5

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 (part 1), Section 27.1 (part 2) and Section 27.1 (part 3).

Firstly:

  1. I believe the book has no explicit definition for an action $\mu$ to be "transitive" and neither does Volume 1. I think this is okay for the book since Proposition 27.6 is not (explicitly) used later on in the book.

    • 1.1. If this wouldn't be okay for the book, then I would ask how, if possible, we could deduce from Tu's definition of principal $G$-bundle that from the action $\mu: P \times G \to P$, we get that $\mu(P_x \times G) \subseteq P_x$, where $P_x := \pi^{-1}(x)$, which is saying something like $\mu$ is fiber-preserving, such that we can define an action $\mu_x: P_x \times G \to P_x$ and then begin to discuss whether or not each $\mu_x$ is transitive.

    • 1.2 Even though I'm not asking (1.1), what I'm about to ask has a similar underlying problem.

  2. Anyway, I assume the definition that an action $\mu$ is "transitive" is the one here, assume that definition is equivalent to the one on Wikipedia and assume that that both definitions are equivalent to "for each $x \in M$, the map $\mu_x : G \to M, \mu_x(g) = \mu(x,g)$, is surjective, where $\mu: M \times G \to M$ is the right action of $G$ on $M$".

Now:

  1. Tu's definition of principal $G$-bundle doesn't say anything about transitive or fiber-preserving, but it may be equivalent to a definition with transitivity (see here). I mean that transitive or fiber-preserving could be somehow deduced from Tu's definition (as stated). Tu's definition is possibly the "Definition 3" in the previous link). I guess the alternative is that Tu made a mistake in the definition of principal $G$-bundle.

  2. I actually notice that for each $U \in \mathfrak U$, while we are given an explicit action $\sigma_U: U \times G \times G \to U \times G$, which is $\sigma_U((x,h),g)=(x,hg)$, we are not given an explicit definition of the action $\zeta_U: P_U \times G \to P_U$, where $P_U := \pi^{-1}(U)$.

    • 4.1. Edit: Oh wait that was kind of wrong. What I meant was to say that $\zeta_U$ is not even declared to exist in the first place. I really think the text is unclear here. I think the text should've said something like "$G$ acts on $U \times G$ (in the way of $\sigma_U$), and then $G$ acts on $\pi^{-1}(U)$ in such a way that $\phi_U$ is invariant". Otherwise, it seems kinda weird that you just say a map is equivariant even though you haven't declared the existence of an action on both domain and range. It just seems that somehow the action $\mu$ on $P$ induces $\zeta_U$'s.

      • 4.1.1. Edit: Probably, there should even be some prior proposition that starts with "given a map $f: N \to M$ and action $\zeta$ by $G$ on $N$ we can define an action $\sigma$ by $G$ on $M$" or that starts with "given a map $f: N \to M$ and action $\sigma$ by $G$ on $M$ we can define an action $\zeta$ by $G$ on $N$" and then the next part would be "that makes $f$ equivariant" and then there might be another proposition or some exercise that says that the defined $\zeta$ or $\sigma$ is unique. I'm thinking of something similar to the pullback metric, from earlier in the book.

      • 4.1.2. Edit: A comment of autodavid: In the definition of principal $G$-bundle, the way in which $G$ is acting on $P$ should make the trivialization maps $G$-equivariant when restricting to a trivialization patch..... Oh okay, there would be some problem because we don't know whether the restrictions are legal. I'm not an expert but I guess Tu implicitly requires that the restrictions to be legal, by talking about equivariance.

  3. I'm expecting something like, for the action $\mu: P \times G \to P$, we get that

    • 5.1. $\mu(P_x \times G) \subseteq P_x$ and $\mu(P_U \times G) \subseteq P_U$ such that we can define, respectively, maps $\mu_x: P_x \times G \to P_x$ and $\mu_U: P_U \times G \to P_U$. These turn out to be actions, probably smooth actions.

    • 5.2. Each $\mu_x$ in (5.1) is transitive. (Well, this is what Proposition 27.6 says.)

    • 5.3. $\zeta_U = \mu_U$: Each $\mu_U$ in (5.1) is the action $\zeta_U$ used to determine whether or not $\varphi_U$ is $G$-equivariant

Questions:

  1. Is this definition of principal $G$-bundle missing some details, such as any notion (explicit or implicit) of fiber-preservation of the action $\mu: P \times G \to P$ or any explicit description of the actions $\zeta_U: P_U \times G \to P_U$?

