Parallel section along a curve

A section along a curve whose covariant derivative along the curve vanishes.
Parallel section along a curve

Let π:EM \pi:E\to M be a smooth vector bundle over a MM, equipped with a \nabla.

Let IRI\subset \mathbb{R} be an interval and γ:IM\gamma:I\to M a smooth curve. A section of EE along γ\gamma is a smooth map s:IEs:I\to E such that πs=γ\pi\circ s=\gamma. Using \nabla, one defines the covariant derivative of ss along γ\gamma by

Dsdt(t)γ˙(t)s~γ(t), \frac{D s}{dt}(t) \coloneqq \nabla_{\dot\gamma(t)} \tilde s \big|_{\gamma(t)},

where s~\tilde s is any smooth extension of ss to a neighborhood of γ(t)\gamma(t) in MM. This is well-defined (independent of the choice of extension) by the locality and C(M)C^\infty(M)-linearity properties of a connection.

A section ss along γ\gamma is called parallel along γ\gamma if

Dsdt(t)=0for all tI. \frac{D s}{dt}(t)=0 \quad \text{for all } t\in I.

Equivalently: given t0It_0\in I and v0Eγ(t0)v_0\in E_{\gamma(t_0)}, there is a unique parallel section ss along γ\gamma with s(t0)=v0s(t_0)=v_0. The resulting identification of fibers is the determined by \nabla along γ\gamma.

Examples

  1. Trivial bundle over an interval. Let E=I×RnIE=I\times \mathbb{R}^n\to I with the standard (componentwise) connection. A section along the identity curve γ(t)=t\gamma(t)=t is a map s(t)=(t,v(t))s(t)=(t,v(t)). The parallel condition is v(t)=0v'(t)=0, so parallel sections are exactly the constant vectors v(t)v0v(t)\equiv v_0.

  2. Tangent bundle of Euclidean space. Take E=TRmE= T\mathbb{R}^m with the standard flat connection. Along any smooth curve γ\gamma, a vector field V(t)Tγ(t)RmRmV(t)\in T_{\gamma(t)}\mathbb{R}^m\cong \mathbb{R}^m is parallel iff its coordinate vector in Rm\mathbb{R}^m is constant in tt.

  3. A rank-one connection written as an ODE. On a trivial real line bundle E=M×RE=M\times \mathbb{R}, any connection can be written locally as =d+α\nabla=d+\alpha for a 1-form α\alpha on MM. Along γ\gamma, a section is a function f(t)f(t), and the parallel condition becomes

    f(t)+α(γ˙(t))f(t)=0, f'(t) + \alpha(\dot\gamma(t))\,f(t)=0,

    so f(t)=f(t0)exp ⁣(t0tα(γ˙(τ))dτ)f(t)=f(t_0)\exp\!\left(-\int_{t_0}^t \alpha(\dot\gamma(\tau))\,d\tau\right).