1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ $\operatorname{Stab}$ 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. $\endgroup$ – conditionalMethod Oct 14 at 14:44
  • 2
    $\begingroup$ There are no $x$ in $M$ for which to write $\operatorname{Stab}(x)$. $\endgroup$ – conditionalMethod Oct 14 at 14:48
  • 1
    $\begingroup$ 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 $\forall x\in M,\ \operatorname{Stab}(x)=\operatorname{****}$. $\endgroup$ – conditionalMethod Oct 14 at 14:54
  • 1
    $\begingroup$ If $M$ is empty, then $\bigcap_{x\in M}whatever$ is a problematic expression. $\endgroup$ – Hagen von Eitzen Oct 14 at 15:18
  • 1
    $\begingroup$ @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? $\endgroup$ – user636532 Oct 14 at 15:21
2
$\begingroup$

$G$ 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.

$\endgroup$
  • 1
    $\begingroup$ Thanks Unit! You actually gave a partial answer to this question. Thanks again! $\endgroup$ – user636532 Oct 14 at 15:29