    • 1.1 Edit: Or any explicit mention of the relationship between $\zeta_U$'s and $\mu$

    • 1.2 Edit: Or mention of some kind of proposition that tells us the $\zeta_U$'s, which may or may not be related to $\mu$, are unique provided $\phi_U$ equivariant and $\sigma_U$ given as such

  2. If the definition is in fact not missing any (explicit or implicit) notion of fiber-preserving (Edit: fiber-preserving or trivializing-open-subset-preserving) of the action $\mu: P \times G \to P$ because we can somehow deduce some kind of notion of fiber-preserving (Edit: fiber-preserving or trivializing-open-subset-preserving) of the action $\mu$ or that any of (5.1),(5.2) or (5.3) is true, then which are true, and how do we deduce these?

  3. Are $\zeta_U$ and $\sigma_U$ necessarily smooth based on Tu's definition (as stated)? If not, then, based on other definitions of (smooth) principal $G$-bundle that you know, are $\zeta_U$ and $\sigma_U$ likely intended to be smooth?

    • I think I was able to prove $\sigma_U$'s are smooth by writing each $\sigma_U$ as a combination of maps, by compositions and multiplication of maps, where the maps include various projection maps and the law of composition on the Lie group $G$.
  4. To clarify, the $\sigma_U$'s are free and transitive right? I think this follows from what I believe is the freedom and transitivity of the left multiplication group action of any group on itself based on its law of composition.


Update: Can we just omit $\mu$ in the definition and then just later make a proposition about $\mu$ in the following way?

I'm thinking we instead first define that for each $U \in \mathfrak U$, $G$ acts on $U \times G$ on the right, still by the given $\sigma_U$ and then we say that $G$ acts on $\pi^{-1}(U)$ by some smooth right action $\zeta_U$ (I guess we don't have to include free or transitive since $\sigma_U$ is free and transitive and then freedom and transitivity are preserved under bijective equivariant or whatever), where $\zeta_U$

  1. satisfies some compatibility condition like $\zeta_U|_{U \cap V} = \zeta_V|_{U \cap V}$ for all $V \in \mathfrak V$

  2. makes $\phi_U$ is $G$-equivariant.

Later, we can make a propositions

  • Lemma A. $\phi_U$ is $G$-equivariant if and only if the $\zeta_U$ is given by $$\zeta_U(e,g) = \phi_U^{-1}(\sigma_U(\phi_U(e),g)) = \phi_U^{-1} \circ \sigma_U \circ ([\phi_U \circ \alpha_U] \times \beta_U) \circ (e,g), \tag{A*}$$ where $\alpha_U: \pi^{-1}(U) \times G \to \pi^{-1}(U)$ and $\beta_U: \pi^{-1}(U) \times G \to G$ are projection maps. (In this case, I guess $\alpha_U$ is the smooth trivial action by $G$ on $\pi^{-1}(U)$.)

    • Exercise A.i. Check that $\zeta_U$ in $(A*)$ is a smooth, right, free and transitive action by $G$ on $\pi^{-1}(U)$.

    • Exercise A.ii. Check that $\zeta_U$ in $(A*)$ satisfies the above compatibility condition.

    • Equivalent Definition A.1. We use Lemma A, Exercise A.i and Exercise A.ii to say instead that $\zeta_U$ is given by ($A*$).

  • Theorem B. $G$ globally acts on $P$ by some (smooth) right, free and transitive global action $\mu$ that turns out to be from collecting all the local actions, the $\zeta_U$'s, together: $\mu(p,g):=\zeta_U(p,g)$ for $p \in \pi^{-1}(U)$ for any $U \in \mathfrak U$, which is well-defined either by the compatibility condition assumption on $\zeta_U$'s in the original definition, where we don't yet know the formula for $\zeta_U$ or by Exercise A.ii, if we use $\zeta_U$ given by ($A*$).

