# Group action on a condensed set and its orbit space

Let $$X$$ be a condensed set, and let $$G$$ be a (discrete) group. Suppose we have an action $$G$$ on $$X$$, which is a group morphism $$a:G \rightarrow \mathrm{Aut}(X)$$, where $$\mathrm{Aut}(X)$$ is the group of condensed isomorphisms. In this case, what would be an appropriate definition of the orbit space of $$a$$?

We may proceed as follows: For each profinite set $$S$$, we get an action $$a_S: G \rightarrow \mathrm{Aut}(X(S))$$, where $$g \mapsto a(g)_S$$, so we can define X/G as \begin{align*} (X/G)(S) = X(S)/G. \end{align*} Where $$X(S)/G$$ is the orbit space of $$a_S$$. This defines a presheaf on the category of profinite sets, but I think it might not be a condensed set without additional hypotheses.

$$\textbf{Question 0:}$$ Is $$X/G$$ a condensed set? (It is not, see Edit.)

$$\textbf{Question 1:}$$ If we have additional hypotheses, can we turn $$X/G$$ into a condensed set?

$$\textbf{Question 2: (Main Question)}$$ Is $$X/G$$ the appropriate definition of the orbit space of a condensed set?

$$\textit{Edit:}$$ I have just checked that even in very good cases, this definition of orbit space is not a condensed set. To see that, take $$S^1 = \mathbb{R}/\mathbb{Z}$$, $$\alpha$$ an irrational real number, and for each $$n \in \mathbb{Z}$$ and each profinite set $$S$$, we have \begin{align*} n: C(S, S^1) &\longrightarrow C(S, S^1) \\ f &\longmapsto f + n \alpha \end{align*} This defines an action in the condensed setting. Let $$Y = S^1/\mathbb{Z}$$. It is not true that $$Y(S \sqcup T) = Y(S) \times Y(T)$$. To see this, start with a map $$f: S \sqcup T \rightarrow S^1$$ and define $$g(s) = f(s) + n\alpha$$ for $$s \in S$$ and $$g(t) = f(t) + m\alpha$$ for $$t \in T$$ with $$n \neq m$$. This way, $$f \neq g$$ in $$Y(S \sqcup T)$$ but the map $$\phi: Y(S \sqcup T) \rightarrow Y(S) \times Y(T)$$ sends $$f$$ and $$g$$ to the same element. Thus, $$\phi$$ is not bijective, concluding the claim.

• Also an important question: when you sheafify, will that be a categorical quotient in the category of condensed sets?
– user473423
Commented Mar 26, 2023 at 6:34
• @Echo, thanks for your comment! I wasn't familiar with the definition of categorical quotient. I was looking for a definition like this! Commented Mar 26, 2023 at 13:53

The answer is no (in general) for question 0 and no for question 2. In the edit of the question I have explained why $$S \mapsto X(S)/G$$ is not a condensed set in general.
In the following sense, the right generalization of the orbit space seems to be the categorical quotient (thanks to Echo's comment, I got to know this definition), and the categorical quotient of an action of $$G$$ on $$X$$ is the sheafification of $$Y:S \mapsto X(S)/G$$.
To see this, let's start by denoting $$Y^+$$ as the sheafification of $$Y$$. Let $$\pi: X \rightarrow Y$$ and $$i: Y \rightarrow Y^+$$ be the projection and the canonical morphism, respectively. For every $$g \in G$$, it is immediate that \begin{align*} i \circ \pi \circ g = i \circ \pi \end{align*} Now, let $$Z$$ be a condensed set with a map $$q:X \rightarrow Z$$, such that for each $$g \in G$$, we have \begin{align*} q \circ g = q \end{align*} Then, for each profinite set $$S$$, we have $$q_S \circ g = q_S$$. We define \begin{align*} \phi_S: Y(S) = X(S)/G &\longrightarrow Z(S) \\ \overline{x} &\longmapsto q_S (x) \end{align*} It is easy to check that $$\phi_S$$ is well-defined and that it defines a natural transformation $$\phi:Y \rightarrow Z$$. Also, note that $$\phi$$ is the only option to commute this diagram. Therefore, by the universal property of the sheafification, there is a unique map $$Y^+ \rightarrow Z$$ commuting the desired diagram.
• If $G$ is a (topological or discrete) group acting on a space $X$, wouldn't the condensed orbit space be $S \mapsto C(S,G)\backslash C(S,X)$? Commented Mar 29, 2023 at 14:03
• Prof @Linus, thank you for your comment! What do you mean by $C(S, G) \ C(S,X)$? Do you mean that $C(S,G)$ acts on $C(S,X)$ in some way? I don't see how to do it. Commented Mar 29, 2023 at 16:25
• Well, $C(S,G)$ is a group and it acts on $C(S,X)$ in a natural way: for $\gamma:S\to G$ and $\phi:S\to X$ put $(\gamma\cdot\phi)(s)=\gamma(s)\phi(s)$. Commented Mar 30, 2023 at 5:10