Delooping and unreduced operads - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T05:57:53Z http://mathoverflow.net/feeds/question/106336 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106336/delooping-and-unreduced-operads Delooping and unreduced operads Ulrich Pennig 2012-09-04T12:23:31Z 2012-09-04T14:14:22Z <p>Suppose I have an operad $P(n)$ in topological spaces in the sense of the book by Markl, Shnider and Stasheff, i.e. there is <em>no</em> space $P(0)$. In agreement with some of the answers to <a href="http://mathoverflow.net/questions/29636/operad-terminology-operads-with-and-without-o0" rel="nofollow"> this</a> question, I will call such an operad <em>unreduced</em>. </p> <p>Let $H$ be an infinite dimensional separable Hilbert space. Then an example for $P$ would be </p> <p>$$P(n) = Iso(H^{\otimes n}, H)$$</p> <p>the linear isometric isomorphisms from $H^{\otimes n}$ to $H$. There is no obvious candidate for $P(0)$ (since we insist on isomorphisms, $P(0) = Hom(\mathbb{C},H)$ will not work). </p> <p>It is still possible to talk about unreduced $E_{\infty}$-operads in this case. But now it is less clear (at least to me) that I can deloop a space, which is an algebra over an unreduced $E_{\infty}$-operad. Hence my question:</p> <blockquote> <p>Let $X$ be a space, which is an algebra over an unreduced $E_{\infty}$-operad $P$, such that $\pi_0(X)$ is a group. Is $X$ an infinite loop space?</p> </blockquote> <p>If this is not true:</p> <blockquote> <p>Are there conditions on the space or the operad (other than that it can be reduced) to ensure that $X$ is an infinite loop space?</p> </blockquote> <p>Using the bar construction by May, I think this would correspond to the question, whether I can drop degeneracies and still get a delooping. </p> http://mathoverflow.net/questions/106336/delooping-and-unreduced-operads/106342#106342 Answer by Peter May for Delooping and unreduced operads Peter May 2012-09-04T13:35:32Z 2012-09-04T13:35:32Z <p>Hey Ulrich, you sent me essentially that question by email, and I answered you just now. (Its earlier here, I just got up). So here publicly is the answer I sent you privately.</p> <p>I know the situation. Let me refer you to Section 7 of a recently posted paper by Guillou and myself where certain operads P, Q, and R are defined and related. The operad R is analogous to your example and embedded in operad Q. The operad Q allows monos as well as isos.</p> <p><a href="http://front.math.ucdavis.edu/1207.3459" rel="nofollow">http://front.math.ucdavis.edu/1207.3459</a></p> <p>If you can play a trick like that, and you have an analog of our Q that acts on your P-algebras, then you are fine. Otherwise I have no ideas.</p> <p>Dealing with isomorphisms intrinsically gets you into that unreduced situation, and then you have no unit elements in your P-algebras, which means you can't do infinite loop space theory. I would bet that there are operads like P, maybe your P itself, which act on spaces that even have non-Abelian fundamental groups. Just a hunch.</p> <p>Sorry not to be more helpful. Of course, I liked your question. </p> http://mathoverflow.net/questions/106336/delooping-and-unreduced-operads/106345#106345 Answer by Jacob Lurie for Delooping and unreduced operads Jacob Lurie 2012-09-04T14:08:17Z 2012-09-04T14:14:22Z <p>The answer to your first question is no: for example, take $X$ to be any connected pointed space, and regard $X$ as a nonunital commutative monoid by saying that the product of any two points of $X$ is equal to the base point. This endows $X$ with the structure of an algebra over any operad $\mathcal{O}$ with $\mathcal{O}(0) = \emptyset$, but $X$ need not be an infinite loop space.</p> <p>However, with a slightly stronger hypothesis, the answer is yes. Let $X$ be a nonunital $E_{\infty}$-space with the property that translation by any point $x \in X$ is a weak homotopy equivalence from $X$ to itself. Then $X$ is weakly homotopy equivalent to an infinite loop space.</p> <p>Assuming that your operads are sufficiently nice (so that "algebras up to coherent homotopy" can be rectified), one has the following more precise statement: the homotopy category of $E_{\infty}$-spaces is equivalent to the subcategory of the homotopy category of "nonunital" $E_{\infty}$-spaces, whose objects are nonunital $E_{\infty}$-spaces $X$ for which $\pi_0 X$ contains a unit element $e_X$ such that translation by (any representative of) $e_X$ is a weak homotopy equivalence, and whose morphisms are maps of nonunital $E_{\infty}$-spaces $X \rightarrow Y$ which carry $e_{X}$ to $e_{Y}$.</p>