  • Corollary C1. $\mu$ is trivializing-open-subset-preserving, i.e. $\mu((U \times G) \times G) \subseteq U \times G$

  • Corollary C2. $\mu$ is fiber-preserving, i.e. $\mu((x \times G) \times G) \subseteq x \times G$


Bounty message: I really believe there's at least one of the following here:

  1. ambiguity or implicit relationship between $\mu$ and $\zeta_U$'s,

  2. implicit rule about uniqueness or existence of an action (in this case $\zeta_U$'s) on domain of a map that makes a map equivariant given an action (the $\sigma_U$'s) on the range

  3. circular reasoning or circular definitions or something that need to be remedied either by some assumption $\mu$ preserves fibers or trivializing open subsets or by first defining smooth compatible local actions, the $\zeta_U$'s on the $P_U$'s, that make $\phi_U$'s equivariant and then later deducing a global action $\mu$ on $P$

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    In your picture (Section 27.1 part 3), Tu wrote $$(x,h)\cdot g = (x,hg).$$ With $\phi_U$ begin fibre-preserving, the fact that first component $x$ is not changed during $(x,h)\cdot g$ means $g$ preserves fibres. Maybe by $hg$ Tu means the multiplication in $G$ so that transitivity is implied. – autodavid Oct 16 '19 at 09:42
  • @autodavid Thanks for commenting. "$hg$" indeed refers to the (smooth) law of composition of the (Lie) group $G$. I understand that the "(x,h)⋅g=(x,hg)" refers to the $\sigma_U$'s. I guess then that each $\sigma_U$ is a fiber-preserving smooth right action. However, it seems that the required fiber-preservation instead of the $\mu$. Do I misunderstand? –  Oct 16 '19 at 10:41
  • @autodavid Actually, before saying $\mu$ is fiber-preserving, I think we must even first say that $\mu$ is trivializing-open-subset-preserving in having $\mu(U \times G \times G) \subseteq U \times G$. I don't know of any explicit relationship between $\mu$ and the $\sigma_U$'s, which is why I think something's missing. –  Oct 16 '19 at 10:45
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    $\phi_U$ is $G$-equivariant, so is $\phi_U^{-1}$. If we define $\phi_U((x',h'))=(x,h)$, then $$(x', h')\cdot g = \phi_U^{-1}((x,h))\cdot g=\phi_U^{-1}((x,h)\cdot g)=\phi_U^{-1}((x,hg)) = (x', \text{something})$$, which is in the same fibre of $P$. (I find it hard to write down because there may be no need to write $\square'$ for things in $P$.) It feels like that a $G$-equivariant bundle isomorphism transfers some behaviors of an action on a bundle to the other. (Edit: so my first comment is not sufficient to imply that the fibres of $P$ are preserved, without referring to equivariance.) – autodavid Oct 16 '19 at 11:42
  • @autodavid Wait, in the first place I think any references to the equivariance of $\phi_U$ might circular: To say $\phi_U$ is equivariant relies on having an action $\zeta_U$ on $P_U$. If you're saying that $\zeta_U=\mu_U$, we must have $\mu_U$ be a well-defined map before we even discuss whether or not $\mu_U$ is smooth map or a group action. Do I misunderstand? If so, then why? If not, then what exactly is your $\zeta_U$ please? Or is it like there exists a unique action $\zeta_U$ such that $\phi_U$ is equivariant (given of course the action $\phi_U$)? –  Oct 16 '19 at 12:29
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    In the definition of principal $G$-bundle, the way in which $G$ is acting on $P$ should make the trivialization maps $G$-equivariant when restricting to a trivialization patch..... Oh okay, there would be some problem because we don't know whether the restrictions are legal. I'm not an expert but I guess Tu implicitly requires that the restrictions to be legal, by talking about equivariance. – autodavid Oct 16 '19 at 13:29
  • @autodavid Replying to your comment without considering Jason DeVito's answer: I knew it! Thanks. You/we/I might be wrong, but hopefully I'm/you're/we're justified in thinking there might (but not necessarily is. Just might) be some kind of implicit requirement. –  Oct 17 '19 at 09:29

1 Answers1

4

I think Tu's definition is equivalent to the one involving free transitive actions on the fibers. His definition of principal $G$-bundle has two parts:

A) we have a fiber bundle $\pi:P\rightarrow M$ with $G$ acting smoothly freely on $P$ AND

B) we are told something more about the action: the fiber-preserving local trivializations are $G$-equivariant, where the action of $G$ on $U\times G$ is given by $(x,h)\cdot g = (x,hg)$.

