Part 8. Differential Forms. Overview. Contents. 1 Introduction to Differential Forms

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Part 8 Differential Forms Printed version of the lecture Differential Geometry on 25. September 2009 Tommy R. Jensen, Department of Mathematics, KNU 8.

1 Part 8 Differential Forms Printed version of the lecture Differential Geometry on 25. September 2009 Tommy R. Jensen, Department of Mathematics, KNU 8.1 Overview Contents 1 Introduction to Differential Forms 1 2 p-forms 3 3 Exterior Derivatives 4 4 Conclusion Introduction to Differential Forms The form of a differential form Properties A differential form is constructed by taking any function f : R 3 R, any differential dx 1,dx 2,dx 3 (or dx,dy,dz), any sum of two existing differential forms, or a special wedge product of two differential forms. Example: is a differential form. (x 2 z + y 3 dxdz 1)(e y dz dxdydx) + (sinx)dz Note: the multiplication used in the formula is the wedge product multiplication and not the usual multiplication

The wedge product A car with the shape of a wedge. 8.4 The wedge product Properties Multiplication rules for differential forms θ,φ,ψ and differentials dx 1,dx 2,dx 3 : 1.

2 The wedge product A car with the shape of a wedge. 8.4 The wedge product Properties Multiplication rules for differential forms θ,φ,ψ and differentials dx 1,dx 2,dx 3 : 1. associative rule: θ(φψ) = (θφ)ψ 2. distributive rules: θ(φ + ψ) = θφ + θψ and (φ + ψ)θ = φθ + ψθ 3. alternation rule: dx i dx j = dx j dx i (i, j = 1,2,3) The usual product is commutative (ab = ba) and does not obey the alternation rule. So we have to be careful about knowing exactly what kind of multiplication we are using. If we want to say clearly that we use the wedge product then we write φ ψ instead of φψ. 8.5 A property of the wedge product Repeats are zero The product of a differential dx i with itself is always zero: dx i dx i = 0, for i = 1,2,3. We can prove this from the alternation rule dx i dx j = dx j dx i. For j = i, we get dx i dx i = dx i dx i, which implies dx i dx i = 0. Example: (x 2 z + y 3 dxdz 1)(e y dz dxdydx) + (sinx)dz = (x 2 z + y 3 dxdz 1)(e y dz + dxdxdy) + (sinx)dz = (x 2 z + y 3 dxdz 1)(e y dz + 0) + (sinx)dz = (x 2 ze y dz + y 3 e y dxdzdz e y dz) + (sinx)dz = (x 2 ze y dz e y dz) + (sinx)dz = ((x 2 z 1)e y + sinx)dz, which happens to be a 1-form! 8.6 2

3 2 p-forms p-forms Definition Let p be one of the numbers 0,1,2,3. A 0-form is an expression f. A 1-form is an expression f dx + gdy + hdz (or f i dx i ). A 2-form is an expression f dxdy + gdxdz + hdydz. A 3-form is an expression f dxdydz. Where f,g,h : R 3 R are any differentiable functions. We can explain a p-form as a sum of products each of which has a factor that is a function, and p factors that are dx i s. Every p-form on R 3 with p > 3 is zero, because of the repeat rule. 8.7 Products of p-forms Let φ = xdx ydy, and ψ = zdx + xdz, and θ = zdy. These are three 1-forms. We calculate the product φ ψ θ. ψ θ = (zdx + xdz)zdy = z 2 dxdy + xzdzdy = z 2 dxdy xzdydz. We note that any product of two 1-forms is a 2-form. φ (ψ θ) = = (xdx ydy)(z 2 dxdy xzdydz) = xz 2 dxdxdy x 2 zdxdydz yz 2 dydxdy + xyzdydydz = x 2 zdxdydz. 8.8 Alternation rule for 1-forms Lemma 6.2 If φ and ψ are 1-forms, then Alternation rule for 1-forms φ ψ = ψ φ. 8.9 Proof of Lemma 6.2 We introduce the Euclidean coordinate functions of φ and ψ : Then φ = f i dx i and ψ = g i dx i. φ ψ = ( f i dx i )( g i dx i ) = ( f 1 dx 1 + f 2 dx 2 + f 3 dx 3 )(g 1 dx 1 + g 2 dx 2 + g 3 dx 3 ) = ( f 1 g 2 dx 1 dx 2 + f 1 g 3 dx 1 dx 3 + f 2 g 1 dx 2 dx f 2 g 3 dx 2 dx f 3 g 1 dx 3 dx 1 + f 3 g 2 dx 3 dx 2 ) = ( g 2 f 1 dx 2 dx 1 g 3 f 1 dx 3 dx 1 g 1 f 2 dx 1 dx 2 g 3 f 2 dx 3 dx 2 g 1 f 3 dx 1 dx 3 g 2 f 3 dx 2 dx 3 ) = (g 2 f 1 dx 2 dx 1 + g 3 f 1 dx 3 dx 1 + g 1 f 2 dx 1 dx 2 + g 3 f 2 dx 3 dx 2 + g 1 f 3 dx 1 dx 3 + g 2 f 3 dx 2 dx 3 ) = (g 1 dx 1 + g 2 dx 2 + g 3 dx 3 )( f 1 dx 1 + f 2 dx 2 + f 3 dx 3 ) = ( g i dx i )( f i dx i ) = ψ φ 3

8.10 Just to remind you 3 Exterior Derivatives The exterior derivative Any 0-form is a function f : R 3 R. We can derive a 1-form d f from f, the differential of f.

