(n+1,r+1)-Theta space of (n,r)-Theta spaces?

2009-11-17T20:59:27Z

I started writing nLab:Theta space. Not done yet, but while I am working on it:

is there a good proposal for what the "$(n+1,r+1)$-$\Theta$-space of all $(n,r)$-$\Theta$-spaces" would be?

Answer by Charles Rezk for (n+1,r+1)-Theta space of (n,r)-Theta spaces?

2009-11-17T23:54:50Z

Let me assume $n=\infty$, to make things easier to write, so "$(\infty,r)$-$\Theta$-space" equals "$r$-$\Theta$-space".

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

$$X[0] = \text{class of objects of $C_r$},$$

and

$$X([m](\theta_1,\dots,\theta_m)) = \coprod_{a_0,\dots,a_m} C(a_0,a_1)(\theta_1)\times \cdots \times C(a_{m-1},a_m)(\theta_m).$$

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.)

The gadget $X$ is almost an $(r+1)$-$\Theta$-space. It satisfies all the "Segal" conditions, and also all the completeness conditions except for the one in bottom dimension. You get an honest $(r+1)$-$\Theta$-space $X'$ from $X$ by applying a suitable localization.

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.)

(n+1,r+1)-Theta space of (n,r)-Theta spaces? - MathOverflow

(n+1,r+1)-Theta space of (n,r)-Theta spaces?

Answer by Charles Rezk for (n+1,r+1)-Theta space of (n,r)-Theta spaces?