User geoffroy horel - MathOverflow

Answer by Geoffroy Horel for Path components of a monoidal category acting on homology

2013-03-31T19:53:13Z

The action of $S$ on $X$ induces an action of $BS$ on $BX$. One way to see it is to do it at the level of $p$ simplices and construct a map $B_pS\times B_pX\to B_pX$ for all $p$. For any category $C$, the $p$ simplices $B_pC$ are the functors $[p]\to C$. Since $S$ acts on $X$, we have a functor $S\times X\to X$. Mapping $[p]$ to this map, we construct a map: $B_pS\times B_pX\to B_p X$ (note that $Fun([p],-)$ preserves product). It is formal to check that this map induce a map of simplicial sets $B_\bullet(S)\times B_\bullet(X)\to B_\bullet(X)$. If you would rather work with topological space, you can use the fact that geometric realization is product preserving and you can turn this map into a map of topological spaces.

This action induces an action map $H_0(BS)\otimes H_p(BX)\to H_p(BX)$ making $H_p(BX)$ into an $H_0(BS)$-module. But $H_0(BS)=\mathbb{Z}[\pi_0(BS)]$, therefore you have an action of the group algebra $\mathbb{Z}[\pi_0(BS)]$ on $H_p(BX)$ or equivalently an action of the group $\pi_0(S)$ on $H_p(BX)$.

What manifolds are boundaries of euclidian spaces ?

2013-02-21T17:42:31Z

I would like to know if there are compact (n-1)-manifolds $N$ that are not spheres but such that there is a manifold with boundary $M$ which satisfies the following two properties:

$\partial M\cong N$
$M-\partial M\cong \mathbb{R}^n$

I am primarily interested in this question in the category of smooth manifolds but I would be interested to know the answer in the topological case as well.

Monoidal model category structure on a functor category.

2013-01-31T10:13:08Z

Let $A$ be a small simplicial category. The category $Fun(A,s\mathrm{Set})$ of simplicial functors from $A$ to simplicial sets can be given the projective model structure in which fibration and weak equivalences are objectwise.

Now assume further that $A$ is a symmetric monoidal category with respect to some binary operarion $\circ: A\times A\to A$. We can define the Day (or convolution) tensor product on $Fun(A,s\mathrm{Set})$ by the following coend: $$F\otimes G(a)=A(-\circ -,a)\otimes_{A\times A}F(-)\times G(-)$$

My question is:

Is it true that the projective model structure is a monoidal model category in the sense that it satisfies the pushout-product axiom and if so is there a place where this is written down ?

Contractibility of the space of collars

2012-10-11T23:23:49Z

I'm looking for a reference or a proof of the following statement :

Let $M$ be a compact smooth manifold with boundary. Then the space of embeddings $\partial M\times[0,1]\to M$ inducing the identity $\partial M\times{0}\to \partial M$ is contractible.

Answer by Geoffroy Horel for Two commuting operad actions

2012-09-22T17:10:02Z

First of all let me mention that the structure you describe is precisely the Boardman-Vogt tensor product of the operads $P$ and $Q$. I'll denote this by $P\otimes_{BV}Q$. A space is an algebra for the BV tensor product of $P$ and $Q$ if it is a $P$-algebra in the symmetric monoidal category of $Q$-algebras (note that you need the fact that the monoidal structure is the cartesian product to show that the category of $Q$-algebras in spaces is symmetric monoidal).

I don't have an answer to your exact question but here is something close to what you ask. Maybe you are aware of all this. If so I apologize :

The category of $P$-algebras in $Q$-algebras is Quillen equivalent to the category of $Ass$-algebras in $Q$-algebras by standard facts about category of algebras over operads. Therefore you can strictify your $P$-algebra structure. You get a space $X'$ which is homotopy equivalent to $X$ with an action of $Ass\otimes_{BV} Q$. To this you can apply the Bar construction. The important fact is that it doesn't matter if you do it in the category of spaces or in the category of $Q$-algebras because the forgetful functor $Q-Alg\to Spaces$ commutes with geometric realization of simplicial objects.

If you are careful about doing everything homotopically, you are going to end up with a space $Y$ which is weakly equivalent to $B_PX$ and which is equipped with a $Q$-action.

The following paper http://arxiv.org/abs/math.AT/0410367 does something similar to show that $THH$ of an $E_2$-algebra is an $A_{\infty}$-algebra.

Something interesting is that you can only strictify one of the two $A_{\infty}$-structure. Indeed $P\otimes_{BV} Q$ is usually equivalent to the little $2$-disk operad (in fact the derived BV tensor product is equivalent to $E_2$). However $Ass\otimes Ass$ is the commutative operad. This paper http://arxiv.org/abs/1102.1311 talks about the BV tensor product of $A_\infty$ and $E_n$.

