3
$\begingroup$

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?

$\endgroup$

2 Answers 2

6
$\begingroup$

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).

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ Commented Jul 1, 2010 at 22:42
  • $\begingroup$ 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... $\endgroup$
    – Tim Perutz
    Commented Jul 2, 2010 at 20:48
  • $\begingroup$ 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... $\endgroup$ Commented Jul 4, 2010 at 16:28
  • $\begingroup$ 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. $\endgroup$
    – Tim Perutz
    Commented Jul 4, 2010 at 18:31
  • $\begingroup$ 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? $\endgroup$
    – Tim Perutz
    Commented Jul 4, 2010 at 19:08
0
$\begingroup$

I was browsing old symplectic questions and spotted this one. I realise that in the intervening decade you "learned a little Lagrangian Floer theory" like the question says, but maybe you'll enjoy another answer anyway.

In the paper https://arxiv.org/abs/1507.05842 by me and YankI Lekili, we give some examples of non-formality for the diagonal for toric manifolds (see corollary 7.3.5). Indeed the simplest is $CP^1$ when Floer cohomology is considered in characteristic 2 (whereas it is formal in characteristic zero, as Tim explains above). Here's how to see it in this simplest case.

The toric Hamiltonian circle action gives you a Lagrangian "moment correspondence" from $T^*S^1$ to $CP^1\times CP^1$ which gives an $A_\infty$ functor from the wrapped Fukaya category to the Fukaya category. This functor sends the cotangent fibre to the diagonal and sends the zero section to the product torus (equator times equator). In particular, it gives an $A_\infty$ map from chains on the based loopspace of $S^1$ (as a ring this is $k[z,z^{-1}]$, and higher products vanish for degree reasons) to quantum cohomology of $CP^1$ (as a ring this is $k[H]/(H^2-1)$, with the higher products to be determined below; note that I'm not working over a Novikov ring, I just set the area parameter equal to 1). Note that I'm using a minimal ($\mu_1=0$) model for the $A_\infty$ structure on the Floer complex of the diagonal, just identifying it with QH with some higher products.

In $T^*S^1$, you can write the zero section as the cone on a morphism from the cotangent fibre to itself. Under the moment correspondence functor, this means you can write the product torus as the cone on an element of quantum cohomology, considered as a morphism from the diagonal to itself. Of course it's nontrivial to compute what that element of QH is because it comes out of an $A_\infty$ functor counting pseudoholomorphic quilts (followed by an $A_\infty$ homotopy transfer to get to a minimal model). However, if we take the field $k$ to be $Z/(2)$ then there are only four elements of QH: $0$, $1$, $H$ and $1+H$.

So you try taking the cone on each of these and computing its self-homs (e.g. in the category of twisted complexes). Only one choice gives the right dimension to match with the Floer cohomology of the product torus (I think* it was $1+H$).

You can get this far only knowing the quantum product on QH, not the higher products. Now compute the ring structure on the cone. For this you need to know $\mu_4$ on QH. Since we can assume by a further homotopy that the $A_\infty$ structure is strictly unital, the only possibilities are $\mu_4(H,H,H,H)=0$ or $H$. To get the correct product structure on the Floer cohomology of the product torus you need this to be H. Therefore Floer cochains on the diagonal are non-formal in characteristic 2.

More generally, corollary 7.3.5 referred to above shows that self-Floer cochains of the diagonal is not formal if QH is not semisimple. It would be fun to explicitly compute the higher products for $CP^2$ in characteristic 3, etc.

*As I say, you can check this is the only possibility using nothing more than linear algebra. But to see why it is geometrically reasonable, remember that, given a Morse function and gradient like vector field, the zero section in $T^*M$ is built out of cotangent fibres at critical points using morphisms coming from Morse flow lines. On the circle you have two flow lines. Now shrink one flowline to bring the two critical points together. Roughly, our moment correspondence functor should be giving Seidel elements: from one flowline (the one which wraps around) you'll get the Seidel element of the $S^1$ action, namely H; from the one which shrinks you'll get the Seidel element of the trivial circle action, namely 1.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .