# Mirror Symmetry A-branes

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 ${\mathbb{P}^n}$ 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 ${S}$ is isotropic if ${S\subset S^\perp}$ and coisotropic if ${S^\perp \subset S}$. The subspace is Lagrangian if it is both isotropic and coisotropic. This extends to manifold language easily by saying an embedded submanifold ${S\subset M}$ is Lagrangian if the tangent subspace ${T_sS\subset T_sM}$ is Lagrangian for every point of ${S}$. 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 ${J: TM \rightarrow TM}$ that behaves similarly to “multiplication by ${i}$“. The condition is that ${J^2=-Id}$, and indeed multiplication by ${i}$ when identifying ${\mathbb{R}^2\simeq \mathbb{C}}$ gives an example of an almost complex structure. In fact, since we’ll always work over ${\mathbb{C}}$, any complex manifold does have multiplication by ${i}$ as a natural almost complex structure.

It is possible that all these things are related by the following. Suppose ${(M, \omega)}$ is a symplectic manifold, ${J}$ an almost complex structre, and ${g}$ a Riemannian metric. These three structures are called compatible if ${\omega(J(-), -)=\langle - , -\rangle_g}$. 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 ${g}$ is Hermitian? If so, then this is definitely what people call Kähler).

Now for the definition of the A-model. Let ${(M, \omega)}$ be a Kähler (in the sense of the previous paragraph) manifold. We define the Fukaya category ${Fuk(M)}$ 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 ${A_\infty}$ 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 ${A_\infty}$ categories ${D^b(X)\rightarrow Fuk(\widehat{X})}$ where ${X}$ 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 ${X}$ is something that has to do purely with the algebraic data of ${X}$. 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.

# The Derived Category 2

I thought we would be able to move on to dg-categories today, but I was wrong. There was too much we didn’t do last time, plus an example might be nice. Recall that last time we took an arbitrary abelian category ${\mathcal{A}}$ and created the derived category ${D(\mathcal{A})}$ by making the category whose objects are complexes and morphisms are equivalence classes up to homotopy equivalence and we “invert” all quasi-isomorphisms (they become isomorphisms).

This post should be taken as just a smattering of remarks on this construction plus focusing in on the actual category we want to consider. Since the first step of the construction was to take the category of complexes we can actually alter this part of the construction in several natural ways and complete the next two steps without any problem. We called the category of complexes ${Kom(\mathcal{A})}$. Instead of using this whole category we might be interested in just the complexes that are bounded below (meaning ${A^i=0}$ for all ${i<<0}$) or bounded above or just plain bounded. These will be denoted ${Kom^+(\mathcal{A})}$, ${Kom^-(\mathcal{A})}$, or ${Kom^b(\mathcal{A})}$ respectively. Finishing the construction of the derived category in the same way with these gives ${D^+(\mathcal{A})}$, ${D^-(\mathcal{A})}$ and ${D^b(\mathcal{A})}$.

For the purposes of this post we will focus on the bounded derived category. The first remark is that if ${\mathcal{A}}$ has enough injectives, then in ${D^b(\mathcal{A})}$ for any ${A^\bullet}$ there is a complex ${I^\bullet}$ of injective objects such that they are isomorphic ${A^\bullet\simeq I^\bullet}$. I.e. any bounded complex is quasi-isomorphic to a complex of injectives and so if all we care about are objects up to isomorphism we may as well just work with complexes of injective objects.

In fact, something much stronger is true. Take ${\mathcal{I}\subset \mathcal{A}}$ to be the full subcategory of injective objects, then we have an equivalence of categories ${K^+(\mathcal{I})\simeq D^+(\mathcal{A})}$. Now let's shift our focus to ${D^b(Coh(X))}$ where ${X}$ is a scheme. Since this is the category that will appear the most we will simply denote it ${D^b(X)}$ and call it "the derived category of ${X}$" instead of the more accurate and cumbersome "the bounded derived category of coherent sheaves on $X$".

If you're paying attention you should be objecting to all of this post at this point. My remarks are useless here because ${Coh(X)}$ might be a nice abelian category, but it certainly doesn't have enough injectives. The fix for this is that in a sense that we'll make precise it is often good enough to take your complex of injectives in a category that has enough injectives as long as when you take cohomology it lands you back in the right place.

