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Dec 9, 2010 at 22:31 vote accept Rex
Dec 9, 2010 at 22:31
Dec 9, 2010 at 19:06 answer added Deane Yang timeline score: 4
Dec 9, 2010 at 10:58 history edited Rex CC BY-SA 2.5
added 275 characters in body; edited title
Dec 9, 2010 at 1:03 comment added Rex @Jose: Locally, $\Omega_P=\Omega_U\oplus\Omega_{GL_n}$. So the section restricted to $U$ is just a $GL_n$ equivariant section $\Omega_{GL_n}\to \Omega_U$, which gives us $\omega^U$. We just need to check what the gluing condition tells us. If $x\in U\cap V$, then the point $(x,A)$ in $U\times GL_n$ is identified with $(x,gA)$ in $V\times GL_n$. This map from $U\cap V\times GL_n\to U\cap V\to GL_n$ induces a pullback at the level of forms. I just put these together. Thanks in advance.
Dec 9, 2010 at 0:51 comment added Rex @Eric: I was just going to tell you that. Thanks.
Dec 9, 2010 at 0:45 comment added Eric O. Korman Just noticed your transition functions are going the opposite way I had them. So yea, the first formula is correct (replace my $g$ with $g^{-1}$). Maybe you made the same mistake I did in getting at the second equation?
Dec 9, 2010 at 0:30 comment added Eric O. Korman I don't see how the first formula is correct. Say $E$ is 1-d with frames $f$ and $e$ with $f = ge$ and $De = \omega\otimes e$. Then $Df = dg \otimes e + g \nabla e = dg\otimes e + g \omega \otimes e = (dg + g\omega) \otimes g^{-1} f = ((dg)g^{-1} + \omega)\otimes f$. So the connection form in the $f$ frame is $(dg) g^{-1} + \omega$. This agrees with the second formula but not the first.
Dec 8, 2010 at 23:41 comment added José Figueroa-O'Farrill The first formula is correct, so clearly there is something wrong with the second. Since you give no details, it's difficult to say where it is that the error has crept up.
Dec 8, 2010 at 23:24 comment added Rex @Eric: The wiki article agrees with the first one.
Dec 8, 2010 at 22:54 comment added Eric O. Korman I'm pretty sure the second one you have is correct, but the first one is not. I've worked it out before and the wikipedia article en.wikipedia.org/wiki/Connection_form agrees.
Dec 8, 2010 at 21:22 history asked Rex CC BY-SA 2.5