most recent 30 from http://mathoverflow.net 2013-05-19T02:53:43Z http://mathoverflow.net/feeds/user/10201 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122617/non-sigma-e-n-algebras Justin Young 2013-02-22T09:07:00Z 2013-02-22T15:44:48Z

<p>Any symmetric operad can be considered as an non-$\Sigma$ operad by throwing away permutations. Does anyone know what sort of structure one gets for algebras over $C_n$ the little n-cubes operad, or some equivalent variant, considered in this way? One thing which is strange is that there is a non-$\Sigma$ operad splitting of the map $Ass \to Com$ (where $Ass$ and $Com$ are the symmetric associative and commutative operads, respectively), and so any non-$\Sigma$ $C_n$ algebra should be a loop space, but I do not see what, if any, extra structure is present.</p>

http://mathoverflow.net/questions/108535/does-the-group-completion-theorem-apply-to-the-james-construction Justin Young 2012-10-01T09:52:02Z 2012-10-01T09:52:02Z

<p>In other words, is the natural map $M \to \Omega B M$, for $M=JX$ the James construction on a space, a group completion? (By "group completion" I mean at the level of homology, I am aware of the space level version.) The versions of the group completion theorem that I have found such as Segal/McDuff have conditions on the monoid $M$, involving some kind of commutativity condition, and it is not clear to me that $JX$ satisfies any of these conditions. Could someone provide a reference or statement of a group completion theorem that clearly applies to $JX$, or is there actually a counterexample?</p>

http://mathoverflow.net/questions/49690/what-do-cohomology-operations-have-to-do-with-the-non-existence-of-commutative-co/107767#107767 Justin Young 2012-09-21T12:26:38Z 2012-09-21T12:26:38Z

<p>One further answer to this is in Mandell's paper "Cochain Multiplications". We can view your hypothetical commutative cochains $C_1^*(X)$ as a functor landing in $E_\infty$ algebras over $\mathbb Z$. In that case, if the functor satisfies a reasonable list of axioms then it follows that it is naturally equivalent to the usual cochain functor. It follows from that that it has non-trivial Steenrod operations on $\mathbb Z/p$ cohomology, and that contradicts commutativity, as the Steenrod operations on a strictly commutative algebra are almost all zero.</p>

http://mathoverflow.net/questions/107175/homotopy-groups-and-homology-groups-for-the-h-mathbb-z-module-dg-module-corresp Justin Young 2012-09-14T13:03:18Z 2012-09-14T17:34:14Z

<p>In Shipley's paper <a href="http://arxiv.org/abs/math/0209215" rel="nofollow">http://arxiv.org/abs/math/0209215</a> she proves a Quillen equivalence between the category of $H\mathbb Z$-modules and dg $\mathbb Z$-modules. So, to a chain complex $C$, she assigns a spectrum $HC$, but is it true in some sense that $\pi_i HC \cong H_i C$? I am aware there are some subtleties involved with homotopy groups of symmetric spectra, but I'm not sure how that plays out in this case.</p>

http://mathoverflow.net/questions/129547/seeking-errata-for-berger-moerdijk-axiomatic-homotopy-theory-for-operads Justin Young 2013-05-06T11:51:53Z 2013-05-06T11:51:53Z

By studying free algebras, one can prove using direct point set arguments, and knowledge of how to build pushouts in $\mathcal P$ algebras from pushouts in pointed spaces, to prove that the pushout in $\mathcal P$ algebras of the inclusion $\mathbb P(S^n_+) \to \mathbb P(D^{n+1}_+)$ along a map $\mathbb P(S^n_+) \to X$ gives a closed inclusion $X \to Y$. This is enough to get the small object argument working in pointed spaces, so one can do a transfer.

http://mathoverflow.net/questions/122672/non-commutativity-of-certain-hopf-spaces/122679#122679 Justin Young 2013-02-24T09:41:28Z 2013-02-24T09:41:28Z

$R$ is an arbitrary ring, but $R=\mathbb R$ suffices as does $R= \mathbb Z$. And yes, the James construction works if you already know about it (which you do).

http://mathoverflow.net/questions/122672/non-commutativity-of-certain-hopf-spaces/122679#122679 Justin Young 2013-02-23T07:25:02Z 2013-02-23T07:25:02Z

Edit: in the above I should specify that tensor algebras are not commutative provided the generators are more than one dimensional, or in odd degree. Remember, to be commutative, odd degree elements must satisfy $2x^2 = 0$.

http://mathoverflow.net/questions/122672/non-commutativity-of-certain-hopf-spaces/122679#122679 Justin Young 2013-02-23T07:15:18Z 2013-02-23T07:15:18Z

In general, if $\tilde H_*(X, R)$ is free then $H_*(\Omega \Sigma X, R)$ is a tensor algebra on $\tilde H_*(X,R)$, and tensor algebras are not commutative. You can prove this using your Serre SS. With $X = S^1$ and $X = S^1 \vee S^1$ you get your result. Hi Prasit!

http://mathoverflow.net/questions/122617/non-sigma-e-n-algebras/122641#122641 Justin Young 2013-02-23T06:54:00Z 2013-02-23T06:54:00Z

Thank you for responding. I do not have any examples in mind other than the obvious free algebras. This question was put to me and I thought it was surprising and interesting, and I had never thought about it, either. I thought that there should be a simple answer, but I was unable to come up with one quickly myself or to find anything in the literature about it.

http://mathoverflow.net/questions/122617/non-sigma-e-n-algebras Justin Young 2013-02-22T15:28:14Z 2013-02-22T15:28:14Z

My ideal answer would be something analogous to the characterization of symmetric $C_n$ algebras as $n$-fold loop spaces. Perhaps a nice description of the free algebras. It is clear you have a loop space for all $n$, and then at $C_\infty$ you get a loop space again, but does anything interesting happen in the middle?

http://mathoverflow.net/questions/108535/does-the-group-completion-theorem-apply-to-the-james-construction Justin Young 2012-10-01T13:47:21Z 2012-10-01T13:47:21Z

I am particularly interested in the case when $X$ is not connected, otherwise there are many proofs out there. For discrete spaces, it seems to be true by direct calculation.