By declaring $\phi_U$ to be $G$-equivariant, it now follows that the $G$ action on $P$ preserves fibers as follows. Suppose $g\in G$, $p\in P$ and $\pi(p)\in U\subseteq M$ where $U$ trivializes $P$. Set $\phi_U(p) = (\pi(p), h)$ and set $\phi_U(pg) = (\pi(pg), h')$. Then $$(\pi(pg), h') = \phi_U(pg) =\phi_U(p)g =(\pi(p),h)g = (\pi(p), hg),$$ from which it follows that $\pi(pg) = \pi(p)$. That is, $pg$ and $p$ are in the same fiber.

Now, to actually answer your questions:

  1. The action isn't explicitly given because it's a general definition. Kind of like when defining groups, you just have some binary operation satisfying some properties. Fiber preservation was handled above.

  2. Both 5.1 and 5.2 and true. I'm a little hazy on what 5.3 is asserting. But the point is the action must look like right multiplication in any trivializing open set.

  3. They are smooth as they are the restrictions of the $G$ action on $P$ to a preserved subset, and the $G$ action on $P$ is assumed to be smooth.

  4. Yes. (As a corollary, Tu didn't need to include "free" in his definition of prinicpal $G$-bundle, since it follows from B above.

  • Thanks Jason DeVito!

    Follow-up question 1. Do you mean the following?

    1A. Suppose $g\in G$, $p\in P$ and $\pi(p) \in U \subseteq M$ where $U$ trivializes $P$.

    1B. $\phi_U(p)g =(\pi(p),h)g = (\pi(p), hg)$

    –  Oct 17 '19 at 09:18
  • Follow-up question 2. For the expression "$\phi_U(pg)$", in the first place, the domain of $\phi_U$ is $\pi^{-1}(U)$? This seems kind of circular. (Thinks for awhile) Oh I think I see what you're doing, you're proving fiber-preserving i.e. $\mu(x \times G \times G) \subseteq x \times G$ assuming trivializing-open-subset-preserving i.e. $\mu(U \times G \times G) \subseteq U \times G$ right? If not, then what exactly are you doing please? If so, then how do you prove trivializing-open-subset-preserving? –  Oct 17 '19 at 09:46
  • Follow-up question 3. Can we just omit $\mu$ in the definition and then just later make a proposition about $\mu$? I'm thinking we instead first define that for each $U \in \mathfrak U$, $G$ acts on $U \times G$ on the right, still by the given $\sigma_U$ and then we say that $G$ acts on $\pi^{-1}(U)$ on the right by some action $\zeta_U$ (I guess we don't have to include free or transitive since $\sigma_U$ is free and transitive and then freedom and transitivity are preserved under bijective equivariant or whatever) in such a way that $\phi_U$ is $G$-equivariant. Later, we –  Oct 17 '19 at 09:49
  • Later, we can make a propositions:

    Lemma A. $\phi_U$ is $G$-equivariant if and only if the $\zeta_U$ is given by $\zeta_U(e,g) = \phi_U^{-1}(\sigma_U(\phi_U(e),g)) = \phi_U^{-1} \circ \sigma_U \circ ([\phi_U \circ \alpha_U] \times \beta_U) \circ (e,g)$, where $\alpha_U: \pi^{-1}(U) \times G \to \pi^{-1}(U)$ and $\beta_U: \pi^{-1}(U) \times G \to G$ are projection maps. (In this case, I guess $\alpha_U$ is the smooth trivial action on $\pi^{-1}(U)$.)

    –  Oct 17 '19 at 09:49
  • Theorem B. $G$ acts on $P$ on the right by some smooth, free and transitive action $\mu$ that turns out to be from collecting all the $\zeta_U$'s together: $\mu(p,g):=\zeta_V(p,g)$ for $p \in \pi^{-1}(V)$ for any $V \in \mathfrak U$, which is well-defined because $\zeta_V(p,g) = \zeta_W(p,g)$ for any $V,W \in \mathfrak U$. Corollary C1. $\mu$ is trivializing-open-subset-preserving.

    Corollary C2. $\mu$ is fiber-preserving.

