Local Form of Covariant Derivative Induced from a Connection one-form

Question

Let $P\rightarrow M$ be a Principal G-Bundle with $E=P\times_\rho V$ the associated vector bundle with $\rho$ a representation of $G$ on $GL(V)$. Also let $\omega$ be a connection one-form on $P$ i.e. $\omega\in\Omega(P; Lie(G))$. We have for some local trivialisation $\phi:P\rightarrow U\times G$,

$ (\phi^{-1})^*\omega=g^{-1}a_Ug+ g^{-1}dg $

with $a_U$ a lie algebra valued one form on $U$, and open set of $M$.

Let $\nabla$ be the covariant derivative induced by the connection $\omega$.

If we have a local trivivialisation $\psi:E\rightarrow U\times V$.

with a section $s\in\Omega^{0}(M; E)$. I am wondering how we derive the formula

$\psi(\nabla s)=(x, ds_U+\rho_*(a_u)s)$

in the local trivialisation. Please just comment if something is unclear. I have seen this formula in multiple sources and can't find a derivation. Appreciate any help that is given.

Answer 1

Put $E = P[\mathbf V;\rho]$ for the associated bundle. The representation $\rho : G \longmapsto \mathsf{GL}(\mathbf V) $ induces also a group action on the tangent space $\mathrm T\mathbf V = \mathbf V \times \mathbf V$ of $\mathbf V$, seen as a manifold: $$ \mathrm T G \times \mathrm T\mathbf V \ni (\mathrm L_g \xi; s,u) \longmapsto a(\mathrm L_g \xi; s, u)= (\rho(g)s,\rho(g)(\rho_*(\xi)s+u)) \in \mathrm T \mathbf V$$ enabling us to define the associated bundle $(\mathrm TP)[\mathrm T \mathbf V; a]$. Recall that $\mathrm T P$ is a $\mathrm T G$-bundle over $\mathrm T M$. One can easily prove that $(\mathrm TP)[\mathrm T \mathbf V; a] = \mathrm T{(P[\mathbf V;\rho])}$.

Given $\xi \in \mathfrak g$, put $P \ni p \longmapsto \xi^*(p)$ for the fundamental vector field on $P$. The representation $\rho$ also enables us to define the fundamental vector field over $\mathbf V$ as follows: $$ \mathbf V \ni s \longmapsto \xi^*(s) = (s, \rho_*(\xi)s) \in \mathrm T\mathbf V$$

Further, there is a natural identification of the vertical subbundle $\mathrm{V}E = \mathrm V( P[\mathbf V])$ of $\mathrm{T} E$ with the associated bundle $P [\mathrm T \mathbf V]$. Roughly this is (leaving the details of the proof) : $$ \mathrm V E \ni [ \xi^*(p); s, u ] ( = [0_p; s,u+\xi^*(s)])\longmapsto [p; s, u+\xi^*(s)] \in P[\mathrm T\mathbf V]$$

Now, define $\Phi : \mathrm T P \times \mathrm T \mathbf V \longmapsto \mathrm T P \times \mathrm T \mathbf V : (\Xi_p; s, u) \longmapsto \Phi(\Xi_p; s, u) = ((\omega\cdot\Xi_p)^*(p); s, u)$. But $((\omega\cdot\Xi_p)^*(p); s, u)$ is a vertical vector in the product space $\mathrm T P \times \mathrm T \mathbf V$. Whence $\Phi$ factors to a map $\hat \Phi$ on the quotient spaces $ \hat\Phi : \mathrm T(P[\mathbf V]) \cong (\mathrm TP)[\mathrm T \mathbf V; a] \longmapsto \mathrm VE \cong P[\mathrm T \mathbf V]$. This is the desired connection on the associated bundle. This map satisfies: $$ \hat\Phi([\Xi_p; s, u]) = [p; s, u + \rho_*(\omega\cdot\Xi_p)s]$$ Taking local coordinates will give that the christoffel symbol in the associated bundle is $$ \Gamma(x) \cdot h = -\rho_*(\omega(x)\cdot h)$$

Answer 2

I think what confuses you is you didn't clarify how the connection $1$-form induces the covariant derivative. From what I know, people tend to use the formula you gave as the definition of the induced covariant derivative after verifying the local covariant derivatives patch together with repsect to different trivializations. In other words, the local formula you gave for the connection $1$-form $\omega$ implies that the local formula you gave for the covariant derivative patches together to a covariant derivative on the whole bundle $E$.

