# In the context of free or faithful group actions, what is the stabilizer subgroup when the set is empty? [duplicate]

My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

I refer to Section 27.1.

Let $$M$$ be a set, possibly empty. Let $$G$$ be a group, possibly a singleton. Let $$G$$ act right on $$M$$ by the action $$\mu: M \times G \to M$$. For each $$x \in M$$, let $$\text{Stab}(x):=\{g \in G | \mu(x,g) = x\}$$ denote a stabilizer subgroup of $$G$$. Let $$1_G$$ be the identity of $$G$$.

I understand definitions of $$\mu$$ to be free as follows:

• Wikipedia: $$\mu$$ is free if for all $$g \in G$$, if there exists $$x \in M$$ such that $$\mu(x,g)=x$$, then we have that $$g=1_G$$.

• jgon in this answer: (same as Wikipedia's, given above)

• Section 27.1: $$\mu$$ is free if for all $$x \in M$$, $$\text{Stab}(x) = \{1_G\}$$

Question 1: For Wikipedia's and jgon's definitions, there is no explicit reference to stabilizers. For Tu's definition, how do I understand $$\text{Stab}(x)$$ for $$M$$ empty and $$G$$ not a singleton?

Question 2: Similarly, for the definition of faithful as

$$\bigcap_{x \in M} \text{Stab}(x) = \{1_G\} \tag{2a}$$

How do I understand $$\mu$$ as never faithful for $$M$$ empty and $$G$$ not a singleton?

My attempt to understand:

• For Question 2, I think I can apply this, by $$M$$ empty assumption to say $$\bigcap_{x \in M} \text{Stab}(x) = G$$. Then I apply $$G$$ not a singleton assumption to get $$\bigcap_{x \in M} \text{Stab}(x) \ne \{1_G\}$$.

• For Question 1, I think we somehow say $$\text{Stab}(x) = G$$ for all $$x \in M = \emptyset$$ by some vacuousness argument. I'm not really sure.

• $\operatorname{Stab}$ is a function from $M$ to subgroups of $G$. If $M$ is empty, the only function is the empty function. For the empty function its name $\operatorname{Stab}$ makes sense, but there are no values $\operatorname{Stab}(x)$ for us to talk about. – conditionalMethod Oct 14 at 14:44
• There are no $x$ in $M$ for which to write $\operatorname{Stab}(x)$. – conditionalMethod Oct 14 at 14:48
• I don't know the intentions of the book. In principle they can talk about $\operatorname{Stab}$ for $M$ empty. It is just the empty function. There is no problem with that. All you said above looks correct, except for $\operatorname{Stab}(x)=G$. there is no $\operatorname{Stab}(x)$ to be computed, and therefore no equation $\operatorname{Stab}(x)=G$ to write. However, the proposition $\forall x\in M,\ \operatorname{Stab}(x)=G$ is true, but so is any other like $\forall x\in M,\ \operatorname{Stab}(x)=\operatorname{****}$. – conditionalMethod Oct 14 at 14:54
• If $M$ is empty, then $\bigcap_{x\in M}whatever$ is a problematic expression. – Hagen von Eitzen Oct 14 at 15:18
• @HagenvonEitzen If $M$ is empty, then whatever(x) for $x \in M$ may or may not be problematic (this is the problem of Question 1). However, based on this, I think $\bigcap_{x \in M} whatever(x)$ is actually not problematic (this is the problem of Question 2). Do I misunderstand? – user636532 Oct 14 at 15:21

$$G$$ acts on $$M$$ faithfully iff the induced homomorphism $$\psi$$ from $$G$$ to the symmetric group on $$M$$ is injective. If $$M$$ is empty then the kernel of this homomorphism is all of $$G$$ (because the symmetric group on the empty set is trivial), so the action is faithful iff $$G$$ is trivial.
$$G$$ acts on $$M$$ freely iff every stabilizer is trivial. If $$M$$ is empty, there are no stabilizers to speak of, so the action is free.