# Connections on principal bundles and vector bundles

In Donaldson and Kronheimer's book on the geometry of four manifolds, a brief review of connections on principal bundles is given. Three equivalent definition are stated:
1) Via horizontal subspaces,
2) Via connections $1$-forms,
3) Via covariant derivatives on an associated vector bundle.
The authors then proceed to sketch very briefly the implications $2) \rightarrow 1)$ and $3) \rightarrow 1)$. I am looking for more detailed proofs.

I have easily found references for the equivalences between $1)$ and $2)$. However I am having a harder time finding references for anything involving $3)$.
Here for instance, the implication $2) \rightarrow 3)$ is proved using local descriptions of connection $1$-forms and covariant derivatives.
On the other hand here, the same implication is proved using, basically the fact that connections are the same thing as parallel transport.
I can't find any proof of $3) \rightarrow 1)$ or $3) \rightarrow 2)$ except the sketch given in Donaldson and Kronheimer's book.

Can anyone help?

Ps: I realize that the same question was asked here. The question was answered. However I have some difficulties navigating in the concepts of the related link (connections on general fibre bundles).

• I don't have my copy handy, but I think this is pretty well explained in Volume 2 of Spivak's Comprehensive Introduction to Differential Geometry. – Jack Lee Jul 27 '14 at 19:41
• Yes indeed it does, thank you! (Chapter 8 addendum 3 for anyone interested) – Lepanais Jul 27 '14 at 20:50
• I would suggest Kolar, Michor, Slovak. Their chapter on connections is very good and thorough. They discuss all three different ways of looking at connections, and also how principal connections induce connections on associated bundles. – ಠ_ಠ May 12 '16 at 8:04

## 1 Answer

Let $M$ be a manifold, $E$ a vector bundle on $M$. Write ${\rm Fr} \, (E)$ for the frame bundle. The following are equivalent:

1. a horizontal distrbution on ${\rm Fr} \, E$
2. equivariant parallel transport on ${\rm Fr} \, E$
3. Linear parallel transport on $E$

So all we need to do is check that linear paralell transport on $E$ is the same as specifying $\nabla_X s$. Suppose that we have linear paralell transport. Then we can define $$(\nabla_X s)_m = \lim_{t \to 0} \frac{T(s_{\gamma(t)})-s_m}{t}$$ where $\gamma$ is the integral curve for $X$ at $m$ and $T(s_{\gamma(t)})$ is the paralell transport of $s_{\gamma(t)}$ to $m$ along $\gamma^{-1}$. This defines a covariant derivative. On the otherhand, if we have a covariant derivative and a curve $\gamma : I \to M$, then we have a covariant derivative on $\gamma^* E$ defined by $$\nabla_T s := \nabla_{\gamma'T} s$$ (recall that $\gamma' : TI \to \gamma^* TM$ is a map of vector bundles.) This formula makes sense because sections of a pullback bundle are germs of sections over the image. But vector bundles on $I$ are smoothly trivial, so $\gamma^* E = I \times V$ for some vector space $V$. It follows from the fundamental theorem of ODEs that there is a paralell section $s(t) \in V$ defined for $t \in [0,\epsilon)$. Using Gronwall's inequality, this section can't blow up in finite time, so it must be defined on all $[0,1]$. This gives us paralell transport. We still need to check that these constructions are inverse to each other. SInce the covariant derivative and paralell transport are determined locally, we just need to check when $M = \mathbb{R}^n$. If you start with a system of parallel sections and define a covariant derivative $\nabla$, then your original sections are paralell for $\nabla$. For the other direction we need to do some work.

Let $V = \mathbb{R} \{ e_1,\dots,e_d \} \cong \mathbb{R}^d$ be a vector space and consider the vector bundle $\mathbb{R}^n \times V$ over $\mathbb{R}^n$. Suppose that $\nabla$ is a connection on $V$. Then we have that $$\nabla_{\partial / \partial x_k} e_j = \sum_i \Gamma_k^{ij} e_i$$ Consider the curve $\gamma(t) = (a_1,\dots,a_k -t,\dots,a_n)$ in $\mathbb{R}^n$. Then we have $$\nabla_{d/dt} e_j = - \sum_i \Gamma_k^{ij} e_i$$ This implies that a section $f(t)$ along $\gamma$ is paralell when $$f' = (\Gamma_k^{pq}) f$$ Now let $\widetilde{\nabla}$ be the connection induced from our paralell sections. Then $$(\widetilde{\nabla}_{\partial / \partial x_k} e_j)_{(a_i)} = \lim_{t \to 0} \frac{T(e_j) - e_j}{t} = \sum_i \Gamma_{k}^{ij}(a_1,\dots,a_n) e_i$$ Therefore $\widetilde{\nabla} = \nabla$. I think this answers your question, but I want to say a little more. The parallel transport of $e_j$ in the negative $x_k$ direction is $$e_j + (\Gamma_k^{ij})e_j t + O(t^2)$$ Therefore, the paralell transport of the frame $I = \{ e_1,\dots,e_d \}$ in the negative $x_k$ direction is $$I + (\Gamma_k^{ij}) t + O(t^2)$$ Therefore the splitting $\sigma : \pi^* TM \to T {\rm Fr} \, E$ is defined by $$(\partial / \partial x_k)_{(m,I)} \mapsto (\partial / \partial x^k, - \Gamma_k^{ij})$$ so the connection form $\omega$ is defined by $$\omega_{(m,I)} (\partial / \partial x_k, a^{ij}) = a^{ij} + \Gamma^{ij}_k$$ Locally, the connection form is defined by a $\mathfrak{gl}_n$-valued 1-form on $M$. This computation demonstrates that this form is exactly the Christoffel symbols $\Gamma^{ij}_k$.