Math 1540 Hastings section Spring 2011 Notes 13 Di¤erential 2-forms on R2:
1 Wedge products
Above we de…ned the 2-form dx dy; etc. These are called the “wedge products”of the one-forms dx and dy: If we have^ general one-forms on R2; say p dx + q dy and r dx + s dy; then their wedge product is (ps qr) dx dy: (1) ^ Once we realize that ps qr is a function from R2 to R we see that this was de…ned earlier, as a 2-form , in equations (3) and (4) of notes 12. The formula (1) can be obtained from ordinary algebra: (p dx + q dy) (r dx + s dy) = pr dx dx + ps dx dy + qr dy dx + qs dy dy; ^ ^ ^ ^ ^ where we used the previously established formulas dx dx = dy dy = 0 and dy dx = dx dy: ^ ^ ^ ^ 2 The exterior derivative of di¤erential forms
We …rst add to our list of di¤erential forms: De…nition 1 A 0-cell on Rn is a function : 0 Rn: f g ! In practice, we will think of a 0-cell as a point in Rn: De…nition 2 A 0-form on Rn is a function F : Rn R: (There is a reason I used F instead of f; as you will see. ! Thus, if is a 0-cell and F is a 0-form, then F ( (0)) is de…ned as a real number. Further on we will use our usual integral notation for the value of a form on a cell, giving F = F ( (0)) : Z (0) We will probably not use ! to denote a 0-form, but instead the name of the function, such as F:
1 De…nition 3 The exterior derivative of a 0-form F is the functional dF : S1;n R given by ! dF ( ) = F ( (1)) F ( (0)) : (2) We know that not all functionals on S1;n are one-forms. For example, the length of the curve is not a 1-form because it cannot be expressed as an integral in terms 2 of 0 = ( 0 ; 0 ). However, considering n = 2; if F : R R is smooth, then 1 2 ! 1 @F @F ( (t) ; (t)) 0 (t) + ( (t) ; (t)) 0 (t) dt @x 1 2 1 @y 1 2 2 Z0 1 d = F ( 1 (t) ; 2 (t)) dt = F ( (1)) F ( (0)) = dF ( ) (3) dt Z0 Hence, dF is a 1-form . Continuing our use of integral notation for 1-forms, we now can write
dF ( ) = dF = F ( (1)) F ( (0)) : (4) Z In fact, comparing the formula above with the formula for a general 1-form with name ! = pdx + qdy; namely,
1 ! ( ) = (p (x (t) ; y (t)) x0 (t) + q (x (t) ; y (t)) y0 (t)) dt; (5) Z0 we see that @F @F dF = dx + dy: (6) @x @y De…nition 4 Suppose that ! = p dx + q dy is a 1-form on R2: Then the exterior derivative of ! is d! = dp dx + dq dy; (7) ^ ^ where dp and dq are the exterior derivatives of the zero forms p and q: We have no di¢ culty knowing this is a form, since the wedge product was de…ned so as always to be a form; that is, an integral involving a determinant. We can easily get a more useful formula. Note that p and q are 0-forms on R2: @p @p We can use equation (6) to give dp = dx dx + @y dy; and a similar expression for dq: The, substituting these in (7) and using the rules for wedge products, we get @q @p d! = dx dy: (8) @x @y ^
2 3 Chains and boundaries.
