(n+1,r+1)-Theta space of (n,r)-Theta spaces? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:04:18Z http://mathoverflow.net/feeds/question/5867 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5867/n1-r1-theta-space-of-n-r-theta-spaces (n+1,r+1)-Theta space of (n,r)-Theta spaces? Urs Schreiber 2009-11-17T20:59:27Z 2009-11-17T23:54:50Z <p>I started writing <a href="http://ncatlab.org/nlab/show/Theta+space" rel="nofollow">nLab:Theta space</a>. Not done yet, but while I am working on it:</p> <p>is there a good proposal for what the "$(n+1,r+1)$-$\Theta$-space of all $(n,r)$-$\Theta$-spaces" would be?</p> http://mathoverflow.net/questions/5867/n1-r1-theta-space-of-n-r-theta-spaces/5886#5886 Answer by Charles Rezk for (n+1,r+1)-Theta space of (n,r)-Theta spaces? Charles Rezk 2009-11-17T23:54:50Z 2009-11-17T23:54:50Z <p>Let me assume $n=\infty$, to make things easier to write, so "$(\infty,r)$-$\Theta$-space" equals "$r$-$\Theta$-space".</p> <p>The totality of $r$-$\Theta$-spaces forms a (large) category enriched over $r$-$\Theta$-spaces, which I'll call $C$. Given this, you can form a presheaf of spaces $X$ on the category $\Theta_{r+1}$, where</p> <p>$$X[0] = \text{class of objects of C_r},$$</p> <p>and </p> <p>$$X([m](\theta_1,\dots,\theta_m)) = \coprod_{a_0,\dots,a_m} C(a_0,a_1)(\theta_1)\times \cdots \times C(a_{m-1},a_m)(\theta_m).$$</p> <p>Here "$[m](\theta_1,\dots,\theta_m)$" represents a typical object in $\Theta_{r+1}$ (so each $\theta_i\in \Theta_r$). The coproduct is over tuples of objects of $C$. The structure maps in the presheaf use the fact that $C$ is a category object. (It's like the way you get a Segal category from a category enriched over spaces.)</p> <p>The gadget $X$ is <em>almost</em> an $(r+1)$-$\Theta$-space. It satisfies all the "Segal" conditions, and also all the completeness conditions <em>except</em> for the one in bottom dimension. You get an honest $(r+1)$-$\Theta$-space $X'$ from $X$ by applying a suitable localization. </p> <p>The gadget $X'$ should be the thing you want. (None of the proofs involved here have been written up, or at least not by me, though we're working on it.)</p>