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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.

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    $\operatorname{Stab}$ 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 '19 at 14:44
  • @conditionalMethod Can we or can we not define $\text{Stab}(x)$ for $M$ empty? I was thinking Tu's definition is assuming $M$ is non-empty. I mean, empty set is never a group or subgroup. –  Oct 14 '19 at 14:47
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    There are no $x$ in $M$ for which to write $\operatorname{Stab}(x)$. – conditionalMethod Oct 14 '19 at 14:48
  • @conditionalMethod So Tu's definition must assume $M$ non-empty? –  Oct 14 '19 at 14:49
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    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 '19 at 14:54
  • @conditionalMethod Thanks. Actually, to clarify, I edited my last comment from "assuming" to "must assume". Then, my question in the last comment is asking about not the intentions (as indicated by the word "assuming") but rather what must be the case for any author of any text to discuss the term $\text{Stab}(x)$ for $M$ non-empty. Anyway, if your comment is still the same despite this clarification: I think if I am correct here, then there's a contradiction with this because... –  Oct 14 '19 at 15:09
  • @conditionalMethod ...oh wait I just saw your edit. I took it to mean that $\text{Stab}(x) = \emptyset$...Never mind I'll ask later: After this is answered, I'll ask it again under Tu's definition of free. –  Oct 14 '19 at 15:10
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    If $M$ is empty, then $\bigcap_{x\in M}whatever$ is a problematic expression. – Hagen von Eitzen Oct 14 '19 at 15:18
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    @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? –  Oct 14 '19 at 15:21
  • You've asked a number of very similar questions back to back, all of which can be resolved just by understanding the notion of vacuous implication. If $M$ is empty, then every group action on $M$ is free, both by wikipedia's definition and Tu's definition, since there are no $m\in M$ to check. Hagen is correct that $\bigcap_{x\in \emptyset}$ is problematic in general - but in the context of subsets of a fixed set $G$, the intersection of an empty collection of subsets of $G$ is $G$ itself. – Alex Kruckman Oct 14 '19 at 17:44

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$G$ 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.

Unit
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