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Let $G$ be a topological group and $(X,\mu)$ be a $G$-set, i.e. $\mu$ defines an action $X \times G \rightarrow X$.

Is it then true that $\mu$ is continuous if and only if for every $x \in X$ the stabilizer subgroup $G_x$ is open in $G$?

If yes, how does one prove this?

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Dear Chuck: Let $\mathbb R$ act on itself by translations. – Pierre-Yves Gaillard Jan 29 '12 at 6:28
@Pierre: OK I had a similar counterexample in mind, but this is much simpler; could you then tell me what MacLane/Moerdijk are on about in Sheaves in Geometry and Logic, exercise I.6.(a) where they ask for a proof of this result? – Chuck Jan 29 '12 at 6:51
I'd use the word "every", in "for every $x\in X$...". The word "any" is ambiguous and best avoided in this sort of context. If you write "Prove that any $x\in X$ is purple", it could reasonably mean "Pick any $x\in X$ and prove that it's purple", or it could mean "Prove that it is the case that any $x\in X$, no matter which one you pick, is purple." – Michael Hardy Jan 29 '12 at 18:01
@Michael - Point taken, edited accordingly – Chuck Jan 29 '12 at 19:56
up vote 10 down vote accepted

Neither implication is true.

1) Let $X$ be a topological space admitting a discontinuous bijection $f: X \rightarrow X$. (E.g. we may take $X = \mathbb{R}$.) Let $G = \mathbb{Z}$ with the discrete topology. There is a unique set-theoretic action of $G$ on $X$ such that $1 \cdot x = f(x)$. Because $1 \cdot = f$ is discontinuous, the action is not continuous. However, since $\mathbb{Z}$ is discrete, all subgroups are open. In particular the point stabilizers $G_x$ are open.

2) Let $G$ be a nondiscrete topological group [Pierre-Yves Gaillard suggests $\mathbb{R}$ in his comment above], and view the group law $G \times G \rightarrow G$ as a left action of $G$ on itself. By definition this action is continuous, but the point stabilizers are $\{e\}$, which is not open since $G$ is not discrete.

Added: Wait! Upon closer reading, it was not explicitly said that $X$ is a topological space, only a $G$-set. In fact, if we view $X$ as a "naked set" and endow it with the discrete topology, then the equivalence becomes true.

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As I said in my comment above, it (at least seems to me) to be asked in Maclane-Moerdijk Ex. I.6.(a) and I was going crazy trying to figure out how to interpret the question differently so craziness compelled me to ask just in case. That said, I still don't know what to make of the exercise... – Chuck Jan 29 '12 at 7:08
The exercise says (almost word-for-word): Let $G$ be a topological group and $\textbf{B}G$ the category of continuous $G$-sets. Let $G^{\delta}$ be the same group with the discrete topology and let $i_G: \textbf{B}G \rightarrow \textbf{B}G^{\delta}$ be the inclusion functor. [what follows is exactly word-for-word] (a) Prove that a $G$-set $(X,\mu)$ is in the image of $i_G$, i.e. that $\mu$ is continuous, iff for each $x \in X$ its isotropy group $I_x = \lbrace g \in G \vert x g = x \rbrace$ is an open subgroup of $G$. – Chuck Jan 29 '12 at 7:21
Oh and thanks for the detailed answer, much appreciated – Chuck Jan 29 '12 at 7:22
@Chuck: okay, based on this it seems my guess is correct. Namely, we are speaking of categories of $G$-sets, not $G$-spaces. (One might well be forgiven for supposing that a "continuous $G$-set" means a topological space together with a $G$-action, but based on the context it seems that it does not mean this!) I suppose this means that we endow the sets with the discrete topology. – Pete L. Clark Jan 29 '12 at 7:23
I see - I was thinking of $X$ as a $G$-space too...I guess this makes sense in the context of the exercise too because we want to think of $\textbf{B}G$ as a subcategory of the $G$-set category and not of $\textbf{Top}$. – Chuck Jan 29 '12 at 7:28

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