1 Introduction Consider the following integral theorems we have used in this course: The Fundamental Theorem of Calculus: Z b f 0 (x) dx = f (b) − f (a) a The Fundamental Theorem for Line Integrals: Z b ∇f (r(t)) · r0 (t) dt = f (r(b)) − f (r(a)) a Green’s Theorem: ZZ D ∂Q ∂P − ∂x ∂y Z dA = P dx + Q dy C Stokes’ Theorem: ZZ Z curl F · dS = S F · dr C The Divergence Theorem: ZZZ ZZ F · dS div F dV = E S All these theorems relate the integral of an expression involving derivatives to an integral over a lower-dimensional set. (In the case of the first two theorems, think of plugging points into a function as integrating that function over a 0-dimensional set.) Using the language of differential forms, we will see that these theorems are special cases of a single theorem valid in higher dimensions. Johns Hopkins University, CTYOnline 2 Differential Forms Differential Forms on R3 Differential forms may be defined on Rn for any n ≥ 1, but in this note we generally focus on the case n = 3. Definition: A differential form on R3 is an expression involving scalar functions and the differentials dx, dy, and dz of one of the four types below. 0-form: f (x, y, z) 1-form: P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz 2-form: P (x, y, z) dy dz + Q(x, y, z) dz dx + R(x, y, z) dx dy 3-form: f (x, y, z) dx dy dz Note that f (x, y, z), P (x, y, z), Q(x, y, z), and R(x, y, z) are scalar functions. Depending on the context, we will often require that these functions be differentiable, twice differentiable, etc., but will usually not explicitly say so. Example: x2 + yz is a differential 0-form, a scalar function. Example: xy dx + sin(x + z) dz is a differential 1-form. Example: (x2 + y 2 + z 2 )−3/2 dx dy dz is a differential 3-form defined for (x, y, z) 6= (0, 0, 0). 2.1 Operations on Differential Forms Differential forms may be added, subtracted, multiplied, differentiated, and integrated. 2.1.1 Addition and Subtraction Two or more k-forms may be added or subtracted by collecting like terms. Example: (2x dx − xy dz) + (dy + 2z dz) = 2x dx + dy + (2z − xy) dz 2 Johns Hopkins University, CTYOnline 2.1.2 Differential Forms Multiplication The product f ω of a 0-form f and a k-form ω may be defined in the obvious way. Example: If f (x, y, z) = x2 and ω = y dx − z dy, then the product of f and ω is f ω = x2 (y dx − z dy) = x2 y dx − x2 z dy A general multiplication operation for differential forms is given by the wedge product: Definition: The wedge product of a k-form ω and an l-form η (with 0 ≤ k + l ≤ 3) is a (k + l)-form ω ∧ η. The wedge product satisfies the following properties: (1) (Distributivity) If ω1 and ω2 are k-forms and η is an l-form, then (ω1 + ω2 ) ∧ η = ω1 ∧ η + ω2 ∧ η (2) (Anticommutativity) If ω is a k-form and η is an l-form, then ω ∧ η = (−1)kl (η ∧ ω) (3) (Associativity) If ω1 , ω2 , ω3 are k1 -, k2 -, and k3 -forms, respectively, then (ω1 ∧ ω2 ) ∧ ω3 = ω1 ∧ (ω2 ∧ ω3 ) (4) If f is a 0-form, ω is a k-form, and η is an l-form, then (f ω) ∧ η = ω ∧ (f η) = f (ω ∧ η) (5) If ω is a k-form, then 0∧ω =ω∧0=0 3 Johns Hopkins University, CTYOnline Differential Forms (6) The following rules hold for basic 1-forms: dx ∧ dx = dy ∧ dy = dz ∧ dz = 0 dy ∧ dz = dy dz = (−1)(dz ∧ dy) dz ∧ dx = dz dx = (−1)(dx ∧ dz) dx ∧ dy = dx dy = (−1)(dy ∧ dx) (dx ∧ dy) ∧ dz = dx dy dz = dx ∧ (dy ∧ dz) The properties of the wedge product allow us to simplify more complicated products. Example: If ω = dx and η = dz dx, then ω ∧ η = dx ∧ (dz dx) = dx ∧ (dz ∧ dx) = dx ∧ ((−1)(dx ∧ dz)) = (−1)(dx ∧ (dx ∧ dz)) = (−1)((dx ∧ dx) ∧ dz) = (−1)(0 ∧ dz) = (−1)(0) =0 