# Annoying principal bundle problem

1. Nov 25, 2009

### Fredrik

Staff Emeritus
Consider a principal bundle $\pi:E\rightarrow M$ with structure group G. Let $I=(a-\epsilon,b+\epsilon)$. We can use a curve $\alpha:I\rightarrow M$ to construct a pullback bundle $Pr_1:IE\rightarrow I$. I'm calling the total space of the pullback bundle IE because that reminds me that it's a subspace of $I\times E$. To be more specific, we have

$$IE=\{(t,p)\in I\times E|\alpha(t)=\pi(p)\}$$

This pullback bundle is a principal bundle too, and $Pr_2$ is a principal bundle isomorphism. If $\omega$ is a connection 1-form on the original principal bundle, then $Pr_2^*\omega$ is a connection 1-form on the pullback bundle. (I was able to prove that, so that's not what I'm going to ask about).

Now let D be the derivative operator that takes a real-valued function on I to its derivative, i.e. Df=f' and Dtf=f'(t). This D is a vector field on I. The "horizontal lift" of this vector field is defined as the unique horizontal vector field $D^H$ that satisfies the condition $\pi_*D^H=D$. "Horizontal" means that it's taken to 0 by the connection 1-form, so we have $(Pr_2^*\omega)_{(t,p)}D^H_{(t,p)}=0$.

I understand all of the above, but there's another detail that looks like it should be easy to prove, and yet I can't figure it out. Define $\gamma$ to be the integral curve of $D^H$ trough (a,p) (where a is the real number mentioned in the definition of I, and p is some arbitrary point in E). This $\gamma$ is a curve in IE, so we can write $\gamma(t)=(\gamma_I(t),\gamma_E(t))$, but why is $\gamma_I(t)=t$?

I'm stuck at a "clearly we have" statement in the book I'm reading, and that statement is only true if $\gamma_I(t)=t$, so apparently the author of this book (Isham) felt that this is so trivial that he didn't even need to include a "clearly we have" statement about that. I'm hoping that someone can tell me why the integral curve $\gamma$ must also be a section.

This stuff is on page 264 of "Modern differential geometry for physicists" by Chris Isham, but for some reason it isn't possible to view page 264 either at Google books or at Amazon. If you happen to own a copy of the book, you might get confused by the fact that my notation is different from his.

Last edited: Nov 25, 2009
2. Nov 25, 2009

### Fredrik

Staff Emeritus
I think I got it. What I want to prove is that $Pr_1\circ\gamma$ is the identity map on I. First note that

$$(Pr_1\circ\gamma)_*D_t=Pr_1_*\gamma_*D_t=Pr_1_*\dot\gamma_{\gamma(t)}=Pr_1_*D^H_{\gamma(t)}=(Pr_1_*D^H)_{Pr_1(\gamma(t))}=D_{Pr_1\circ\gamma(t)}$$

So if we write $\sigma=Pr_1\circ\gamma$, we have

$$\sigma_*D_t=D_{\sigma(t)}$$

but $\sigma$ is a function from I into I (and I is an open interval in the field of real numbers), so the chain rule tells us that we also have

$$\sigma_*D_t f=D_t(f\circ\sigma)=(f\circ\sigma)'(t)=f'(\sigma(t))\sigma'(t)=D_{\sigma(t)}f\sigma'(t)$$

i.e.

$$\sigma_*D_t=\sigma'(t)D_{\sigma(t)}$$

and a comparison with the first result shows that $\sigma'(t)=1$ for all t, which means that $\sigma=id_I+C$, where C is a constant. The constant can be determined from the condition $\gamma(0)=(a,p)$, which implies that

$$Pr_1\circ\gamma(0)=Pr_1(a,p)=a[/itex] So we have C=a, which isn't what I expected. Here's what Isham does: He defines $\alpha^H=Pr_2\circ\gamma$ and then says that "clearly we have" $\pi(\alpha^H(t))=\alpha(t)$, but if I'm right, this isn't just "not clear", it's wrong. We have [tex]\gamma(t)=(Pr_1(\gamma(t)),Pr_2(\gamma(t)))=(\sigma(t),\alpha^H(t))$$

and therefore

$$\pi(\alpha^H(t))=\pi\circ Pr_2\circ\gamma(t)=\pi(\alpha^H(t))=\alpha(\sigma(t))=\alpha((id_I+a)(t))=\alpha(t+a)$$

3. Nov 27, 2009

### Fredrik

Staff Emeritus
New question:

I ran into some difficulties on page 266 too. Can someone explain (6.2.14) to me? He mentions (6.1.26). There's no preview for that page, but he writes that gauge transformation like this:

$$A_\mu(x)\mapsto\Omega(x)A_\mu(x)\Omega(x)^{-1}+\Omega(x)\partial_\mu\Omega(x)^{-1}$$

Both A and $\Omega$ are matrix-valued functions.