Differential Geometry of Manifolds Connections and Covariant Differentiation, guaranteed and verified 100% Pass

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Connections and Covariant Differentiation

We saw an example earlier of a vector field being a contravariant vector. This
is true in general for vector fields on a manifold. Let 𝜒(𝑀) be the space of all
smooth vector fields on an 𝑛-dimensional manifold, 𝑀. If 𝑋 ∈ 𝜒(𝑀), then in local
coordinates 𝑥 1 , … , 𝑥 𝑛 we can write:
𝑛
𝜕
𝑋 = ∑ 𝐴𝑖 (𝑥 ) .
𝜕𝑥 𝑖
𝑖=1
𝜕 ⃗⃗⃗
𝜕Φ
⃗
⃗⃗ 𝑛 𝑘
Recall that if Φ: 𝑈 ⊆ ℝ → 𝑀 ⊆ ℝ , then 𝑖 means 𝑖 .
𝜕𝑥 𝜕𝑥
If 𝑥̅ 1 , … , 𝑥̅ 𝑛 are another set of local coordinates, then by the Chain Rule we can
write:
𝑛
𝜕 𝜕𝑥̅ 𝑗 𝜕 𝜕𝑥̅ 𝑗 𝜕
=∑ =
𝜕𝑥 𝑖 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑗 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑗
𝑗=1


Thus:

𝑛 𝑛 𝑛 𝑛 𝑛
𝑖( )
𝜕 𝑖
𝜕𝑥̅ 𝑗 𝜕 𝑖
𝜕𝑥̅ 𝑗 𝜕
𝑋 = ∑𝐴 𝑥 = ∑∑𝐴 = ∑ (∑ 𝐴 )
𝜕𝑥 𝑖 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑗 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑗
𝑖=1 𝑖=1 𝑗=1 𝑗=1 𝑖=1


So we know:

𝑛
𝑗
𝜕𝑥̅
𝐴̅ 𝑗 = ∑ 𝐴𝑖 𝑖
𝜕𝑥
𝑖=1


and 𝑋 is a contravariant vector.

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However, what happens if we differentiate a vector field? Is that also a tensor?

We know that a vector field:

𝑛
𝜕
𝑋 = ∑ 𝐴𝑖 (𝑥 )
𝜕𝑥 𝑖
𝑖=1

is a contravariant vector, so if we change coordinates we get:


𝑛
𝜕𝑥̅ 𝑗 𝜕𝑥̅ 𝑗
𝐴̅ 𝑗 (𝑥̅ ) = ∑ 𝐴𝑖 =𝐴𝑖
𝜕𝑥 𝑖 𝜕𝑥 𝑖
𝑖=1


Now let’s differentiate this equation with respect to 𝑥̅ 𝑘 .



𝜕𝐴̅ 𝑗 𝑖
𝜕 𝜕𝑥̅ 𝑗 𝜕𝑥̅ 𝑗 𝜕𝐴𝑖
=𝐴 ( )+ 𝑖
𝜕𝑥̅ 𝑘 𝜕𝑥̅ 𝑘 𝜕𝑥 𝑖 𝜕𝑥 𝜕𝑥̅ 𝑘


𝜕 𝜕𝑥̅ 𝑗 𝜕𝐴𝑖
Applying the Chain Rule to
𝜕𝑥̅ 𝑘
( 𝜕𝑥 𝑖 ) and 𝜕𝑥̅ 𝑘 we get:

𝜕𝐴̅ 𝑗 𝑖
𝜕 2 𝑥̅ 𝑗 𝜕𝑥 𝑙 𝜕𝑥̅ 𝑗 𝜕𝐴𝑖 𝜕𝑥 𝑙
=𝐴 +
𝜕𝑥̅ 𝑘 𝜕𝑥 𝑙 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑘 𝜕𝑥 𝑖 𝜕𝑥 𝑙 𝜕𝑥̅ 𝑘


𝜕𝐴̅ 𝑗 𝑖
𝜕 2 𝑥̅ 𝑗 𝜕𝑥 𝑙 𝜕𝑥̅ 𝑗 𝜕𝑥 𝑙 𝜕𝐴𝑖
=𝐴 + .
𝜕𝑥̅ 𝑘 𝜕𝑥 𝑙 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑘 𝜕𝑥 𝑖 𝜕𝑥̅ 𝑘 𝜕𝑥 𝑙