    –  Oct 17 '19 at 09:50
  • Exercise D. Check that $\phi_U^{-1} \circ \sigma_U \circ ([\phi_U \circ \alpha_U] \times \beta_U)$ is a smooth, right, free and transitive action by $G$ on $\pi^{-1}(U)$. –  Oct 17 '19 at 09:52
  • For follow-up question 3: I edited post to include all of these in the "Update" at the end. –  Oct 17 '19 at 10:08
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    Follow up question 1: yes. Follow up question 2: I'm not assuming anything circular. I am assuming 1. $G$ acts on $P$ by some action. 2. For any $U\subseteq M$, $G$ acts on $U\times G$ by right multiplication on the second factor. 3. For any $U\subseteq M$ for which $P$ trivializes over $U$, there is a trivialization $\phi_U:\pi^{-1}(U)\rightarrow U\times G$ which is $G$-equivariant. Then, using all this, I proved that the $G$ action on $P$ must be fiber preserving. Question 3: You cannot omit $\mu$ because "$G$-equivariant" is meaningless without an initial action of $G$ on $P$. – Jason DeVito - on hiatus Oct 17 '19 at 14:23
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    But you could follow a different approach where you have a trivializing cover with all transition maps $\phi_U\circ \phi_V^{-1}: (U\cap V)\times G\rightarrow (U\cap V)\times G$ equivariant. Then you can use this to define a $G$ action on $P$, and you will get a principal bundle that way as well. – Jason DeVito - on hiatus Oct 17 '19 at 14:24
  • Follow-up question 4: You are proving $\mu$ is fiber-preserving without first proving $\mu$ is trivializing-open-subset-preserving? Follow-up question 5: Let $U \in \mathfrak U$. I understand $\phi_U$ is $G$-equivariant for the $\sigma_U$ on $U \times G$ and some $\zeta_U$ on $\pi^{-1}(U)$. What's the relation between $\zeta_U$ and $\mu$ please? Follow-up question 6: Do you disagree with autodavid " we don't know whether the restrictions are legal. I'm not an expert but I guess Tu implicitly requires that the restrictions to be legal, by talking about equivariance. " ? –  Oct 17 '19 at 20:20
  • Follow-up question 7A: Why is $G$-equivariant meaningless without action of $G$ on $P$ if for each $U \in \mathfrak U$, I have an action $\zeta_U$ on $\pi^{-1}U$. Follow-up question 7B: (possibly related to follow-up question 5) Why is $G$-equivariant meaningful with action $\mu$ of $G$ on $P$ when for each $U \in \mathfrak U$, $\zeta_U$ is an action not on $P$ but on $\pi^{-1}U$. It seems like there's some relation between $\mu$ and $\zeta_U$, but I just can't seem to figure it out. (Maybe it's not something to be figured out but something to be given in the definition...) –  Oct 17 '19 at 20:24
  • Leucippus, why did you reject my edit? https://math.stackexchange.com/posts/3396183/revisions –  Oct 18 '19 at 03:55
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    Question 4: Yes, I am proving that $\mu$ is fiber preserving, from which trivializing-open-subset-preserving follows. Question 5: $\zeta_U$ is the restriction of $\mu$ to $\pi^{-1}(U)$. Question 6: I agree that the definition could be clearer by asking that the restrictions be legal, but I proved above that it follows from Tu's definition that the restrictions are legal (if I'm understanding what autodavid meant, and of course, assuming my proof is right!) Question 7A. If you have a bunch of actions on each $\pi^{-1}(U)$, you need them to be compatible on $\pi^{-1}(U)\cap \pi^{-1}(V)$ – Jason DeVito - on hiatus Oct 18 '19 at 14:16
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    (continued). If they are compatible, then you basically already have a $G$ action on $P$. 7B: I think I answered this in 5. – Jason DeVito - on hiatus Oct 18 '19 at 14:24
  • Thanks Jason DeVito. For compatibility: I edited the post. Is my proposed definition in the "Update" part missing anything or is incorrect somewhere? For Follow-up Question 5: I believe your proof is right (regarding your parenthetical remark in your answer to Follow-up question 6), but my concern is that your proof might be circular. I think the very idea of $\zeta_U$ being the restriction of $\mu$ means to have $\mu$ be trivializing-open-subset-preserving in the first place. I mean that it sounds like the book is saying –  Oct 19 '19 at 02:31
  • "If $A, B, C$ are square $n \times n$ matries such that $AB=BC$, then $B=A^{-1}BC$" and then I ask how we know $A$ is invertible and then you say something like (my analogy weakens after this point) "If $A$ weren't invertible, then we couldn't have '$B=A^{-1}BC$' ". –  Oct 19 '19 at 02:33