Answer by Geoffroy Horel for Gamma spaces and monoidal categories II

2012-03-02T15:29:16Z

You can construct a $\Gamma_{epi}$-category from your category (where $\Gamma_{epi}$ denotes the category of finite pointed sets with epimorphisms). You cannot extend it to a $\Gamma$-category if your symmetric monoidal category doesn't have a unit (if you could, this would give you a unit). When you apply nerve you get a $\Gamma_{epi}$-space and this is probably the same as a non unital $E_{\infty}$-space (i.e. a space over the operad $E_{\infty}$ where you forgot about the $0$-th space).

However there is a result of Lurie in Higher Algebra that says that homotopy units on non unital $E_{\infty}$-spaces can be strictified so that the result is a unital $E_{\infty}$-space. You seem to have a homotopy unit in your category so this result might be relevant.

Does the category of topological symmetric spectra satisfy the monoid axiom ?

2010-11-09T18:20:12Z

In the paper "Symmetric spectra"by Hovey, Smith and Shipley, they say that they don't know if the monoid axiom holds for topological symmetric spectra. This paper was written in 1998 so I am wondering what is the situation on this question today.

Comment by Geoffroy Horel

2013-04-27T01:24:04Z

The geometric realization of a simplicial set is a CW-complex so it is certainly compactly generated. But you don't even need that. It's perfectly fine to define homology of a simplicial set by applying the free abelian group functor degreewise and then take the alternating sum of the faces as differential and take homology of the resulting chain complex.

Comment by Geoffroy Horel

2013-04-17T18:11:22Z

If $E$ is an infinite dimensional vector space, a finite configuration of points $X$ of $E$ is contained in a finite dimensional subspace $V$. You can apply the Mayer Vietoris long exact sequence for the inclusion $V-X\to E$. It tells you that $E-X$ has the same connectivity as $V-X$ i.e. $dim(V)-1$. Since you can do that for arbitrarily large $V$'s you can show that $E-X$ is contractible. If $Gvect(f)$ is homeomorphic to an infinite dimensional vector space then you are fine but I'm not sure that it is the case.

Comment by Geoffroy Horel

2013-04-17T13:44:36Z

Is GVect(f) infinite dimensional ? If this is the case, you are right. If not, I think claim 2 is not true.

Comment by Geoffroy Horel

2013-04-14T15:26:19Z

@Dan Petersen: You're right. Sorry for the confusion.

Comment by Geoffroy Horel

2013-04-14T01:10:05Z

I think the paper of Bodigheimer and Cohen. ``Rational cohomology of configuration spaces of surfaces'' answers that question.

Comment by Geoffroy Horel

2013-04-01T18:32:04Z

Done. Let me know if you would like to see more details.

Comment by Geoffroy Horel

2013-03-31T00:26:36Z

$BS$ acts on $BX$, therefore you have an action map $H_0(BS)\otimes H_p(BX)\to H_p(BX)$. $H_0(BS)=\mathbb{Z}[\pi_0(BS)]$, therefore you have an action of the group algebra $\mathbb{Z}[\pi_0(BS)]$ on $H_p(BX)$ or equivalently an action of the group $\pi_0(S)$ on $H_p(BX)$.

Comment by Geoffroy Horel

2013-02-21T22:52:39Z

Thanks for a great answer Ricardo. I have accepted Ryan's answer but it's nice to have details.

Comment by Geoffroy Horel

2013-02-21T22:50:10Z

Great answer, thanks !

Comment by Geoffroy Horel

2013-02-19T00:31:27Z

You might want to look at this paper : arxiv.org/abs/1104.0181 by Jon Francis. He shows that Hochschild cohomology of an e_n algebra is the Lie algebra of some derived algebraic group. Of course this doesn't really answer the question.

Comment by Geoffroy Horel

2013-02-03T17:54:50Z

Thanks Ricardo, this answers my question.

Comment by Geoffroy Horel

2012-10-12T00:18:42Z

Ryan. I'm not sure I have some flexibility on what topology I can use. I have edited the question to take your comment into account.

Comment by Geoffroy Horel

2012-03-03T02:31:48Z

What you do is adding a disjoint unit or equivalently writing you non unital monoid as the augmentation kernel of an augmented monoid. I can tell you what happens for commutative algebras in the category of chain complexes. I don't know how relevant it is to this situation. So if A is an augmented algebra and I is its augmentation ideal, then the iterated bar $B^n(A)$ is equivalent to $k\oplus B^n(I)[n]$. You can find this result in a paper of Po Hu.

Comment by Geoffroy Horel

2011-11-27T20:11:29Z

I'm French. I believe your translation is correct. I don't know why the term "produit direct" is not used (maybe just to avoid monotony) but the adjective direct obviously modifies the noun "compose" exactly as in "direct product" in english. I guess what is meant should be clear if you look at the proof of the lemma.

Comment by Geoffroy Horel

2011-11-16T21:23:44Z

If $E$ is an $E_\infinity$ ring spectrum (actually $E_2$ is enough) then the $E$-homology of a 2-fold loop space is a commutative $E_*$-algebra. This is merely a stable version of the facts you mention so this is probably not an acceptable answer.