a year old. Recall what a symplectic form is on a smooth manifold is. It is just a closed non-degenerate 2-form. A smooth manifold plus symplectic form is called a symplectic manifold. The cotangent bundle always has a canonical symplectic form on it. An example that may be less well-known is that any smooth complex projective variety is symplectic because the Fubini-Study Kähler form on restricts to a symplectic form.

If we just think about vector spaces for a second, then given a symplectic form, we say that a subspace is isotropic if and coisotropic if . The subspace is Lagrangian if it is both isotropic and coisotropic. This extends to manifold language easily by saying an embedded submanifold is Lagrangian if the tangent subspace is Lagrangian for every point of . If you want to get used to these definitions, a quick exercise would be to check that the zero section of the cotangent bundle is Lagrangian with respect to the canonical symplectic structure.

My second semi-cheat is to ask you to recall the definition of an almost complex structure from close to two years ago. The way to think about it is that it is a bundle map that behaves similarly to “multiplication by “. The condition is that , and indeed multiplication by when identifying gives an example of an almost complex structure. In fact, since we’ll always work over , any complex manifold does have multiplication by as a natural almost complex structure.

It is possible that all these things are related by the following. Suppose is a symplectic manifold, an almost complex structre, and a Riemannian metric. These three structures are called compatible if . I am far out of my depth here, but I’m pretty sure such a manifold is called Kähler if this happens, but maybe some slight more conditions are needed (e.g. does this automatically imply that is Hermitian? If so, then this is definitely what people call Kähler).

Now for the definition of the A-model. Let be a Kähler (in the sense of the previous paragraph) manifold. We define the Fukaya category to have as objects the Lagrangian submanifolds. The morphisms require a bit of technicality to define, but essentially are a way to intersect the submanifolds. It involves all the structures above and is called Floer cohomology. Recall that we’re merely sketching an idea here! Somehow this should be an or dg-category if you remember from last time, and this just comes from the fact that the morphisms have to do with cohomology classes of intersections.

If you’ve been following this at all, then you should be in utter amazement. We can state mirror symmetry now as an equivalence of categories where is a Calabi-Yau. Why is this amazing (for those not following along)? Look at the left side of this equivalence. The bounded derived category of coherent sheaves (in the Zariski topology!!) on is something that has to do purely with the algebraic data of . I mean, the Zariski topology is algebraic, the definition of coherent is very algebraic, the construction of the derived category is algebraic, etc.

The right hand side seems to have forgotten all of the algebraic data. You forget that it is a variety and instead think of it as a smooth manifold. You consider a bunch of structure that helps you study the smooth structure. You consider Lagrangian submanifolds. The Fukaya category is almost entirely analytic in nature. But now the conjecture of Kontsevich mirror symmetry is that the two are always equivalent. That’s it for today. There should be one more post in this series in which I try to sketch the conjecture in the case of an elliptic curve.