# Exterior derivative on principal bundle [closed]

In Nakahara's Geometry, Topology and Physics on page 375, he constructs a Lie-algebra-valued one-form $\omega$ on a principal bundle $P$ by "lifting" a Lie-algebra-valued one-form $\mathcal A_i$ on an open covering $\{U_i\}$ on the base manifold $M$. Given a $\mathfrak{g}$-valued one-form $\mathcal{A}_i$ on $U_i$ and a local section $\sigma_i:U_i\to\pi^{-1}(U_i)$, he defines the connection one-form $\omega$ like this:

$$\omega_i := g_i^{-1}\pi^*\mathcal A_i g_i + g_i^{-1}\mathrm{d}_Pg_i$$

Here, $\mathrm{d}_P$ denotes the exterior derivative on $P$ and $g_i$ is the canonical local trivialization such that $\phi_i^{-1}(u)=(p,g_i)$ and $u=\sigma_i(p)g_i$. Later he proves that $\sigma_i^*\omega_i=\mathcal A_i$, where he uses the fact that $\mathrm{d}_Pg_i({\sigma_i}_*X)=0$ because of $g\equiv e$ along ${\sigma_i}_*X$.

I don't understand the notation used here, especially the exterior derivative. As I understand it, the exterior derivative maps $p$-forms to $(p+1)$-forms, so for $\mathrm{d}_P$ to produce a one-form on $P$, it's input must be a 0-form, i.e. a function on $P$.

What's the function that serves as the "input" of $\mathrm{d}_P$? In other words, how does $g_i^{-1}\mathrm{d}_Pg_i$ produce a Lie-algebra-valued one-form?

## closed as off-topic by Alex Degtyarev, Ben McKay, Stefan Kohl, Franz Lemmermeyer, user21574 Jun 25 '16 at 14:41

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• It appears that you are speaking about a connection on a bundle, which is a well known and well established concept. – Alex Degtyarev Jun 25 '16 at 9:54
• @AlexDegtyarev Yes, this definition is part of a theorem about connections on principal bundles. – Andy Miles Jun 25 '16 at 10:53

## 1 Answer

The expression $g^{-1} dg$ means the left invariant Maurer-Cartan 1-form on the Lie group $G$. It is only suggestive notation, but if $G$ is a subgroup of the general linear group $GL(n,\mathbb{R})$, then it has a precise meaning. The function $g : G \to \mathbb{R}^{n \times n}$ is the identity map, and then $dg : TG \to \mathbb{R}^{n \times n}$, and $g^{-1} dg$ is a 1-form on $G$ valued in $\mathfrak{g}$. So $g$ in his notation is the projection to $G$ from the local trivialization.

A clarification: in Nakahara, $g_i^{-1} d_P g_i$ really denotes the pullback of the Maurer-Cartan form through the map $g_i:\pi^{-1}(U_i)\to G$

• Thanks Ben! One last question, isn't the Maurer-Cartan 1-form defined as $(g^{-1})_*$, the pushforward of the action of $g^{-1}$? Only this way you get a map $T_gG\to T_eG\simeq\mathfrak g$. Shouldn't it be $g\mathrm{d}_P g^{-1}$ instead of $g^{-1}\mathrm{d}_P g$? – Andy Miles Jun 25 '16 at 10:58
• Both are Maurer-Cartan 1-forms. One is left invariant, the other right invariant. You can easily check which is which. – Ben McKay Jun 25 '16 at 12:09