User steve - MathOverflow

Homology in the $A_\infty$ World

2013-03-26T19:03:00Z

This question is turning out to be a little long so let me start off with the headline. Given a differential graded algebra $A$, we can recover $A$ from its homology $HA$ if we know "the" $A_\infty$-structure of $HA$. Vaguely, I would like to know:

Question 0: Can this be done "functorially" in some sense?

Ok, now for the long version.

For a dg algebra $A$, the association $A\mapsto HA$ gives a functor from the category of dg algebras into the category of graded algebras. Kadeishvili's theorem states that there is a unique(ish) $A_\infty$-structure on the homology $HA$ with certain nice properties. In this way we can think of the association $A\mapsto HA$ as having values in the category of $A_\infty$-algebras. Unfortunately, there seems to be two problems when trying to make this into a functor:

Given a dg morphism $f:A\to B$ the induced graded morphism $Hf:HA\to HB$ may not be an $A_\infty$-morphism
One can get an $A_\infty$-morphism $f_*:=q\circ f\circ j:HA\to HB$ where $j:HA\to A$ and $q:B\to HB$ are $A_\infty$-quasi-isomorphisms, but then the association $f\mapsto f_*$ is no longer functorial.

This brings us to:

Question 1: Is there any way to fix this? Eg., can we somehow view homology as an $A_\infty$-functor or some other sort of "functor up to homotopy"?

Similarly, for a dg $A$-module $M$ there is an $A_\infty$-$HA$-module structure on $HM$ having nice properties.

Question 2: Can we recover the category of dg $A$-modules from the category of $A_\infty$-$HA$-modules, i.e., is there a functor $A$-mod$\to$ $HA$-$A_\infty$-mod (or better yet in the other direction) giving some sort of equivalence?

References also would be much appreciated.

Motivic DT-Invariants for the Algebro-Geophobic

2012-06-25T00:54:02Z

I am looking for as gentle of possible of an introduction to Kontsevich-Soibelman's theory of motivic DT-invariants. Specifically I am interested in the algebraic aspects of the theory and the relation with cluster categories. Obviously there is Kontsevich and Soibelman's 150 page paper on the subject, but words like etale and stack tend to make me panic. I am much more comfortable with 3-Calabi-Yau categories than 3 dimensional Calabi-Yau varieties.

Massey Products vs. $A_\infty$-Structures

2012-03-26T22:14:36Z

Does anyone know a good reference for a proof of the fact that given a dga $A$, an $A_\infty$-structure on $HA$ is ''the same'' as coherent choices for all of the higher Massey products of $HA$? More concretely the fact I am looking for is something like the following.

When defining the Massey product $\langle x_1,\dots, x_n\rangle$ there are multiple non-canonical choices that need to be made, which in turn give multiple cycles that could be called the Massey product of $x_1,\dots, x_n$. If $M(x_1,\dots, x_n)$ is the set of all possible resulting Massey products of $x_1,\dots, x_n$, and $\mu_n$ is the $n$-th $A_\infty$ structure map (on $HA$), then $$\mu_n(x_1\otimes\cdots\otimes x_n)\in M(x_1,\dots, x_n)$$ for all $n$ and $x_i$.

Is $Sym^g$ of a Riemann Surface of genus $g$ Calabi-Yau?

2011-06-22T20:55:00Z

The $g$-fold symmetric product of a Riemann surface of genus $g$ naturally has both a symplectic structure as well as a complex structure. Is it in fact Calabi-Yau? If so, is anything known about a mirror for it (in the sense of mirror symmetry)?

My motivation for this comes from Mirror Symmetry and Heegaard-Floer Homology (neither of which I've done any serious work in so the following might be complete garbage). Heegaard-Floer homology is essentially the study of the Floer homology of various special Lagrangian tori in $Sym^g S$ for $S$ a surface of genus $g$. If $Sym^g S$ admits any sort of sensible mirror, under HMS one should be able to extract a lot of information about the Heegaard-Floer homology of 3-manifolds admitting a genus $g$ Heegaard splitting by looking at morphisms of sheaves on the mirror side.

So modulo the CYness of $Sym^g S$ a related question is what work has been done in this direction?

When is a Homology Class Represented by a Submanifold?

2010-04-13T01:15:48Z

Possible Duplicate:
Cohomology and fundamental classes

Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi_*[N]\in H_*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every homology class of $M$ can be represented by a submanifold in this manner, however for some special cases it is.

For example, for $M$ an oriented (and closed maybe?) 4-manifold every homology class can be represented by a submanifold. Another example is when $M$ an Euclidean configuration space.

My questions are:

1) Under what circumstances can every homology class of $M$ be represented by a submanifold and

2) What are some examples of manifolds who have homology classes not representable in this manner?

