Calculus on Manifolds Stokes Theorem on Manifolds, guaranteed and verified 100% Pass

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Stokes’ Theorem on Manifolds

Stokes’ Theorem for Compact Oriented Manifolds:
If 𝑀 is a compact oriented 𝑘-dimensional manifold with boundary and 𝜔
is a (𝑘 − 1)-form on 𝑀, then

∫ 𝑑𝜔 = ∫ 𝜔.
𝑀 𝜕𝑀


We saw earlier that this theorem is true when 𝑀 is a 𝑘-chain. To get this result for a
compact manifold, we take a partition of unity {𝜓𝛼 }𝛼∈𝐼 for 𝑀 subordinate to an open
cover 𝒰. So for any point 𝑥 ∈ 𝑀:


0 = 𝑑 (1) = 𝑑 (∑ 𝜓𝛼 ) = ∑ 𝑑𝜓𝛼
𝛼∈I 𝛼∈I
Thus

∑ 𝑑𝜓𝛼 ∧ 𝜔 = 0.
𝛼∈I
Since 𝑀 is compact, 𝒰 has a finite subcover. The above sum is finite and


∑ ∫ 𝑑𝜓𝛼 ∧ 𝜔 = 0.
𝛼∈I 𝑀
Therefore:


∫ 𝑑𝜔 = ∑ ∫ 𝜓𝛼 (𝑑𝜔) = ∑ ∫ (𝑑𝜓𝛼 ∧ 𝜔 + 𝜓𝛼 ∧ 𝑑𝜔)
𝑀 𝛼∈I 𝑀 𝛼∈I 𝑀




= ∑ ∫ 𝑑 (𝜓𝛼 𝜔) = ∑ ∫ 𝜓𝛼 𝜔 = ∫ 𝜔.
𝛼∈I 𝑀 𝛼∈I 𝜕𝑀 𝜕𝑀

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Corollary 1: If 𝑀 is a compact manifold without boundary (e.g. a sphere
or torus) and 𝜔 is a 𝑘 − 1 form, then:

∫ 𝑑𝜔 = 0.
𝑀


Proof:

∫ 𝑑𝜔 = ∫ 𝜔
𝑀 𝜕𝑀
But 𝜕𝑀 = 𝜙, so:

∫ 𝜔 = 0.
𝜕𝑀



Corollary 2: If 𝑀 is a compact manifold (with or without boundary) and
𝜔 is a 𝑘 − 1 form that is closed (i.e. 𝑑𝜔 = 0), then:

∫ 𝜔 = 0.
𝜕𝑀


Proof:

∫ 𝜔 = ∫ 𝑑𝜔 = ∫ 0 = 0.
𝜕𝑀 𝑀 𝑀



Notice that if 𝑀 is not compact, then Stokes’ Theorem need not hold. For example, if
𝑀 = 𝐷 − (0, 0), where 𝐷 = {(𝑥, 𝑦) ∈ ℝ2 |𝑥 2 + 𝑦 2 ≤ 1} and
−𝑦 𝑥
𝜔 = 𝑥 2 +𝑦2 𝑑𝑥 + 𝑥 2 +𝑦2 𝑑𝑦 , then 𝑑𝜔 = 0. But we have:


∫ 𝜔 = 2𝜋 ≠ ∫ 𝑑𝜔 = 0.
𝜕𝑀 𝑀