Let $X$ be a smooth manifold and suppose I have a smooth vector bundle $E\to X$ which admits a connection $D$. Then on an open set $U\subset X$ where $E$ is trivial, once I choose a frame, say $e_1,...e_n$, I get a connection matrix by the rule $D(e_i)=\sum_{j=1}^{n}e_j\otimes\omega_{ji}^U$ (a column vector of 1 forms). If $V \subset X$ is another open set where $E$ is trivial and I choose a frame over $V$, say $f_1,...f_n$, then similarly I get $\omega^V$. Let $e_i=\sum_{j=1}^{n}g_{ji}f_j$. Then we have $\omega^V=g\omega^U g^{-1}-dgg^{-1}$.

On the other hand, we can form the associated principal $GL_n$ bundle $P\stackrel{\pi}{\to} X$, with $GL_n$ acting on the right. A connection on $P$ is a $GL_n$ invariant splitting of $\pi^*\Omega_X\to \Omega_P$. Locally, after we choose trivialisations for $P$ using the frames for $E$ chosen earlier, this becomes equivalent to giving a $GL_n$ equivariant splitting of $\Omega_U\to \Omega_U\oplus\Omega_G$, which is equivalent to giving a map $\Omega_{G,e}\to\Omega_{U,x}$ for each point $x\in U$, this gives the $\omega^U$. If $x\in U\cap V$, then the point $(x,A)$ in $U\times G$ is identified with the point $(x,gA)$ in $V\times G$. The gluing gives the following relation $\omega^V=g\omega^U g^{-1}+dgg^{-1}$. This is different from the one above. Is that OK?

  • $\begingroup$ 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. $\endgroup$ Commented Dec 8, 2010 at 22:54
  • $\begingroup$ @Eric: The wiki article agrees with the first one. $\endgroup$
    – Rex
    Commented Dec 8, 2010 at 23:24
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Dec 8, 2010 at 23:41
  • $\begingroup$ 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. $\endgroup$ Commented Dec 9, 2010 at 0:30
  • $\begingroup$ 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? $\endgroup$ Commented Dec 9, 2010 at 0:45

1 Answer 1


I don't think there's any problem here. The question you want to ask is: Given a section $u$ of $E$, whether you get a consistent answer for $\nabla u$, using either frame. In other words, does $\nabla (u^ie_i) = \nabla (\hat{u}^i\hat{e}_i)$ using your change of frame formulas? I think you'll see what's going on if you work this out.

Let me elaborate a little bit. I never write change of frame formulas for the connection $1$-forms in isolation, because I am always confused by whether I should be acting on the frame by $g$ or $g^{-1}$. And whether $G$ is supposed to be acting on the right or left. But I know I will always get the right answer if I think about applying the connection to local section $u$ and expand it using two different local frames. Everything always automatically works out correctly when I do it this way.

  • $\begingroup$ My problem is on the one hand I get $\omega^V=g\omega^Ug^{-1}-dgg^{-1}$ and on the other hand I get $\omega^V=g\omega^Ug^{-1}+dgg^{-1}$. I have checked the calculation several times and so I think I am goofing up with some definition?? $\endgroup$
    – Rex
    Commented Dec 9, 2010 at 19:44
  • $\begingroup$ Let me try to extend Deane Yang's answer (hope it is OK): Suppose you are looking at the same connection and the same trivialization in these two different ways. Do you get the same form $\omega^U$? $\endgroup$
    – t3suji
    Commented Dec 9, 2010 at 20:08
  • $\begingroup$ Your calculation is correct. Please focus on an actual section of $E$ and expand it using two different frames. You'll see why everything is actually OK. $\endgroup$
    – Deane Yang
    Commented Dec 9, 2010 at 21:25
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    $\begingroup$ Both formulas appear correct to me. $\endgroup$
    – Deane Yang
    Commented Dec 9, 2010 at 21:58
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    $\begingroup$ I encourage you to read the second paragraph of my answer (which I just added). $\endgroup$
    – Deane Yang
    Commented Dec 9, 2010 at 22:36

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