Contentssnovikov/math437.pdf · Diﬀerential 2-forms in Rn. Change of coordinates and restriction of 2-forms to 2-surfaces in Rn. Case n= 2. Integration. Case of any k-forms in Rn.

S.P.Novikov — Math 437Spring 2015

Differential forms

Textbook: D.Bachman, A geometric approach to differential forms, Birkhauser.Additional material: M.Spivak, Calculus on Manifolds.

Contents

Program 3

Homeworks 5Homework 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Homework 1 — Solutions . . . . . . . . . . . . . . . . . . . . . 6Homework 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Homework 2 — solutions . . . . . . . . . . . . . . . . . . . . . 8Homework 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Homework 3 — Solutions . . . . . . . . . . . . . . . . . . . . . 9Homework 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Homework 4 — Solutions . . . . . . . . . . . . . . . . . . . . . 11Homework 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Homework 5 — Solutions . . . . . . . . . . . . . . . . . . . . . 13Homework 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Homework 6. Solutions. . . . . . . . . . . . . . . . . . . . . . 16Homework 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Homework 7. Solutions. . . . . . . . . . . . . . . . . . . . . . 17Homework 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Homework 8. Solutions. . . . . . . . . . . . . . . . . . . . . . 20Homework 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Homework 9. Solutions. . . . . . . . . . . . . . . . . . . . . . 24Homework 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Homework 10. Solutions. . . . . . . . . . . . . . . . . . . . . . 28

Midterm Tests. 33First Midterm Test. . . . . . . . . . . . . . . . . . . . . . . . . 33Second Midterm Test. . . . . . . . . . . . . . . . . . . . . . . 34Second Midterm Test. Solutions. . . . . . . . . . . . . . . . . 34

1

Lectures 1–30 38Introductory lecture: History of integration and calculus . . . 38Lecture 2. Differential 1-forms . . . . . . . . . . . . . . . . . . 41Lecture 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46Lecture 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Lecture 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54Lecture 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Lecture 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Lecture 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Lecture 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Lecture 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Lecture 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75Lecture 12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Lecture 13. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82Lecture 14. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84Lecture 15. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86Lecture 16. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Lecture 17. Linear Algebra (Minicourse). . . . . . . . . . . . . 93Lecture 18. Linear Algebra - II. . . . . . . . . . . . . . . . . . 97Lecture 19. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Lecture 20. Volume and Differential Forms. Symplectic Man-ifolds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105Lecture 21. Duality Operator and Maxwell’s Equations. . . . . 110Lecture 22. Stokes Formula. . . . . . . . . . . . . . . . . . . . 115Lecture 23. Algebraic Boundary. . . . . . . . . . . . . . . . . . 118Lecture 24. Differential Forms and Homotopy Poincare Lemma.122Lecture 25. Examples of important closed differential forms. . 125Lecture 26. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129Lecture 27. Curvature. . . . . . . . . . . . . . . . . . . . . . . 133Lecture 28. Curvature. . . . . . . . . . . . . . . . . . . . . . . 137Lecture 29. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143Lecture 30. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

2

Program

1.Coordinates. Cartesian coordinates in a domain U ⊂ Rn. Important ex-amples (polar, cylindrical, spherical, hyperbolic, complex).

2.Differential 1-forms and vector fields, Jacoby matrix, change of coordi-nates, pull-back for functions and 1-forms, restriction to curves and integra-tion. Orientation.

3.Important examples, closed 1-forms, Newton–Leibnitz property for in-tegration, exact and closed 1-forms, 1-dimensional (co)homology of domainsU ⊂ R2. The differential of the angle function. Topology of planar domain,number of ”cuts” necessary to make it simply connected.

4.Complex coordinates z, z in R2. Cauchy 1-forms f(z) dz, the Cauchy–Riemann equation ∂f/∂z = 0. What is the value of the integral

∮|z|=1

zn dz?

Is the form f(z) dz closed if ∂f/∂z = 0?5.What is d(X) if X is a 1-form? Differential 2-forms in Rn. Change

of coordinates and restriction of 2-forms to 2-surfaces in Rn. Case n = 2.Integration. Case of any k-forms in Rn.

6.Product of 1-forms as a 2-form. Product of differential forms. Associa-tivity and skew commutativity for k-forms in U ⊂ Rn.

7.The De Rham operator d acting on differential forms, its definition andalgebraic properties. De Rham operator and multipication of forms. DeRham Operator and pull-back operation. Closed k-forms and cohomology(definition for the domains in Rn). Important examples of the closed formsin R2 and R3, angle form in R2; area form for S2 in spherical coordinatesand as a restriction of the closed 2-form in R3.

8.Change of coordinates for k-forms in u ⊂ Rn. Pull-back operation fork-forms for arbitrary smooth maps φ : U → V .

9.Integration of n-forms in Rn. Integration of k-forms along the orientedk-surfaces U → Rn, U ⊂ Rk.

10.The Stokes formula for integration∫· · ·∫

D

dΩk+1

=

∫· · ·∫

∂D

Ωk

,

cases k = 0, 1.11.First pair of Maxwell Equations and De Rham operator d in the 3-

space and in 4-space. Electromegnetic field as a 2-form in R4. Electric andMagnetic fields in R3.

3

12.Minicourse in Linear Algebra. Vectors and covectors. Inner products,Gram Matrices. Nondegenerate inner product. Classification of symmetricInner Products. Euclidean and Minkowski cases. Volume element and GramMatrix. Groups GLn(R) and GLn(C), On, SOn. How many connected piecesdo they have? Same question for group O1,1.

13.Nondegenerate Symplectic Inner Product in R2n and 2-form, its pow-ers and volume element, Pfaffian (square root of determinant) of the skewsymmetric matrix and powers of 2-form.

15. Riemanian and Pseudoriemanian metrics. Euclidean and Minkowskicases, metrics of eulideasn plane, 2-sphere and pseudosphere in the polarcoordinates. Arc length of curves, definition of geodesics as (locally) minimalarc curves. Arc length of timelike curves in Minkowski space. Arc length ofthe ”Fermat metric” in R2. The Fermat principle for the propagation of lightand ”Snells law” in the water/air case.

15. Duality of forms in euclidean metric, the case of 3-space. The op-erators div and curl in terms of differential forms. Product of 1-forms andvector product in R3. Vector product and commutator of skew symmetricmatrices.

16.Differential forms and homotopy of mappings. Poincare’ lemma.17.Curves in the euclidean plane. Curvature and Gauss map. Curvature

as a pull-back of the form dϕ. Integral of curvature along the closed curveand degree of Gauss map.

18.Surfaces in R3. Riemannian metric and curvature form. Mean andGauss curvature. Gauss curvature and Gauss map. Total (integral ) ofGauss curvature for the convex body in R3.

19. Hopf invariant for the maps S3 → S2 and Kelvin-Whitehead integralalong S3.

4

Homeworks

Homework 1

1. Calculate the Jacoby matrix for a shift, rotations, and affine transfor-mations of Rn.

2. Find the restriction of the 1-forms dx and dy to the lines

a) y = λx;

y

x

b) x2 + y2 = 1.

3. Is (ρ(x, y), φ(x, y)) a pair of “Cartesian” type coordinates in R2? Is ittrue for R2 \ 0?

4. In which domain (ρ, φ) may be treated as a pair of “Cartesian” coor-dinates? Is it ok for the domain 0 < φ < 2π?

0

5. Is ρ a “good” Cartesian coordinate for the domain R2 \ 0?

6. Find the restriction of the form

x dy − y dxx2 + y2

to the circle x2 + y2 = 1 (in the coordinates (ρ, φ)).

5

Homework 1 — Solutions

1. Shift: x 7→ x+ b, J = 1.

Rotation: x 7→ Ax, J = A.

Affine transformation: x 7→ Ax+ b, J = A.

2. a) Use x as the parameter. y = λx, x = x, dx→ dx, dy → λ dx.

b) x2 + y2 = 1, x = cos t, y = sin t, use t as the parameter.

dx→ d(cos t) = − sin t dt,

dy → d(sin t) = cos t dt.

3. (ρ, φ) is not a pair of Cartesian coordinates in R2 and R2 \ 0.

4. The domain U ⊂ R2 \ 0 should not contain a closed path around 0.

0

5. No, the level ρ = const > 0 is a circle S1 ⊂ R1.

6.x dy − y dx

ρ2= dφ.

6

Homework 2

1. Calculate the change of coordinates (ρ1, φ1)→ (ρ2, φ2),

ρ1(ρ2, φ2), φ1(ρ2, φ2) = ?

•

•

(0, 0)

(2, 3)

1

2

2. Prove that the 1-form α dφ1+β dφ2 is closed and not exact in a domainV like:

(0, 0)

(2, 3)

V

3. Find a domain in S2 ⊂ R3 in which dθ is a well-defined 1-form.

4. Find a domain in S2 in which dφ is a well-defined 1-form.

5. The cylindrical coordinates (z, ρ, φ) in R3 are defined by z = z, x =ρ cosφ, y = ρ sinφ. In which domain V ⊂ R3 the form dφ is a well-defined 1-form?

7

6. Calculate the integral∮γ

(αz2 + β/z) dz, z = x+ iy, α, β = const

along a closed contour γ.

Homework 2 — solutions

1.x = ρ1 cosφ1,

y = ρ1 sinφ1,

x− 2 = ρ2 cosφ2,

y − 3 = ρ2 sinφ2.

Conclusion:ρ1 cosφ1 = 2 + ρ2 cosφ2,

ρ1 sinφ1 = 3 + ρ2 sinφ2.

Solve these equations:

ρ21 = (2 + ρ2 cosφ2)2 + (3 + ρ2 sinφ2)

2,

cosφ1 =2 + ρ2 cosφ2

ρ1,

sinφ1 =3 + ρ2 sinφ2

ρ1.

2. d(α dφ1 + β dφ2) = 0 (obvious).∮C1

Ω = 2πα,

∮C2

Ω = 2πβ.

3. dθ is OK in the domain S2 \ poles.

4. dφ is OK in S2 \ poles.

5. Cylindrical coordinates in R3. dφ is OK in R3 \ line(x = 0, y = 0).

6. ∮C

(αz2 +

β

z

)dz = 2πiβ

(Cauchy)∮Czn dz = 0, n = −1.

8

Homework 3

1. Prove that the 1-form x dy− y dx is invariant under all rotations of R2

around (0, 0).

2. Prove the same for x dx+ y dy.

3. Introduce “hyperbolic coordinates” (χ, ψ): x = χ coshψ, y = χ sinhψ.In which domain U ⊂ R2 are they defined?

4. Calculate the area 2-form dx∧ dy of R2 in the polar coordinates (ρ, φ):dx ∧ dy = C dρ ∧ dφ, C = ?.

5. Calculate the area 2-form dx ∧ dy in the complex coordinates z, z,dx ∧ dy = C dz ∧ dz, C = ?.

6. Calculate the volume 3-form of R3 in the spherical coordinates (r, φ, θ),

z = r cos θ,

x = r sin θ cosφ,

y = r sin θ sinφ,

dx ∧ dy ∧ dzthe volume 3-form

= (?) dr ∧ dθ ∧ dφ.


1. x dy − y dx = dφ · ρ2 — invariant under rotation.

2. x dx+ y dy = ρ dρ — invariant under rotation: ρ 7→ ρ, φ 7→ φ+ const,dρ 7→ dρ, dφ 7→ dφ.

x = χ coshψ, y = χ sinhψ, x2 − y2 = χ2 > 0

9

x

y

x2−y2>0

3. dx ∧ dy = ρ dρ ∧ dφ.

4.

dx ∧ dy =dz ∧ dz−2i

,

dz = dx+ i dy,

dz = dx− i dy.

5. R3: dx ∧ dy ∧ dz = r2 sin θ dr ∧ dθ ∧ dφ.For the unit sphere S2 (r = 1) we have Area = sin θ dθ∧dφ, 0 6 θ 6 π.

10

Homework 4

1. Prove that d(f dz) =∂f

∂zdz ∧ dz in R2.

2. Calculate the area form in the hyperbolic coordinates x = χ coshψ,y = χ sinhψ.

3. Introduce hyperbolic coordinates in a domain of R3

x = χ sinh θ cosφ,

y = χ sinh θ sinφ,

z = χ cosh θ.

Find a domain where these coordinates are well defined.

4. Find linear transformations of R2 such that ⟨Aη,Aζ⟩ = ⟨η, ζ⟩, where⟨η, ζ⟩ = η1ζ1 − η2ζ2.

5. How many components does the group of such transformations have(see no. 4)?

6. Find the value of the integral∮|z|=1

f(z)

zndz,

where f(z) = a0 + a1z + . . .+ aNzN .


1. df = fx dx+ fy dy = fz dz + fz dz by definition.

Here ∂z =12(∂x − i∂y), ∂z = 1

2(∂x + i∂y). For the form f(z) dz we have

d(f dz) = fz dz ∧ dz (calculation).

2-3. x = χ coshψ, y = χ sinhψ,

dx ∧ dy = (dχ coshψ + χ sinhψ dψ) ∧ (dχ sinhψ + χ coshψ dψ)

= χdχ ∧ dψ(cosh2 ψ − sinh2 ψ) = χdχ ∧ dψ.

11

x

y

x2−y2>0

4. Linear transformations A preserving inner product ⟨η, ζ⟩ = η1ζ1−η2ζ2:

a) hyperbolic rotations

A =

(coshψ sinhψsinhψ coshψ

);

b) time reflection

T =

(−1 00 1

), T 2 = 1;

c) space reflection

P =

(1 00 −1

), P 2 = 1.

5. Four components: TαP βA, α, β = 0, 1.

6. ∮C

|z|=1

f(z)

zndz = 2πiak, k = n− 1,

f(z) = a0 + a1z + . . .+ aNzN ,∮

C

dz

z= 2πi,

∮C

dz

zn= 0, k = −1.

12

Homework 5

1. Find components of the group GL2(R).

2. Same for O2.

3. Same for O1,1 (preserving the quadratic form ds2 = (dx0)2 − (dx1)2 orthe inner product ⟨Aη,Aζ⟩ = ⟨η, ζ⟩, ⟨η, ζ⟩ = η0ζ0 − η1ζ1).

4. Prove that the Mobius band cannot be oriented, M ⊂ R3.

a a

5. Prove that the following 2-form in R3 \ 0

Ω = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy

coincides with the area form on the unit sphere S2 (x2 + y2 + z2 = 1)and is invariant under rotations from SO3.

6. Prove that the 2-form Ω/r3 is closed in R3 \ 0 and∫∫S2:x2+y2+z2=1

Ω

r3= 0.


1. The eigenvalues are real. Choose a basis consisting of eigen vectors (ora Jordan basis).(

λ1 00 λ2

)= A0 = A, At =

(λ1(t) 00 λ2(t)

)

We come to

(1 00 1

)or

(−1 00 −1

).

For a Jordan cell we have

A0 =

(λ 10 λ

)−→

(1 10 1

)−→

(1 00 1

)13

or (λ 10 λ

)−→

(−1 10 −1

)−→

(−1 00 −1

)Rotation family: (

−1 00 −1

)−→

(1 00 1

).

2. O2: A =

(cosφ sinφ− sinφ cosφ

)and

(−1 00 1

)= P , A and PA — the

whole group O2, two components.

3. O1,1: A =


), P =

(−1 00 1

), T =

(1 00 −1

). Compo-

nents: A, PA, TA, PTA.

4. An orientation of the Mobius band does not exist because it has onlya single side.

5-6.Ω = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy

Substitute the spherical coordinates. We have

Ω = r3 sin θ dθ ∧ dφ.

This is the area form on S2, which is SO3-invariant.

