A sanity check for the Fukaya category of a cotangent bundle

Yesterday I gave a seminar about Fukaya categories and I didn’t have chance to do quite as much explicit computation as I’d hoped. I thought I'd write a blog post with a basic calculation to show you the kind of things that are involved in doing computations in Fukaya categories. I will show (using Abouzaid's description of the zero section in terms of the cotangent fibre) that the zero section and the cotangent fibre have \(rank(HF) = 1\), in the special case of \(T^*S^1\). This is such a trivial result in the end (you could do the computation just by looking at the intersection and seeing it's a single point) that you should think of this post as more of a sanity check.

The first 3 hours of what I said was essentially covering Auroux's Introduction to Fukaya categories, so if you look at that first, you should be able to figure out what I'm talking about. I hope that all my grading conventions agree with those of Abouzaid – if you notice a discrepancy, please let me know!

Abouzaid proves that a cotangent fibre in \(T^*M\) generates the wrapped Fukaya category of \(T^*M\). In fact, if \(L\) is an exact Lagrangian in \(T^*M\) he gives a prescription for writing down a twisted complex for \(L\) using only cotangent fibres. In the case when \(L\) is the graph of an exact 1-form \(dF\) where \(F\) is a Morse function, there's a very nice description: for each critical point \(p\) of \(F\) you write down \(T_p^*M\) shifted in degree by minus the Morse index \(i_p\) of \(p\) and the differential in the twisted complex between \(T_p^*M[-i_p]\) and \(T_q^*M[-i_q]\) is an element of \(CF^{1+i_q-i_p}(T_p^*M[-i_p],T^*_qM[-i_q])\). Abouzaid computed that this Floer complex is quasiisomorphic to the cubical chains on the space \(P(p,q)\) of paths between \(p\) and \(q\) with the grading reversed, so the morphism should live in \(C_{i_p-i_q-1}(P(p,q))\). What cubical chain do we take? We consider the moduli space of downward Morse flowlines from \(p\) to \(q\), cubulate it, and consider that as a \((i_p-i_q-1)\)-dimensional cubical chain on path space. The usual combinatorics of the boundaries of Morse moduli spaces translates into a relation satisfied by the different components of the differential: namely the Maurer-Cartan equation. This means that the object we write down is a twisted complex.

So here's our example.

Take \(T^*S^1\). The obvious Morse function \(F\) on \(S^1\) has 2 critical points \(p,q\) and two flowlines. The corresponding twisted complex should be \[T_p[-1] \to T_q\] where \(T_p\) and \(T_q\) are the cotangent fibres at \(p\) and \(q\). The morphism should be an element of \(CF^0(T_p,T_q)\), which is quasiisomorphic to 0-chains on the space of paths from \(p\) to \(q\). Pick one of the two flowlines as a path connecting \(q\) to \(p\) and postcompose with this path to identify the path space with the loop space based at \(p\). The based loop space of \(S^1\) has \(\mathbb{Z}\) components (indexed by winding number) each of which is contractible so there's a model for chains on the based loop space which is the group ring of \(\mathbb{Z}\), in other words Laurent polynomials \(\mathbb{Z}[t,t^{-1}]\) (\(t\) is a formal variable of degree zero whose exponent is the winding number).

The flowlines from \(p\) to \(q\) give two points in the based loop space; by the way we identified paths with loops, we know that one of these is contractible. The other one wraps once around the circle, so in total the morphism should be \(1+t\). Therefore the twisted complex representing \(S^1=\mathrm{graph}(dF)\) is \[T[-1] \stackrel{1+t}{\to} T\] (I'm no longer distinguishing between different cotangent fibres: they're all quasiisomorphic so I don't need to).

Now let's check that \(HF(S^1,T)\) has rank 1 (which it should because the zero-section and the cotangent fibre intersect transversely once!). \(HF\) means taking Homs from this twisted complex to \(T\): \[HF(S^1,T) = \mathrm{Hom}(T[-1]\stackrel{1+t}{\to}T,T)\] Remember this just means: take the direct sum \[\mathrm{Hom}(T[-1],T)\oplus \mathrm{Hom}(T,T)\] with the differential \[a\oplus b \mapsto (m_1(a)+(1+t)b) \oplus m_1(b)\] (this is precisely the lower/upper-triangular thing you'd write down in the case of taking cones of chain complexes). The differential \(m_1\) is zero for this model of chains on the based loop space (I am basically saying that the based loop space of \(S^1\) is homotopy equivalent to a bunch of points, so it's formal and we can take its cohomology as the chain model) so this differential is just \[a\oplus b \mapsto (1+t)b \oplus 0\] When is \(a\oplus b\) closed? When \((1+t)b=0\). But multiplication by \((1+t)\) is injective on Laurent polynomials (we can invert \((1+t)\) in the bigger ring of formal Laurent series), so this means \(b=0\).

When is a closed element \(a\oplus 0\) exact? When \(a=(1+t)b\) for some \(b\). But a Laurent polynomial \[a_{-n}t^n + ... +a_mt^m\] is divisible by \(1+t\) if and only if the alternating sum of the coefficients \(a_i\) vanishes. This is one equation on the coefficients, hence the cokernel of \(b\mapsto(1+t)b\) is rank 1.

Therefore \(HF(S^1,T)\) has rank 1.

While this seems like a lot of effort for a computation we can do just by inspection (there's only one intersection point!) it's also something you can imagine working for you much more generally if you knew enough about chains on the based loop space.

Exercise: do the same computation for some other manifolds with particularly simple Morse functions, e.g. \(S^n\).

N.B. If you're trying the exercise, note that Abouzaid proves the cotangent fibre is quasiisomorphic to cubical chains on the based loop space: this is important – composition of loops is basically associative on the nose for cubical chains. For singular chains I think you would need a nontrivial \(A_{\infty}\) structure (like Stasheff found). Now there's a nice, tractable, model for cubical chains on the based loop space, due to Adams and Hilton from the 1950s; if you're interested you can read about it (in the simply-connected case) here.