# The B-model

Sorry for the delay, but I’ve been incredibly busy and this topic is basically lowest priority right now. Today we will finish describing the “B-side” of mirror symmetry. The typical way this is done is with ${A_\infty}$-categories. I’m going to do it using dg-categories, because I’m much more comfortable with the language. Since we are working over ${\mathbb{C}}$ it turns out these two things are exactly the same. I won’t go into why that is the case (probably because I don’t have an understanding why they are). The reason it is called the “B-side” is that we are constructing what is called the B-model. The statement of mirror symmetry will be something along the following lines: the B-model on ${X}$ is equivalent to the A-model on ${\widehat{X}}$ where ${\widehat{X}}$ is that mirror pair we looked at here.

The two different “models” have something to do with string theory that I definitely don’t understand. As I said above we will need the notion of a dg-category. Without being incredibly careful with this definition (we’ll see later that this essentially adds no information to our derived category) a dg-category, ${T}$, (over ${\mathbb{C}}$) is a collection of objects ${Ob(T)}$ such that for any pair of objects and any integer we have ${T(x,y)^n}$ a ${\mathbb{C}}$-module. This should be thought of as morphisms in degree ${n}$. There is a composition of morphisms, and it should act like multiplication in a graded ring, so ${T(x,y)^n\times T(y,z)^m\rightarrow T(x,z)^{n+m}}$ is bilinear and associative.

There needs to be an identity map in degree zero for any object which we notate ${e_x\in T(x,x)^0}$. So far nothing should feel strange to you, but our final condition is that there is a differential ${d: T(x,y)^n\rightarrow T(x,y)^{n+1}}$ with the property that ${d^2=0}$ (i.e. ${Hom(x,y)}$ is a complex with respect to the differential) satisfying a graded Leibniz rule ${d(fg)=d(f)g+(-1)^mfd(g)}$.

If you want something to help you wrap your head around this here is an analogy. A groupoid is a category in which every morphism is invertible, so you can think of any group ${G}$ as a groupoid by taking the category to have only one object and ${Hom(*,*)=G}$. In a similar way, a dg-category or “differential graded” category can be formed by taking any differential graded algebra ${A=\bigoplus A^n}$ and then forming the category with exactly one object and ${T(*,*)^n=A^n}$. All those definitions are just basically turning the definition of a differential graded algebra into a category in a consistent way.

Let ${X}$ be a smooth projective variety over ${\mathbb{C}}$. Then we have already constructed a triangulated category ${D^b_{Coh}(X)}$, the derived category. This is not a dg-category, but we can ask if there is a dg-enhancement. It is a bit technical to describe what this means, but roughly an enhancement is asking for a dg-category ${T}$ where the homotopy category ${H^0(T)}$ is equivalent to ${D^b(X)}$. Here ${H^0(T)}$ means to take the same objects and define the morphisms to just be the ${0}$-th cohomology ${H^0(Hom(x,y))}$.

There are varying levels of strength of what it means to be a unique enhancement mostly stemming from the fact that there is a choice of equivalence ${H^0(T)\rightarrow D^b(X)}$ that may or may not be respected. Again, we’ll skip over these technicalities because the hope is to take away from these posts a general flavor of what mirror symmetry says rather than trying to describe all the technical details (the Clay Math book called “Mirror Symmetry” is 929 pages for goodness sake!). Thanks to Lunts and Orlov when our variety is projective we get that there always exists a dg-enhancement and it is unique in the strongest sense.

A priori a dg-category is big with lots of information and behaves nicely (is relatively easy to work with). A triangulated category has very little structure and behaves rather poorly (is difficult to work with). This enhancement theorem says that it doesn’t really matter which one we work with if we care about ${D^b(X)}$ which should be surprising. Somehow ${D^b(X)}$ is just the ${0}$-th part of the big dg-category so for instance having an equivalence of the triangulated categories doesn’t seem strong enough to be able to extend it to an equivalence of the whole dg-structure, but it is.

To conclude this half of the series, suppose we have a Calabi-Yau threefold ${X}$, then for our purpose we will call the B-model of ${X}$ denoted by ${B(X)}$ to be the unique dg-enhancement of ${D^b(X)}$. Mirror symmetry will eventually be an equivalence (as dg-categories) with ${A(\widehat{X})}$. This means we need to move on to the ${A}$-model.

## 2 thoughts on “The B-model”

1. Does this have anything to do with the fact that $A_\infty$ algebras can be strictified into associative algebras (and $E_\infty$-algebras can be strictified into commutative ones in char. zero)? There are a whole bunch of general results on when you can do this: basically, to get a “homotopy invariant” theory, you have to work with an operad on which the symmetric group acts reasonably freely. See Berger and Moerdijk’s paper “Axiomatic homotopy theory for operads.”

2. hilbertthm90 says:

Here is my (poorly informed) notion of some correspondences: $A_\infty$-categories are horizontal categorifications of $A_\infty$-algebras and can be thought of as categories homotopically enriched over chain complexes.
This means that if the composition is in the $A_\infty$-category is strictly associative, then it is a dg-category. Now I said these two things should be equivalent, but this seems to say that dg-categories are just a special type of A-infinity-category. It turns out that every $A_\infty$-category is $A_\infty$-equivalent to a dg-category.
My guess would be that if you strictify an $A_\infty$-algebra to a dg-algebra and you take horizontal categorifications of both you get the $A_\infty$-equivalence between the A-infinity category and the dg-category. I.e. if you make a square with strictification, horizontal categorifications, and sending an A-infinity category to an equivalent dg-category it should commute (these “functors” may not even be well-defined?). Again, that is definitely just a guess on my part.