4 8.10 Just to remind you 3 Exterior Derivatives The exterior derivative Any 0-form is a function f : R 3 R. We can derive a 1-form d f from f, the differential of f. The operator d can also derive a 2-form from a 1-form: Definition Let φ = f i dx i be a 1-form on R 3. The exterior derivative of φ is the differential form dφ = d f i dx i. xxxxxxxxxxxx 8.11 Then dφ is itself a 2-form. Definition Let ψ = f dxdy + gdxdz + hdydz be a 2-form. Then the exterior derivative of ψ is the 3-form dψ = d f dxdy + dg dxdz + dh dydz A formula for exterior derivative We have φ = f i dx i, and dφ = d f i dx i. From Corollary 5.5, d f i = 3 j=1 We combine to get a formula f i x j dx j. dφ = f i d f i dx i = dx j dx i i=1 i=1 j=1 x j = f 1 dx 2 dx 1 + f 2 dx 1 dx 2 + f 1 dx 3 dx 1 + f 3 dx 1 dx f 2 dx 3 dx 2 + f 3 dx 2 dx 3 = f 1 dx 1 dx 2 + f 2 dx 1 dx 2 f 1 dx 1 dx 3 + f 3 dx 1 dx 3 f 2 dx 2 dx 3 + f 3 dx 2 dx 3 = ( f 2 f 1 )dx 1 dx 2 + ( f 3 f 1 )dx 1 dx 3 + ( f 3 f 2 )dx 2 dx

5 Example It is easier to apply the definition directly: Let Then φ = xyzdx y 3 dz. dφ = d(xyz) dx d(y 3 ) dz = (xd(yz) + yzdx) dx 3y 2 dy dz = (xydz + xzdy + yzdx) dx 3y 2 dydz = xydzdx + xzdydx 3y 2 dydz = xzdxdy xydxdz 3y 2 dydz Some rules for exterior derivatives Theorem 6.4 Let f,g be functions and let φ,ψ be 1-forms. Then 1. d( f g) = d f g + f dg = gd f + f dg. 2. d( f φ) = d f φ + f dφ = d f φ + f dφ. 3. d(φ ψ) = dφ ψ φ dψ. (1) is the same as Lemma 5.6. Proving (2) is easier than (3). So we prove (3) Proof of Theorem 6.4 part (3) First assume φ = f dx and ψ = gdy. Then d(φ ψ) = d( f gdxdy) = d( f g) dxdy = ( f g) dzdxdy (Corollary 5.5) z = ( f g z + g f z )dxdydz, dφ ψ = d( f dx) gdy = (d f dx) gdy = gd f dxdy = g f z dxdydz, φ dψ = ( f dx) d(gdy) = f dx g dz dy z = f g g dxdzdy = f z z dxdydz Proof of Theorem 6.4 part (3) So we have shown when φ = f dx and ψ = gdy. d(φ ψ) = dφ ψ φ dψ If we have φ = f i dx i and ψ = g j dx j instead, then d(φ ψ) = ( f i dx i ) (g j dx j ) = (d( f i dx i ) (g j dx j ) ( f i dx i ) d(g j dx j )) = d( f i dx i ) (g j dx j ) ( f i dx i ) d(g j dx j ) = d( f i dx i ) g j dx j f i dx i d(g j dx j ) = d f i dx i g j dx j f i dx i d g j dx j = dφ ψ φ dψ

How to remember the formula (3) Short form of (3)

side of φ changes the sign from + to just as if d

6 How to remember the formula (3) Short form of (3) in Theorem 6.4 If we write d(φψ) = dφψ φdψ without wedges, then we only have to remember that moving d to the other side of φ changes the sign from + to just as if d were a 1-form Conclusion The End END OF THE LECTURE! 8.19 Next time: Mappings

Lecture Notes: Divergence Theorem and Stokes Theorem

Lecture Notes: Divergence Theorem and Stokes Theorem Lecture Notes: ivergence heorem and tokes heorem Yufei ao epartment of omputer cience and Engineering hinese University of Hong Kong taoyf@cse.cuhk.edu.hk In this lecture, we will discuss two useful theorems