The form A = Ω/r3 is invariant under x 7→ λx, y 7→ λy, z 7→ λz. So,

A comes from the map R3 \ 0 ϕ→ S2, (x, y, z) 7→ (x/r, y/r, z/r) ∈ S2.So, A = ϕ∗(Area). d(Area) = 0, hence d(ϕ∗Area) = 0, Q.E.D.

14

Homework 6

1. What is closed 1 - form? What is closed 2 - form? Are the following formsclosed?

a)dz

z2 − ab) x dy ∧ dz c) x dy + y dx

2. Prove that in R2 we have (orthonormal basis)

⟨η, ζ⟩ = η1 ζ1 + η2 ζ2 = |η| |ζ| cosφ

ξ

η

3. Prove that symmetric nondegenerate inner product ⟨η, ζ⟩ in R2 (bi-linear) can be reduced to the form

⟨η, ζ⟩ = η1 ζ1 ± η2 ζ2

+ = euclidean space R2

− = Minkovski space R1,1

4. Classify skew-symmetric inner products in R2 , R3 :

⟨η, ζ⟩ = −⟨ζ, η⟩

Nondegeneracy∀ η ∃ ζ : ⟨η, ζ⟩ = 0

5. Gramm matrix of inner product G:

e1, . . . , en − basis , gij = ⟨ei, ej⟩

Nondegeneracy: det gij = 0 (prove).

6. New basise = A e′ (ei = aji e

′j)

⟨ei, ej⟩ = gij = G , G = AG′At

Prove!

15

Homework 6. Solutions.

2. High school⟨η, ζ⟩ = |η| |ζ| cosφ

ϕ

Bilinearity - important

⟨η, ζ⟩ = η1 ζ1 + η2 ζ2

3.O1,1 : ds2 = (dx0)2 − (dx1)2 R1,1

⟨η, ζ⟩ = η0 ζ0 ± η1 ζ1

4. Skew-symmetric inner product

⟨η, ζ⟩ = −⟨ζ, η⟩

Define a complement to any subspace. After appropriate normaliza-tion of basis vectors it becomes (

0 1−1 0

)

5. Gramm matrix for inner product G:

⟨ei, ej⟩ = gij , e = A e′ , G = AG′At

Corollary: detG = |detA|2 detG′

16

Homework 7

1. Prove that eAt is orthogonal if At = −A . Find differential equation foreAt .

2. Prove thatdet eAt = eTr(At)

3. Prove that rotation in Rn splits into rotations in n/2 orthogonal planesfor n = 2k . Use result of Problem 1 and skew-symmetric matrices.

4. Calculate d (Ωn−1) in Rn and its relation with operation

div ζ =n∑i=1

∂ζi∂xi

for vector fields.

5. Calculate dΩ1 in R3 (1 - form) and its relation with operation curl (ζ)for vector field ζ .

6. For 2× 2 - matrix B(t) ∈ O(2) in R2 prove equation (Frenet)

B(t) :e1 e1(t)

→e2 e2(t)

,de1dt

= k e2(t) ,de2dt

= − k e1(t)


1. At = −A , eA ∈ On ?

Proof: (eA)t

= eAt

= e−A =(eA)−1

OK.

Differential equation for eAt = ψ :

ψ = Aψ = ψA

2.det eAt = eTrAt ?

17

Proof: Diagonal matrices

A =

a1 . . . 0...

. . ....

0 . . . an

, eAt =

ea1t . . . 0...

. . ....

0 . . . eant

det eAt = ea1t · . . . · eant = ea1t+···+ant = e(TrA) t

Set of the “diagonalizable” matrices is dense in space of matrices n× nover C . So the identity

det eAt = eTrAt

is true everywhere (it is polynomial).

OK.

3. Rotations in R2n : A = eB , Bt = −B :There exist orthonormal coordinates such that (this statement was not

proved in this course)

B =

0 a1 . . . 0 0−a1 0 . . . 0 0...

.... . .

......

0 0 . . . 0 an0 0 . . . −an 0

, eB =

A1 . . . 0...

. . ....

0 . . . An

,

Aj = exp

(0 aj−aj 0

)− 2× 2 orthogonal matrix

4. dΩn−1 in Rn , Ωn−1 =n∑i=1

ai dx1 ∧ . . . ∧ dxi ∧ . . . ∧ dxn

dΩn−1 =∑ ∂ai

∂xi(−1)i dx1 ∧ . . . ∧ dxn

18

OK.

5. Ω = a dx + b dy + c dz

dΩ = ay dy∧dx + az dz∧dx + bx dx∧dy + bz dz∧dy + cx dx∧dz + cy dy∧dz =

= (bx − ay) dx ∧ dy + (cy − bz) dy ∧ dz + (cx − az) dx ∧ dz

∗ : Λ1(R3)→ Λ2

(R3)

∗ dΩ = (cy − bz) dx − (cx − az) dy + (bx − ay) dz =

= curl (a, b, c) = (cy − bz, az − cx, bx − ay)

6. Frenet:e1e3

→ e1(t)e2(t)

a) e1 ⊥ e1(t) : ⟨e1, e1⟩ = 1 ⇒ e1 ⊥ e1

e1 = k n (definition)

e2 = − k n (skew symmetry)

A =

(0 k−k 0

) ∣∣∣∣ A(t) = 1 + B t + . . .

B = A|t=0 = −B

19

Homework 8

1. Find 0 - cohomology group H0(GLn(R)) and H0(GLn(C)) . H0(On) = ?H0(On,1) = ?

2. Prove that all 1 - forms in S2 invariant under rotations SO3 , are trivial(equal to 0) (Also true for Sn and SOn+1).

3. Prove that for every rotation A ∈ SO3 acting on S2 , the form closedA∗Ω and Ω are cohomologous (i.e. A∗Ω − Ω = du).

4. Prove that all closed forms in

T2 : (x1, x2) ≃ (x1 + 2π, x2) ≃ (x1, x2 + 2π)

are cohomologous to their shifts

h∗Ω − Ω = du h = h(a,b)

h : (x1, x2) → (x1 + a, x2 + b)

h∗ Ω = Ω(x1 − a, x2 − b)

5. Prove that forms dx1, dx2 are linearly independent in H1(T2) and 2 -form dx1 ∧ dx2 = du (nontrivial) in T2 .


1. H0(GLn(R)) = ? , H0(On) = ? , H0(On,1) = ?

U =N∪i=1

Ui , Ui ∩ Uj = ∅

f : U → R , d f = 0 ↔ f is locally constant

So H0(U) = RN (number of pieces)

GLn(R) − 2 pieces, On,1 − 4 pieces, On − 2 pieces.

20

2. 1 - form ω in Λ1(S2) :

Rotations A : S2 → S2 (R3 → R3) .

Let x ∈ S2 , A(x) = x ⇒ Aα(x) = x

xAα − rotations in tangent plane.

No vectors in R2 invariant under all rotations. So, for any invariant 1 -form we have

Λ1inv(S

2) = 0

3. Let A ∈ SO3 , A : S2 → S2 .

∃At such that A0 = A , A1 = I (Homotopic! At ∼ A0 = I)

We have for closed forms A∗t Ω − Ω = dut (every 2-form in the sphere

S2 is closed, homotopy F (x, t) = At is given).

OK.

4. Like in the Problem 3, we have for shifts A : T2 → T2 that there existshomotopy

At : T2 → T2 ×R such that A0 = A , A1 = I

So Ω(x, y) − Ω(x− a, y − b) = du, Ω is k - form in T2 , dΩ = 0 .

5. dx1 and dx2 in T2 (closed 1 - forms).

x = x1

y = x2

∣∣∣∣∣∣(x, y) ≃ (x + 2πm, y + 2πn)

m, n ∈ Z

21

y

x

∮ 2π

0dx1 = 2π ,

∮ 2π

0dx1 = 0

(x− cycle) (y − cycle)

∮ 2π

0dx2 = 0 ,

∮ 2π

0dx2 = 2π

(x− cycle) (y − cycle)

OK.

5a. dx1 ∧ dx2 − closed 2 - form∫∫T2

dx1 ∧ dx2 = 4π2 = 0

So dx1 ∧ dx2 = 0 in H2(T2) .

22

Homework 9

1. Calculate restriction of euclidean metric to S2 ⊂ R3 given as

z2 = 1 − x2 − y2 , z =√

1− x2 − y2 , u = x , v = y

2. Calculate in the spherical coordinates restriction of the 2-form in R3

Ω = r−3 (x dy ∧ dz − y dx ∧ dz + z dx ∧ dy)

to the sphere

z =√1− x2 − y2 , u = x , v = y

and prove that it is an area form on the sphere

3. Prove that for the closed curve with nonzero tangent vector C ⊂ R2 wehave ∮

C

k ds = 2π(integer)

4. Consider 2 vectors η, ζ ∈ R3 , x, y, z − orthogonal coordinates. Provethat vector product η×ζ is “dual” to AB−BA (skew symmetric matrices)

A = ∗ η , B = ∗ ζ

1 2 3

1 2 3η = (a, b, c) , ∗ η =

0 c −b−c 0 ab −a 0

123

1 → (23) = − (32)2 → (31) = − (13)3 → (12) = − (21)

∗ (η × ζ) = [∗ η , ∗ ζ] = AB − BA (?)

23


1. z2 = 1 − x2 − y2 (S2) ,

zy

xz =

√1− x2 − y2

ds2 = dx2 + dy2 + dz2 =

= dx2 + dy2 +

(x dx + y dy√1− x2 − y2

)2

g11 = 1+x2

1− x2 − y2, g22 = 1+

y2

1− x2 − y2, g12 =

xy

1− x2 − y2

gij|(0,0) = δij

2. Area form:d2σ =

√det gij dx ∧ dy

det gij =

(1 +

x2

1− x2 − y2

)(1 +

y2

1− x2 − y2

)− x2y2

(1− x2 − y2)2=

1

1− x2 − y2

dx , dy , dz = − x dx + y dy√1− x2 − y2

, r2 = x2 + y2 + z2

(x dy ∧ dz − y dx ∧ dz + z dx ∧ dy)|r=1 = Ω |r=1 =

= (x dy − y dx) ∧ −x dx− y dy√1− x2 − y2

+√

1− x2 − y2 dx ∧ dy =

= dx ∧ dy

(y2 + x2√1− x2 − y2

+1 − x2 + y2√1− x2 − y2

)=

=1√

1− x2 − y2dx ∧ dy =

dx ∧ dyz

24

3.

∮C

k ds = 2πm (number of rotations of τ(s)) , m ∈ Z

C

k ds = G∗ (dφ) , G : C → S1

G (s+ T ) = G(s) + 2πm , m ∈ Z

x(s) =

x(s+ T ) = x(s)y(s+ T ) = y(s)

∣∣∣∣ τ = velocity ˙x(s)

Assumption τ = 0 , |τ | = | ˙x| = 1

k∆s = ∆φ

4. η , ζ ∈ R3 , (η × ζ)12 = η1ζ2 − ζ1η2 = ∗ (η × ζ)3

(12) → 3 , (13) → −2 , (23) → 1

Orthonormal basis. Product of 1 - forms

ηi dxi ∧ ζj dx

j =∑

ij ηi ζj dxi ∧ dxj

η ζ

∗ (η ∧ ζ) = (η × ζ) (vector product)

∗ η = η1 dx2 ∧ dx3 − η2 dx

1 ∧ dx3 + η3 dx1 ∧ dx2

∗ ζ = ζ1 dx2 ∧ dx3 − ζ2 dx

1 ∧ dx3 + ζ3 dx1 ∧ dx2

∗ η =

0 η3 −η2−η3 0 η1η2 −η1 0

= A , ∗ ζ =

0 ζ3 −ζ2−ζ3 0 ζ1ζ2 −ζ1 0

= B

25

Calculate AB and BA :

AB =

−η3ζ3 − η2ζ2 η2ζ1 η3ζ1η1ζ2 −η3ζ3 − η1ζ1 η3ζ2η1ζ3 η2ζ3 −η2ζ2 − η1ζ1

BA =

−ζ3η3 − ζ2η2 ζ2η1 ζ3η1ζ1η2 −ζ3η3 − ζ1η1 ζ3η2ζ1η3 ζ2η3 −ζ2η2 − ζ1η1

AB − BA =

0 η2ζ1 − ζ2η1 η3ζ1 − ζ3η1η1ζ2 − ζ1η2 0 η3ζ2 − ζ3η2η1ζ3 − ζ1η3 η2ζ3 − ζ2η3 0

(skew symmetric).

Conclusion: AB − BA = ∗ (η × ζ) .

− Tr (AB) = 2 ⟨η , ζ⟩ Inner Product!

26

Homework 10

1. Calculate Riemannian metric

− ds2 =(dx20 − dx2 − dy2

)∣∣M2⊂R2,1

for the surface M2 in Minkowski space R2,1 with indefinite metric

dx20 − dx2 − dy2 , M2 =x20 − x2 − y2 = 1

Prove that this metric is positive.Express it in terms of “pseudospherical” coordinates:

x0 = ρ coshχ , x1 = ρ sinhχ cosφ , x2 = ρ sinhχ sinφ

x1 = x , x2 = y , ρ2 = x20 − x2 − y2 = 1

− ds2 = gij dx′i dx′j , gij = gij(θ, φ) , x′1 = θ , x′2 = φ

2. Find volume element in R2,1 in coordinates (ρ, χ, φ) , d3σ = ? In whicharea of R2,1 these coordinates are OK?

3. Let metric be given in U ⊂ R2 as

ds2 = dz dz · g (|z|2)

Find area element d2σ for such metric, g (|z|2) > 0 . Write it in theform (?) · dz ∧ dz .

4. Let 2 - sphere be given in the “Riemann” form

z (in C = S2 \∞) , w (in C = S2 \ 0)

with z = 1/w for z = 0 , w = 0 . Calculate metric

ds2 =dz dz

(1 + |z|2)2

in the local coordinate w = 1/z , z = dx + i dy .

27

5. Geodesics:= line with extremal (minimal) length among the paths join-ing P and Q ∈ M

P Qx(t)

l(γ) =

∫ Q

P

|x(t)| dt , ds2 = gij(x) dxi dxj

Fermat Principle: Let c(x) > 0 = speed of light at the point x .

“time” =

∫ Q

P

|dx|c (x(t))

, |dx| = |x| dt

Minimal time principlefor the propagation of light:Find geodesics of metric

ds2 =dx2 + dy2

c2(x)

P

Q

O

1c

c2X

beach

(they are straight lines for y > 0 and y < 0).

c = c1 (Air) y > 0 , c = c2 (Water) y < 0

∫ Q

P

dx

c(x)=

∫ P

0

dx

c(x)+

∫ Q

0

dx

c(x)


1. − ds2 =((dx0)2 − dx2 − dy2

)∣∣M2 , x0 = t

28

M2ρ=1 : t2 − x2 − y2 = 1 , ρ2 = t2 − x2 − y2

t = coshχ , x = sinhχ cosφ , y = sinhχ sinφ

(χ, φ) − coordinates:

dt = sinhχ dχ , dx = coshχ cosφ dχ − sinhχ sinφ dφ ,

dy = coshχ sinφ dχ + sinhχ cosφ dφ

dt2 − dx2 − dy2 = dχ2(sinh2 χ − cosh2 χ

)−

− dφ2(sinh2 χ sin2 φ + sinh2 χ cos2 φ

)= −

(dχ2 + sinh2χ dφ2

)- hyperbolic (Lobachevski) plane. (Sphere metric is dθ2 + sin2θ dφ2).

ds2 = gij dxi dxj , dnσ =

√det gij dx

1 ∧ . . . ∧ dxn

Gramm matrices are (gij = gji) following.