The exact statement is that if ${\mathcal{A}\subset \mathcal{B}}$ is a thick subcategory and any ${A}$ can be embedded in some ${I\in \mathcal{A}}$ which is injective as an object of ${\mathcal{B}}$ then we get an equivalence of categories ${D^b(A)\simeq D^b_\mathcal{A}(\mathcal{B})}$ where the notation ${D^b_\mathcal{A}(\mathcal{B})}$ means the full subcategory of ${D^b(\mathcal{B})}$ of complexes with cohomology in ${\mathcal{A}}$.

Now to not get in trouble at some point let's suppose our scheme ${X}$ is always a smooth projective variety over a field ${k}$ (this is the only case we'll consider in mirror symmetry so it's fine for these purposes). If we just take for a moment the category ${QCoh(X)}$ of quasi-coherent sheaves on ${X}$, then it is a standard theorem in algebraic geometry that any quasi-coherent sheaf ${\mathcal{F}}$ has an injective resolution ${0\rightarrow \mathcal{F}\rightarrow \mathcal{I}^\bullet}$ by quasi-coherent sheaves that are injective as ${\mathcal{O}_X}$-modules. If you think of the definition of the sheaf cohomology ${H^i(X, \mathcal{F})}$ in relation to the cohomology of the injective resolution complex it immediately shows that we can view ${D^b(QCoh(X))}$ as the full subcategory ${D^b_{QCoh}(Sh_{\mathcal{O}_X}(X))}$ of the bounded derived category of sheaves of ${\mathcal{O}_X}$-modules on ${X}$.

Even though this can't be done directly by some general theorem as in the quasi-coherent case we actually do get a theorem that parallels that case (again, warning: this cannot be done the same way since coherent things will not be injective). There is a natural functor ${D^b(X)\rightarrow D^b(QCoh(X))}$ and it induces an equivalence of ${D^b(X)}$ with the full subcategory ${D^b_{Coh}(QCoh(X))}$. This means that to work with the derived category of ${X}$ we can always replace (up to isomorphism) a complex of coherent sheaves by a complex of injective quasi-coherent sheaves whose cohomology is coherent.

There are so many interesting things we could talk about at this point (a derived form of Serre duality, how derived functors relate to this, when can you recover a variety from its derived category, Fourier-Mukai transforms, and so on…), but my goal is to keep trudging towards a statement of mirror symmetry and stopping to talk about all these things would lead us too far astray. Although, talking about a few of them might help some of you who might be in unfamiliar territory here feel a little more comfortable.

The last thing we'll look at today is an example. This might be completely unenlightening because I won't take the time to properly define everything, but we'll do it anyway. What is ${D^b(\mathbb{P}^n)}$ or even just ${D^b(\mathbb{P}^1)}$? We will say that the derived category ${D^b(X)}$ is generated by a collection of objects ${S}$ if the smallest full subcategory that contains ${S}$ and is closed under taking shifts and cones (we haven't defined that, but recall that the derived category satisfies this condition called being triangulated and those conditions just make sure the generated category is still triangulated) is ${D^b(X)}$ itself.

For some intuition, what this means is that any object of ${D^b(X)}$ can be reached by doing the two main operations of ${D^b(X)}$ a finite number of times using only the objects of ${S}$. In the case of projective space we get that ${D^b(\mathbb{P}^n)}$ is generated by just ${n+1}$ objects ${S=\{\mathcal{O}(n), \mathcal{O}(n-1), \ldots, \mathcal{O}(1), \mathcal{O}\}}$.

# The Derived Category 1

I promised some sort of an explanation of Kontsevich mirror symmetry. Going in there are two things to know. First, it is a conjecture so it hasn’t been established yet. Second, it isn’t even really a well-formulated conjecture. There are cases where there is a firm conjecture, but there are other cases where part of the conjecture is to figure out what the conjecture should be.

The other thing I should point out is that this is sometimes called “homological mirror symmetry”. The general form is that a certain derived category should be equivalent to a certain other category. This means that for at least the first two posts of this series I will only be describing two different categories and it won’t be at all obvious why this should be related to the mirror symmetry I alluded to last time.