Let me try to explain how to obtain the formula you mentioned from the usual definition of connections on principal bundles. For simplicity, let's assume the representation $\rho: G \to GL(V)$ is faithful. So we can regard the principal bundle $\pi: P \to M$ as a $G$-subbundle of the frame bundle of $E=P \times_{\rho} V$. A connection $A$ on $P$ can be thought as a splitting of the following sequence of equivariant bundles on $P$: \begin{equation}\tag{*} 0 \longrightarrow VP \longrightarrow TP \longrightarrow \pi^*TM \longrightarrow 0 \end{equation} where $VP = \ker(\pi_*:TP \to TM)$ is the vertical bundle. Note there is a canonical trivialization $\alpha: P \times \mathfrak{g} \to VP$ given by the derivative of $G$-right multiplication on $P$. The composition of the splitting $TP \to VP$ and $\alpha: VP \to \mathfrak{g}$ is the connection $1$-form $\omega_A \in \Omega^1(P, \mathfrak{g})$ you mentioned. This connection $1$-form is characterized by two properties:

(1) $\omega_A: TP \to \mathfrak{g}$ is $G$-equivariant, i.e. $r^*_g \omega_A = g^{-1} \omega_A g$, where $g \in G$ and $r_g$ is the right $g$-multiplication on $G$.

(2) The restriction of $\omega$ to each vertical fiber is identified with the Maurer-Cartan form under the indentification $\alpha$.

Given a local trivialization $\phi: P_U \to U \times G$ as you mentioned, the pull-back $(\phi^{-1})^*\omega_A$ decomposes accordingly from $TU \oplus TG$. Then (1) implies that that the restriction on $TU$ takes the form $g^{-1}a_Ug$ for some $a_U \in \Omega^1(U, \mathfrak{g})$, and (2) implies that the restriction on $TG \simeq G \times \mathfrak{g}$ takes the form $g^{-1}dg$. That's how to obtain the first formula you mentioned.

In order to deduce the local formula of the covariant derivative, I have to clarify what is the induced covariant derivative. Otherwise I can just take your local formula as the definition. Then the first formula implies it's independent of the choice of the local trivializations.

The splitting of the sequence (*) gives a decomposition $TP = H_A \oplus VP$ with $H_A$ a $G$-invariant subbundle referred to as the horizontal space. A section $s \in C^{\infty}(M, E)$ corresponds to an equivariant function $s^{\dagger}: P \to V$ from writing $s(x) = [p, s^{\dagger}(p)] \in P_x \times_\rho V = E_x$. Given $X \in T_xM$, one gets a unique horizontal lift $v_X \in H_A|_p$ for each $p \in \pi^{-1}(x)$. Then one can define the covariant derivatie $\nabla_X s$ by \begin{equation} (\nabla_X s)^{\dagger} := ds^{\dagger}(v_X). \end{equation} It's easy to check that $ds^{\dagger}(v_X): P \to V$ is $G$-equivariant, and $\nabla_X s$ satisfies the Leibniz's rule. In this way, a covariant derivative $\nabla$ on $E$ has been induced from a connection $A$ on $P$.

Now we verify your formula with a local trivialization $\psi: E_U \to U \times V$ given. After fixing a basis of $V$, the local trivialization $\psi$ provides a local frame of $E_U$, thus a local section $\sigma: U \to P$ (recall we have regarded $P$ as a subbundle of the frame bundle of $E$). Given a local section $s: U \to E_U$, the corresponding equivariant function $s^{\dagger}: P_U \to V$ is related by \begin{equation} s^{\dagger} \circ \sigma = \psi \circ s. \end{equation} Given $x \in T_xM$, since $\sigma$ is a local section, the horizontal lift of $x$ at $\sigma(x)$ is \begin{equation} v_X = d\sigma(X) - \alpha \circ \omega_A(d \sigma(X)):=d\sigma(X) - \xi_X. \end{equation} Then we can compute \begin{equation}\tag{**} (\nabla_X s)^{\dagger} = ds^{\dagger} v_X = ds^{\dagger}(d\sigma(X)) - ds^{\dagger} \xi_X. \end{equation} We note that $\xi_X \in VP$, differentiating the equivariance equation $s^{\dagger}(p\cdot g) = g^{-1} s^{\dagger}(p)$ with respect to $g = \exp(t\xi_X)$ at $t=0$ gives \begin{equation} ds^{\dagger} \xi_X = \xi_X s^{\dagger} = -\alpha^{-1}(\xi_X) s^{\dagger}. \end{equation} Substituting back to (**) and using $s^{\dagger} \circ \sigma = \psi \circ s$, we get \begin{equation} (\nabla_X s)^{\dagger}|_{\sigma(x)} = d(\psi \circ s) X + \sigma^*\omega_A(X). \end{equation} Write $s^{\psi} = \psi \circ s$ for the corresponding section under the trivialization. Then its covariant derivative is \begin{equation} \psi(\nabla s) = (\nabla s)^{\dagger} \circ \sigma = d s^{\psi} + \sigma^*\omega_A. \end{equation} When the trivialization $\psi: E_U \to U \times V$ is induced from the trivialization $\phi: P_U \to U \times G$, it's clear that $\sigma^*\omega_A = a_U$. This gives the second formula you mentioned.