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From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

From Math 2220 Class 40

• Dr. Allen Back
• Dec. 3, 2014

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

Differential forms are sums of wedge products of the “basis 1-forms” dx, dy, and dz. They are kinds of tensors generalizing

• rdinary scalar functions and vector fields. They have a

“skew-symmetry” property so that for 1 − forms (e.g. dx or dy) ω ∧ η = −η ∧ ω. This implies dx ∧ dx = 0 for example. Marsden and Tromba drop the ∧ symbol, writing dz dx for dz ∧ dx, etc.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

InR3 all differential forms have an interpretation as either an

• rdinary function or a vector field.

0-forms are scalar functions - f (x, y, z) on R3. 1-forms are linear combinations P dx + Q dy + R dz where P, Q, and R are functions of three variables (x, y, z). Think of this as associated with the vector field (P, Q, R). 2-forms are linear combinations P dy ∧ dz + Q dz ∧ dx + R dx ∧ dy. Think of this as associated with the vector field (P, Q, R). 3-forms are of the form f (x, y, z) dx ∧ dy ∧ dz. Think of this as associated with the scalar function f (x, y, z).

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

There is an operation exterior differentiation d which takes any k form to a k + 1 form. For ordinary functions f (x, y, z) , it looks like the expression for differentials: df = fx dx + fy dy + fz dz.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

d2 = 0; i.e. for any (C 2) k-form ω d(dω) = 0 So e.g. d(dx) = 0. In the end this is because of the independence of order for mixed partials.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

Exterior derivative obeys the following anti-commutativity rule for wedge products: d (η ∧ ω) = (dη) ∧ ω + (−1)kη ∧ (dω). if η is a k-form and ω is an l-form. ω ∧ η = (−1)klη ∧ ω as well.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

For differential forms, the operations of grad, div, and curl are all special cases of exterior differentiation d ! grad is d from 0-forms (functions) to 1-forms as long as we identify the 1-form P dx + Q dy + R dz with the vector field (P, Q, R). curl is d from 1-forms to 2-forms as long as we identify the 1 and 2-forms with their associated vector fields. div is d from 2-forms (vector fields in disguise) to 3-forms (functions in disguise.)

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Algebra of Differential Forms

Let D = [0, 1] × [0, 1] ⊂ R2 be a square, Φ : D → R3 a parameterization of a piece of surface S, and ω a 1-form on R3. There is a pullback operation Φ∗ taking a 2-form (or more generally any k-form) on the image of Φ to a 2-form on R2. Φ(u, v) = (u3 + v3, u − v, u + v), Φ∗(xy dx∧dy) = (u3+v3)(u−v) (3u2 du + 3v2 dv)∧(du − dv).

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

Integration of a differential k-form over a rectangle in Rk is the same as an ordinary k-fold multiple integral. We use chains (essentially formal sums of parameterizations, with coefficients like ±1 to express orientation) to reduce the integration of any differential k-form (over e.g. a path, surface,

• r region in R3) to an ordinary multiple integral.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

Let D = [0, 1] × [0, 1] be a square, Φ : D → R3 a parameterization of a piece of surface S, and ω a 1-form on R3. Assume Φ(∂D) = ∂(Φ(D)) ∂S. Then

• Φ(∂D)

ω =

Φ(D)

d ω Stokes in R3 follows from

• ∂D

Φ∗ω =

D

d Φ∗ω

• n the square in R2. (A simple version of Green.)

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

Here Φ∗ refers to the pullback operation; e.g. for Φ(u, v) = (u3 + v3, u − v, u + v), Φ∗(xy dx∧dy) = (u3+v3)(u−v) (3u2 du + 3v2 dv)∧(du − dv).

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

Integration of differential forms is basically defined in terms of pullback. If c : [a, b] → R3 with c(t) = (x(t), y(t), z(t)), then c∗dx = x′(t)dt, etc. and so the integral of the 1-form η = P dx + Q dy + R dz over C corresponding to c is essentially defined by

• C

η =

• C

P dx + Q dy + R dz =

• C

c∗(P dx + Q dy + R dz) = b

a

(Px′ + Qy′ + Rz′) dt in line with one of our older classical notations/computation styles for line integrals.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

Now let’s look at the differential form version of the surface integral

• S

(x, y, z) · ˆ n dS

• ver part of the paraboloid z = 1 − x2 − y2 with z ≥ 0.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

The associated 2-form to the vector field (x, y, z) is µ = x dy ∧ dz − x dz ∧ dx + z dz ∧ dy. For the parameterization c(r, θ) = (r cos θ, r sin θ, 1 − r2), we have dy = cos θ dr − r sin θ dθ, dz = −2r dr etc., so

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Integration and Stokes for Differential Forms

• S

µ = =

• D

c∗(x dy ∧ dz − x dz ∧ dx + z dx ∧ dy) = 1 2π r cos θ(cos θ dr − r sin θ dθ) ∧ (−2r dr) + . . . which (thinking about ˆ n in terms of cross products) can be seen to agree with our usual!

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Integration of a conservative vector field cartoon.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Green’s Theorem cartoon.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Stokes’ Theorem cartoon.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Both sides of Stokes involve integrals whose signs depend on the orientation, so to have a chance at being true, there needs to be some compatibility between the choices. The rule is that, from the “positive” side of the surface, (i.e. the side chosen by the orientation), the positive direction of the curve has the inside of the surface to the left. As with all orientations, this can be expressed in terms of the sign of some determinant. (Or in many cases in terms of the sign of some combination of dot and cross products.)

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Problem: Let S be the portion of the unit sphere x2 + y2 + z2 = 1 with z ≥ 0. Orient the hemisphere with an upward unit normal. Let F(x, y, z) = (y, −x, ez2). Calculate the value of the surface integral

• S

∇ × F · ˆ n dS.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Gauss’ Theorem field cartoon.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

The surface integral side of Gauss depends on the orientation, so there needs to be a choice making the theorem true. The rule is that the normal to the surface should point outward from the inside of the region. (For the 2d analogue of Gauss (really an application of Green’s)

• C
• F · ˆ

n =

• inside

(Px + Qy) dx dy we also use an outward normal, where here C must of course be a closed curve.

From Math 2220 Class 40 V1df Algebra of Differential Forms Integration and Stokes for Differential Forms Stokes and Gauss Traditionally

Stokes and Gauss Traditionally

Problem: Let W be the solid cylinder x2 + y2 ≤ 3 with 1 ≤ z ≤ 5. Let F(x, y, z) = (x, y, z). Find the value of the surface integral

• ∂W
• F · ˆ

n dS.