Free loop space objects and actions

The free loop space object of an object $X$ in an $(\infty,1)$-category $\mathcal{C}$ can be defined as the pullback $\mathcal{L}X= X\times_{X\times X} X$. Unlike the based loop space, this is not generally a group object in $\mathcal{C}$: there is no way to compose loops with different basepoints. However, if we regard $\mathcal{L}X$ as an object of the slice $\mathcal{C}/X$ (via either projection; they are equivalent), then as sketched here, it should be a group object in $\mathcal{C}/X$ (which "puts together" all the based loop spaces over all possible basepoints), and moreover it should act canonically on all objects of $\mathcal{C}/X$ by "transport".

Where can I find this group structure and action defined formally and coherently?

• Consider the map $\ast\coprod\ast\rightarrow \ast$ in pointed spaces (choose an arbitrary pointing on the left). Its Cech conerve is a cogroupoid $\ast\rightrightarrows S^1 ...$ which gives $S^1$ the structure of a cogroup in pointed spaces. Therefore, if $\mathcal{C}$ is cotensored over spaces, $\mathrm{Map}(S^1, X)\rightarrow \mathrm{Map}(\ast, X)$ is a group. Commented Sep 20, 2017 at 8:00
• @PavelSafronov Can you supply a citation, and can you define the action as well? Commented Sep 20, 2017 at 8:08
• Let $\mathcal{D}$ be an $\infty$-category with finite limits. Given a pointed object $A$, $\Omega A$ is a group. Given a morphism $B\rightarrow A$, $\Omega A$ acts on $B\times_A \ast$. You can apply this to $\mathcal{D}=\mathcal{C}_{/X}$, $A=(p_1\colon X\times X\rightarrow X)$ pointed by the diagonal and $B=(p_1\colon X\times Y\rightarrow X)$ with $(\mathrm{id}\times f)\colon X\times Y\rightarrow X\times X$. Unfortunately, I don't know a reference and this might be folklore. Commented Sep 20, 2017 at 9:13

Suppose $\mathcal{D}$ is an $\infty$-category with finite limits. Given a pointed object $\ast\rightarrow A$, its Cech nerve is a simplicial object $M_\bullet$. By Higher Topos Theory, Proposition 6.1.2.11, it is actually a groupoid and, moreover, since $M_0\cong \ast$, it is a group. This gives $M_1\cong \Omega A$ the structure of a group.

Given a morphism $B\rightarrow A$ consider the pullback diagram

$$\require{AMScd} \begin{CD} B\times_A \ast @>>> B\\ @VVV @VVV \\ \ast @>>> A \end{CD}$$

Taking the Cech nerves horizontally you get a morphism of simplicial objects $M'_\bullet\rightarrow M_\bullet$. Using that $M'_\bullet$ is a groupoid, it is easy to see that $M'_\bullet\rightarrow M_\bullet$ is a left action object in the sense of Higher Algebra, Definition 4.2.2.2. So, the group $M_1\cong \Omega A$ acts on $M'_0\cong B\times_A \ast$.

Let's restrict to the case $\mathcal{D} = \mathcal{C}_{/X}$. Consider the object $A=(p_1\colon X\times X\rightarrow X)$. The final object in $\mathcal{C}_{/X}$ is $\mathrm{id}\colon X\rightarrow X$ and $A$ has a pointing by the diagonal map $\Delta\colon X\rightarrow X\times X$. Its based loop space is $\Omega_\Delta A = (\mathcal{L} X\rightarrow X)$.

Note that using that $\mathcal{C}$ is cotensored over spaces, you can obtain the Cech nerve of $X\rightarrow X\times X$ in $\mathcal{C}_{/X}$ by applying $\mathrm{Map}(-, X)$ to the Cech conerve of $S^0\rightarrow \ast$ in pointed spaces which gives $\Sigma S^0\cong S^1$ the structure of a cogroup in pointed spaces.

Given a morphism $f\colon Y\rightarrow X$ in $\mathcal{C}$, let $B = (p_1\colon X\times Y\rightarrow X)$ with $B\rightarrow A$ given by $(\mathrm{id}\times f)\colon X\times Y\rightarrow X\times X$. Then $B\times_A \ast = (X\times Y)\times_{X\times X} X\cong Y$ which by above gives an action of $\mathcal{L} X\rightarrow X$ on $Y\rightarrow X$.

Note that you could instead consider $\tilde{B} = (p_1\circ(f\times \mathrm{id})\colon Y\times X\rightarrow X)$. This will give the trivial action of $\mathcal{L} X\rightarrow X$ on $Y\rightarrow X$.

In algebraic geometry you can think of it as follows. Let's say $X$ is a (derived) scheme. $\widehat{X\times X} = [X/\mathcal{L} X]$ is the classifying space of the group scheme $\mathcal{L}X\rightarrow X$ and $\widehat{X\times Y} = [Y/\mathcal{L} X]$.