Penner's formula for volume of the Moduli Space

2012-01-28T18:06:12Z

In his paper "Weil-Petersson Volumes" Penner gives the following formula for the integral of a top-dimensional cohomology class $\omega$ on the moduli space $\mathcal M_g^s$ of $s$-punctured riemann surfaces of genus $g$:

$$\int_{\mathcal M_g^s}\omega=\sum_G\frac{1}{|Aut(G)|}\int_{D(G)}\phi^*\omega$$

where the sum is taken over all isomorphism classes of ribbon graphs $G$, $\phi:\widetilde{\mathcal T_g^s}\to\mathcal M_g^s$ is the natural map from the decorated Teichmüller space to the moduli space, and $D(G)$ is more or less the pre-image under $\phi$ of the orbicell in $\mathcal M_g^s$ corresponding to $G$.

My question is to what extent can this formula be generalized. Specifically,

1) Is a similar formula for evaluating forms of arbitrary degree?

2) Is this formula the ``shadow'' of a general formula for evaluating cohomology classes on (smooth) orbifolds?

I myself cannot understand Penner's paper well enough to see where the top-dimensional assumption on $\omega$ enters the picture. References to the literature would be much appreciated as well.

Answer by Steve for An algorithm for constructing the AR-quiver of a path algebra corresponding to a change in the orientation.

2012-01-29T02:13:29Z

The algorithm for constructing the AR-quiver of any orientation of $A_n$ is the same as the algorithm for constructing the AR-quiver of the orientation you describe. Start out by figuring out all the irreducible maps between the indecomposable projectives. Then write out the cokernels, and then the cokernels of the new maps, and so on until you get to all of the indecomposable injectives.

For example, consider the $A_3$ of the form $1\xleftarrow{} 2\xrightarrow{} 3$. You have two simple projectives $P_1$ and $P_3$ and the non-simple $P_2=(k\xleftarrow{}k\xrightarrow{} k)$. Both $P_1$ and $P_3$ map into $P_2$ with respective cokernels $I_3=(0\xleftarrow{} k\xrightarrow{} k)$ and $I_1=(k\xleftarrow{} k\xrightarrow{} 0)$. The cokernel of $P_2\to I_1\oplus I_3$ is the simple injective $I_2$. So the AR-quiver is like a fish swimming to the right.

Do Denominator Vectors Determine the Cluster Variable

2011-09-14T15:13:50Z

Given a cluster algebra $A=A(\mathbf{x},Q)$, the Laurent Phenomenon states that all the cluster variables of $A$ are Laurent polynomials in the elements of $\mathbf{x}$. Thus, any cluster variable $y$ can be written $$y=\frac{p(x_1,\dots,x_n)}{x_1^{d_1}\cdots x_n^{d_n}}$$ where $p$ is a polynomial and $d_i$ are positive integers. We call $d(y):=(d_1,\dots,d_n)$ the denominator vector of $y$.

If $Q$ is mutation equivalent to a simply laced Dynkin diagram, all cluster variables are uniquely determined by their denominator vector. I would like to know to what extent this holds in general. That is:

Is it true that for any cluster algebra, the clusters are determined by their denominator vectors? If not, what classes of cluster algebras have this property? I am particularly interested in surface cluster algebras.

Answer by Steve for Why are operads useful?

2011-08-09T18:09:05Z

Here are a couple 2-3 line answers to your question:

1) They allow you to treat various algebraic problems uniformly. For example, Commutative, Associative, and Lie algebras all have their own cohomology theories (Harrison, Hochschild, and Chevalley-Eilenberg respectively). These can all be seen as instances of a single operad cohomology.

2) One can use operads to construct cohomology classes for the Mapping Class group and Out(Fn) (and others). The idea is to use graphs "colored" by operads, and construct a chain complex out of these colored graphs that computes the desired cohomology.

and perhaps the most classic answer:

3) They allow you to classify loop spaces and infinite loop spaces. For connected spaces, these are exactly classified as algebras over various operads.

Which cluster algebras have been categorified?

2011-05-18T18:13:41Z

In "Tilting Theory and Cluster Combinatorics" Buan, Marsh, Reineke, Reiten, and Todorov constructed cluster categories for mutation finite cluster algebras (without coefficients), and Amiot gives a construction of a cluster category given a quiver with potential whose Jacobian algebra is finite dimensional (in particular, this gives a cluster category for cluster algebras coming from unpunctured surfaces with nonempty boundary).

My question is simply in what other instances have cluster categories been constructed? In particular, what about cluster algebras with coefficients? What about other surface cluster algebras? What about tame cluster algebras?

Why does (Ribbon) Graph (co)Homology Compute (co)Homology of MCG?

