26
votes
Has anything precise been written about the Fukaya category and Lagrangian skeletons?
I noticed this question has been bumped up to the front page, and the
most recent answer is about 8 years old: the subject has moved on
since then, and more has been written. Here is my understanding ...
15
votes
Has anything precise been written about the Fukaya category and Lagrangian skeletons?
A comment re. Jonny's nice answer: there was indeed a time when that was the envisioned strategy of proof. However our present approach does not require the arborealization. Because: now we know ...
13
votes
Accepted
Meaning of A-infinity relations
For your first question. Suppose $(A,d,m,m_3,m_4\dots)$ is an $A_\infty$-algebra. The operation $m_3$ gives a homotopy between $m(-,m(-,-))$ and $m(m(-,-),-)$, which I will abusively denote as $a(bc)$ ...
13
votes
Geometric Langlands: From D-mod to Fukaya
To answer [a paraphrase of] your second question first: yes the Kapustin-Witten perspective on geometric Langlands has I think been taken very seriously by a segment of the math community. I find it ...
8
votes
Geometric Langlands: From D-mod to Fukaya
One answer to your initial question is that the $D$-modules are supposed to actually do something - they're supposed to analogize to automorphic forms under the sheaf-functions dictionary. Therefore ...
7
votes
Meaning of A-infinity relations
I'll be informal but try to give a topologist's intuitive interpretation, which gives the original source of the idea. Historically, $A_{\infty}$ structures start with Stasheff's work determining ...
6
votes
Wrapped Fukaya categories of Stein manifolds
By now we understand quite well generators of such categories. The co-core disks for any relative skeleton of the sector associated to the superpotential will do. (The argument for this is geometric,...
5
votes
Geometric Langlands: From D-mod to Fukaya
More a comment than an answer:
I think the lagrangian-to-sheaf dictionary is more “immediately applicable” than is commonly supposed. In particular, it’s possible to immediately apply the dictionary ...
5
votes
Generating Fukaya category vs split-generating Fukaya category
Yes, generation is stronger than split-generation, because you don't need to split off summands after forming iterated mapping cones with degree shifts.
The simplest example I have in mind is when $M$ ...
5
votes
What is Known about the $K$-Theory of Fukaya Categories?
For $X$ Weinstein, it's a result of Oleg Lazarev that the map $H_n(X) \to SH(X)$ factors as
$$H_n(X) \twoheadrightarrow K_0(Fuk(X)) \to HH_\bullet(Fuk(X)) = SH_\bullet(X)$$
Here the map to $K_0(X)$ ...
5
votes
Accepted
Mirror symmetry for singular Lagrangian torus fibrations
Yes, there is a concrete program on how to handle singular fibers in the SYZ fibration and several steps of this program are already completed.
You can watch the videos of Abouzaid lectures on the ...
4
votes
Meaning of A-infinity relations
One interpretation I prefer is obtained by defining $A_{\infty}$ structures on graded space $L$ as a degree $-1$ square zero derivation $d$ on (co)free coalgebra on $L[-1]$ aka its bar construction. ...
3
votes
Accepted
How to understand geometrically, the count of pseudoholomorphic discs by (multi)section perturbation of the kuranish structure on the moduli space?
You are right that the solutions of the perturbed equation do not satisfy the $\bar\partial_J$ equation for any $J$ anymore. Please note that this is a feature: if they would still be properly $J$-...
2
votes
Accepted
Are all exact Lagrangian spheres, vanishing cycles?
No, this is not necessarily true. In fact, this is never the case if $E = B^4$, and $F$ has connected boundary and genus $g \ge 2$.
Since $H_1(E) = 0$, the vanishing cycles span $H_1(F)$.
An Euler ...
1
vote
Bridgeland stability to Fukaya stability on elliptic curve; geometric proof of no slope decreasing homs
Until someone better comes along here is some info about mirror symmetry on elliptic curve.
The rational line of slope $m/n$ for coprime $m,n$ is the appropriate stable vector bundle.
Then if you have ...
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