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I am trying to learn a little Lagrangian Floer theory and I was hoping someone could explain the following calculation. Consider CP^n x CP^n with the form (omega,-omega) and the diagonal Lagrangian L. Now the FH*(L,L) is isomorphic to the quantum cohomology of CP^n as a ring. How about the higher A-infinity structure on the Floer cochains, CH*(L,L)? Can we compute this in a reasonable way? I'd be happy to even understand this for CP^1, though I suspect this might be obvious somehow. Is there a way to extract this from the Gromov Witten invariants of CP^n?

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Here's an argument that the diagonal Lagrangian correspondence $\Delta$ in $\mathbb{C}P^n \times \mathbb{C}P^n$ is formal. That is, its Floer cochains $CF^\ast(\Delta,\Delta)$, as an $A_\infty$-algebra over the rational Novikov field $\Lambda=\Lambda_\mathbb{Q}$ (say), are quasi-isomorphic to the underlying cohomology algebra $HF^\ast(\Delta, \Delta)\cong QH^\ast(\mathbb{C}P^n; \Lambda)$ with trivial $A_\infty$ operations $\mu^d$ except for the product $\mu^2$.

Be critical; I might have slipped up!

Write $A$ for $QH^\ast(\mathbb{C}P^n; \Lambda)=\Lambda[t]/(t^{n+1}=q)$. Here $q$ is the Novikov parameter. I claim that $A$ is intrinsically formal, meaning that every $A_\infty$-structure on $A$, with $\mu^1=0$ and $\mu^2$ the product on $A$, can be modified by a change of variable so that $\mu^d=0$ for $d\neq 2$.

Suppose inductively that we can kill the $d$-fold products $\mu^d$ for $3\leq d\leq m$. Then $\mu^{m+1}$ is a cycle for the Hochschild (cyclic bar) complex $C^{m+1}(A,A)$. The obstruction to killing it by a change of variable (leaving the lower order terms untouched) is its class in $HH^{m+1}(A,A)$. But $A$ is a finite extension field of $\Lambda$ (and, to be safe, we're in char zero). So, as proved in Weibel's homological algebra book, $HH^\ast(A,A)=0$ in positive degrees, and therefore the induction works. Taking a little care over what "change of variable" actually means in terms of powers of $q$, one concludes intrinsic formality.

You made a much more geometric suggestion - to invoke GW invariants. If you want to handle $\Delta_M\subset M\times M$ more generally, I think this is a good idea, though I can't immediately think of a suitable reference. One can show using open-closed TQFT arguments that $HF(\Delta_M,\Delta_M)$ is isomorphic to Hamiltonian Floer cohomology $HF(M)$. One could do this at cochain level and thereby show that the $A_\infty$ product $\mu^d$ of $HF(\Delta_M,\Delta_M)$ corresponds to the operation in the closed-string TCFT of Hamiltonian Floer cochains arising from a genus zero surface with $d$ incoming punctures and one outgoing puncture (and varying conformal structure). Via a "PSS" isomorphism with $QH(M)$, these operations should then be computable as genus-zero GW invariants (or at any rate, the cohomology-level Massey products derived from the $A_\infty$-structure should be GW invariants).

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I am not sure how this argument works... I am surely making a mistake but imagine an A-infinity algebra on a vector space of dimension 2, with a unit in degree zero and with a generator e in degree 1, such that the multiplication is as stated, say, e^2=1. HH*(A) vanishes, but as I understand the situation, there are possibly some non-trivial higher infinity structures namely one can have a higher multiplication e(tensor n-times)-->1. This is Koszul dual to the curved algebra k[x], x^n, whose HH* is the Jacobian ring... What am I missing? – Daniel Pomerleano Jul 1 '10 at 22:42
Daniel, thanks for your comment. My reference for the algebra is Seidel's "Homological mirror symmetry for the quartic surface", section 3. Strictly, the grading situation is not quite the same as his (namely, the grading of $A$ is periodic). So that's a place to check for loopholes. I don't know enough about Koszul duality to understand the significance of your observation... – Tim Perutz Jul 2 '10 at 20:48
The HH*(A) is equal to the HH* of the category of matrix factorizations over k[x] with curving x^n, this fact is in Dyckerhoff's paper on the subject(I believe section 4.5). which is the Jacobian ring. I think this grading thing might be a real issue... but I'm not sure. I have no opinion, however, on the original question except that on Y= P^1 x P^1, my limited intuition says that the diagonal and anti-diagonal should generate so maybe one in this case can infer what the categorical structure is by understanding just the HH*(Fuk(Y))(which we know) and working backwards... – Daniel Pomerleano Jul 4 '10 at 16:28
I can't find a fault in the following statement... Let $A$ be a mod 2 graded $A_\infty$-algebra over a field, with $\mu^1=0$. Suppose that $HH^d(A,A)=0$ for each $d>2$. Then every $A_\infty$-structure $A'$ on $A$ (with $\mu^1_{A'}=0$ and $\mu^2_{A'}=\mu^2_A$) can be trivialised by a "gauge transformation", i.e. an $A_\infty$-homomorphism $A \to A'$ whose leading-order term is the identity. – Tim Perutz Jul 4 '10 at 18:31
Looking at Dyckerhoff's Theorem 4.7 - the $A_\infty$-structure you describe seems to correspond to the superpotential $w=x^2 + x^n$ (not $x^n$). The Jacobian ring of $w$ (which by Dyckerhoff's Cor. 5.5 is the Hochschild cohomology of the dg category of matrix factorisations of $w$) is the ground field! Does this resolve the problem? – Tim Perutz Jul 4 '10 at 19:08

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