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In Foundations of Differential Geometry by Kobayashi and Nomizu they give the definition for "A (differentiable) principal fibre bundle over $M$ with group $G$".

$M$ is a (differentiable) manifold, $G$ is a Lie group, $P$ is a (differentiable) manifold. There is a (differentiable) right action of $G$ on $P$.

(1) $G$ acts freely on $P$

(2) $M$ is the quotient space of $P$ by the equivalence relation induced by $G$, $M=P/G$, and the canonical projection $\pi:P\to M$ is differentiable;

(3) $P$ is locally trivial, that is, every point $x$ of $M$ has a neighborhood $U$ such that $\pi^{-1}(U)$ is isomorphic with $U\times G$ in the sense that there is a diffeomorphism $\psi:\pi^{-1}(U)\to U\times G$ such that $\psi(u) = (\pi(u), \varphi(u))$ where $\varphi$ is a mapping of $\pi^{-1}(U)$ into $G$ satisfying $\varphi(ua) = (\varphi(u))a$ for all $u\in \pi^{-1}(U)$ and $a\in G$.

Note that my questions are similar to the one that prompted this answer but I think my questions are even more basic.

(A) In condition (2) what does it mean $M=P/G$? It can't be normal set equality since $M$, $P$, and $G$ are defined separately. Does it mean $M$ and $P/G$ are homeomorphic? Or does it mean $M$ and $P/G$ are diffeomorphic? If it the latter then how do we know $P/G$ has a differential structure on it? Does $P$ being a differentiable manifold and $G$ being a Lie group imply $P/G$ inherits a certain differentiable structure?

(B) Similarly, in (2), it says $\pi: P \to M$ is differentiable. But, by my understanding, $\pi$ is actually a map $\pi: P\to P/G$. Suppose the morphism between $P/G$ and $M$ is called $f: P/G \to M$. If $f$ is a diffeomorphism then I could take the statement to mean $f \circ \pi: P \to M$ is differentiable. But if $f$ is only a homeomorphism it isn't obvious to me that $f\circ \pi$ would be diffeomorphism.

(C) I paraphrased slightly. I want to make sure the definition I've given above exactly matches the definition in the text. Let me know if I've inadvertently made some changes. Namely, is it correct that the right action of $G$ on $P$ should be differentiable? This wasn't explicitly stated in the definition in the book. I also added in parenthetical indications that all manifolds are differentiable manifolds, but I think it is the normal convention in the book that manifold means differentiable manifold.

(D) Is $G$ just a group that is also differentiable manifold such that group multiplication and group inverse are differentiable? Or are there additional conditions on $G$ like compactness or something?

I think my questions are very basic. I'm a beginner when it comes to topology etc. It seems like Kobayashi and Nomizu is a bit of an advanced text and they are using some shorthand notation (like $P/G = M$) that I'm just not familiar with yet. I'm just needing the details spelled out in a more beginner-friendly way.

Jagerber48
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  • To begin with: Do you know what a Lie group is? This is where the definition of a principal bundle (in the KN book) starts. Do you know what the quotient topology is and how to define one given a group action on a topological space? – Moishe Kohan Apr 02 '24 at 17:40
  • @MoisheKohan I understand a Lie group to be a group $G$ that is also a (differentiable/smooth) manifold and the group multiplication $\cdot:G\times G \to G$ is a differentiable/smooth map and the group inverse operation ${}^{-1}:G\to G$ is also differentiable/smooth. I understand that if we have a group action of $G$ on $P$ you can define a quotient $P/G$ where equivalence classes are the orbits of points of $P$ under the group action $G$. That is $[p]=[q]$ if there is $g\in G$ such that $qg=p$. I assume $P/G$ inherits the usual topological quotient topology. – Jagerber48 Apr 02 '24 at 17:46

1 Answers1

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(A) $M=P/G$ means that $M$ is homeomorphic to $P/G$ equipped with the quotient topology. The fact that $M$ is a smooth manifold and that the quotient map $\pi: P\to M$ gives you more, namely, a relation between the smooth structures of $P$, $G$ and $M$. Yes, if one has a smooth, free and proper action $G\times P\to P$, one can prove that the quotient $P/G$ has a natural smooth structure. But this is not the route taken by Kobayashi and Nomizu, they simply postulate that $M=P/G$ is a smooth manifold.

(B) If you want to be very formal, you take a homeomorphism $f: P/G\to M$ and require the composition $f\circ \pi$ to be differentiable. But this just introduces an unnecessary notation since once you have a homeomorphism $f: P/G\to M$, you can identify $P/G$ and $M$ as sets via $f$, making the extra map unnecessary.

(D) $G$ is just a Lie group, no compactness assumption. In fact, some of the most important examples of principal bundles, such as the frame bundle, have noncompact group $G$. As I explained in other answers, with a bit different axiomatics, one replaces the unnecessary compactness assumption with the properness assumption on the action of $G$ on $P$. With the KN definition, properness is automatic.

Moishe Kohan
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  • Thank you, this is very helpful. Just one last question. From $M=P/G$ we only have that $f: P/G \to M$ is a homeomorphism. Is it a further constraint on $f$ to require that $f\circ \pi: P \to M$ is a diffeomorphism? Or is that only a constraint on $\pi$? That is, do we need to specify that "$M$ is homeomorphic to $P/G$ such that $f\circ \pi$ is/can be a diffeomorphism". I hope this question makes sense. – Jagerber48 Apr 03 '24 at 14:46
  • @Jagerber48: It cannot be a diffeomorphism unless $G=1$. – Moishe Kohan Apr 03 '24 at 14:56
  • oh right. I should have said differentiable (like in the definition) instead of diffeomorphism in my last comment. – Jagerber48 Apr 03 '24 at 15:07
  • @Jagerber48 This is the condition 2 in the definition. – Moishe Kohan Apr 03 '24 at 15:09
  • Ok, so I'll take condition (2) to mean both (a) that there is a homeomorphism $f: P/G \to M$ and (b) that $f\circ \pi: P \to M$, where $\pi: P \to P/G$, is differentiable. I guess the answer to my question is that yes, there are indeed two conditions on $f$. First that it is a homeomorphism, and second, that when composed with $\pi$, the result is differentiable. – Jagerber48 Apr 03 '24 at 19:58
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    @Jagerber48: Right. Most of the time, as I said, we can identify $P/G$ and $M$ as sets with $f=id$. When you first learn the subject, this is enough. But one can consider different principal bundles over the same base $M$ and then things become a bit trickier since, as sets, $P'/G$ and $P/G$, are clearly different... – Moishe Kohan Apr 03 '24 at 20:07