Today we’ll discuss what the derived category of an abelian category is. In your head you can just keep thinking that our category is the abelian category of coherent sheaves on some sufficiently nice scheme or variety, since this is what will come up in mirror symmetry. I see absolutely no reason why I shouldn’t do the construction in the more general case of an aribtrary abelian category since it doesn’t introduce any extra difficulties (and for algebraists you can just think of ${R}$-modules or something).

The notation in the literature varies greatly. I’m going to follow Huybrechts. Let ${\mathcal{A}}$ be an abelian category. One basic related category is to form ${Kom(\mathcal{A})}$ which is the category of complexes. The objects are chain complexes ${A^\bullet}$ and morphisms ${f: A^\bullet \rightarrow B^\bullet}$ are just maps ${A^i\rightarrow B^i}$ for all ${i}$ that commute with the morphisms in the complex. This is again an abelian category and there is an embedding ${\mathcal{A}\rightarrow Kom(\mathcal{A})}$ as a full subcategory by ${A\mapsto (\cdots \rightarrow 0 \rightarrow A \rightarrow 0 \rightarrow 0 \rightarrow \cdots)}$ where ${A}$ sits in degree ${0}$.

Given any object ${(A^\bullet, d^\bullet)}$ in ${Kom(\mathcal{A})}$ (i.e. any complex of objects from ${A}$) we can take cohomology ${H^i(A^\bullet)=ker(d^i)/im(d^{i-1})}$. The cohomology lies in ${\mathcal{A}}$. Given a morphism of complexes ${f:A^\bullet \rightarrow B^\bullet}$ there is an induced map on each cohomology ${H^i(f): H^i(A^\bullet)\rightarrow H^i(B^\bullet)}$. If ${f}$ induces an isomorphism for all ${i}$ we say that ${f}$ is a quasi-isomorphism.

Loosely speaking the derived category ${D(\mathcal{A})}$ is a category (not necessarily abelian) in which quasi-isomorphisms become isomorphisms. One way to construct it is by proving the existence of a functor ${Q: Kom(\mathcal{A})\rightarrow D(\mathcal{A})}$ that takes quasi-isomorphisms to isomorphisms and ${Q}$ satisfies a certain universal property. This isn’t so easy to work with so we’ll describe ${D(\mathcal{A})}$ in a different way.

Recall that two morphisms ${f,g: A^\bullet \rightarrow B^\bullet}$ are homotopy equivalent if there is a collection of maps ${h^i: A^i\rightarrow B^{i-1}}$ such that ${f^i-g^i=h^{i+1}\circ d_A^i+d_B^{i-1}\circ h^i}$. We can form ${K(\mathcal{A})}$ often called the homotopy category of ${\mathcal{A}}$ where the objects are still just complexes, but a morphism of complexes is an equivalence class where the relation is being homotopy equivalent.

The functor ${Q}$ factors through ${K(\mathcal{A})}$. To describe the derived category ${D(\mathcal{A})}$ I’ll just tell you the objects and the morphisms. The objects are still just complexes, i.e. the objects of ${K(\mathcal{A})}$. The morphisms are trickier to describe. A morphism ${A^\bullet \rightarrow B^\bullet}$ is a choice of (equivalence class of) quasi-isomorphism ${C^\bullet \rightarrow A^\bullet}$ plus a map (in ${K(\mathcal{A})}$) ${C^\bullet \rightarrow B^\bullet}$.

This may look weird at first, but recall that all we are really trying to do is make all our quasi-isomorphisms actual isomorphisms. Thus when giving a morphism ${A^\bullet \rightarrow B^\bullet}$ you may need to pick something isomorphic ${C^\bullet}$ before defining the map, and then since ${Q}$ factors through ${K(\mathcal{A})}$ everything must be done only up to homotopy equivalences.

The derived category is no longer abelian, but it is what is known as a triangulated category. We will not talk about the general definition of one of these, since it is a sort of ad hoc, difficult, poorly behaved definition. Some people might tell you that the reason for the bad behavior is actually a result of forgetting too much stuff. You are actually decategorifying (an ${\infty}$-delooping, whatever that means) a stable ${(\infty, 1)}$-category which is a very nicely behaved structure and you may as well just work with the nicer thing.

