User geoffroy horel - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T08:42:52Z http://mathoverflow.net/feeds/user/10707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125966/path-components-of-a-monoidal-category-acting-on-homology/126131#126131 Answer by Geoffroy Horel for Path components of a monoidal category acting on homology Geoffroy Horel 2013-03-31T19:53:13Z 2013-04-01T18:31:09Z <p>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.</p> <p>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 $H_p(BX)$.</p> http://mathoverflow.net/questions/122557/what-manifolds-are-boundaries-of-euclidian-spaces What manifolds are boundaries of euclidian spaces ? Geoffroy Horel 2013-02-21T17:42:31Z 2013-03-29T23:55:24Z <p>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:</p> <ul> <li><p>$\partial M\cong N$</p></li> <li><p>$M-\partial M\cong \mathbb{R}^n$ </p></li> </ul> <p>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.</p> http://mathoverflow.net/questions/120395/monoidal-model-category-structure-on-a-functor-category Monoidal model category structure on a functor category. Geoffroy Horel 2013-01-31T10:13:08Z 2013-02-01T04:18:53Z <p>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.</p> <p>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: $$F\otimes G(a)=A(-\circ -,a)\otimes_{A\times A}F(-)\times G(-)$$</p> <p>My question is:</p> <p>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 ?</p> http://mathoverflow.net/questions/109422/contractibility-of-the-space-of-collars Contractibility of the space of collars Geoffroy Horel 2012-10-11T23:23:49Z 2012-10-12T01:08:38Z <p>I'm looking for a reference or a proof of the following statement :</p> <p>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.</p> http://mathoverflow.net/questions/106249/two-commuting-operad-actions/107854#107854 Answer by Geoffroy Horel for Two commuting operad actions Geoffroy Horel 2012-09-22T17:10:02Z 2012-09-22T19:35:35Z <p>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).</p> <p>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 :</p> <p>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.</p> <p>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.</p> <p>The following paper <a href="http://arxiv.org/abs/math.AT/0410367" rel="nofollow">http://arxiv.org/abs/math.AT/0410367</a> does something similar to show that $THH$ of an $E_2$-algebra is an $A_{\infty}$-algebra.</p> <p>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 <a href="http://arxiv.org/abs/1102.1311" rel="nofollow">http://arxiv.org/abs/1102.1311</a> talks about the BV tensor product of $A_\infty$ and $E_n$.</p> http://mathoverflow.net/questions/89935/gamma-spaces-and-monoidal-categories-ii/90047#90047 Answer by Geoffroy Horel for Gamma spaces and monoidal categories II Geoffroy Horel 2012-03-02T15:29:16Z 2012-03-02T15:29:16Z <p>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). </p> <p>However there is a result of Lurie in <em>Higher Algebra</em> 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.</p> http://mathoverflow.net/questions/45472/does-the-category-of-topological-symmetric-spectra-satisfy-the-monoid-axiom Does the category of topological symmetric spectra satisfy the monoid axiom ? Geoffroy Horel 2010-11-09T18:20:12Z 2010-12-04T16:44:10Z <p>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.</p> http://mathoverflow.net/questions/125966/path-components-of-a-monoidal-category-acting-on-homology/126131#126131 Comment by Geoffroy Horel Geoffroy Horel 2013-04-27T01:24:04Z 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. http://mathoverflow.net/questions/127843/contractibility-of-a-configuration-space Comment by Geoffroy Horel Geoffroy Horel 2013-04-17T18:11:22Z 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. http://mathoverflow.net/questions/127843/contractibility-of-a-configuration-space Comment by Geoffroy Horel Geoffroy Horel 2013-04-17T13:44:36Z 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. http://mathoverflow.net/questions/127495/cohomology-of-configuration-spaces Comment by Geoffroy Horel Geoffroy Horel 2013-04-14T15:26:19Z 2013-04-14T15:26:19Z @Dan Petersen: You're right. Sorry for the confusion. http://mathoverflow.net/questions/127495/cohomology-of-configuration-spaces Comment by Geoffroy Horel Geoffroy Horel 2013-04-14T01:10:05Z 2013-04-14T01:10:05Z I think the paper of Bodigheimer and Cohen. Rational cohomology of configuration spaces of surfaces'' answers that question. http://mathoverflow.net/questions/125966/path-components-of-a-monoidal-category-acting-on-homology/126131#126131 Comment by Geoffroy Horel Geoffroy Horel 2013-04-01T18:32:04Z 2013-04-01T18:32:04Z Done. Let me know if you would like to see more details. http://mathoverflow.net/questions/125966/path-components-of-a-monoidal-category-acting-on-homology Comment by Geoffroy Horel Geoffroy Horel 2013-03-31T00:26:36Z 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 $H_p(BX)$. http://mathoverflow.net/questions/122557/what-manifolds-are-boundaries-of-euclidian-spaces/122576#122576 Comment by Geoffroy Horel Geoffroy Horel 2013-02-21T22:52:39Z 2013-02-21T22:52:39Z Thanks for a great answer Ricardo. I have accepted Ryan's answer but it's nice to have details. http://mathoverflow.net/questions/122557/what-manifolds-are-boundaries-of-euclidian-spaces/122568#122568 Comment by Geoffroy Horel Geoffroy Horel 2013-02-21T22:50:10Z 2013-02-21T22:50:10Z Great answer, thanks ! http://mathoverflow.net/questions/121832/analogue-of-cyclic-homology-for-e-n-algebras Comment by Geoffroy Horel Geoffroy Horel 2013-02-19T00:31:27Z 2013-02-19T00:31:27Z You might want to look at this paper : <a href="http://arxiv.org/abs/1104.0181" rel="nofollow">arxiv.org/abs/1104.0181</a> 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. http://mathoverflow.net/questions/120395/monoidal-model-category-structure-on-a-functor-category/120487#120487 Comment by Geoffroy Horel Geoffroy Horel 2013-02-03T17:54:50Z 2013-02-03T17:54:50Z Thanks Ricardo, this answers my question. http://mathoverflow.net/questions/109422/contractibility-of-the-space-of-collars Comment by Geoffroy Horel Geoffroy Horel 2012-10-12T00:18:42Z 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. http://mathoverflow.net/questions/89935/gamma-spaces-and-monoidal-categories-ii/90047#90047 Comment by Geoffroy Horel Geoffroy Horel 2012-03-03T02:31:48Z 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. http://mathoverflow.net/questions/82023/what-does-compose-direct-mean-in-mathematical-french Comment by Geoffroy Horel Geoffroy Horel 2011-11-27T20:11:29Z 2011-11-27T20:11:29Z I'm French. I believe your translation is correct. I don't know why the term &quot;produit direct&quot; is not used (maybe just to avoid monotony) but the adjective direct obviously modifies the noun &quot;compose&quot; exactly as in &quot;direct product&quot; in english. I guess what is meant should be clear if you look at the proof of the lemma. http://mathoverflow.net/questions/81090/applications-of-eckmann-hilton-argument-to-topology Comment by Geoffroy Horel Geoffroy Horel 2011-11-16T21:23:44Z 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.