Suppose I define a "principal $G$-bundle" as follows:

A principal $G$-bundle is a fiber bundle $F \to P \overset{\pi}{\to} X$ with a left group action of $G$ on $F$ that is free and transitive, together with a trivializing cover whose transition maps are $G$-valued.

However, it seems that many references define "principal $G$-bundle" via a right action of $G$ on $P$ (not $F$).

How does my definition induce a natural right action of $G$ on $P$? Can this be done without saying the phrase "identify $F$ with $G$"?

The reason I would like to avoid this identification is two-fold. First, I would like to keep the fiber $F$ and the group $G$ separate in my head -- at least for now -- in part because not all $G$-bundles are principal. Second, and more importantly, I am concerned that any identification of $F$ with $G$ will involve an arbitrary choice of base-point of $F$, and I would rather not make such unnecessary choices if possible.

Ultimately, I would like to say that the specified trivializations in my definition of "principal $G$-bundle" are $G$-equivariant with respect to the actions on $P$ and $F$. I would like to deduce this as a consequence of the definition of the $G$-action on $P$, rather than taking this equivariance as the definition of the action.

Aside: As usual, this question is a refinement of a previous, less focused question of mine.

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    $\begingroup$ I am really interested in wether it is possible to deduce $G$-equivariance of the trivializations. Otherwise I would have suggested, that $G$-equivariance and the (natural) right action (induced by the left action with $g^{-1}$) on $F$ induces a right action on $P$. $\endgroup$ – gofvonx Oct 27 '13 at 10:33

A clean way to do it (without making an identification you do not like), is to assume that we have two commuting actions of $G$ on $F$ (left and right actions). Below are the details.

Let $g_{\alpha\beta}: U_\alpha\cap U_\beta\to G$ be the transition maps satisfying the cocycle condition. Let $F$ be any left $G$-space (the action at this point need not be transitive). Then one forms the bundle $E\to X$ (with the fibers diffeomorphic to $F$ in the smooth setting) by taking the quotient of the disjoint union $$ \tilde E=\sqcup_{\alpha} U_\alpha\times F $$ by the natural equivalence relation: $(x,v)\in U_\alpha \times F \sim (x,w)\in U_\beta\times F$ whenever $g_{\alpha\beta}v=w$ (here we are using the left action).

Now, suppose in addition that we have a second action, a right action of $G$ on $F$, this time transitive, commuting with the left action used earlier. For instance, if $F$ is identified with $G$ itself (I know that you do not want to make this identification explicit, so this is only an illustration of the construction), then we can use the action of $G$ on itself via right multiplication for the right action and via left multiplication for the left action.

Using this right action, the group $G$ also acts naturally (on the right) on $\tilde E$. This action commutes with the left action used to make the identification and, hence, projects to a (right) $G$-action on $E$. Since we assumed that the right action of $G$ on $F$ is transitive, then the resulting right action of $G$ on $E$ is transitive on the fibers, as required by the usual definition of a principal bundle. Hope it helps to clear the confusion.

  • $\begingroup$ This answer is very helpful. Thank you. $\endgroup$ – Jesse Madnick Nov 7 '13 at 8:55
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    $\begingroup$ I'm wondering whether we can go the other way. That is, given a free right action of $G$ on $P$, transitive on fibers, can we find a left action of $G$ on $F$ and a trivializing cover whose transition maps are $G$-valued? $\endgroup$ – Jesse Madnick Nov 15 '13 at 22:50

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