Recall that a k-cell in Rn is a smooth function : Ik Rn: ! n De…nition 5 A k-chain (on R ) is a set of pairs (ai; i) of numbers ai and k-cells n i (on R ): = (a1; 1) ; :::; (aN ; N ) : f g Given a k-chain ; and a k-form !; we de…ne the integral of ! over as follows:
N ! = i=1ai !: (9) Z Z i P This is a real number, and we can consider ! as a functional on the set of all k-chains. R As an example, consider the following four 1-cells on R2:
1 (t) = (1; t) ; 2 (t) = (0; t) ; 3 (t) = (t; 1) ; 4 (t) = (t; 0) ; (10) for 0 t 1.(t I1.) Let 2 = (1; 1) ; ( 1; 2) ; ( 1; 3) ; (1; 4) (11) f g Notice that 1 maps [0; 1] onto the right edge of the unit square, with the orientation given by the unit tangent vector (0; 1) : This points upward. Also, 2 maps [0; 1] onto the left edge of the unit square, also with upward orientation. By pairing 1 with 2; when we compute ! for some 1-form ! = pdx + qdy; we get a term
R !: Z 2 This is the value of ! on the 1-form (0; 1 t) ; which goes along the left side of the unit square from top to bottom. We have not changed the image of 2; just its 1 orientation. Similar considerations apply to ( 1; 3) and to (1; 4) ; and putting these together, we see that we have computed a line integral around the boundary of I2 going in the counterclockwise direction. Next we wish to de…ne the “boundary”of a k-cell, for k = 1 and 2: We start with a 1-cell : I1 Rn: ! De…nition 6 The boundary of a 1-cell is the 0-chain @ = (1; (1)) ; ( 1; (0)) (12) f g 1 Contrast this with the usual de…nition of 2; which would change its image. 3 3.1 The fundamental theorem of calculus. Suppose that F : R1 R (which equals R1 ). Then as we saw in De…nition 2, F is a 0-form on R1. And! in De…nition 3 we saw that the exterior derivative of F is
dF ( ) = dF = F ( (1)) F ( (0)) : (13) Z This is a 1-form, as seen in equation (3), making allowances for the fact that there we were discussing a 0-form on R2. Also, if is a 0-cell, we de…ned the boundary of ; @ , in equation (12). Using (9) we get F = F F: (14) Z@ Z (1) Z (0) We have to interpret the right side correctly. Recall that a 0-cell is a point, and a 0-form is a functional on the set of zero cells. As we saw earlier, the usual integral notation for forms gives
F = F ( (1)) ; F = F ( (0)) : Z (1) Z (0) Using (13) and (14) we get dF = F: (15) Z Z@ This is purely from the notation; no deep mathematics is involved. However, if F is continuously di¤erentiable, with derivative F 0 = f; then we can consider the 1-form f dx; which by de…nition is given by
1 f dx = f ( (t)) 0 (t) dt: Z Z0 (We are still in one dimension.) Using the change of variable theorem and the fundamental theorem of calculus, we get
(1) f = f (x) dx = F ( (1)) F ( (0)) : Z Z (0)
From this we see that the 1-form dF is the same as the 1-form F 0 dx: This result does require nontrivial mathematics, namely the fundamental theorem of calculus.
4 3.2 Boundary of a 2-form, Green’s theorem for the unit square. We can now de…ne the “boundary”of a 2-cell.
