Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have two related questions.

Suppose I have a principal $G$-bundle $P\xrightarrow{\pi} M$. The usual construction of an associated vector bundle goes as follows. Fix some representation $\rho : G\rightarrow GL(V)$ of $G$ on the vector space $V$. Now define a right action of $G$ on the product manifold $P\times V$ by $(x,v)\mapsto (xg,\rho(g^{-1})\cdot v)$. Let $P\times_G V$ denote the orbits of $P\times V$ under this action. Then $P\times V\xrightarrow{q} P\times_G V$, where $q$ is the quotient map by $G$, is a principal $G$-bundle and $P\times_G V\xrightarrow{\pi_G} M$, where $\pi_G:[(x,v)]\mapsto \pi(x)$, is a vector bundle with typical fiber $V$.

Now suppose I have a principal connection $\Phi$ on $P\xrightarrow{\pi} M$, understood as an equivariant projection onto the vertical bundle of P that acts as the identity on vertical vectors. The (again) "standard" way to induce a linear connection on $P\times_G V\xrightarrow{\pi_G} M$ is to first define a connection $\Phi'=\Phi\times 1_V$ on $P\times V\xrightarrow{\pi\circ pr_1} M$. Then one shows that there's a unique linear connection $\tilde{\Phi}$ on the associated bundle $P\times_G V\xrightarrow{\pi_G} M$ such that $q_*\circ\Phi'=\tilde{\Phi}\circ q_*$.

Both of these constructions make me uneasy because they seem to depend on a choice of global trivialization of the bundle $P\times V\xrightarrow{\pi} P$, even though the results of the constructions appear independent of this choice. My first question is: is there a way to think of associated bundles that doesn't require you to mention a product structure?

What I am imagining is something like the following. Suppose I have a trivial vector bundle $E\xrightarrow{\wp} P$ with typical fiber $V$ over the total space of a principal $G$-bundle $P\xrightarrow{\pi} M$, and suppose I have a free right action of $G$ on $E$ satisfying $\wp(xg)=\wp(x)g$ for any $x\in E$. It certainly follows, with no additional conditions, that there's a unique manifold structure on $E/G$ such that $E\xrightarrow{q} E/G$ is a principal $G$-bundle. But is there a natural sense in which (1) this right action induces a representation $\rho$ of $G$ on $V$ that is unique up to automorphisms of $V$ and (2) $E/G\xrightarrow{\pi_G} M$, where $\pi_G$ is the unique map such that $\pi_G\circ q=\pi\circ \wp$, forms a vector bundle over $M$ isomorphic to the bundle $P\times_G V\xrightarrow{\pi_G} M$ defined above, for the representation $\rho$?

This leads to the second question, which is (largely) independent of the first. Suppose I am given (1) a principal bundle $P\xrightarrow{\pi} M$ with structure group $G$, (2) a trivial vector bundle $E\xrightarrow{\wp} P$ with typical fiber $V$, and (3) a right action of $G$ on $E$, all such that $E\xrightarrow{q} E/G$ is a principal G-bundle, $E/G\xrightarrow{\pi_G} M$ is a vector bundle, and $(\wp,\pi_G)$ is a principal bundle homomorphism from $E\xrightarrow{q} E/G$ to $P\xrightarrow{\pi} M$. (I get all of this from the standard associated bundle construction.) Does a principal connection on $P\xrightarrow{\pi} M$ induce a canonical linear connection on $E/G\xrightarrow{\pi_G} M$ that can be characterized without mentioning a trivialization?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.