In fact, property (2) and the first part of property (6) imply that any wedge product of basic forms with a repeated dx, dy, or dz is 0. For example, (dy dz) ∧ dz = 0 because of the repeated dz, and dx ∧ (dy dx) = 0 because of the repeated dx. Example: If ω = dx + y dz and η = x dy − yz dz, then ω ∧ η = (dx + y dz) ∧ (x dy − yz dz) = dx ∧ (x dy) + dx ∧ (−yz dz) + (y dz) ∧ (x dy) + (y dz) ∧ (−yz dz) = x (dx ∧ dy) − yz (dx ∧ dz) + xy (dz ∧ dy) − y 2 z (dz ∧ dz) = x(dx ∧ dy) + yz (dz ∧ dx) − xy (dy ∧ dz) = −xy dy dz + yz dz dx + x dx dy 4 Johns Hopkins University, CTYOnline Differential Forms Example: If ω = y 2 dx − dz and η = −z dz dx + x dx dy, then ω ∧ η = (y 2 dx − dz) ∧ (−z dz dx + x dx dy) = (y 2 dx − dz) ∧ (−z dz ∧ dx + x dx ∧ dy) = −y 2 dx ∧ dz ∧ dx + xy 2 dx ∧ dx ∧ dy + z dz ∧ dz ∧ dx − x dz ∧ dx ∧ dy = −x dz ∧ dx ∧ dy = x dx ∧ dz ∧ dy = −x dx ∧ dy ∧ dz = −x dx dy dz 2.1.3 Differentiation The derivative of a k-form ω is a (k + 1)-form denoted by dω. Because there are no nonzero 4-forms on R3 , if ω is a 3-form, then dω = 0. The derivative satisfies the following properties: (1) If f is a 0-form, then df = ∂f ∂f ∂f dx + dy + dz ∂x ∂y ∂z (2) (Linearity) If ω1 and ω2 are k-forms, then d(ω1 + ω2 ) = dω1 + dω2 (3) If ω is a k-form and η is an l-form, then d(ω ∧ η) = (dω ∧ η) + (−1)k (ω ∧ dη) (4) If ω is a k-form, then d(dω) = 0 5 Johns Hopkins University, CTYOnline Differential Forms Remark: By property (1), the basic 1-forms dx, dy, and dz are the derivatives of the 0forms x, y, and z, respectively. It follows from property (4) that d(dx) = d(dy) = d(dz) = 0 We may use properties of the derivative to find the derivative of any k-form. Example: If ω = dx dy, then dω = d(dx dy) = d(dx ∧ dy) = (d(dx)) ∧ dy − dx ∧ (d(dy)) = 0 ∧ dy − dx ∧ 0 =0 Similarly, the derivatives of dy dz and dz dx are 0. This simplifies the computation of derivatives, as the next examples show. Example: If f is a 0-form and ω is a basic 1- or 2-form (dx, dy, dz, dx dy, dy dz, or dz dx), then d(f ω) = d(f ∧ ω) = (df ) ∧ ω + f ∧ (dω) = (df ) ∧ ω + f ∧ 0 = (df ) ∧ ω Example: If ω = (xy + z 2 ) dy dz, then by the result of the previous example, 6 Johns Hopkins University, CTYOnline Differential Forms dω = d(xy + z 2 ) ∧ (dy dz) = (y dx + x dy + 2z dz) ∧ (dy dz) = y dx ∧ dy ∧ dz + x dy ∧ dy ∧ dz + 2z dz ∧ dy ∧ dz = y dx ∧ dy ∧ dz = y dx dy dz Example: If ω = y 2 dx + xz 2 dz, then dω = d(y 2 dx) + d(xz 2 dz) = d(y 2 ) ∧ dx + d(xz 2 ) ∧ dz = (2y dy) ∧ dx + (z 2 dx + 2xz dz) ∧ dz = 2y dy ∧ dx + z 2 dx ∧ dz + 2xz dz ∧ dz = −2y dx ∧ dy − z 2 dz ∧ dx = −z 2 dz dx − 2y dx dy 2.1.4 Integration A k-form may be integrated over a k-dimensional subset of R3 , assuming that the form and the subset are “nice” enough. We will consider integrals of k-forms for k = 0, 1, 2, 3 separately. Integrals of 0-forms A 0-form f may be “integrated” over an oriented 0-dimensional subset of R3 . In this context, a 0-dimensional subset of R3 is simply a finite set of points, an orientation of such a subset is a labeling of each point in the set as either “positive” or “negative,” and integration is simply function evaluation. Suppose P is a point in a 0-dimensional subset S of R3 . If P is “positive,” then P contributes f (P ) to the integral of f over S, and if P is “negative,” then P contributes −f (P ) to the integral. Example: The integral of f (x, y, z) = xyz over the set {(1, 1, 1), (2, 2, 2)}, where (1, 1, 1) is “positive” and (2, 2, 2) is “negative,” is f (1, 1, 1) − f (2, 2, 2) = 1 − 8 = −7 7 Johns Hopkins University, CTYOnline Differential Forms If both points are