  • I mean before you choose $\zeta_U$ as the restriction of $\mu$, you need the restriction possible. I really think either (A) that $\mu$ comes from $\zeta_U$'s in the way that I described in "update" or (B) that we need to assume the restrictions on $\mu$ are possible by assuming $\mu$ is trivializing-open-subset-preserving or fiber-preserving or something and then we choose $\zeta_U$'s as the restrictions on $\mu$ –  Oct 19 '19 at 02:40
  • I mean, to say a map is $G$-equivariant, you need, in the first place, actions on both domain and range. To choose $\zeta_U$ as a restriction of $\mu$ you need, in the first place, $\mu$ as trivializing-open-subset-preserving. Your proof that $\mu$ is trivializing-open-subset-preserving follows from $\mu$ being fiber-preserving which is proved assuming $\phi_U$ is $G$-equivariant. What am I misunderstanding? The only potential error I see here is "To choose $\zeta_U$...$\mu$ as trivializing-open-subset-preserving." –  Oct 19 '19 at 02:58
  • Jason DeVito how can you choose $\zeta_U$ as the restriction of $\mu$, when in the first place we don't know if the restriction is possible? The circular logic seems to be in using $\phi_U$'s equivariance to say restriction of $\mu$ is possible while we rely on such restriction of $\mu$ to talk about whether or not $\phi_U$ is equivariant. Oh also, I'm assuming you restrict $\mu$ both in domain and range so not just $\mu_1: P_U \times G \to P$ but actually $\mu_2: P_ \times G \to P_U$ because for some reason $\mu(P_U \times G) \subseteq P_U$, for the original $\mu: P \times G to P$ –  Oct 22 '19 at 07:17
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    Maybe I should have written point B in the answer differently. There exist local trivializations $\phi_U: \pi^{-1}(U)\rightarrow U\times G$ which are $G$-equivariant, but I am not asserting that all trivializations have this property. If you'd like, asserting the existence of these then forces the $G$ action on $P$ to preserve fibers. That is, if the $G$ action on $P$ does not preserve fibers, then the equivariant trivializations don't exist. So asserting their existence forces the $G$ action on $P$ to preserve the fibers. – Jason DeVito - on hiatus Oct 22 '19 at 15:05
  • Jason DeVito you're still assuming a relationship between $\mu$ and the $\zeta_U$'s (and thus the book indeed has a missing part of the definition in that $\zeta_U$' is supposed to be $\mu$)? I'm thankful for all the comments, but I don't think your last comment answered the questions in my comments prior. Could you please elaborate? – Selene Auckland Oct 23 '19 at 15:07
  • Also, the book says all the local trivializations are $G$-equvariant right? – Selene Auckland Oct 23 '19 at 15:08
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    I will try to elaborate, but maybe I am missing the core of what your question really is (despite the great lengths you've gone to in order to clarify it!). I really feel as though we are dancing around a sticky logic point, rather than making progress towards understanding what a principal bundle is. We begin with a smooth action of $G$ on $P$ which we know essentially nothing about. Also, for any $U\subseteq B$, we can define an action of $G$ on $U\times G$ by just right multiplying the second factor. Now, for $U\subseteq B$ nice enough, we assert the existence of a map .... – Jason DeVito - on hiatus Oct 23 '19 at 18:20
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    $\phi_U:\pi^{-1}(U)\rightarrow U\times G$ which is $G$ equivariant. You're right that we haven't yet specified an action of $G$ on $\pi^{-1}(U)$, but there is only one reasonable possibility - we use the $G$ action we already had on $P$. (Compare with me saying "$\mathbb{Z}$ acts on $\mathbb{R}$ by translations by integer steps. Thus, it also acts on $\mathbb{Z}$ by translations." I don't think anyone would have a problem with that sentence, and yet, to be correct I should have mentioned that the action of $\mathbb{Z}$ on itself I was considering is the restriction... – Jason DeVito - on hiatus Oct 23 '19 at 18:26
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    of the action on $\mathbb{R}$ to $\mathbb{Z}$). Now that we've settled on using the $G$ action we had on $P$, $G$-equivariance now forces the $G$ action on $P$ to be fiber preserving. As to your second question - it does seem as though the book is asserting all trivializations are $G$-equivariant, but this is definitely false. If you, say, consider the action of a Lie group $G$ on itself, you get a principal $G$-bundle over a base $B$ consisting of a single point. Then any diffeo $\rho:G\rightarrow G$ gives a trivializtion of this bundle, but very few diffeos are equivariant. – Jason DeVito - on hiatus Oct 23 '19 at 18:30