De…nition 7 The boundary of a 2-cell is the 1-chain
@ = (1; (1; t)) ; ( 1; (0; t)) ; (1; (t; 0)) ; ( 1; (t; 1)) : (16) f g One example is the 1-chain I gave earlier (equations (10) and (11) ), which is the boundary of the 2-cell (x; y) = (x; y) : Suppose that n = 2; and pdx + qdy is a 1-form on R2: We can ask: What is
@ pdx + qdy ? This is the integral over a chain, so we have to use equation (9). In this section we will only consider the case where (x; y) = (x; y) : Thus, the image R of is the unit square I2: Then
@ (t) = (1; (1; t)) ; ( 1; (0; t)) ; (1; (t; 0)) ; ( 1; (t; 1)) : f g In that case, (9) becomes
pdx + qdy = pdx + qdy pdx + qdy Z@ Z 1 Z 2 pdx + qdy + pdx + qdy; Z 3 Z 4 where 1; 2; 3; and 4 are 1-forms in (10). I will evaluate one of these integrals and then give the …nal answer. Since 1 (t) = (1; t) ; the standard line integral formula, as in (5) above, gives
1 1 pdx + qdy = (0 + q (1; t) (1)) dt = q (1; t) dt: Z 1 Z0 Z0 Evaluating the other integrals similarly, we get
1 pdx + qdy = (q (1; t) q (0; t) + p (t; 0) p (t; 1)) dt (17) Z@ Z0 We will now evaluate d!; Z
5 where ! = pdx + qdy and where we again choose to be the 2-form (x; y) = (x; y) : Recall from equation (8) that
@q @p d! = dx dy: @x @y ^ We then apply the de…nition of the 2-form in De…nition 17 from Notes 12 (equation (3) of those notes). Recall that we are taking (x; y) = (x; y) : Hence
@ 1 @ 1 1 0 det @x @y = det = 1 @ 2 @ 2 0 1 @x @y ! and using Fubini’stheorem,
@q @p 1 1 @q 1 1 @p dx dy = dx dy dy dx: @x @y ^ @x @y 0 0 0 0 Z Z 1 Z Z Z = ((q (1; s) q (0; s)) (p (s; 1) p (s; 0))) ds Z0 which is the same as the right side of equation (17). This gives the formula
d! = !: (18) Z Z@ This is Green’stheorem for the unit square. Compare with equation (15).
4 Green’stheorem for a general 2-cell on R2:
4.1 Brute force method It is often useful to see more than one proof of an important theorem. Green’s theorem is simple enough, because it is set in R2; to give a direct computational proof, in which both sides of equation (18), for a general 1-form ! and general 2-cell ; are evaluated from their de…nitions and shown to be equal. Most mathematicians prefer a proof which is more easily generalized to higher k and n; but I will give this special case proof …rst. Proof. Recall that a general 1-form on R2 can be written as
! = p (x; y) dx + q (x; y) dy:
6 Suppose that is a 2-cell. Then the boundary @ was given in equation (16):
@ = (1; (1; t)) ; ( 1; (0; t)) ; (1; (t; 0)) ; ( 1; (t; 1)) : f g
Here (x; y) = ( 1 (x; y) ; 2 (x; y)) : Then @ ! is, by de…nition, the sum of four standard line integrals, the …rst two of which add up to R pdx + qdy pdx + qdy = Z (1;t) Z (0;t) 1 @ 1 @ 1 p ( 1 (1; t) ; 2 (1; t)) (1; t) p ( 1 (0; t) ; 2 (0; t)) (0; t) dt 0 @y @y Z 1 @ 2 @ 2 + q ( 1 (1; t) ; 2 (1; t)) (1; t) q ( 1 (0; t) ; 2 (0; t)) (0; t) dt: @y @y Z0 These terms involve integrals along the left and right sides of I2: From now on we will just discuss the terms with p: The terms with q work out in the same way. Also, the terms in p and q along the top and bottom of I2 work out similarly and will not be given here. So we consider
1 @ 1 @ 1 p ( 1 (1; t) ; 2 (1; t)) (1; t) p ( 1 (0; t) ; 2 (0; t)) (0; t) dt: (19) @y @y Z0
We now consider the corresponding terms in d!: We gave d! in equation (7): d! = dp dx +Rdq dy: ^ ^ Further, in (??) we saw that
@p @p dp = dx + dy; @x @y and so using the rules for wedge product, the term in d! involving p is @p (x; y) dx @y ^ dy: Hence, using the de…nition of a 2-form, the terms involving p in d! are