instead “positive,” then the integral is f (1, 1, 1) + f (2, 2, 2) = 1 + 8 = 9 Integrals of 1-forms Let ω = P dx + Q dy + R dz be a 1-form, and let C be an oriented simple curve in R3 . We define the integral of ω over C as a line integral: Z Z ω= P dx + Q dy + R dz C C We can think of a 1-form ω as a function from the set of oriented, smooth, simple curves R in R3 to the set of real numbers. This function takes a curve C to the real number C ω. A similar observation holds for k-forms for any value of k. In this way, differential forms are generalizations of scalar functions. Example: Let ω = y dx + x dy + dz, and let C be any oriented, smooth, simple closed curve. Then Z Z y dx + x dy + dz = 0 ω= C C because F(x, y, z) = hy, x, 1i is a conservative vector field and C is closed. Integrals of 2-forms Let ω = P dy dz + Q dz dx + R dx dy be a 2-form, and let S be an oriented smooth surface in R3 parametrized by 8 Johns Hopkins University, CTYOnline Differential Forms r(u, v) = hx(u, v), y(u, v), z(u, v)i, (u, v) ∈ D Letting F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i, we define the integral of ω over S as a surface integral: ZZ ZZ F · dS ω= S S ZZ F(r(u, v)) · (ru × rv ) du dv = D ZZ = D ∂(z, x) ∂(x, y) ∂(y, z) + Q(r(u, v)) + R(r(u, v)) P (r(u, v)) du dv ∂(u, v) ∂(u, v) ∂(u, v) where ∂y ∂(y, z) ∂u = ∂(u, v) ∂z ∂u ∂y ∂v , ∂z ∂v ∂z ∂(z, x) ∂u = ∂(u, v) ∂x ∂u ∂z ∂v , ∂x ∂v ∂x ∂(x, y) ∂u = ∂(u, v) ∂y ∂u ∂x ∂v ∂y ∂v Example: Let ω = x dy dz + 2 dz dx + 2z dx dz and F(x, y, z) = hx, 2, 2zi, and let S be the surface of the unit sphere in R3 with positive (outward) orientation. Then by the Divergence Theorem, the integral of ω over S is 9 Johns Hopkins University, CTYOnline Differential Forms ZZ ZZ F · dS ω= S S ZZZ = div F dV E ZZZ = 3 dV E = 3 × (Volume of the unit sphere) = 4π Integrals of 3-forms Let ω = f dx dy dz and let E be a simple solid region in R3 . The integral of ω over E is given by the triple integral of f over E: ZZZ ZZZ ω= f dV E E Example: The integral of ω = 4x dx dy dz over the unit cube E in R3 is ZZZ Z 1Z 1Z 1 ω= 4x dx dy dz = 2 E 0 0 0 10 Johns Hopkins University, CTYOnline 3 Differential Forms General Stokes’ Theorem It turns out that there is a very simple formula relating an integral of a differential form ω to an integral of dω. To state this result, we first unify sets of points, curves, surfaces, and 3D regions in R3 using the language of manifolds. Definition: An oriented k-manifold in R3 is one of the following: an oriented set of points, an oriented smooth simple curve, an oriented smooth surface, or a simple solid region in R3 . The number k is the dimension of the manifold. The boundary (if any) of the manifold is denoted by ∂M . The boundary ∂M of an oriented k-manifold M (k = 1, 2, 3) inherits an orientation from M . We handle separately the cases k = 1, 2, and 3. Boundary of an Oriented 1-Manifold Let C be an oriented simple curve in R3 , beginning at the point P0 and ending at the point P1 . If P0 = P1 , then ∂C is empty. Otherwise, ∂C is the set {P0 , P1 }, where P0 is “negative” and P1 is “positive.” Boundary of an Oriented 2-Manifold Let S be an oriented surface with boundary curve ∂S. (Note that not every surface has a “nice” boundary, or even a boundary at all; we make the assumption that ∂S is a smooth, simple, closed curve.) Recall that choosing an orientation of S is equivalent to choosing a continuous field of unit normal vectors on S. The induced orientation of ∂S is such that if you travel along ∂S in the direction of its orientation in such a way that the unit normal vectors at nearby points of S are pointing from your feet toward your head, then the surface S is to your left. This is nothing new; the induced orientation on ∂S is the same that is used when applying the version of Stokes’ Theorem we have seen previously in this course. 