Exterior derivatives and Stokes' theorem
Relate interior variation to oriented boundary integrals, while distinguishing locally exact forms from global potentials.
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The question for today
What you’ll understand
- Compute the exterior derivative of a coordinate one-form.
- Apply Stokes' theorem with a declared orientation.
- Distinguish closed forms from globally exact forms on domains with holes.
The fundamental theorem of calculus converts a derivative integrated over an interval into values at its endpoints. Stokes' theorem extends the same pattern to differential forms on oriented manifolds: interior differentiation corresponds to integration over the boundary. Its power comes from a consistent notion of orientation and a derivative that does not require a metric. The theorem also exposes a global question: a form can look like a derivative in every small patch without being the derivative of a single globally defined potential.
Exterior differentiation raises degree
For a smooth function f, the exterior derivative is its differential df. Extend the operation to forms using linearity, a graded product rule, and the fact that differentiating coordinate differentials gives zero. If a one-form on the plane is Pdx+Qdy, its derivative is . The minus sign comes from moving dy past dx, not from a separate convention invented for planar curl. Exterior differentiation is intrinsic and commutes with smooth pullback.
Applying d twice gives zero. In coordinates the mixed second derivatives cancel by antisymmetry for smooth coefficients. A form ω with dω=0 is called closed; a form equal to dη is called exact. Every exact form is closed, but the reverse implication is global and can fail. The Poincare lemma supplies a local converse on sufficiently small coordinate neighborhoods, or a global converse for positive-degree closed forms on a star-shaped open subset of Euclidean space.
Boundary orientation carries the sign
For a compact oriented smooth n-manifold M with boundary and a smooth (n−1)-form ω defined up to the boundary, Stokes' theorem identifies the integral of dω over M with the integral of ω over its boundary. A compact-support version handles appropriate noncompact settings. The boundary orientation is induced by the outward-normal-first convention: an outward transverse vector followed by a positively oriented boundary basis must give the orientation of M. No Riemannian unit normal is required to define this convention.
For the standard dx∧dy orientation of a planar region, this convention makes the outer boundary counterclockwise and inner-hole boundaries clockwise. Reversing the orientation of the region reverses both sides of Stokes' theorem. If the region is closed and has no boundary, the integral of every exact top-degree form is zero. Green's theorem, the classical curl theorem, and the divergence theorem emerge after translating their vector fields into forms, sometimes using a metric to make that translation.
A hole can obstruct a global potential
On the punctured plane, the one-form is smooth and closed. Restricted to an angular coordinate patch it is dθ. Around the positively oriented unit circle, however, its integral is 2π. If α were the differential of a globally defined real-valued smooth function, its integral around every closed curve would be zero. The nonzero circulation therefore proves that local angle functions cannot be assembled into one global real angle.
There is no contradiction with Stokes' theorem. The filled disk is not contained in the punctured domain where α is smooth. Applying the theorem to an annulus instead includes an inner boundary whose opposite orientation cancels the outer circulation. This is a reliable way to debug apparent paradoxes: check smoothness on the entire integration region, include every boundary component, and state orientations before simplifying integrals. A suggestive arrow plot cannot replace these domain checks.
Work it through
Verify Stokes on a unit square
Use the standard orientation on M=[0,1]² and the one-form . Compare its interior derivative integral with its counterclockwise boundary integral.
Differentiate to obtain . Its integral over the square is its area, exactly one. The square is piecewise smooth; the corresponding piecewise-smooth Stokes version applies to its edges.
On the bottom and top edges, dy=0, so their contributions vanish. On the right edge x=1 and y increases from zero to one, giving integral one.
On the left edge x=0, so the contribution vanishes even though its orientation runs downward. Adding the four oriented edge contributions gives 0+1+0+0=1, matching the interior integral.
Both integrals equal 1. Reversing the orientation would make both equal −1.
The living notebook
Explore the model
Interactive lab · fields
Predict a circulation sign
Set source strength q to zero and choose positive rotation Ω in the softened planar field lab. The remaining vector field is Ω(−y,x). Its Euclidean dual one-form has derivative 2Ω dx∧dy, so predict the sign of circulation around a counterclockwise circle. The arrows illustrate a planar field; the lab does not numerically integrate boundaries or solve differential-form cohomology.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forDomain and orientation determine how a local derivative becomes a global boundary statement.
Pause & reason
Check your understanding
Your turn
Try it yourself
On the counterclockwise boundary of the unit square, compute the integral of using exterior differentiation, then identify the nonzero edge.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Remember that dy∧dx=−dx∧dy, and check the direction of the top edge.
Compare with the solution
We have , whose square integral is −1. Directly, only the top edge contributes: y=1 and x decreases from one to zero, giving −1. The bottom edge has y=0 and the vertical edges have dx=0.
Keep the idea
Take it with you
- Exterior differentiation is intrinsic and satisfies d²=0.
- Stokes requires suitable smoothness, a valid domain, and induced boundary orientation.
- Closed forms may lack global potentials when topology obstructs patching.
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