  • For the analogy with $\mathbb Z$ and $\mathbb R$: We can prove that for $\mu: \mathbb R \times \mathbb Z \to \mathbb R$, $\mu(a,b)=a+b$ that $$\mu(\mathbb Z \times \mathbb Z) \subseteq \mathbb Z. \tag{RZ1}$$ This set inclusion, tagged as $(RZ1)$, is something we can prove based on our knowledge the fact, call this $(RZ2)$, that $\mathbb Z$ is a subgroup of $\mathbb R$. $(RZ1)$ is the very basis of our ability to define $\mu_{\mathbb Z}: \mathbb Z \times \mathbb Z \to \mathbb Z$ with $\mu_{\mathbb Z}(a,b)=\mu(a,b)=a+b$. What's the analogue of $(RZ2)$ here? I think there shouldn't be – Selene Auckland Oct 25 '19 at 05:53
  • any analogy unless we have analogue of $(RZ1)$ = analogue of $(RZ2)$ = the assumption that $\mu(P_U \times G) \subseteq P_U$. So you indeed agree that it's either missing from the definition OR somehow understood, and then you're almost definitely leaning towards the "somehow understood"? – Selene Auckland Oct 25 '19 at 05:55
  • For the singleton $B$ why don't we say, instead of that Tu is wrong, that Tu would not consider that as a principal $G$-bundle then i.e. Tu's definition is different from yours? Actually, what's your definition of principal $G$-bundle please? – Selene Auckland Oct 25 '19 at 06:30
  • Detour: I notice each of the collections ${Orbit(p)}{p \in P}$ and ${P_x}{x \in M}$ partition P. Is this relevant? I was thinking if $Orbit(p) = P_x$ for all $p \in P_x$ (or even just for some $p \in P_x$, I'm not sure all $p \in P_x$ is required), then we can use $\mu(Orbit(p) \times G) \subseteq Orbit(p)$ to deduce $\mu(P_x \times G) \subseteq P_x$ – Selene Auckland Oct 25 '19 at 07:25
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    I do agree that it would have been better for Tu to just say "fiber preserving" when describing the $G$-action. The point of my "translating $\mathbb{R}$" analogy was not whether or not we can make it rigorous (as you did with your $RZ1$ and $RZ2$), but that the phrase "Thus is also acts on $\mathbb{Z}$ by translations" is missing the same ingredient Tu is missing: we haven't specified any action on $\mathbb{Z}$ at this point, so it's wrong to talk about $G$ acting on $\mathbb{Z}$. I would argue that both in my analogy and in Tu's book, the writer is hoping the reader will make.... – Jason DeVito - on hiatus Oct 25 '19 at 13:03
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    the obvious connection between the two actions: that one is the restriction of the other. I am definitely leaning towards the "somehow understood", but would strengthen it to simply "understood". When one talks about a group action on a set and then, without more context, talks about the group acting on a subset of the original set, it seems to me the only reasonable meaning is the restriction of the action. As far as the singleton case is concerned, I just chose that as its the simplest example where things go wrong.... – Jason DeVito - on hiatus Oct 25 '19 at 13:06
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    In fact, given any prinicipal bundle $P\rightarrow M$, there is a trivialization which is not $G$ equivariant (assuming, e..g, $G$ is positive dimensional): start with your favorite equivariant trivialization $\pi^{-1}(U)\rightarrow U\times G$ and compose it with a map $U\times G\rightarrow U\times G$ which is identity on the $U$ factor and a non-equivariant diffeo on the on the second factor. My definition of principal bundle would be Tu's, though I would probably explicitly state the action preserves fibers. Finally, for your detour: once your believe the $G$ action is fiber-preserving... – Jason DeVito - on hiatus Oct 25 '19 at 13:09
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    and transitive and free, the two partitions you list are identical. (By the way, I just noticed I didn't preface my latest block of comments with @Selene - sorry about that!) – Jason DeVito - on hiatus Oct 25 '19 at 13:11
  • I believe we are finished with this. Thanks Jason DeVito. You don't have to be sorry for anything. – Selene Auckland Oct 29 '19 at 06:06
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    Sure thing - until next time! – Jason DeVito - on hiatus Oct 29 '19 at 14:37