11 Johns Hopkins University, CTYOnline Differential Forms Boundary of an Oriented 3-Manifold Let E be a simple solid region in R3 with boundary surface ∂E. Then we take ∂E to be oriented outward. Now we are ready to state the General Stokes’ Theorem. Main Theorem (General Stokes’ Theorem): Let M be an oriented k-manifold (k = 1, 2, 3) in R3 with boundary ∂M , and let ω be a (k − 1)-form. Then Z Z ω= dω ∂M M We will see below that depending on the value of k, the statement of the General Stokes’ Theorem is equivalent to either the Fundamental Theorem for Line Integrals, (usual) Stokes’ Theorem, or the Divergence Theorem. The benefit of combining these theorems into the General Stokes’ Theorem is not only that it is pretty, but also that the General Stokes’ Theorem generalizes to higher dimensions if we make the small effort to define differential forms and oriented manifolds in Rn . Unfortunately, the proof of the General Stokes’ Theorem in Rn is beyond the scope of this course. 3.1 The Fundamental Theorem of Calculus Using differential forms on R, we will see that the Fundamental Theorem of Calculus from single-variable calculus is a special case of the General Stokes’ Theorem. On R, there are only two kinds of differential forms, 0-forms and 1-forms. A 0-form is a scalar function f (x), and a 1-form is an expression of the form f (x) dx. The derivative of 12 Johns Hopkins University, CTYOnline Differential Forms a 0-form f (x) is simply df = f 0 (x) dx Given a 0-form f (x) and a closed interval [a, b], the General Stokes’ Theorem says that Z Z f= df {a,b} [a,b] The integral on the left side is the integral of the 0-form f over the 0-dimensional set {a, b}, where a is “negative” and b is “positive.” Its value is simply f (b)−f (a). The integral on the right is the integral of the function f 0 (x) on the interval [a, b]. Rewriting the equality, we get Z b f (b) − f (a) = f 0 (x) dx a which is the Fundamental Theorem of Calculus! 3.2 The Fundamental Theorem for Line Integrals We now return to R3 . Let C be an oriented simple curve in R3 with endpoints P0 and P1 . Assume for simplicity that P0 6= P1 (otherwise, C has no boundary). Let F(x, y, z) be a conservative vector field. It follows that there exists a scalar function f (x, y, z) such that ∂f ∂f ∂f , , F(x, y, z) = ∂x ∂y ∂z Applying the General Stokes’ Theorem to the 0-form f and the curve C, we obtain Z Z f= df {P0 ,P1 } C The integral on the left side is taken over the 0-dimensional set {P0 , P1 }, where P0 is “negative” and P1 is “positive.” Its value is f (P1 ) − f (P0 ). Rewriting the integral on the right side, we get 13 Johns Hopkins University, CTYOnline Z Z df = C C Differential Forms ∂f ∂f ∂f dx + dy + dz ∂x ∂y ∂z Z F · dr = C The statement of the General Stokes’ Theorem applied to f and C therefore simplifies to Z f (P1 ) − f (P0 ) = F · dr C which is the Fundamental Theorem for Line Integrals! 