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I am trying to check why curvature 0 implies that the horizontal distribution is involutive.

Let $\pi:P\to U$ be a principal $G:=GL_n$ bundle. Assume that $P$ is trivial and $\pi$ admits a section. Thus, $P\cong U\times G$. A connection on $P$ is a $G$ equivariant splitting of the short exact sequence of bundles $0\to P\times\mathfrak{g} \to TU\oplus TG\to \pi^*TU\to 0$. Let us denote this splitting by $s$. If $(v,u)$ is a tangent vector at the point $(x,g)$, then $s(v,u)=w_x(v)+(R_{g^{-1}})_*(u)$, where $w\in\Gamma(X,\Omega_U\otimes\mathfrak{g})$.

Let $x_1,x_2,...x_n$ be coordinates on $U$ and consider vector fields $X_i$ corresponding to them. Use these to construct horizontal $G$ invariant vector fields on $P$ which are given by $((X_i)_x, -w_x((X_i)_x))$ at the point $(x,Id)$. This should give an involutive distribution on $P$.

Using that $[X_1,X_2]=0$, when we compute the Lie bracket, we get $[(X_1,-w(X_1)),(X_2,-w(X_2))]=-[w(X_1),X_2]-[X_1,w(X_2)]+[w(X_1),w(X_2)]$. =$X_2(w(X_1))-X_1(w(X_2))+[w(x_1),w(X_2)]$. We need to show that this is 0.

The condition that curvature is 0 is given by $dw+w\wedge w=0$, i.e. $dw_{ij}+\sum_{k=0}^{n}w_{ik}\wedge w_{kj}=0$. This means that $dw(X_1,X_2)+[w(x_1),w(X_2)]=0$, which is same as saying that $X_1(w(X_2))-X_2(w(X_1))+[w(x_1),w(X_2)]=0$

There seems to be some mismatch.

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    $\begingroup$ I like José's answer below. I would add that this kind of sign discrepancy usually arises because you are assuming that an action is a right action but really is a left action or vice versa. You should of course also doublecheck to make sure all your definitions of things like exterior differentiation are obeying the right sign convention. $\endgroup$
    – Deane Yang
    Commented Jan 5, 2011 at 2:06
  • $\begingroup$ Dear Deane, Thanks for the comment and you are absolutely correct. The problem is that $w(X_1)$ and $w(X_2)$ are right invariant vector fields and not left invariant. Thus, $[w(X_1),w(X_2)]=w(X_2)w(X_1)-w(X_1)w(X_2)$, which makes everything all right. $\endgroup$
    – Rex
    Commented Jan 5, 2011 at 22:15

1 Answer 1

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I realise that you are asking for someone to find where your sign mistake is, but in my opinion that's not what MO is for. Instead, let me sketch why flatness of the connection is equivalent to the integrability of the horizontal distribution.

Let $\pi: P \to M$ be a principal $G$-bundle and $\mathcal{H}\subset TP$ a connection, with connection 1-form $\omega$. In other words, $\mathcal{H} = \ker \omega$. I take it as read that these objects all satisfy the standard $G$-equivariance conditions.

Let $h: TP \to \mathcal{H}$ denote the horizontal projection along $\mathcal{V}:=\ker\pi_*$. Then the curvature 2-form $\Omega$ is defined by $$ \Omega(X,Y) := d\omega(hX,hY) $$ for vector fields $X,Y$ on $P$.

Using the definition of $d\omega$ $$ \Omega(X,Y) = (hX)\omega(hY) - (hY)\omega(hX) - \omega([hX,hY]) = - \omega([hX,hY]) $$ where we have used that $\omega(hX)=0$ for all $X$. Therefore $\Omega\equiv 0$ if and only if for all vector fields $X,Y$, $[hX,hY] \in \ker \omega$, which is equivalent to saying that the Lie bracket of any two horizontal vector fields is again horizontal.

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  • $\begingroup$ Thanks Jose. I realize that in this form it is a bad question to ask at MO, I apologize for not being able to resist the temptation. The $w_P$ that you construct is an element of $\Gamma(\Omega_P\otimes\mathfrak{g})$. Curvature is defined as $dw_P+w_p\wedge w_p$. In my case I started with a connection on a, say trivial, vector bundle $E$. This already gives me a connection form $w_E$ and I assumed that $dw_E+w_E\wedge w_E=0$. It is not clear to me why this should imply that $dw_P+w_p\wedge w_p=0$. $\endgroup$
    – Rex
    Commented Jan 5, 2011 at 18:16
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    $\begingroup$ Hi Rex. This is now a separate question. You are asking how the curvatures of a principal and associated bundles are related. Once you pull them back to the base via a local section, say, they are related by the representation of $G$ defining the associated bundle. Hence if $\omega_P$ is flat then so will be $\omega_E$, but the converse need not hold unless $\rho$ is faithful (at the level of the Lie algebra). For example, this is the case if $P$ is the bundle of frames of $E$. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jan 5, 2011 at 18:48
  • $\begingroup$ Jose thanks for that comment. I am interested in the case of a vector bundle $E$ and the bundle of frames, $P$, of $E$. In this case could you please explain or give me a reference why $w_E$ is flat implies $w_P$ is flat. All the notes I found on the internet work out the case of a principal bundle. $\endgroup$
    – Rex
    Commented Jan 5, 2011 at 20:27
  • $\begingroup$ I was basically trying to understand why $w_E$ flat implies that $E$ is a local system. I was trying to do this by going to the frames bundle. Could you suggest a better way to see this. $\endgroup$
    – Rex
    Commented Jan 5, 2011 at 20:39

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