2011-01-13T16:33:33Z

The title says it all. I am looking for an explanation or reference for why the homology of the ribbon graph complex computes the cohomology of the mapping class groups of surfaces.

I've seen explanations of this using operads, but my understanding is that the operad viewpoint is more recent and not how the above question was originally understood.

Group Extensions and Line Bundles on $BG$

2010-12-06T21:24:37Z

I am sure the answer to this question is well-known, but

It is well known that the group cohomology $H^2(G,\mathbb Z)$ classifies group extensions $0\to \mathbb Z\to E\to G\to 1$ and that for a topological space $X$ elements of $H^2(X,\mathbb Z)$ are in natural bijection with complex line bundles on $X$.

My question is thus:

What is the direct correspondence between extensions of $G$ and line bundles on $BG$?

That is, given an explicit line bundle $L$ how does one construct an explicit group extension $E$ such that the two give the same cohomology class and vice versa?

Index of a Morse function via the Hessian tensor

2010-08-06T22:10:30Z

For a smooth function $f:M\to \mathbb{R}$ one usually defines the degeneracy and index of a critical point $p\in M$ in terms of the eigenvalues of the Hessian matrix $(\partial^2 f/\partial x_i\partial x_j)$.

On the other hand, if we have a Riemannian metric $g$ we can define the Hessian tensor $H(f,g)=\nabla df$. From what I understand, one can recover the Hessian matrix from this tensor and define the index and degeneracy of a critical point as above.

My question is: can we cut out the middle man, i.e., is there a natural way to define the index and degeneracy of a critical point from the Hessian tensor without using the Hessian matrix?

Defining Quotient Bundles

2010-06-07T22:54:35Z

This is an extremely elementary question but I just can't seem to get things to work out. What I am looking for is a natural definition of the quotient bundle of a subbundle $E'\subset E$ of $\mathbb R$ (say) vector bundles over a fixed base space $B$. Every source I find on this essentially leaves the construction to the reader. I would like to glue together sets of the form $U\times E_x/E'_x$ where $x\in U$ is a locally trivial neighborhood by some sort of transition function derived from those corresponding to $E$ and $E'$, but this doesn't actually make sense in any meaningful way.

While I am tagging this as differential geometry, I would like a construction that works in the topological category (i.e., does not invoke Riemannian metrics) and avoids passing to the category of locally free sheafs.

Sorry if this is a repost (I'm sure it is, but I can't seem to find anything) and thanks in advance.

Comment by Steve

2013-04-05T03:47:26Z

@Aaron: I left the setup intentionally vague so that it can be (hopefully) modified to give a correct answer. I guess what I am really asking is weather or not something non-canonical can be done canonically in some sense.

Comment by Steve

2013-03-28T15:26:04Z

@Fernando: yes, I believe so but I would like to know what the precise statements are.

Comment by Steve

2012-07-20T20:31:54Z

The answer to your first question is yes if A is a hereditary category (meaning the higher Ext groups vanish), but it is not true in general.

Comment by Steve

2012-06-26T14:36:20Z

@Jim: this sounds like what I was looking for. Do you know of any references for these aspects of the theory?

Comment by Steve

2012-03-27T15:44:25Z

@John: this is more or less the kind of result I was looking for. Do you know if we can say anything about $\mu_n(x_1,\dots,x_n)$ if the product $\langle x_1,\dots, x_n\rangle$ is not defined? For example, does this force $\mu_n(x_1,\dots,x_n)=0$?

Comment by Steve

2012-01-29T04:59:54Z

This isn't exactly what you are looking for, but I believe you can algorithmically go between the AR-quivers of orientations related by source/sink mutations (also called APR mutations). This corresponds to flipping the direction of all arrows incident to a source or a sink.

Comment by Steve

2012-01-29T02:27:48Z

Right. By ``similar formula of arbitrary degree'' I'm asking for the integral over a cycle of the appropriate dimension.

Comment by Steve

2012-01-28T21:51:17Z

The link appears to be broken. Do you know the title of the work it is supposed to link to?

Comment by Steve

2011-08-10T21:55:32Z

@Chris: try section 4.2 of "Koszul duality for Operads" by Ginzburg and Kapranov (arxiv.org/abs/0709.1228). The arXiv version doesn't seem to have any pictures though...

Comment by Steve

2011-06-22T22:19:37Z

True, it's not simply connected but that's not so important for what I want (elliptic curves aren't simply connected but are still CY enough for MS)

Comment by Steve

2011-06-22T22:18:33Z

Ahh.. the final nail in the coffin. Thanks

Comment by Steve

2011-03-06T20:31:10Z

@John: that is absolutely true. Thanks for pointing that out

Comment by Steve

2010-04-13T12:43:45Z

The original post seems to be closer to what I was looking for. Thanks for the link and sorry for the repost.