3.3 Green’s Theorem In this section, we will consider differential 1-forms and 2-forms on R2 . These, respectively, take the form P (x, y) dx + Q(x, y) dy and f (x, y) dx dy Let ω = P dx + Q dy be a 1-form, and let C be a smooth, simple, closed curve in R2 bounding a region D. We give C the positive (counterclockwise) orientation. Then by the General Stokes’ Theorem, ZZ Z dω ω= C D 14 Johns Hopkins University, CTYOnline Differential Forms To see that this is Green’s Theorem, we must simplify dω: dω = d(P dx) + d(Q dy) = d(P ∧ dx) + d(Q ∧ dy) = (dP ) ∧ dx + (dQ) ∧ dy = ∂P ∂Q ∂Q ∂P dx + dy ∧ dx + dx + dy ∧ dy ∂x ∂y ∂x ∂y = ∂P ∂P ∂Q ∂Q dx ∧ dx + dy ∧ dx + dx ∧ dy + dy ∧ dy ∂x ∂y ∂x ∂y = ∂Q ∂P dy ∧ dx + dx ∧ dy ∂y ∂x =− = = ∂P ∂Q dx ∧ dy + dx ∧ dy ∂y ∂x ∂Q ∂P − ∂x ∂y ∂Q ∂P − ∂x ∂y dx ∧ dy dx dy Looks familiar, right? The statement of the General Stokes’ Theorem above becomes Z ZZ ∂Q ∂P P dx + Q dy = − dx dy ∂x ∂y C D which is Green’s Theorem! 3.4 Stokes’ Theorem We again return to R3 , and consider a smooth oriented surface S with smooth, simple, closed boundary C with orientation induced from the orientation on S. Let ω = P dx + Q dy + R dz be a 1-form. Then the statement of the General Stokes’ Theorem 15 Johns Hopkins University, CTYOnline Differential Forms applied to S and ω is: Z ZZ ω= dω C S As before, let’s simplify dω: dω = d(P dx) + d(Q dy) + d(R dz) = d(P ∧ dx) + d(Q ∧ dy) + d(R ∧ dz) = (dP ) ∧ dx + (dQ) ∧ dy + (dR) ∧ dz = ∂P ∂P ∂P dx + dy + dz ∂x ∂y ∂z + + = ∧ dx ∂Q ∂Q ∂Q dx + dy + dz ∂x ∂y ∂z ∂R ∂R ∂R dx + dy + dz ∂x ∂y ∂z ∂R ∂Q − ∂y ∂z dy dz + ∧ dy ∧ dz ∂R ∂P − ∂z ∂x dz dx + If we let F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, x)i, then Z Z ω= C P dx + Q dy + R dz C Z F · dr = C 16 ∂Q ∂P − ∂x ∂y dx dy Johns Hopkins University, CTYOnline Differential Forms and ZZ ZZ dω = S S ZZ = S ∂R ∂Q − ∂y ∂z dy dz + ∂P ∂R − ∂z ∂x ∂R ∂Q ∂P ∂R ∂Q ∂P − , − , − ∂y ∂z ∂z ∂x ∂x ∂y dz dx + ∂Q ∂P − ∂x ∂y dx dy · dS ZZ curl F · dS = S Putting everything together, the statement of the General Stokes’ Theorem applied to S and ω is Z ZZ F · dr = curl F · dS C S which is Stokes’ Theorem! 3.5 The Divergence Theorem Let E be a simple solid region in R3 with boundary surface S, and let ω = P dy dz + Q dz dx + R dy dz be a 2-form. The statement of the General Stokes’ Theorem in this case is ZZ ZZZ ω= dω S E 17 Johns Hopkins University, CTYOnline Differential Forms We again simplify dω: dω = d(P dy dz) + d(Q dz dx) + d(R dy dz) = d(P ∧ dy ∧ dz) + d(Q ∧ dz ∧ dx) + d(R ∧ dy ∧ dz) = (dP ) ∧ dy ∧ dz + (dQ) ∧ dz ∧ dx + (dR) ∧ dy ∧ dz = ∂P ∂P ∂P dx + dy + dz ∂x ∂y ∂z + + = ∧ dy ∧ dz ∂Q ∂Q ∂Q dx + dy + dz ∂x ∂y ∂z ∂R ∂R ∂R dx + dy + dz ∂x ∂y ∂z ∧ dz ∧ dx ∧ dx ∧ dy ∂P ∂Q ∂R dx ∧ dy ∧ dz + dy ∧ dz ∧ dx + dz ∧ dx ∧ dy ∂x ∂y ∂z = ∂Q ∂R ∂P + + ∂x ∂y ∂z dx dy dz If we let F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i, then ZZ ZZ ω= P dy dz + Q dz dx + R dy dz S S ZZ F · dS = S and ZZZ ZZZ dω = E E ∂P ∂Q ∂R + + ∂x ∂y ∂z ZZZ = div F dV E 18 dx dy dz Johns Hopkins University, CTYOnline Differential Forms The statement of the General Stokes’ Theorem in this case becomes ZZ ZZZ F · dS = div F dV S E which is the Divergence Theorem! 4 Conclusion We have now seen that the fundamental integral theorems of single-variable and multivariable calculus are all special cases of a simply-stated theorem, the General Stokes’ Theorem, that is valid in all dimensions. Using differential forms, which can be defined in all dimensions, we are no longer limited by operations like the cross product that are only valid in R3 . As an exercise, in R4 , you might try to: (1) define differential forms; (2) define oriented curves and surfaces; (3) define integrals of 1-forms and 2-forms over oriented curves and surfaces; (4) verify the General Stokes’ Theorem for a particular 1-form ω and a particular oriented surface S with smooth, simple boundary. The surface S might be, for example, the image of the unit disk in R2 under a one-to-one linear function T : R2 → R4 . 19