1 1 @ 1 @ 1 R @p @x @y ( 1 (x; y) ; 2 (x; y)) det dxdy @y @ 2 @ 2 Z0 Z0 @x @y ! 1 1 @p @ 1 @ 2 @ 1 @ 2 = ( 1 (x; y) ; 2 (x; y)) dxdy: (20) @y @x @y @y @x Z0 Z0
7 We wish to evaluate this integral. We note that @ @p @ @p @ p ( (x; y) ; (x; y)) = ( (x; y) ; (x; y)) 1 (x; y) + ( (x; y) ; (x; y)) 2 (x; y) @x 1 2 @x 1 2 @x @y 1 2 @x @ @p @ @p @ p ( (x; y) ; (x; y)) = ( (x; y) ; (x; y)) 1 (x; y) + ( (x; y) ; (x; y)) 2 (x; y) : @y 1 2 @x 1 2 @y @y 1 2 @y @ @p (Be sure to notice the di¤erence between @y p ( 1 (x; y) ; 2 (x; y)) and @y ( 1 (x; y) ; 2 (x; y)) :) @ 1 @ 1 Multiply the …rst of these equations by @y ; the second by @x and subtract the second resulting equation from the …rst. The …rst terms on the right cancel, and we are left with @ 1 @ @ 1 @ p ( 1 (x; y) ; 2 (x; y)) p ( 1 (x; y) ; 2 (x; y)) @y @x @x @y @p @ 2 @ 1 @ 2 @ 1 = ( 1 (x; y) ; 2 (x; y)) @y @x @y @y @x The term on the right is the integrand in (20), and so we are left to evaluate 1 1 @ 1 @ @ 1 @ p ( 1 (x; y) ; 2 (x; y)) p ( 1 (x; y) ; 2 (x; y)) dxdy: (21) @y @x @x @y Z0 Z0 Considering the …rst term, we integrate the x integral by parts, getting 1 @ 1 @ @ 1 x=1 (x; y) p ( 1 (x; y) ; 2 (x; y)) dx = (x; y) p ( 1 (x; y) ; 2 (x; y)) x=0 0 @y @x @y j Z 1 2 @ 1 p ( 1 (x; y) ; 2 (x; y)) dx @y@x Z0 @ 1 @ 1 = p ( 1 (1; y) ; 2 (1; y)) (1; y) p ( 1 (0; y) ; 2 (0; y)) (0; y) @y @y 1 2 @ 1 p ( 1 (x; y) ; 2 (x; y)) dx (22) @y@x Z0 The …rst two terms are the integrand in (19), and so are the integrands in the line integrals involving p along the left and right sides of I2. This means that the line integrals along the left and right sides of I2 involving p are equal to two of the terms in d!. The last term, (22), will cancel a similar term obtained from the second integrand in (21), where we do the y integral …rst, again using integration by parts. R The rest of the second integral in (21) will contribute the p terms for line integrals over the top and bottom of I2: All q terms are similar, giving the result. In the next set of notes we will give a less computational proof which can more easily be modi…ed to apply to higher dimensions.
8 5 Homework due April 6
1. Suppose that (x; y) = 3 x 1 2 ; y for (x; y) I2. Suppose that ! = 2 2 dx dy: Find ! from the de…nition of this integral, and explain your answer in terms^ of area. R 2. Last week you were asked to …nd 2-cells on R3 whose images were the unit sphere and the unit hemisphere. While there are many possible answers, here are two that work: Sphere: (x; y) = (sin x cos 2y; sin x sin 2y; cos x) for (x; y) I2 2 Hemisphere: (x; y) = sin x cos 2y; sin x sin 2y; cos x for (x; y) I2: 2 2 2 2 Find the boundaries of each of these 2-cells, describing the image and orienta- tion of each section.
3. Find a 2-cell on R2 which is 1 : 1 and whose image is the region in the (u; v) plane between concentric circles with radii 1 and 2 and center the origin. (If you wish, the domain of can be some more convenient rectangle than the unit square.) Find the boundary @ :
4. (10 pts.) Continuing from (3), let ! = 3x2dx + xy2dy; and verify Green’s theorem in this case by explicitly calculating both sides of equation (18). You can use a computer (for example, with Maple or Mathematica) to evaluate the integrals. Or you can Google “evaluate integrals”and go the …rst link, Wolfram Mathematica online integrator.
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