Now whether or not you actually believe this (I’m on the fence myself) there is a much nicer, easier to understand way to say this for the derived category of coherent sheaves on a projective variety. We’ll talk about this next time, because it is actually a necessary part of the statement of Kontsevich Mirror Symmetry.

# Quick Update

I will be doing the Mirror Symmetry thing. I had someone “like” the post which I’ll count as a vote for it and someone explicitly vote for it. That’s two whole people! For all I know that is half of my regular readers, so by majority rule I have to do it.

I haven’t gotten around to a post because I’ve been incredibly busy. I’ve been doing my usual research and I’m taking two classes this quarter. In addition to all that I’ve been teaching a “mini-course” in our AG club. So for the next several weeks when I have down time I’ll probably think about what I’m going to say there and/or I’ll be typing up notes for that. I should point out that mirror symmetry has two parts, the derived category side and the Fukaya category side. The derived side is something that is really close to what I do, so it is without a doubt worthwhile to blog about that half, and I plan to soon. I may be a bit sketchier on the side that isn’t as important to me.

I’ll just leave you with a quick taste of what some people mean by “Mirror Symmetry” which is a bit different than Kontsevitch Mirror Symmetry which is what I’ll be explaining at some point. Suppose you have a Calabi-Yau threefold. By this I’ll mean a smooth, projective variety of dimension three over $\mathbb{C}$ with $\omega_X\simeq \mathcal{O}_X$ and $H^1(X,\mathcal{O}_X)=H^2(X,\mathcal{O}_X)=0$. By general Hodge theory there is a symmetry in the Hodge numbers $h^{pq}(X)=\dim_{\mathbb{C}}H^q(X, \Omega^p)$, namely that $h^{pq}(X)=h^{qp}(X)$. Also, you can use the fact that it is Calabi-Yau to check that all Hodge numbers are completely determined (independently of $X$) as either 0 or 1 except $h^{11}$ and $h^{12}$. Fun exercise in Serre duality!

Suppose $X$ is a Calabi-Yau threefold. There is some specified $h^{11}$ and $h^{12}$ (the only two unknown Hodge numbers). A mirror pair for $X$ is another Calabi-Yau threefold with $h^{11}$ and $h^{12}$ swapped. In my brief encounter with Kontsevich Mirror Symmetry (which says something about an equivalence of categories) this will follow as a special case.

Since I’m in the mood I may as well say some things that immediately pop into mind when seeing this as someone that has recently been thinking in the arithmetic world. If we are over an algebraically closed field of characteristic 0, then there is a result that says $h^{11}>0$. In particular, there cannot be a rigid CY 3-fold if mirror symmetry is true, since $h^{12}$ gives the space of deformations of $X$. But in positive characteristic there are tons of rigid CY 3-folds! Interesting.

I’ll leave you with that little taste of what is to come.

# Mirror Symmetry

Well, I keep putting off writing a new post because I’m not sure what I’m going to do it on. I had an idea. I work with Calabi-Yau varieties a lot, and so inevitably the term “mirror symmetry” appears all over the place. I’m mostly interested in arithmetic properties of Calabi-Yau’s where mirror symmetry doesn’t apply, so I know absolutely nothing about it. Since I’m curious what is meant when people use this term I thought I’d do a series of posts trying to explain the main idea of mirror symmetry.

In fact, a week or so ago Matt Ballard (who graduated the year I started grad school at the same school) put up on the archive a really nice introduction to the subject. This means I even have a nice reference to work with now. Here’s the problem. To type up something fairly reasonable on the subject is going to be a major undertaking. I have very little old blog material that is relevant, so I’ll basically be starting from scratch. It is also going to be quite time consuming since I know nothing about it.

This is my dilemma. I’d love to learn about it and blog about it, but I’m not sure it is worth the time and effort it will take. I’m at that point of grad school where I probably shouldn’t go off on a wild tangent for a long time just because I want to know what this term means when it has basically nothing to do with my research. On the other hand, one of the purposes of this blog is to keep me doing things that aren’t directly related to my research so that I maintain some sort of breadth.

I’m just throwing the idea out there for you all. What do you think? Do you have an interest in hearing about (homological) mirror symmetry, because if there seems to be interest, then it is probably worth doing.