S2 L2 R2

dθ2 + sin2 θ dφ2 dχ2 + sinh2 χ dφ2 dρ2 + ρ2 dφ2

2. Area elements.

ds2 = dχ2 + sinh2 χ dφ2 , L2 : G =

(1 00 sinh2 χ

)d2σ = sinhχ dχ ∧ dφ

ds2 = dθ2 + sin2 θ dφ2 , S2 : G =

(1 00 sin2 θ

)d2σ = sin θ dθ ∧ dφ

29

ds2 = dρ2 + ρ2 dφ2 , R2 : G =

(1 00 ρ2

)d2σ = ρ dρ ∧ dφ

Volume element in R2,1 .Coordinates ρ, χ, φ (correction), ρ2 = t2 − x2 − y2 .

t = ρ coshχ , x = ρ sinhχ cosφ , y = ρ sinhχ sinφ

d3σ = ?

dt2 − dx2 − dy2 =

= dρ2[cosh2 χ − sin2 χ cos2 φ − sinh2 χ cos2 φ

]− dχ2 − ρ2sinh2 χ dφ2 =

= dρ2 − ρ2(dχ2 − sinh2 χ dφ2)

G =

1 0 00 ρ2 0

0 0 ρ2sinh2χ

,√detG = ρ2sinhχ

d3σ = ρ2sinhχ dρ ∧ dχ ∧ dφ

3. ds2 = dz dz · g , g = g(|z|2) , dz dz = dx2 + dy2

G =

(g 00 g

),

√detG = g

d2σ = g dx ∧ dy =1

2ig dz ∧ dz

4. ds2 =dz dz

(1 + zz)2(S2 − round sphere)

30

z =1

w, z =

1

z

ds2 =d(1/w) d(1/w)

(1 + 1/ww)2=

dw dw · 1/w2w2

(1 + ww)2/(ww)2=

dw dw

(1 + ww)2

Same!

5. ds2 =dx2 + dy2

c2(x, y), x(t) , y(t) − curve

Arc length:

l(γ) =

∫ b

a

√x2 + y2 dt

c (x(t), y(t))

a b

t

“Geodesics” = i.e. curve with minimal arc length

Example: Fermat Principle Let

c(x, y) =

c2

c1

y=0

a

b

Find geodesic joining a and b.

31

c1a

b

c2

X

y

y=0Aψθ

a = (x1, y1) , b = (x2, y2)

Time from a to A = |a− A|/c1 = T1

Time from A to b = |b− A|/c2 = T2

Total time = T1 + T2 (depends on A), A = (A, 0) .

T1 + T2 =√(x1 − A)2 + y21 / c1 +

√(x2 − A)2 + y22 / c2 = T (A)

minA T (A) = ?

dT

dA= 0 ?

0 =1

c1

(A − x1)√(x1 − A)2 + y21

+1

c2

(A − x2)√(x2 − A)2 + y22

↑sin θ

↑− sinψ

A − x1 < 0 , A − x2 > 0 .

c2c1

=sinψ

sin θ

32

Midterm Tests.

First Midterm Test.

1. Differential 0-forms and 1-forms, define pull-back map, change of coor-dinated, restriction to curve, integration. Newton-Leibnitz formula forintegration. Calculate differential 1-form dz/z in polar coordinates

2. Formulate necessary and sufficient condition for 1-form X = udx+vdyin the planar domain U ⊂ R2 to be closed. Find condition for closedform to be exact in terms of contour integration? Is the form dz/(z2−1)exact in the domain U ⊂ R2 where 2 points ±1 removed?

Define 2-forms and De Rham differential operator d from 1-forms to2-forms for the planar domains.

3. Define exterior (grassmanian) product of any number of 1-forms. For-mulate main properties of this product. Calculate product X1∧X2∧X3

of 3 forms in R3:

X1 = 2ydx+3dy+6dz, ,X2 = 6xdx+5dy+12dz, ,X3 = (y+3x)dx+4dy+9dz

33

Second Midterm Test.

1. Differential 1-forms and change of coordinates. Differential 1- and 2-forms in R2, R3, define the exterior product of 1-forms and De RhamOperator d for 1-forms. What is a closed form? What is an exact form?Calculate 2-form d[(xdy − ydx)/(x2 + y2)s]. Prove that it is closed fors = 1 only in R2minus0.

2. By definition, Electromagnetic Field F is a differential 2-form in R4 orpair (E,B) of 1- and 2-forms in R3 depending on time (i.e. the 4-spaceis presented as R3×R). Formulate the 1st pair of Faraday Laws in R3

(Non-relativistic form) and in R4 = (x0 = ct, x1, x2, x3) (Relativisticform) in terms of differential forms and De Rham Operator. ApplyStokes Formula to express the integral of Electric Field along the closedcontour C in R3 through the magnetic field inside of contour (its timederivative).

3. Prove that for every symmetric indefinite nondegenerate inner productin R2 there exists a basis e, e′ such that Gramm Matrix has a (”light-like”) form

(e, e) = (e′, e′) = 0, (e, e′) = 1

Second Midterm Test. Solutions.

I. First problem is collection of definitions and calculation of dΩ for somespecific 1 - form Ω .

Definitions:

1. 1 - form in R1 , R2 , R3 , . . . :∑i

ui(x) dxi

Change of coordinates x(x′) (pull-back, restriction, . . . )

f(x) → f (x(x′)) ,

34

dxi → ∂xi

∂x′kdx′k ,

matrix form (Jacoby matrix):

dx →(∂x

∂x′

)dx′

2. 2 - form in R2 , R3 , . . . :

Ω =∑ij

aij(x) dxi ∧ dxj

De - Rham operatord : Ω → dΩ

Multiplication of forms ∧ :(f dxi ∧ dxj

)∧(g dxk ∧ dxl

)= fg dx1 ∧ dxj ∧ dxk ∧ dxl

Operator d :

df =∂f

∂xidxi , d

(f dxi ∧ . . . ∧ dxj

)= df ∧ dxi ∧ . . . ∧ dxj

3. Calculate dΩ for

Ω =(x2 + y2

)−s(x dy − y dx)

Exact form: Ω = dω .

Closed form: dΩ = 0 .

Properties

ω − exact ⇒∮C

ω = 0 (closed contour)

ω − closed ⇒∮C

ω = 0 C ⊂ R2

35

D

C

if domain D is simply connected(ω is defined and smooth in D).

II. Second Problem:

Electromagnetic field = 2 - form F in R4 (space-time)

R4 = (x0 = ct, x, y, z)

First part of Maxwell Equations (Faraday Law)

dF = 0

3 − dimensional form in R3 × R :(x, y, z) (t)

E = Ex dx + Ey dy + Ez dz

F = E ∧ dx0 + B , x0 = ct

B =∑

i,j=1,2,3

bij dxi ∧ dxj

d(4)E = 0 ⇔ “Pair” :

a) d(3)B = 0 ,

b) d(3)E + ∂B/∂x0 = 0

(d(4) − in R4 , d(3) − in R3).

D

C

Stokes:∮C

E =1

c

∂

∂t

∫D

B

Second pair of Maxwell Equations:

36

d (∗F ) = J (4) (“4 − current”)

↑ ↑involves non− geometricalinner term

product

This law is nongeometrical(interaction with matter)

First pair dF = 0 is purely geometrical.

III. Third problem: R2 with indefinite inner product (symmetric). Findbasis e, e′ with Gramm Matrix

(e , e′) = 1 , (e , e) = (e′ , e′) = 0 ,

(0 11 0

)= G

Proof. Start with basis

e1 , e2 , G′ =

(1 00 −1

)Put

e = e1 + e2 , e′ = e1 − e2

(e , e) = (e′ , e′) = 0 , (e , e′) = 2

Put

e =e√2, e′ =

e′√2

37

Lectures 1–30

Introductory lecture: History of integration and cal-culus

1. XVII–XVIII centuries:

x∫a

f(y) dy = g(x), g′(x) = f(x), . . .

2. XIX century: idea of a differential 1-form (Green, Cauchy, . . . ) andfirst topological ideas (1820)

0 =

∮C

f(z) dz

C

∂f

∂z= 0 inside C −→ complex analytic function

Integration of differential forms is the best possible type ofintegration:

(i) It is well-defined without use of any other structures (Riemannianmetric, . . . ).

(ii) It is invariant under changes of coordinates both in the space(manifold) and in the body. Admissible changes of coordinatesform a very broad class (even not one-to-one along the body).

(iii) They can be differentiated (d). This differential is invariant.

(iv) They have remarkable “Stokes Property” connecting (body) and(its boundary). It connects them with topology — “De RhamCohomology” known since XIX century.

38

(v) They can be described in the beautiful “differential” language aswell as in the classical vector (tensor) language.

3. Electromagnetism, Faraday laws, Gauss formula and differential forms,topological quantities, Maxwell equations (mid XIX century)

(linking number)

4. Vector (tensor) calculus (in the 3-space), Maxwell, Kelvin, Stokes,Poincare.

Stokes-type formulas:

b∫a

f ′(x) dx = f(b)− f(a)a− +

b

body and boundary

(simplest Newton–Leibnitz formula).

5. Poincare, differential forms and topology (vector (tensor) calculus inmultidimensional spaces) (1895). Hamiltonian systems, Riemann sur-faces.

6. E.Cartan and differential forms (1920s). Discovery of differential formsin the modern notations. Topology.

7. De Rham Theorem (1930s). Proof of the Cartan–Poincare conjecture.

8. Differential forms in modern mathematics. Complex geometry.

Differential forms are special tensors (“contravariant”, skew symmetric).Why are they especially important?

1. They provide a universal integration and differentiation (independentof any other structure like Riemannian metric and so on).

39

2. They present the best bridge between Analysis and Topology. Cauchystarted to use these ideas in early XIX century.

3. They describe the laws of Electromagnetism (Faraday laws and Maxwellequations).

4. In the special case of Complex Geometry everything important canbe written in terms of differential forms including Riemannian metric(“Kahler metric”).

5. The differential notations of Cartan are extremely convenient.

40

Lecture 2. Differential 1-forms

Question: What is a differential 0-form?Answer: Differential 0-forms are functions f(x), 0-bodies are points, 0-chains are collections of points and numbers:

Q =[•

(P1,n1)•

(P2,n2). . . •

(Pk,nk)

], nj ∈ Z,

Q = “0-chain” = (P1, n1;P2, n2; . . . ;Pk, nk).

The “value” (or the “integral”) of the 0-form f along a 0-chain Q is

⟨f,Q⟩ =k∑j=1

njf(Pj).

Example. Q is a single point P with n1 = ±1,

•P1,n1=±1

⟨f,Q⟩ = ±f(Q).

What is a differential 1-form?In the space with coordinates (x1, . . . , xn), R2 = (x, y), R1 = (x),

R3 = (x, y, z) (n = 1, 2, 3), (co)-“vector field” X = (η1, . . . , ηn), ηj =ηj(x

1, . . . , xn).n = 1: η(x) has a single component;n = 2: (η1(x, y), η2(x, y)), x

1 = x, x2 = y.A differential 1-form is

X =n∑j=1

ηj dxj.

n = 1: X = η(x) dx;n = 2: X = η1(x, y) dx+ η2(x, y) dy;n = 3: X = η1(x, y, z) dx + η2(x, y, z) dy + η3(x, y, z) dz, x

1 = x, x2 = y,x3 = z.Question: How to integrate a differential 1-form?Question: How to write it down in any other system of coordinates?

η(x) dx = η(x(y))dx

dydy, x = x(y).

41

x←→ yx

y

What is a one-dimensional “oriented” body?

n = 1a b

X

n = 2

ab

x(t), y(t)

Oriented linet

n = 1.

Oriented parametrized curve (body)

A

B

(x(t), y(t))

y

x

γt = a (point A)

t = b (point B)

Integration of a 1-form along an “oriented” curve (“body”) γ (case n = 2)

B∫A

(η1(x, y) dx+ η2(x, y) dy

)= I

along the curve γ.

Definition. (see n = 2)

Step 1. Restrict 1-form to the body. Substitute x = x(t), y = y(t), a 6 t 6 b

I =

b∫a

(η1(x(t), y(t))

dx

dt+ η2(x(t), y(t))

dy

dt

)dt =

b∫a

φ(t) dt.

42

Step 2. Calculate the ordinary integral along the line. “Parametrized curve”x(t), y(t): R1 → R2 (or R1 → Rn).

Vector notation: X = (η1, η2). Restriction to the curve γ = (x(t), y(t)),γ = (dx/dt, dy/dt) for n = 2:

⟨X, γ⟩ = η1(x(t), y(t))dx

dt+ η2(x(t), y(t))

dy

dt.

Integral along the curve γ reduces to the ordinary integral

I =

B∫A

⟨X, γ⟩ dt,

where ⟨ , ⟩ is the “inner product” in orthonormal coordinates.We assume here that the space is Euclidean. In fact, our integral does

not depend on the Euclidean geometry.

Examples :n = 1. X = η(x) dx, x = t

I =

b∫a

η(x) dx.

n = 2. x = cos t, y = sin t (circle)

t

X =x dy − y dxx2 + y2

,

a very interesting differential 1-form (angle). It will be considered later.Calculate the restriction to x2 + y2 = ρ2.

Examples from physics:

1. The speed of particles is not a differential 1-form. One cannot integrateit without Euclidean geometry.

43

2. Electric field E = (E1, E2, E3) is a differential 1-form (n = 3).

3. Magnetic field is not a differential 1-form (it is in fact a 2-form).

4. The gradient of a function is a differential 1-form (n is arbitrary):

df =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz + . . .

Integration (see the proof below).

Theorem 1.B∫A

df = f(B)− f(A)

(along any smooth path γ : R1 → Rn joining A and B)

A Bγ

We will prove it later.What is the “de Rham” d-operator? It is d : f → df .

Proof of Theorem 1. Let n = 1. We have A = a, B = b,

b∫a

f ′(x) dx = f(b)− f(a), df = f ′(x) dx.

Let n > 1. We have

df = fx dx+ fy dy + . . . , fx =∂f

∂x, . . .

A B

x(t), y(t) a 6 t 6 bt = a

t = b

44

B∫A

df =

b∫a

(fx(x(t), y(t))

dx

dtdt+ fy(x(t), y(t))

dy

dtdt)

(let g(t) = f(x(t), y(t))

=

b∫a

dg(t) = g(b)− g(a) = f(B)− f(A).

The theorem is proved.

45

Lecture 3

0-forms = functions f(x), I = f(P ).1-forms = X =

∑i ηi(x) dx

i (x = x1, . . . , xn).The integral of a 1-form X along 1-body (parametrized curve) γ =

x(t) = (x1(t), . . . , xn(t)), a 6 t 6 b is defined by formula:

I =

b∫a

∑i

ηi(x(t))dxi

dtdt =

∫γ

X =

∫⟨X, γ⟩ dt, γ = (x1(t), . . . , xn(t)),

X = η1(x(t)), . . . , ηn(x(t)), ⟨X, γ⟩ =∑i

ηixi (“inner product”).

Example. X = df =∑

i∂f∂xi

dxi.

I =

B∫A

df = f(B)− f(A)

(any path γ).

B : t = bA : t = a

Change of coordinates

1. Along the curve γ

γ = x(t), t = t(τ), dt/dτ > 0.

Claim: ∫γ(t)

X =

∫γ(t(τ))

X

because on the line we have

b∫a

f(t) dt =

d∫c

f(t(τ)) dτ · dtdτ

46

t

b

a

c d τ

But it is also true for “non-monotonic” changes (sometimes we may havedt/dτ < 0):

t

b

a

c d τ

2. Change of coordinates in the space x↔ x′

x1(x′1, . . . , x′

n),

x2(x′1, . . . , x′

n),

. . .

xn(x′1, . . . , x′

n),

Let n = 1.η(x) dx = η′(x′) dx′,

dx′ · dxdx′

= dx,

dx(x(x′)) =dx

dx′dx′

one-to-one change!

Definition.η(x) dx = η′(x(x′)) dx′

47

is a definition of the same 1-form in the new coordinate x′. η and η′ representthe same 1-form, so

b∫a

η(x(t)) dt =

b∫a

η′(x(x′(t)) dt

because it is the same integral.

η′ = η · dxdx′

, for n = 1,

in the same point x(x′), x′ ↔ x.For every n > 1 we define the same change of coordinates∑

i

η′i(x) dx′i =

∑j,k

ηj∂xj

∂x′idx′

i,

η′i = ηj∂xj

∂x′i

[−→η ′ = −→η · ∂x∂x′

].

matrix multiplication

n = 2.η1(x, y) dx+ η2(x, y) dy = η′1 dx

′ + η′2 dy′

x(x′, y′); y(x′, y′) ↔ one-to-one change.

dx→ ∂x

∂x′dx′ +

∂x

∂y′dy′,

dy → ∂y

∂x′dx′ +

∂y

∂y′dy′.

Jacoby matrix

J =

∂x

∂x′∂x

∂y′

∂y

∂x′∂y

∂y′

.

Requirement: det J = 0.

48

Change is one-to-one, invertible as a smooth map x′(x, y), y′(x, y)∂x′

∂x

∂x′

∂y

∂y′

∂x

∂y′

∂y

= J−1.

−→x = (x1, . . . , xn), −→x ′ = (x′1, . . . , x′n). −→x (−→x ′), −→x ′(−→x )−→x (−→x ′(−→x )) ≡ −→x .

∂xi

∂x′j∂x′j

∂xk= δik =

1 0 . . . 0

0. . . . . .

......

. . . . . . 00 . . . 0 1

.

the unit matrix

∂x

∂x′∂x′

∂x≡ 1 (matrix multiplication).

φ(x(y))∂φ

∂y=∂φ

∂x

∂x

∂y(matrix multiplication).

Remark 1. Change of coordinates for ordinary vectors (like speed of parti-cles). γ(t) : x1(t), . . . , xn(t). Speed: γ(t) = (x1, . . . , xn). New coordinatesx(x′).Speed:

dx

dt=∂x

∂x′dx′

dt

( ∂xi∂x′j

)= J

γ(x)

= Jγ′

(x′)

⇒ γ′

(x′)

= J−1γ(x)

1-form:

η dx = η′ dx′, dx =∂x

∂x′dx′, η

∂x

∂x′dx′ = η′ dx′,

so we have

η∂x

∂x′= η′ or ηJ⊤ = η′

(ηi∂xi

∂x′j= η′j

)It is not the same law: the equality

J−1 = J⊤ ⇔ J ∈ On

(orthogonal)!

49

is true for Orthogonal matrices only.)

Remark 2. For 1-forms this formula makes sense even if the change is noton-to-one

η′ = Jη x = x(x′(t))

x = (x1, . . . , xn), x′ = (x′1, . . . , x′n).May be even det J = 0 in some points (even everywhere). This formula

makes sense even if dim x = dim x′, γ : Rk

(x′)→ Rn

(x)

X =∑

ηi(x) dxi, γ∗X =

∑i,j

ηi(x(t))∂xi

∂tjdtj, t = (t1, . . . , tk).

Example. Let k = 1, n = 2,

γ : R1

t

−→ R2

(x,y)

, x(t), y(t)

1-form η1 dx+ η2 dy = X in R2.“Restriction” of a 1-form to γ = R1 is also a partial case of this formula

γ∗X = η1(x(t), y(t))dx

dtdt+ η2(x(t), y(t))

dy

dtdt.

50

Lecture 4

Coordinates in Rn. Let (x1, . . . , xn) — Cartesian coordinates: −→x = (x1, . . . , xn).−→x 1 = −→x 2 ⇔ the points are distinct. Let U ⊂ Rn be an open domain withthe same cartesian coordinates.

Examples (changes of coordinates).

a) Shift −→x = −→x ′ +−→a (−→a is a vector),

x′i → x′

i+ ai = xi.

b) Rotation (let n = 2). (x1, x2) = (x, y)

x′

x

y′y

α

x = x′ cosα + y′ sinα,

y = −x′ sinα + y′ cosα

∣∣∣ (x′, y′)→ (x, y)

c) Reflection x→ −x, y → y, x′ = −x, y′ = y.

d) Linear change

x = ax′ + by′

y = cx′ + dy′

∣∣∣ (xy

)= A

(x′

y′

).

e) Affine change (xy

)= A

(x′

y′

)+−→β

∣∣∣ A =

(a bc d

).

51

Polar coordinates (n = 2). x = ρ cosφ, y = ρ sinφ, (x′, y′) = (ρ, φ),ρ2 = x2 + y2, x/ρ = cosφ, y/ρ = sinφ.

Spherical coordinates (n = 3).

z = r cos θ,

x = r sin θ cosφ,

y = r sin θ cosφ.

(r, θ, φ)

x

y

z

θ

φ

“Formal” complex coordinates

z = (x+ iy),

z = (x− iy).(z, z), i2 = −1.

Let n = 2. A differential 1-form is u(x, y) dx + v(x, y) dy. Change ofcoordinates (x, y)↔ (x′, y′):

u(x, y)→ u(x(x′, y′), y(x′, y′)),

v(x, y)→ v(x(x′, y′), y(x′, y′)).

For functions:f(x, y)→ f(x(x′, y′), y(x′, y′))

— same functions in the new coordinates.For 1-forms:

u dx+ v dy = u′ dx′ + v′ dy′ ?

dx→ ∂x

∂x′dx′ +

∂x

∂y′dy′,

dy → ∂y

∂x′dx′ +

∂y

∂y′dy′.

For every function f(x, y) we have:

df → ∂f

∂xdx+

∂f

∂ydy → ∂f

∂x

∂x

∂x′dx′ +

∂f

∂x

∂x

∂y′dy′ +

∂f

∂y

∂y

∂x′dx′ +

∂f

∂y

∂y

∂y′dy′.

52

Conclusion. In every coordinates we have

df =∂f

∂xdx+

∂f

∂ydy =

∂f

∂x′dx′ +

∂f

∂y′dy′.

(For all n > 1: df =n∑i=1

∂f

∂xidxi.)

Operator d: functions → 1-forms, does not depend on coordinates. Itcommutes with all C∞-maps φ : Rk → Rn (x′1, . . . , x′k)

φ→ (x1, . . . , xn),−→x = φ(−→x ′).

The rule is:φ∗ : f(−→x )→ f(−→x (−→x ′)),

φ∗ : df(−→x )→ df(−→x (−→x ′)),

dxi →∑j

∂xi

∂x′jdx′

j.

Terminology: we call it“induced map” (or pull-back map) φ∗ acting onfunctions and forms.

Examples.Shifts −→x = −→x ′ +−→a .Rotations −→x = A−→x ′, A⊤ = A−1 (orthogonal maps) AA⊤ = 1.General linear maps (A is not orthogonal).Polar coordinates (n = 2) (x′, y′) = (ρ, φ), ρ2 = x2 + y2.

Remark. Properties of Cartesian coordinates in any open U ⊂ Rn (x1, . . . , xn):

a) every coordinate domain xj = c is ??? open domain in Rn−1;

b) −→x 1 = −→x 2 ⇒ points are distinct.

53

Lecture 5

Coordinates in the plane R2:Cartesian (x, y), x ∈ R, y ∈ R,Polar (ρ, φ), x = ρ cosφ, y = ρ sinφ.

Changes: shifts, linear transformations, affine transformations well- de-fined for all points in R2 (or Rn). We have also ρ > 0 in R2.

Polar coordinates are well-defined in domains not containing a closedpath surrounding 0 ∈ R2.

Examples.

0 Yes

0 Yes

0 Yes

54

0

Yes

0 No

How does dφ look in the coordinates (x, y)?

dρ = d√x2 + y2 =

2x dx+ 2y dy

2√x2 + y2

=x dx+ y dy

ρ.

dφ =?x = ρ cosφ ⇒ dx = dρ cosφ− ρ sinφdφ,y = ρ sinφ ⇒ dy = dρ sinφ+ ρ cosφdφ.

x dy = ρ cosφ(dρ sinφ+ ρ cosφdφ),

y dx = ρ sinφ(dρ cosφ− ρ sinφdφ),x dy − y dx = ρ2 cos2 φdφ+ ρ2 sin2 φdφ

= ρ2 dφ.

x dy − y dxρ2

= dφ.

55

Spherical coordinates in R3:

z = r cos θ,

x = r sin θ cosφ,

y = r sin θ cosφ,

r2 = x2 + y2 + z2.

(r, θ, φ)

x

y

z

θ

φ

r = 1 is the unit sphere S2. (θ, φ) are coordinates in S2.0 6 θ 6 π,

0 6 φ 6 2π

∣∣∣∣∣ Are (θ, φ) Cartesiancoordinates in S2?

In which open domains U ⊂ S2 are they ok?

θ = 0

θ = π

ℓ S2 − ℓ = U

Complex coordinates in R2: x+ iy = z, x− iy = z, (z, z).The differential of a function:

df(x, y) = fx dx+ fy dy.

Make “change”

dx =1

2(dz + dz), dy =

1

2i(dz − dz).∣∣∣∣∣∣∣∣

∂z

∂x

∂z

∂y

∂z

∂x

∂z

∂y

∣∣∣∣∣∣∣∣ =∣∣∣∣1 i1 −i

∣∣∣∣ = −2i = 0.

56

“Jacobian” = 0.Let Adx+B dy = X be a differential 1-form. We can write???

1

2A(dz + dz) +

1

2iB(dz − dz) = 1

2(A− iB) dz +

1

2(A+ iB) dz.

Let f = u(x, y) + iv(x, y) be a function.

df = du+ i dv = ux dx+ uy dy + i(vx dx+ vy dy) = (ux + ivx) dx+ (uy + ivy) dy

= Adx+B dy =1

2(A− iB) dz +

1

2(A+ iB) dz.

Definition.A− iB

2=∂f

∂z,

A+ iB

2=∂f

∂z,

df =∂f

∂zdz +

∂f

∂zdz,

∂f

∂z= (ux + ivx) + i(uy + ivy) = (ux − vy) + i(vx + uy).

Condition (“analyticity”):

∂f

∂z= 0 ⇒ ux = vy, vx = −uy.

Example. f(z) = zn, df = nzn−1 dz.

Taylor series: ∑n>0

cn(z − a)n.

It is convergent in some disc (Abel)

z = a

Laurent series: ∑n∈Z

cn(z − a)n.

57

It is convergent in an annulus

z = a

The expression like f(z, z) dz+g(z, z) dz defines arbitrary complex-valuesdifferential 1-form in R2.

58

Lecture 6

Coordinates:

Polar (ρ, φ)R2

ρ = 0?which domains U ⊂ R2 \ 0?dφ is a closed 1-form in R2 \ 0.

Spherical (r, θ, φ)R3 ⊃ S2 (r = 1),r = 0.Sphere S2 \ N,S

N

S

θ = 0, π, φ?dφ is a closed 1-form in S2\N,S.

Claim. a) For any closed path γ around 0 in R2 \ 0 we have

∮γ

dφ = 2π. 0γ

b) For S2 \ N,S and γ around N,S we have

∮γ

dφ = 2π

N

S

γ

59

Proof of a).b∫

a

dφ = ∆φ(along γ)

(no matter which path if γ does not cross 0).

0a

bγ

0a

b

γ ∆φ does not dependon the path

Proof of b): Same.

Complex coordinates in R2:

z = x+ iy,

z = x− iy,

∣∣∣ dz = dx+ i dy,

dz = dx− i dy,

∣∣∣ dx =1

2(dz + dz),

dy =1

2i(dz − dz).

Let f = u(x, y) + iv(x, y) and f(x, y) ∈ C.

df = fx dx+ fy dy = Adz +Bdz.

Definition 1. We call A = ∂f/∂z abd B = ∂f/∂z partial derivatives alongthe complex directions.

∂

∂z=

1

2

( ∂∂x− i ∂

∂y

),

∂

∂z=

1

2

( ∂∂x

+ i∂

∂y

).

∣∣∣∣∣∣∣∣ operators ∂z = ∂, ∂ = ∂z

For every 1-form X = U dx+ V dy we can write X = U dz + V dz.

Definition 2. Complex analytic function f = u+ iv is such that:

∂f/∂z = 0 ⇔ ux + vy = 0, vx − uy = 0.

60

Examples: z = ρeiφ.f = 1, z, z2, . . . (any polynomial in z);f = P (z)/Q(z)—a rational function (except “poles”Q = 0), f = 1/z, 1/z2, . . .;f = log z—multivalued, log z = log |z|+ i arg z = log(ρ) + iφ;f = ez, . . .

Important observation (topology):∮γ

f(z) dz = 0 if∂f

∂z≡ 0 inside γ.

γ

the contour γ is “simply con-nected” (no holes inside).

Explanation. Let f(z) = zn. We have

0

γ : |z| = 1ρ = 1

∮γ

dz = 0,

∮γ

z dz = 0, . . . ,

∮γ

zn dz = 0.

z = x+ iy = ρ(cosφ+ i sinφ) = ρeiφ.

Let γ be ρ = 1.∮γ

dz = ?

∮γ

z dz =

∮γ

dw = 0, . . .? dw/dz = z.

The proof will appear later. It follows from the fact that for every smoothfunction f(x, y) we have

∮df ≡ 0.

61

γ

no holes!

← closed path, simply connected inside

What is a closed 1-form?

Definition. a) A 1-form X = u dx+ v dy is closed iff for every point P ∈ R2

there exists a small neighbourhood P ∈ U ⊂ R2 such that X = u dx+v dy =df in U . Same definition for all n > 2.

b) “An exact 1-form” X is such that X = df in the whole domain Vwhere it is defined.

Examples. a) The domain V is R2 \ 0, X = dφ. It is closed but not exactbecause φ is multivalued. The obstruction is

∮γdφ = 2π = 0.

0γ

b) The domain V is S2 \ N,S, X = dφ is closed, but not exact.c) The domain V ⊂ R2 \ 0, where φ is one-valued smooth function, dφ

is exact in V .

62

Lecture 7

Our program and comparison with Bachman:

1. We assume that multivariable calculus is known (partial derivatives,differential of a function).

2. Special coordinates (polar, spherical, complex z = x + iy). Complexcoordinates are missing in Bachman.

3. Theory of differential 1-forms. This theory is briefly done in Bachman.Something appears later only as partial cases of n-forms. This theory(including elementary topology) is fundamental especially complex 1-forms in R2 and Riemann surfaces (Cauchy, Riemann, Poincare).

4. Contravariant property of the category of functions and forms underC∞-maps f : Rm → Rn appears only in the Chapter “Manifold” inBachman, it is called a ”pull-back map” f ∗

From 1-forms to 2-forms:Bachman discusses “product” of 1-forms. We will discuss it later, and startfrom Calculus:Problem. Let a differential 1-form be given in R2 (or Rn): X = u dx+ v dy(in some domain V ).a) Is it locally the differential of some function f :

(x, y) ∈ Uε ⊂ V, X = df, u =∂f

∂x, v =

∂f

∂y?

b) Is it globally differential X = df of any one-valued smooth function in thewhole domain V ?

Examples. 1) X = λ dx+ µ dy, f = λx+ µy, λ, µ = const.

2) X = dφ =x dy − y dxx2 + y2

.

a) V1 = R2 \ 0, dφ is not df for a 1-valued f in V ;

b) V2 = •0

V2 , X = df , f = φ.

Closed 1-forms = locally differentials. Exact 1-forms = globally differen-tials. Closed forms modulo exact ones = Cohomology group H1(V,R) (Thefirst “de Rham” cohomology).

63

What do we know from Calculus? X = u dx+ v dy ((u, v) = “a covectorfield” in V ⊂ R2),

u =∂f

∂x, v =

∂f

∂y(locally) in Uε ⊂ V.

Uε

Claim. Nearby every point from V (n = 2) the condition

∂u

∂y=∂v

∂x

is necessary and sufficient for the “local” existence of f , ∇f = (u, v), ordf = X.

n > 2: X =∑

i ui dxi:

∂ui∂xj

=∂uj∂xi

.

“Global obstruction” for the existence of a single-valued f = Topology.

Example.

γ V ∮γ

dφ = 2π.

γ1 γ2

dφ1 dφ2 ∮γ1

dφ1 = 2π,

∮γ1

dφ2 = 0,

∮γ2

dφ1 = 0,

∮γ2

dφ2 = 2π.

64

“The cohomology with real coefficients” H1(V,R) is generated by dφ1, dφ2

as a linear space over the field R. Similar answer we have for H1(V,C) usingthe complex-valued 1-forms.

Consider the formal expression

dX =(∂u∂y− ∂v

∂x

)dx ∧ dy (n = 2),

dX =∑i<j

(∂ui∂xj− ∂uj∂xi

)dxi ∧ dxj (n > 2).

Claim. dX = 0 if and only if X is locally exact (i.e. “closed” 1-form).

Formal expressions like a dx ∧ dy,∑i<j

aij(x) dxi ∧ dxj (n > 2)

are called “differential 2-forms”.We extend aij to all pairs (i, j) by the requirement:

aij = −aji, ajj = 0.

The operator d : 1-forms→ 2-forms will be defined.The product of 1-forms X ∧ Y will also be defined by the rule

X = u dx+ v dy,

Y = w dx+ t dy,

∣∣∣ X ∧ Y = (ut− vw) dx ∧ dy.

65

Lecture 8

From 1-forms to 2-forms: let a differential 1-form X = u dx+ v dy be given.It is exact if X = df , i.e. fx = u, fy = v. It is closed if locally there exists afunction f such that fx = u, fy = v.

Example. X = dφ in R2 \ 0.

dφ =x dy − y dxx2 + y2

.

Lemma 1. A differential 1-form

X =n∑i=1

ui dxi

in a domain U ⊂ Rn is closed iff the following equations are satisfied:

∂ui∂xj

=∂uj∂xi

.

(Calculus.)

Introduce a 2-form

dX = Ω =∑i6j


)dxi ∧ dxj,

where dxi ∧ dxj = −dxj ∧ dxi by definition.More generally,

Ω =∑i<j

aij(x) dxi ∧ dxj.

We have an operator

d : (1-forms)→ (2-forms)

such that

d(∑

i

ui dxi)=∑i<j


)dxi ∧ dxj.

66

We also haved : (0-forms)→ (1-forms),

df(x) =∑i

∂f

∂xidxi.

Properties of the operator d:

1. d · d ≡ 0;

2. df = exact 1-forms, dX = exact 2-forms;

3. ker d: closed 1-forms dX = 0. Closed 0-forms = constant functionsdf = 0;

4. Cohomology H1(U,R) = closed 1-forms modulo exact ones.

Examples:

a) Simply connected domain U ⊂ R2: H1 = 0;

b)

U

1 hole, H1 = R;

U

2 holes, H1 = R2.

Define now multiplication of 1-forms X ∧Y . It is bilinear and has a formfor n = 2

X = u dx+ v dy,

Y = w dx+ t dy,

∣∣∣ X ∧ Y = ?

X ∧ Y = ut dx ∧ dy + vw dy ∧ dx = (ut− vw) dx ∧ dy.

67

For general n > 2 we define:

X =∑

ui dxi, Y =

∑vj dx

j,

X ∧ Y =∑i<j

(uivj − viuj) dxi ∧ dxj.

So we have can formulate axioms for multiplication: 1) dxi∧dxj = −dxj∧dxi;2) numbers (functions, 0-forms) commute with everything, ui dx

j = dxj · ui.Extension to products of more than 2 1-forms and to arbitrary m-forms

(m > 2) follows from the Associativity Axiom (dx∧dy)∧dz = dx∧ (dy∧dz).

Motivation: Let 2 vectors be given in R2

η

ξ

Area =

(ξ1 ξ2η1 η2

)

determinant

= ξ ∧ η = −η ∧ ξ.

n = 3. Volume η, ζ, ξ ∈ R3

η

ξζ Volume =

∣∣∣∣∣∣η1 η2 η3ζ1 ζ2 ζ3ξ1 ξ2 ξ3

∣∣∣∣∣∣ .

Volume = η ∧ ζ ∧ ξ = −ζ ∧ η ∧ ξ = ζ ∧ ξ ∧ η.Same is true in Rn for n vectors. Volume is a multilinear skew symmetric

function of n vectors. We write it as η(1), . . . , η(n)

Volume = η(1) ∧ η(2) ∧ . . . ∧ η(n)

V (η(1), η(2), . . . , η(n))

linear in every variable, skew symmetric (changes sign by (−1)σ for anypermutation σ). We write it as

(η(1)i dxi) ∧ (η

(2)j dxj) ∧ . . . ∧ (η

(n)k dxk).

68

Lecture 9

(0-forms)(f)

−→d

(1-forms)(X)

−→d

(2-forms)(Ω)

f(x), x ∈ U ⊂ Rn(X =

∑i

ui dxi) (

Ω =∑i<j

aij dxi ∧ dxj

)n = 2:

X = u dx+ v dy, Ω = a dx ∧ dy.

Axioms.

1. dxi ∧ dxj = −dxj ∧ dxi;

2. functions commute with everything, u(x) dx = dx · u(x);

3. associativity (extension to k-forms for k > 2):

dx ∧ (dy ∧ dz) = (dx ∧ dy) ∧ dz;

4. multiplications of k-forms is bilinear (“distributive”).

We have “algebra of R(C)-valued differential forms” Λ(U,R(C)) in thedomain U ⊂ Rn with Cartesian coordinates (x1, . . . , xn). A 2-form is∑

aij(x) dxi ∧ dxj, i < j,

a 3-form is ∑i<j<k

aijk(x) dxi ∧ dxj ∧ dxk,

. . .k-form is ∑

i1<...<ik

ai1,...,ik(x) dxi1 ∧ . . . ∧ dxik ,

n-form isa(x) dx1 ∧ . . . ∧ dxn.

Herea(x) = a1,2,...,n(x).

69

Examples.k = 1: X =

∑ui(x) dx

i, n = 2: X = u dx+ v dy;k = 2: Ω =

∑i<j aij dx

i ∧ dxj, n = 2: Ω = a dx ∧ dy;k = 2, n = 3:

2-form = a12 dx ∧ dy + a13 dx ∧ dz + a23 dy ∧ dz.

The “vector” (a23,−a13, a12) = −→η is “an axial vector” in the physical litera-ture.

Operations. d : f → df = fxi dxi.

n = 2. d : X → dX = (uy − vx) dx ∧ dy, a12 = uy − vx.n > 2.

dX =∑i<j


)dxi ∧ dxj.

Example. 2-form a dx ∧ dy + b dy ∧ dx = (a− b) dx ∧ dy.

Change of coordinates: what do we already know?Functions x′ → x or x(x′), x1(x′1, . . . , x′n), . . . , xn(x′1, . . . , x′n). f(x(x′))

is the pull-back of the function f for the map x′ → x (map φ : U ′ → U),f(x(x′)) = φ∗f(x′).

1-forms. φ : U ′ → U , x′ → x, x(x′).

φ∗X =∑i

ui dxi φ∗−→

pull-back

∑i,j

ui(x(x′))∂xi

∂x′jdx′

j.

Or u(x)φ∗−→ u(x(x′)) as a function

dxjφ∗−→

∑i

∂xj

∂x′idx′

i.

Partial cases of “pull-back”:

1. Change of coordinates x↔ x′, one-to-one (may be local)

det( ∂x∂x′

)= det J = 0,

J = Jacoby matrix =( ∂xi∂x′j

).

70

2. Restriction to some surface (curve) (number of x′ is less than n).

General definition of the pull-back is given

f(x)→ f(x(x′)) = φ∗f(x′),

φ∗ : df → ∂f

∂xj(x(x′))

∂xj

∂x′ldxl = d(φ∗f) (sum in j, l).

Conclusions. The differential d of a function (0-form) commutes with thepull-back operation: φ∗(df) = d(φ∗f). By definition, the exterior multipli-cation of forms commute also with pull-back map. So we have

Claims. 1. φ∗d = dφ∗ for all k-forms (proof for k = 1 will be given later).2. φ∗(X ∧ Y ) = φ∗(X) ∧ φ∗(Y ). Product of forms is “natural” (commuteswith the pull-back for maps φ : U ′ → U including restrictions and changesof coordinates).

71

Lecture 10

Remarks.

1. Differential of function f is 1-form df =∑fxi dx

i as we write it inAnalysis. Vectors in Analysis we write as fist order differential opera-tors of the form

∑aj(x)∂/∂xj = w.

The scalar product of a 1-form and a vector attached to some point isby definition“the directional derivative” of a function

∇wf =∑i

∂f

∂xiai(x) = ⟨df, w⟩.

2. The homology H1(U) of a planar domains U in fact appear a lot inComplex Analysis as number of “cuts” necessary to make domain sim-ply connected.

Number of cuts =rank of H1(U).

Remaining domain should be simply connected (every closed contour= boundary of a “ball”).

The homology is generated by “cycles”. A cycle c is ∼ 0 iff for everyclosed 1-form X, dX = 0, we have∮

c

X = 0.

3. Cartesian coordinates in U ⊂ Rn (x1, . . . , xn) “inherited” from Rn are(x1, . . . , xn) in U . Other Cartesian coordinates = set (y1, . . . , yn) ofC∞-functions U →

yiR such that −→y 0 = −→y 1 ⇔ the points are distinct.

0-forms = functions,1-forms:

∑ui dx

i in U ⊂ Rn or u dx+ v dy in U ⊂ R2,

72

2-forms:∑

i<j aij dxi ∧ dxj, U ⊂ Rn, a dx ∧ dy in R2,

. . . ,k-forms: ∑

i1<i2<...<ik

ai1,...,ik(x) dxi1 ∧ . . . ∧ dxik , dxi ∧ dxj = −dxj ∧ dxi.

n-form in Rn

a dx1 ∧ . . . ∧ dxn, a = a1,2,...,n(x),

an object of integration in Rn (ordinary)∫· · ·∫

D⊂Rn

a dx1 ∧ dx2 ∧ . . . ∧ dxn.

Algebraic operations: forms = linear space, multiplication X ∧ Y

a) functions commute with everything f(x) ∧X = X ∧ f(x);

b) dxi ∧ dxj = −dxj ∧ dxi, dx ∧ dy = −dy ∧ dx;

c) associativity (X ∧ Y ) ∧ Z = X ∧ (Y ∧ Z).

Symbols dxi1 ∧ . . . ∧ dxik for i1 < . . . < ik generate the space of k-formslinearly (coefficients are functions ai1,...,ik(x)).

The differential operation: df .d : functions→ 1-forms, f 7→ df ;d : 1-forms→ 2-forms, u dx+ v dy 7→ (uy − vx) dx ∧ dy.

Definition of the operator d for all k-forms (it is a linear operator):

1. d(dxi1 ∧ . . . ∧ dxik) = 0;

2. d(f) = df for functions;

3. d(f dxi1 ∧ . . . ∧ dxik) = df ∧ dxi1 ∧ . . . ∧ dxik .

d : k-forms −→ (k + 1)− forms.

For the case k = 1 we already know

0-formsd−→ 1− forms

d−→ 2− forms.

73

d d = 0

H1(U) = closed 1-formsdX=0

/exact 1-formsX=df

.

Integration of k-forms along a parametrized k-surface

φ : Dk

(t1,...,tk)−→ U ⊂ Rn

(x1,...,xn), xi(t1, . . . , tk), −→x (−→t ).

Step 1. Take a k-form X =∑ai1,...,ik dx

i1 ∧ . . . ∧ dxik . Pull it back to Dk

(restriction to Dk), i.e.

φ∗X =∑

ai1,...,ik(x(t))∂xi1,...,ik

∂t1,...,kdt1 ∧ . . . ∧ dtk = b(t) dt1 ∧ . . . ∧ dtk.

Step 2. Integrate the restricted k-form along the body Dk ⊂ Rk (t1, . . . , tk)(ordinary integration).

Is this program well-defined?Is it invariant under changes of coordinates?Is it well-defined in the category of manifolds U ⊂ Rn and C∞-maps?

φ : URn

x1,...,xn

−→ VRm

y1,...,ym

, φ : x 7→ y, yi = yi(x1, . . . , xn).

We need to prove that

a) operations (addition and multiplication) of k-forms are “natural”, i.e.they commute with change of coordinates;

b) the differential d—does it commute with φ∗? Or with change of coor-dinates in particular?

We define:

1. k = 0. φ∗f(x) = f(y(x)) for functions;

2. k = 1. φ∗df = d(φ∗f)?

d(φ∗f) = df(y(x)) =∂f

∂yi∂yi

∂xjdxj = d(φ∗f)

φ∗(df) = φ∗( ∂f∂yi

dyi)=∂f(y(x))

∂yj∂yj

∂xidxi = d(φ∗f) — ok.

Conclusion: φ∗d = dφ∗.

74

Lecture 11.

Diffrential forms: change of coordinates and pull-back map:

φ : U ′x′→→Ux

f−→ R, f is a function,

Theorem.

1. φ∗f(x′) = f(x(x′)) (0-forms).

2. φ∗(X ∧ Y ) = φ∗(X) ∧ φ∗(Y ) (commute with external product).

3. φ∗(dx) = dφ∗(x). More general: φ∗ commutes with operator d, map-ping k-forms to (k + 1) forms:

d(dxi1 ∧ . . . ∧ dxik) = 0,

d(fdxi1 ∧ . . . ∧ dxik) = df ∧ dxi1 ∧ . . . ∧ dxik .

Example. n = 2, Ω = dx ∧ dy.

φ : (x′)→ (x).

φ∗Ω =

(∂x

∂x′dx′ +

∂x

∂y′dy′)∧(∂y

∂x′dx′ +

∂y

∂y′dy′)

=

=

(∂x

∂x′∂y

∂y′− ∂x

∂y′∂y

∂x′

)dx′ ∧ dy′ = det

(∂x∂x′

∂x∂y′

∂y∂x′

∂y∂y′

)dx′ ∧ dy′.

Formal properties of d in the given system of coordinates:

1. df =∑i

∂f∂xidxi.

2. d(dxi1 ∧ . . . ∧ dxik) = 0

3. d(fdxi1 ∧ . . . ∧ dxik) = df ∧ dxi1 ∧ . . . ∧ dxik .

Corollaries:

1. d(df) = 0.

75

Proof:

d(df) = d

(∑i

∂f

∂xidxi

)x

Key point

=∑ij

∂2f

∂xj∂xj·x

Symmetricin (i,j)

dxj ∧ dxixSkew symmetricin (i,j)

= 0 (!)

2. For 0-forms we have:

φ : U ′ φ−→ Uf−→ R, x′ → x,

φ∗f(x′) = f(x(x′)),

d(φ∗f(x′)) = φ∗(df) =∂f

∂xi∂xi

∂x′jdx′j.

3. Operator d has the foillowing property:

d(Ω1xk−form

∧ Ω2) = dΩ1 ∧ Ω2 + (−1)kΩ1 ∧ dΩ2.

Proof: a) Let k = 0, Ω2 = dxi. d(fdxi) = df ∧ dxi. O.K. d(dxi) = 0.

b) Let k and l be arbitrary: (fdxi1 ∧ . . . ∧ dxik) ∧ (gdxj1 ∧ . . . ∧ dxjl).Prove that d d = 0?

d(adxi1 ∧ . . . ∧ dxik) = da ∧ dxi1 ∧ . . . ∧ dxik

d(d(adxi1 ∧ . . . ∧ dxik)) = d(da ∧ dxi1 ∧ . . . ∧ dxik) = 0,

becaused(da) = 0, d(dxi1 ∧ . . . ∧ dxik) = 0.

Prove that d(f ·X) = df ∧X + fdX (X is 1-form): X = udx

d(f · udx) = d(fu)∧ dx =∂f

∂y∧ udx+ f · ∂u

∂ydy ∧ dx = df ∧X + f · dX.

Prove that d(X ∧ f) = dX ∧ f −X ∧ df .

X = udx, X ∧ f = udx ∧ f = ufdx,

76

d(u · dx · f) = d(fu) ∧ dx =∂u

∂y∧ dx · f + u · ∂f

∂ydy ∧ dx = df ∧X + f · dX,

dX ∧ f = du ∧ dx · f =∂u

∂y· fdy ∧ dx

X ∧ df = udx ∧ ∂f∂ydy = u

∂f

∂ydx ∧ dy = −u∂f

∂ydy ∧ dx.

O.K.

Sod(Ωk ∧ Ωl) = dΩk ∧ Ωl + (−1)kΩk ∧ dΩl.

Examples:

n = 2 : d(udx) = uydy ∧ dx = −uydx ∧ dyd(vdy) = vxdx ∧ dy.

n = 3 : udx, vdy, wdz, (u, v, w)− “vector” X

curl X : dX =∂u

∂ydy ∧ dx+ ∂u

∂zdz ∧ dx+ ∂v

∂xdx ∧ dy + ∂v

∂zdz ∧ dy + ∂w

∂xdx ∧ dz =

=

(∂v

∂x− ∂u

∂y

)dx ∧ dy +

(∂w

∂y− ∂v

∂z

)dy ∧ dz +

(∂u

∂z− ∂w

∂x

)dz ∧ dx

Curl X – “vector” with components:

(wy − vz)1↔(23)

dy ∧ dz + (uz − wx)2↔(31)

dz ∧ dx+ (vx − uy)3↔(12)

dx ∧ dy.

This formula is true in any coordinate system (as a 2-form).Association of 2-form with vector is O.K. for n = 3 in orthogonal positive

coordinates and is invariant under rotations only (det = +1).

77

Lecture 12.

Claim:

d(k

Ω1 ∧ (l

Ω2) = dΩ1 ∧ Ω2 + (−1)kΩ1 ∧ dΩ2.

Let k = 1, l = 0.Proof: Let Ω1 = X = udx, Ω2 = f .

1) d(fX) = df ∧X + f · dX = d(fudx) = df ∧ udx+ f(du ∧ dx),

2) d(Xf) = dX · f?−X ∧ df = d(udx · f) = du ∧ dx · f +u df ∧ dx = dX · f −X ∧ df.

O.K.Claim: d d ≡ 0.Proof:

1) d(df) = 0 =∑ij

∂2f

∂xj∂xjdxj ∧ dxi.

1) d(df ∧ dxi1 ∧ . . . ∧ dxik) = d(df) ∧ dxi1 ∧ . . . ∧ dxik = 0.

O.K.Examples: n-forms in Rn:

Ω = a dx1 ∧ . . . ∧ dxn = (−1)σa dxi1 ∧ . . . ∧ dxin ,

where

σ =

(1 . . . ni1 . . . in

).

Integration: ∫· · ·∫

D⊂Rn

adx1 ∧ . . . ∧ dxn

denotes ordinary integral.Change of variables for n-form in Rn: x(x′)

Ω = a dx1 ∧ . . . ∧ dxn ⇒ a J dx′1 ∧ . . . ∧ dx′n,

where denotes the Jacobian:

J = det

(∂x

∂x′

).

78

Proof for n = 2: dx1 ∧ . . . ∧ dxn = dx ∧ dy.

dx = αdx′ + βdy′, dy = γdx′ + δdy′, where det

(α βγ δ

)= J = det

(∂x

∂x′

),

dx ∧ dy = αδ dx′ ∧ dy′ + βγ dy′ ∧ dx′ = (αδ − βγ) dx′ ∧ dy′ = J dx′ ∧ dy′.

Proof for any n: Let

ηj =∂xj

∂x′1dx′1 + . . .+

∂xj

∂x′ndx′n.

Consider the product

η1 ∧ . . . ∧ ηn = Sum of terms:∂x1

∂x′j1. . .

∂xn

∂x′jndx′j1 ∧ . . . ∧ dx′jn =

= det

(∂xi

∂x′j

)dx′1 ∧ . . . ∧ dx′n.

O.K.What is the image (pull-back) of the form dxi1 ∧ . . . ∧ dxik? As above,

x = x(x′), dxi → ηi,

ηi1 ∧ . . . ∧ ηiki1<...<ik

=∑

j1<...<jk

J i1...ikj1...jk

dx′j1 ∧ . . . ∧ dx′jk

For any k ≤ n we have

Ω =∑

i1<...<ik

ai1,...,ik dxi1 ∧ . . . ∧ dxik .

Let φ : x′ → x, i.e. x = x(x′).

φ∗Ω =∑

i1<...<ik

∂xI

∂x′Jdx′j1 ∧ . . . ∧ dx′jk ,

whereI = (i1 < . . . < ik), J = (j1 < . . . < kk),

∂xI

∂x′J= det

(∂xi1 . . . ∂xik

∂x′j1 . . . ∂x′jk

)are k × k minors in the Jacoby matrix.

79

Examples.

k = 0 : φ∗a(x′) = a(x(x′)),

k = 1 : φ∗X = φ∗(uidxi) = ui(x(x

′))∂xi

∂x′jdx′j, (sum in i, j),

k = 2, i < j : φ∗(uijdxi ∧ dxj) = uij(x(x

′))

(∂xi

∂x′ldx′l)∧(∂xj

∂x′sdx′s

)=

= uij(x(x′))J

(i jl s

)dx′l ∧ dx′s,

where J(i jl s

)is 2-minor

(i jl s

)in Jacoby matrix,

. . .

k = n φ∗(adx1 ∧ . . . ∧ dxn) = a(x(x′)) · J · dx′1 ∧ . . . ∧ dx′n.

a) Restriction to l-surface in Rn: (x1, . . . , xl)→ Rn.b) Change of variables l = n: U ′ → U , one-to-one in U .1-space: 0, 1 forms:

f, udx.

2-space: 0, 1, 2 – forms:

f(x), udx+ vdy, a dx ∧ dy.

2-forms are like scalar functions, but under change of variables we have:

a→ a · det(∂x

∂x′

).

3-space: 0, 1, 2, 3 – forms:

f(x), udx+vdy+wdz, a12 dx∧dy+a13 dx∧dz+a23 dy∧dz, b dx∧dy∧dz.

3-forms are like scalar functions:

φ : b(x)→ b(x(x′))J , x′ → x.

2-forms are like 1-forms (“vectors”):

a12 → a3σ=(123), +

, a13 → −a2σ=(132), −

, a23 → a1σ=(231), +

.

80

What is “Curl”? d : 1-forms→ 2-forms.

d : uidxi →

∑i<j

(∂uj∂xi− ∂ui∂xj

)dxi ∧ dxj,

a12 = (vx − uy), a13 = (wx − uz), a23 = (wy − vz),

Here the “star operator” ∗ from k-forms into n− k-forms: ∗ :∧k →

∧n−k isused.

It depends on Riemannian metric. This formula is written in the or-thonormal coordinates in euclidean space

81

Lecture 13.

Forms in 3-space: (x, y, z) = (x1, x2, x3).

1. Let X = udx+ vdy + wdz.

We have

d(udx+vdy+wdz) = (vx−uy) dx∧dy+(wx−uz) dx∧dz+(wy−vz) dy∧dz.

Denote:

a12 = (vx − uy), a13 = (wx − uz), a23 = (wy − vz),

To define “Curl” denote:

a12 → a3σ=(123), +

, a13 → −a2σ=(132), −

, a23 → a1σ=(231), +

.

The “vector” (a1, a2, a3) is called the “Curl” of X: dX = curlX.

2. Let u = (a1, a2, a3).

d(a12 dx ∧ dy + a13 dx ∧ dz + a23 dy ∧ dz) =(∂a1

∂x1+∂a2

∂x2+∂a3

∂x3

)dx ∧ dy ∧ dz.

We write:

÷u =3∑i=1

∂ai

∂xi.

d d = 0, therefore

curl (df) = 0, div curl (X) = 0.

Examples: 2-forms in R3 are:

a) “Vorticity” of vector field: X =∑uidx

i is by definition a 2-form Ω = dXwritten as a vector field in euclidean space and orthonormal coordinates.

b) Magnetic field H is a closed 2-form. A =∑aidx

i is called “Vector po-tential”.

dA = H ↔ dH = 0 (H is closed).

82

Constant 2-forms in Rn = skew symmetric matrices: (aij = −aji) = A.eAt is a rotation.

Let n = 3,

A =

0 a b−a 0 c−b −c 0

.

Claim: eAt ∈ SOn for all t → At = −A.Proof:

eAt = 1 + At+O(t2),

< eAtη, eAtξ >=< η, ξ > +t (< Aη, ξ > + < η,Aξ >) +O(t2) =< η, ξ >

For t = 0 we have: < Aη, ξ >= − < η,Aξ >. The basis is orthonormal,therefore At = −A.

“Lie Algebra” of SOn consists of constant 2-forms. eAt ∈ SOn.For n = 3 it looks like 3-vectors (Euler).Consider 4-space: x0 = ct, x1, x2, x3. Let F = Fij dx

i ∧ dxj be a 2-form(“Electromagnetic Field”). dF = 0 – Faraday laws.

F = E1 dx1 ∧ dx0 + E2 dx

2 ∧ dx0 + E3 dx3 ∧ dx0︸︷︷︸

electric field

+∑

α,β=1,2,3α<β

Hαβ dxα ∧ dxβ

︸︷︷︸magnetic field

.

We assume that (x0, x) = (ct, x) = (time, space).In 3-space R3(x1, x2, x3) we have 1-form E and 2-form H, depending on

t as of parameter.

d(4)F = 0→ d(3)H = 0

and∂Eα∂xβ

dxβ ∧ dxα ∧ dx0 + ∂Hαβ

∂x0dx0 ∧ dxα ∧ dxβ = 0,

therefore∂Eα∂xβ

− ∂Hαβ

∂x0= 0.

Finally in the 3-space R3 we have:

d(3)E − H

c= 0, where x0 = ct,

∂

∂x0=

1

c

∂

∂t.

83

Lecture 14.

Differential forms and physics.

1. Non-relativistic physics. Space R3, (x, y, z) = x, time t is a param-eter.

Electric charge: e.

Electric field: E = (E1, E2, E3) is a 1-form∑α

Eαdxα, α = 1, 2, 3.

In R3: x1 = x, x2 = y, x3 = z, E = E(x, t).

Newtonian equations: mxi = eEi + other forces.

Magnetic field is a 2-form Bαβ, α, β = 1, 2, 3.

B(x, t) =∑α<β

Bαβ dxα ∧ dxβ,

Magnetic (Lorentz) force acting on the charged particle moving withspeed (v1, v2, v3) is fα = e/cBαβv

α where α, β = 1, 2, 3.

Faraday law: d(3)B = 0 (is closed), or

∂B12

∂x3− ∂B13

∂x2+∂B23

∂x1= 0,

B1 = B23, B2 = −B13, B3 = B12, B = (B1, B2, B3).

div B = 0 ↔ ∂B1

∂x1+∂B2

∂x2+∂B3

∂x3= 0.

2. Relativistic physics.

x0 = ct, c ∼= 300000kmsec (speed of light in vacuum).

4-space: (x0, x1, x2, x3) = (ct, x1, x2, x3).

Electromagnetic field:

F =∑j<j

Fij dxi ∧ dxj, i, j = 0, 1, 2, 3,

F = E ∧ dx0 +B = E ∧ cdt+B =

= E1 dx ∧ dx0 + E2 dy ∧ dx0 + E3 dz ∧ dx0++B12 dx ∧ dy +B13 dx ∧ dz +B23 dy ∧ dz.

84

The 1st pair of Maxwell Laws: dF = 0 in the 4-space.

dF = d(3)E ∧ dx0 + d(3)B +∂B

∂x0∧ dx0.

Here d(3) denotes d in the 3-space R3(x, y, z), and d(4) denotes d in R4.

x0 = ct,

therefore

∂B

∂x0=

1

c

∂B

∂t=

1

c

∑α<β

∂Bαβ

∂tdxα ∧ dxβ =

=1

c

∂B12

∂tdx ∧ dy + 1

c

∂B13

∂tdx ∧ dz + 1

c

∂B23

∂tdy ∧ dz.

d(3)E =

(∂E2

∂x− ∂E1

∂y

)dx ∧ dy +

(∂E3

∂x− ∂E1

∂z

)dx ∧ dz+

+

(∂E3

∂y− ∂E2

∂z

)dy ∧ dz, x1 = x, x2 = y, x3 = z.

Conclusion:

(a) d(3)B(x, t) = 0

(b) d(3)E + 1c∂B∂t

= 0.

Corollary:

(a) B = d(3)A (A is vector-potential).

B12 =∂A2

∂x− ∂A1

∂y, B13 =

∂A3

∂x− ∂A1

∂z, B23 =

∂A3

∂y− ∂A2

∂z.

(b) 1cB = −d(3)E in R3.

MagneticField B

metal wire

Electric Field E

C

Current appeaqrs in Crotate it!

(Flux of B) = Circulation of E

85

Lecture 15.

Integration of differential forms. Consider the n-forms in the n-space U ⊂ Rn, x1, . . . , xn.

Ω = f(x) dx1 ∧ . . . ∧ dxn.

By definition,∫· · ·∫

Dn⊂U

f(x) dx1 ∧ . . . ∧ dxn =

∫ (. . .

∫ (∫f(x)dx1

)dx2 . . .

)dxn,

is the ordinary integral.

Let Dn be n-cube (0 ≤ xi ≤ 1).

x1

x2

x11n=10 n=2 n=3

Dn = I1x1× . . .× I1

xn.

What do we know: Let us have a 1 to 1 change of variables: x =x(x′). Change of variables for n-form in Rn: x(x′)

Ω = f(x) dx1 ∧ . . . ∧ dxn =

Ω′

f(x(x′)) det

(∂xi

∂x′j

)dx′1 ∧ . . . ∧ dx′n .

Let D ∼= D′ ⊂ (x′) (i.e. we have a change of coordinates). Then∫D

Ω =

∫D′

Ω′.

Integration of k-forms in n-space along k-surface.

φ : Ux′→ Rn

x, U ⊂ Rk, x = x(x′).

86

Let

Ω =∑I

aI(x)dxI , aI = ai1...ik , dxI = dxi1 ∧ . . .∧ dxik , i1 < . . . < ik.

Then

φ∗Ω =∑I

aI(x(x′))

(∂xI

∂x′

)dx′1 ∧ . . . ∧ dx′k,

where (∂xI

∂x′

)= det

(∂xi1 . . . xik

∂x′1 . . . x′k

).

Examples.

k = 1 : φ∗ Ω =∑i

∂xi

∂tai(x(t)) dt, x′ = t,

k = 2, : φ∗ Ω =∑i1<i2

ai(x(t))

(∂xi1 , xi2

∂u, v

)du ∧ dv, x′ = (u, v),

where

(∂xi1 , xi2

∂u, v

)=

(∂xi1

∂u

∂xi2

∂v− ∂xi1

∂v

∂xi2

∂u−),

. . .

k = n φ∗ Ω = a1...n(x(x′)) · J

( xx′

)x

Jacobian

· dx′1 ∧ . . . ∧ dx′n.

Claim: Consider domain Dk with the boundary ∂Dk. Then:∫Dkx

Domain

dΩ =

∫∂Dkx

Boundary (oriented?)

Ω.

Example (XVII century) – Newton-Leibnitz.

k = 1 :

∫D′df =

∫∂D′

f = f(b)− f(a)

87

Consider a curve in n-space Rn.

A

B

D’D′ :

xi(t), i = 1, . . . , nt = a, x(a) = At = b, x(b) = B

.

∫D′

∂f

∂xidxi =

def

b∫a

∂f

∂xi(x(t))

∂xi

∂tdt =

b∫a

(∂f

∂xi(x(t))

∂xi

∂tdt

)φ∗(df)

=

b∫a

φ∗(df) =

=

b∫a

Φ(t) dt = f(b)− f(a), where Φ(t) =

(∂f

∂xi(x(t))

∂xi

∂t

)x=x(t)

.

What is orientation for k > 1?

For linear space over R orientation is a choice of basis up to linearchange with det > 0.

Orientation in Rn or in U ⊂ Rn is given by Cartesian coordinates(x1, . . . , xn) up to change of coordinates x = x(x′) such that J > 0.

Another point of view: orientation provided at the point P ∈ U ⊂Rn is given by the basis (∂/∂x1, . . . , ∂/∂xn) in the “tangent” spaceTP (Rn) generated by ∂/∂xi.

Consider a submanifold in Rn defined by an equation (globally!):

Mn−1 = f(x1, . . . , xn) = 0, such that (df)Mn−1 = 0.

∆)

f)

∆)

f)

P

Dn

n−1M

(u)externalnormal

“Tangent vectors” to the submanifold M are (∂/∂u1, . . . , ∂/∂un−1) if(u1, . . . , un−1) are local coordinates in Mn−1 near the point P .

88

Take vectors(∇f, ∂

∂u1, . . . ,

∂

∂un−1

), where ∇f =

n∑i=1

(∂f

∂xi

)∂

∂xi.

Compare this basis the chosen oriented basis(

∂∂x1, . . . , ∂

∂xn

).

(∇f, τu) =(∇f, ∂

∂u

)+←−−→

(∂

∂x1, . . . ,

∂

∂xn

)= τx.

Induced orientation at Mn−1 is such that determinant is + (> 0),Mn−1 = ∂Dn.

89

Lecture 16.

Orientation:

Mn−1 − hypersurface in Rn given by equation f (x1, . . . , xn) = 0 ,such that df = 0 in all points x where f(x) = 0 .

n−1M

n−1M

P

Dn

Dn=

∇ f =

(∂f

∂x1, . . . ,

∂f

∂xn

),

∇ f = 0 on M.

Orientation in Rn with given coordinates (x1, . . . , xn) .

∂

∂x1, . . . ,

∂

∂xn− basis in TP (Rn)

Let∂f

∂x1= 0 at P ∈ Mn

(u1, . . . , un−1) = (x2, . . . , xn)

- local coordinates in M near P . “Implicit function theorem”:

x1 = Φ(x2, . . . , xn) if∂f

∂x1= 0

∣∣∣∣P

near P on the surface given by the equation f (x1, . . . , xn) = 0 .

Basis (∇ f

∣∣∣∣P

,∂

∂x2, . . . ,

∂

∂xn

)= A

(∂

∂x1, . . . ,

∂

∂xn

)det A > 0 ↔ x2, . . . , xn given (+)

A

(∂

∂x1, . . . ,

∂

∂xn

)=

(∇ f

∣∣∣∣P

,∂

∂x2, . . . ,

∂

∂xn

)90

Definition:

det A > 0 ↔ (x2, . . . , xn) are oriented coordinates in Mn−1

det A < 0 ↔ (−x2, . . . , xn) are oriented coordinates in Mn−1

“Orientation of ∂ D induced by orientation of Rn ”

Example:

1. n = 1

∆

f

∆

fa b

_ +

D∂ D = b − a .

2.

u

v 1

2

Ddu=0 du=0

dv=0

dv=0

Orientation of D inducesorientation on ∂ D .

Theorem. ∫∫D

dΩ =

∫∂D

Ω

Proof for 1 - form Ω . Let x1 = u , x2 = v and

Ω = f(u, v) dv , dΩ =∂f

∂udu ∧ dv

∫∫D

(fu du ∧ dv) =

∫∂D

f dv =

∫ 1

0

f(1, v) dv −∫ 1

0

f(0, v) dv

∫∫D

fu du ∧ dv =

∫ 1

0

dv

(∫ 1

0

fu(u, v) du

)=

∫ 1

0

dv [f(1, v) − f(0, v)]

So we have∫∫D

fu du ∧ dv =

∫∂D

f dv =

∫∫D

dΩ

91

Theorem is proved for n = 2 .

For any n ≥ 2 we have

D = cube In , x = x1, . . . , xn , 0 ≤ xi ≤ 1

∫∂D

f dx2 ∧ · · · ∧ dxn =

∫fx1 dx

1 ∧ · · · ∧ dxn

u = x1 , v = (x2, . . . , xn)∫∂D

f dv =

∫∫D

fu du ∧ dv

∫∂D

f dv =

∫In−1

dv

(∫ 1

0

fu du

)=

=

∫In−1

dv (f(1, v) − f(0, v)) =

∫∫D

fu du ∧ dv.

So ∫∂D

Ω =

∫∫D

dΩ

Theorem is proved.

92

Lecture 17. Linear Algebra (Minicourse).

Linear space over field k (= R, C)

L : basis (e1, . . . , en) in L

Every vector

e =∑i

ηi ei[e = (η1, . . . , ηn)

]Linear Map

LA−→ M (e′1, . . . , e

′n)

A(ei) = aji e′j , aji is m× n matrix

A(e) =∑i

ηi A(ei) =(ηi aji

)e′j (sum in i, j)

For “components” (ηi)

A(e) =(ηi aji

)e′j

(η1, . . . , ηn) → (η′1, . . . , η′n) , η′j = aji ηi

Adjoint operator At : spaces of “covectors” L∗ , M∗ .

L∗ =e1, . . . , en

, M∗ =

e′1, . . . , e′n

⟨ei, ek⟩ = δik , ⟨e′j, e′p⟩ = δjk

⟨Atζ, η⟩ = ⟨ζ, Aη⟩

ζ = uj e′j , η = ηi ei , Atζ =

(aji uj

)(vectors η , covectors ζ).

93

“Kernel” of operator A : L → M

S ⊂ L , A(S) = 0

rank A = dim L − dim S

“Cokernel” of A :T = M/A(L)

Change of basis

(e1, . . . , en) , ei = rji e′′j (sum over j) , e′′j ∈ L , e = R(e′′)

e′′R−→ e

A−→ e′ ∈ M (change in L)

LR−→ L

A−→ MQ−1

−−→ M (change in M)

A ⇒ Q−1AR

Case L = M (eigenvalue problem)

M = L , Q = R

A ⇒ R−1AR (!)

Theory of inner products

Inner Product ⟨η1, η2⟩ = number, η1, η2 ∈ L .

1. Bilinear⟨η + ζ, η1⟩ = ⟨η, η1⟩ + ⟨ζ, η1⟩

(same in the variable η1).

2. ⟨λη1, η2⟩ = λ ⟨η1, η2⟩3. ⟨η1, λη2⟩ = ⟨η1, η2⟩ · λor ⟨η1, λη2⟩ = ⟨η1, η2⟩ · λ (“hermitian”), k = C , λ ∈ C .

94

Gramm matrix G = (gij)

⟨ei, ej⟩ = gij

Change of basis: ei = aji e′j

gij = ⟨api e′p , aki e′k⟩ = api akj g

′pk

ConclusionG = AtGA

η1 = ηi1 ei , η2 = ηj2 ej , ⟨η1, η2⟩ = ηi1 ηj2 gij

Nondegenerate Inner Product

det G = 0

k = R : sign of detG is invariant under change of basis e = Ae′ .

Proof: detG′ = (detA)2 detG .

Symmetric Inner Product

⟨η1, η2⟩ = ⟨η2, η1⟩

Theorem. Every symmetric Inner Product can be written in some basis as

G =

0 0 00 I 00 0 −I

, η2 =

k+p∑j=k+1

η2j −n∑

j=k+p+1

η2j

“rank G ” = n − dimension of “Kernel”

G : L → L∗

Gt : (L∗)∗ → L∗ , L∗∗ = L

95

Gt = G − (“symmetric”)

Nondegenerate case:

G =

1 . . . 0 0 . . . 0...

. . ....

.... . .

...0 . . . 1 0 . . . 00 . . . 0 −1 . . . 0...

. . ....

.... . .

...0 . . . 0 0 . . . −1

Type (p, q) − “signature” , p + q = n

Euclidean case q = 0 (or p = 0) : Rn .

Lorentzian case p = 1 , q = n− 1 : R1,n−1

Hermitian Case is needed in Quantum Theory.

96

Lecture 18. Linear Algebra - II.

Theory of inner products:

L − linear space, w, v ∈ L .

⟨w, v⟩ = number

bilinear:⟨λw, v⟩ = ⟨w, λv⟩ = λ ⟨w, v⟩

nondegenerate:∀ v ∃w : ⟨v, w⟩ = 0

Inner ProductG : L → L∗

e1, . . . , en − basis L , e1, . . . , en − basis L∗

⟨ei, ej⟩ = gij , G(ei) = gij ej (sum)

Gramm Matrix

det G = 0 ↔ nondegenerate Inner Product

Symmetric:

⟨v, w⟩ = ⟨w, v⟩ , gij = gji , Gt = G

Lemma. Let M ⊂ L is subspace such that restriction G|M is nondegen-erate. The orthogonal complement space M⊥ exists and unique;

∀ v ∈ M , w ∈ M⊥ : ⟨v, w⟩ = 0

Proof: Choose basis e1, . . . , ek ∈ M , ek+1, . . . , en (dimM = k) inL . We have

GM = (gij) , i, j ≤ k , det GM = 0

97

Find all vectors w ∈ M⊥

w =∑i≤k

αi ei +∑j>k

βj ej = e +∑j>k

βj ej , e ∈ M

Linear Equations

⟨w,M⟩ = 0 ↔ ⟨w, e1⟩ = 0 , . . . , ⟨w, ek⟩ = 0

Or for the variables (α1, . . . , αk, β1, . . . , βn−k)∑αi ⟨ei, e1⟩ +

∑βj ⟨ej, e1⟩ = 0

. . .

∑αi ⟨ei, ek⟩ +

∑βj ⟨ej, ek⟩ = 0

- k linear homogeneous equations.

Rank of system = k (minor gij , i, j = 1, . . . , k is = 0).

So we have linear space Rn−k of solutions

M⊥ ≃ Rn−k , dim M⊥ = n − k

(any field k instead of R).Lemma is proved.

Proof of Theorem:

⟨v, w⟩ − symmetric inner product

∃ v1 = e1 such that ⟨v1, v1⟩ = 0

Normalize v1 such that ⟨v1, v1⟩ = ± 1 .

Take orthogonal space, dim = n − 1 .

Iterate procedure. We are coming to

e1 = v1 , . . . , ⟨ei, ej⟩ =

+1 or− 1 or0

98

So we can reduce our matrix G to the form

G =

I 0 00 −I 00 0 0

pqr

, p + q + r = n

Theorem is proved.

Nondegeneracy: r = 0 . “Signature” (p, q) , p pluses, q = n − pminuses.

Cases:

q = 0 : Euclidean Inner Product

ds2 =∑ (

dxi)2

, G = I

q = n − 1 : Minkowki Inner Product

(+ − − . . . −) (Relativity)

ds2 =(dx0)2 − n−1∑

α=1

(dxα)2 , G =

(1 00 −I

)

Isometries

A : Rn → Rn

Let A be Linear Map At = A−1 (orthogonal)

⟨Av,Aw⟩ = ⟨v, w⟩ = general case

Euclidean space R2 (G = 1) .

A = Pα ·(

cosφ sinφ−sinφ cosφ

), P =

(1 00 −1

), α = 0, 1

Minkowski case: R1,1(x0, x1) .

A = Pα · T β ·(

coshψ sinhψsinhψ coshψ

)99

P =

(1 00 −1

), T =

(−1 00 1

), α, β = 0, 1

“Lorentz Transformations”:

A =


),

x0 = ctx1 = x

x0 = coshψ x′0 + sinhψ x′1

x1 = sinhψ x′0 + coshψ x′1,

x′0 = ct′

x′1 = x′

ct = a ct′ + b x′

x = b ct′ + a x′

, a2 − b2 = 1

“Physical Parametrization”:

a =1√

1− v2, b =

v√1− v2

, v = w/c (“speed”)

t = a t′ +b

cx′ , x = c b t′ + a x′

Assumption: v ≪ 1 (w ≪ c). So we have

t ≃ t′ , x ≃ x′ + w t

a ≃ 1 , b ≃ 0 , c b ≃ w (speed)

- Galilean Transformation.

100

Lecture 19.

Theorem. Every skew-symmetric inner product can be reduced to the form

G =

0 1 . . . 0 0 0 . . . 0−1 0 . . . 0 0 0 . . . 0...

.... . .

......

.... . .

...0 0 . . . 0 1 0 . . . 00 0 . . . −1 0 0 . . . 00 0 . . . 0 0 0 . . . 0...

.... . .

......

.... . .

...0 0 . . . 0 0 0 . . . 0

(“canonical basis”). det G = 0 ⇒ n = 2k .

Proof. Find vectors v, w such that ⟨v, w⟩ = 0 . Normalize them suchthat ⟨v, w⟩ = 1 . Take subspace M = Span v, w . Restriction of innerproduct to M is

GM =

(0 1−1 0

), det GM = 0

Take orthogonal complement M⊥ and so on.

Theorem is proved.

Remark. Skew-symmetric inner product is a 2 - form

Ω =∑

gij dxi ∧ dxj , gij = − gji

Let detG = 0 . We choose a basis (e1, . . . , ek, e′1, . . . , e

′k) and coor-

dinates (q1, . . . , qk, p1, . . . , pk) such that

Ω =k∑i=1

dqi ∧ dpi

G =

(0 Ik−Ik 0

)

Riemannian Geometry:

101

At every point x = (x1, . . . , xn) ∈ U positive quadratic form is given

ds2 = gij dxi dxj , gij = gji

a) ds2 > 0 − Riemannian Geometry: Rn .

b) det gij = 0 − Pseudoriemannian Geometry of the type (p, q) (sig-nature) : Rp,q .

Case p = 1 : − Lorentzian Geometry (Relativity).

Euclidean (Pseudoeuclidean) Space

gij(x) = const

Arc length:γ b

a

γ =xi(t)

, i = 1, . . . , n , a ≤ t ≤ b

- curve (smooth).

Length

l(γ) =

∫ b

a

√gij(x(t)) xi xj dt =

∫ b

a

|x| dt

|x|2 = gij xi xj

“Pseudoriemannian Case”: (Lorentzian, R1,n−1)

timelike vector − |x|2 > 0 , gij xi xj > 0

lightlike vector − |x|2 = 0 , gij xi xj = 0

spacelike vector − |x|2 < 0 , gij xi xj < 0

Symplectic Geometry:

x1, . . . , xn , Ω = gij dxi ∧ dxj , gij = − gji (G)

102

det gij = 0 , n = 2k , dΩ = 0

Lorentzian Geometry R1,1 :

(x0, x1) , x0 = ct , x1 = x

lightlike

spacelike

timeliket

x

timelike vector − ⟨ζ, ζ⟩ > 0

lightlike vector − ⟨ζ, ζ⟩ = 0

spacelike vector − ⟨ζ, ζ⟩ < 0

World - line of any object = curve (x0(τ), x1(τ)) .

“Real object” (nonzero mass): (x0)2 − (x1)2 > 0 .

Zero mass (light-like): (x0)2 − (x1)2 = 0 .

t

x

γγ’ "

Statement: length (γ′) > length (γ′′).

Axiom:

“Living time of object” =1

clength (World line)

Volume in euclidean geometry

103

e

e

1

2gij = ⟨ei, ej⟩ (G)

Claim:vol (K) =

√detG

Why:

ei = λji ej , ⟨ej, ek⟩ = δjk (orthonormal)

e’2 K’

e’1

vol (K ′) = 1 , vol (K) = det(λji)

= detΛ

G = ΛΛt , detG = (detΛ)2

Conclusion: vol (K) =√detG .

Definition: in any pseudoriemannian geometry we define

Dvol D =

∫D

√|detG| dx1∧ . . .∧dxn

detG = g2 , Ω = g dx1∧ . . .∧dxn

In orientable manifolds volume is integral of n - form (differential) .

Change of coordinates:

x (x′) , J =

(∂xi

∂x′j

),

G′ = J G J t , detG′ = detG · |J |2

104

Lecture 20. Volume and Differential Forms. Sym-plectic Manifolds.

Inner Product in Rn ∋ v, w : ⟨w, v⟩

Bilinear⟨λw, v⟩ = ⟨w, λv⟩ = λ ⟨w, v⟩

Nondegenerate∀ v ∃w : ⟨v, w⟩ = 0

Gramm matrix (basis e1, . . . , en)

G = (gij) = ⟨ei, ej⟩

Nondegeneracy: det G = 0 .

Inner Product = Map G : L → L∗

⟨ei, ej⟩ = δij , G(ei) = gij ej

Symmetric Inner Product: gij = gji , ⟨v, w⟩ = ⟨w, v⟩ .Theorem 1. There exists basis e1, . . . , en such that gij is diagonal and

gij =

+1 or− 1 or0

, G =

I 0 00 −I 00 0 0

pqr

p+ q + r = n .

Nondegenerate: r = 0 .

(p, q) = “signature” .

Euclidean: q = 0 (or p = 0) : ds2 =∑

(dxi)2

Lorentzian: p = 1 (or q = 1) : ds2 = (dx0)2 −∑

(dxα)2

(x0 = ct) .

“Orthogonal Maps”:

105

⟨Av,Aw⟩ = ⟨v, w⟩

Skew - Symmetric Inner Product: gij = − gji , ⟨v, w⟩ = −⟨w, v⟩ .Theorem 2. There exists basis

(e1, . . . , ek, e′1 = ek+1, . . . , e

′k = e2k, e

′′1, . . . , e

′′s)

such that

gij = ⟨ei, ej⟩ =

0 Ik 0−Ik 0 00 0 0

kks

For s = 0 we have n = 2k . “Canonical Coordinates”:

(q1, . . . , qk, p1, . . . , pk)

(ei) (e′i)

Define “symplectic 2 - form” in that basis

Ω =k∑i=1

dqi ∧ dpi , (s = 0)

det gij = 1

Property:

Ωk = Ω ∧ . . . ∧ Ω︸︷︷︸k times

= k! dq1 ∧ dp1 ∧ . . . ∧ dqk ∧ dpk =

= k! (−1)k dq1 ∧ . . . ∧ dqk ∧ dp1 ∧ . . . ∧ dpk

So for the volume element

dnσ =√gij dq

1 ∧ . . . ∧ dqk ∧ dp1 ∧ . . . ∧ dpk

106

we have1

k!Ωk = dnσ

For any coordinates Symplectic 2 - form is

Ω =∑i<j

gij dxi ∧ dxj (basis e)

det G = 0 , n = 2k

Ω =k∑i=1

dqi ∧ dpi

in canonical basis (e) .

Linear Change: matrix J :

e = J e , det G =(det J

)2det G

So we have:Volume element

dnσ =√

det G dx1 ∧ . . . ∧ dxn =(det J

)dx1 ∧ . . . ∧ dxn

dnσ = dq1∧ . . . ∧dqk∧dp1∧ . . . ∧dpk =1

k!Ωk =

1

k!Ω∧ . . . ∧Ω

det G =(det J

)2Volume element

dnσ =(det J

)dx1 ∧ . . . ∧ dxn =

1

k!Ωk

Result:det G =

(det J

)107

det J − Pfaffian

Theorem. A Pfaffian

det J =√det (gij)

is a polynomial of the skew-symmetric matrix G .

Proof. We have for the volume element

dnσ =√det G dx1 ∧ . . . ∧ dxn =

1

k!Ω ∧ . . . ∧ Ω︸︷︷︸

k times

because multiplication of forms is invariant under changes of coordinates.Elements of Ω ∧ . . . ∧ Ω are polynomials.

Theorem is proved.

Inner Product of vectors

⟨v, w⟩ =∑

gij viwj , v = vi ei , w = wj ej , v, w ∈ L

Inner product of covectors

v∗, w∗ ∈ L∗ , v∗ = vi ei , w∗ = wj e

j , ⟨v∗, w∗⟩ =∑

viwj g∗ij

and (g∗ij)

= G∗ = G−1

Why?

LG−→←−G∗ L∗

G∗ = G−1 − definition .

Example.

Symplectic Inner Product (2 - form) : gij = − gji .

Ω = gij dxi ∧ dxj , i, j = 1, . . . , n

Ω =∑

dqj ∧ dpj , j = 1, . . . , k

108

Inner Product of gradients is given by matrix (g∗ij) = G−1 .

Example. Hamiltonian System (H) .

df(x)

dt= ∇f , ∇H = g∗ij

∂f

∂xi∂H

∂xj

109

Lecture 21. Duality Operator and Maxwell’s Equa-tions.

In presence of Riemannian (or Pseudoriemannian) metric (gij = gji) dualityoperator is defined for k - forms Ω ∈ Λk(Un) :

∗ Ω is a k − form

∗ ∗ Ω = ±Ω

Define it first for Euclidean Metric: gij = δij using orthonormal basis(e1, . . . , en) , ⟨ei, ej⟩ = δij . Dual basis of 1 - forms is

dx1, . . . , dxn : dxi (ej) = δij

Definition:

∗(dxi1 ∧ · · · ∧ dxik

)= dxj1 ∧ · · · ∧ dxjn−k (−1)σ

where i1 < . . . ik , j1 < · · · < jn−k ,

σ =

(1 . . . . . . . . . . . . ni1 . . . ik j1 . . . jn−k

)j1

i1 ik

jn−k

dx

*(dx)(e ,...,e )

(e ,...,e )

(Dual means “orthogonal”?)

We have:∗ ∗ (Ω) = (−1)k(n−k) Ω

Examples:

k = 0 : ∗ (1) = dx1 ∧ · · · ∧ dxn = dnσ (volume)

∗ (dnσ) = 1

k = 1, n = 2

110

1

2

∗ (1) = (2)

∗ (2) = −(1)∗2 = −1

k = 1 , n = 3

∗ (1) = (23)∗ (2) = −(13)∗ (3) = (12)

k = 2 , n = 3

∗ (12) = (3)∗ (13) = −(2)∗ (23) = (1)

∗2 = 1

k = 1 , n = 4

∗ (0) = (123)∗ (1) = −(023)∗ (2) = (013)∗ (3) = −(012)

k = 3 , n = 4

∗ (012) = (3)∗ (013) = −(2)∗ (023) = (1)∗ (123) = −0

∗2 = − 1

k = 2 , n = 4

∗ (01) = (23)∗ (02) = −(13)∗ (03) = (12)

∗ (12) = (03)∗ (13) = −(02)∗ (23) = (01)

∗2 = 1

Claim: This map is well - defined on the linear space of k - forms in Euclideanspace and do NOT depend on orthogonal basis.

Proof for k = 1 , n = 2 :

111

(e ,e )21

e

e

1

2

∗ (e1) = e2

∗ (e2) = −e1

Take any unit vector v = a e1 + b e2 . We have

w = ∗ (v) = a e2 − b e1

So : |v| = |w| and ⟨v, w⟩ = 0 .

OK.

Now define ∗ operator for Pseudoeuclidean (Lorentzian) metric

gij =

1 0 . . . 00 −1 . . . 0...

.... . .

...0 0 . . . −1

, ds2 = (dx0)2 −n∑

α=1

(dxα)2

e0, . . . , en − basis (x0 = ct, x1, . . . , xn)

dx0, . . . , dxn − dual basis ⟨dxi, ej⟩ = δij

Definition.

∗ (i1, . . . , ik) = (−1)s (j1, . . . , jn+1−k) (−1)σ

σ =

(0 . . . . . . . . . . . . ni1 . . . ik j1 . . . jn+1−k

)s = number of times we have “negative” squares among the indices (i1, . . . , ik) .

We have: either s = k (all ik > 0) or s = k − 1 (i1 = 0).

Examples.

∗ (0) = (12 . . . n) , ∗ (1) = (02 . . . n) , ∗ (2) = − (013 . . . n) , . . .

∗ (0 i1 . . . ik−1) = (−1)k−1 (j1, . . . , jn+1−k) (−1)σ

112

σ =

(1 . . . . . . . . . . . . ni1 . . . ik−1 j1 . . . jn+1−k

)So we have for n = 3 : (Minkowski Space)

(e0 e1 e2 e3) , ds2 = (dx0)2 −3∑

α=1

(dxα)2

∗ (1) = dx0 ∧ · · · ∧ dx3 , ∗ (dx0 ∧ · · · ∧ dx3) = − 1

Minkowski space (x0, x1, x2, x3) :

∗ (0) = (123) , ∗ (1) = (023) , ∗ (2) = − (013) , ∗ (3) = (012)

∗ (01) = − (23) , ∗ (02) = (13) , ∗ (03) = − (12)

∗ (12) = (03) , ∗ (13) = − (02) , ∗ (23) = (01)

∗ (012) = (3) , ∗ (013) = − (2) , ∗ (023) = (1) , ∗ (123) = (0)

Electric FieldE = Ex dx + Ey dy + Ez dz

Magnetic Field

B = B12 dx ∧ dy + B13 dx ∧ dz + B23 dy ∧ dz

Let us put

Bx = B23 , By = B31 , Bz = B12

Electromagnetic Field

F = E ∧ dx0 + B

∗F = Ex dy∧dz + Ey dz∧dx + Ez dx∧dy + (Bx dx + By dy + Bz dz)∧ dx0

113

Maxwell’s Equations:

1st pair (Faraday Law):

dF = 0 ⇒ dB = 0 , dE =1

c

∂B

∂t, (x0 = c t)

2nd pair:

d ∗ F = ∗ (4− Current)

“4 - Current” = J (4) (1 - form)

4 - Current:

(ρ, −Jx, −Jy, −Jz) = J (4) =(ρ, − J (3)

)ρ dx0 − Jx dx − Jy dy − Jz dz = J (4)

Claim:

d ∗ J (4) = 0

∗ J (4) = ρ dx∧ dy ∧ dz − dx0 ∧ (Jx dy ∧ dz + Jy dz ∧ dx + Jz dx ∧ dy)

1

c

∂ρ

∂t= − div J (3)

114

Lecture 22. Stokes Formula.

∫D

dΩk =

∫∂D

Ωk

D - any body, ∂D - its boundary (orientation is induced).

D

n τD

τ - tangent to ∂D ,n - external normal,(n, τ) - orientation of ∂D ,τ - orientation of ∂D .

Examples.

k = 0 (Newton - Leibnitz)

n na b

(−) (+)

Ω0 = f(x),∫ b

a

d f = f(b) − f(a),

D = [a, b], ∂ D = (a ∪ b)

k = 1

D

I

II

III

VI

x

y(x, y) - orientation in R2 .Cube D - n = 2 : square I2 .

(I) : n = y , τ = −x (y, −x) ∼ +(II) : n = x , τ = y (x, y) ∼ +(III) : n = −y , τ = x (−y, x) ∼ +(IV) : n = −x , τ = −y (−x, −y) ∼ +

Proof of theorem for I2 :Let D = I2 and Ω1 = f(x, y) dx .

115

We have:

dΩ1 =∂f

∂ydy ∧ dx = − fy dx ∧ dy∫∫

D

dΩ1 = −∫ 1

0

∫ 1

0

dy dx fy = −∫ 1

0

dx (f(x, 1) − f(x, 0))∫∂D

Ω1 = −∫ 1

0

f(x, 1) dx +

∫ 1

0

f(x, 0) dx =

∫ 1

0

dx [f(x, 0) − f(x, 1)]

x

y

0 1

1 1,1

OK. This is proved.

Proof for D = In , any n ≥ 2 :same as for n = 2 : ∫

∂D

Ωn−1 =

∫D

dΩn−1

LetΩn−1 = f(x1, . . . , xn) dx1 ∧ · · · ∧ dxn−1

dΩn−1 =∂f

∂xndxn ∧ dx1 ∧ · · · ∧ dxn−1 = (−1)n−1 dx1 ∧ · · · ∧ dxn ∂f

∂xn

We have∫D

dΩn−1 = (±)∫ 1

0

. . .

∫ 1

0

dx1 . . . dxn−1

∫ 1

0

∂f

∂xndxn =

= (±)∫ 1

0

. . .

∫ 1

0

dx1 . . . dxn−1(f(x1, . . . , xn−1, 1) − f(x1, . . . , xn−1, 0)

)∫∂D

Ωn−1 = ±(∫

xn=1

Ωn−1 −∫xn=0

Ωn−1

)

116

x =1n

x =0n

In−1 n−1(x ,...,x )1

Ωn−1 = f(x1, . . . , xn−1, xn) dx1∧· · ·∧dxn−1

We see that results are the same (check sign!).

Theorem is proved for D = In , k = n− 1 .

Now we prove this theorem for every manifold U of any dimension M ,U ⊂ RM , Ωn−1 - (n-1)-form in U .

φ : In → U

- “singular cube” (smooth map).

We need to proveφ∗ Ω = Ωn−1

(pull-back) in cube In .

By definition ∫Ω =

∫φ∗Ω

φ : In−1 → U In−1

So our theorem follows from the result above.

OK.

117

Lecture 23. Algebraic Boundary.

Stokes formula: ∫D

dΩ =

∫∂D

Ω

We proved it for cube In , 0 ≤ xj ≤ 1 , j = 1, . . . , n . But everyconvex body is isomorphic to cube up to change of coordinates. We calculatenow boundary of cubes and simplices with orientation.

∂ In =n∪i=1

(In−1xi=1 ∪ In−1

xi=0

)x=0 x=1

I

x=0 x=1

y=1

y=0 x

yI

1

2

Orientation of cube is given by coordinates (x1, . . . , xn) , basis e1, . . . , en ,where ej = ∂/∂xj .

Calculate orientation of In−1xi=0 and In−1

xi=1 induced by (x1, . . . , xn) orien-tation of In :

We have (n = xi) for In−1xi=1 and (n = −xi) for In−1

xi=0 . We have τ =(x1, . . . , xi, . . . xn) (−1)? . For (n, τ) we have

(xi, x1, . . . , xi, . . . xn) (−1)? = (−1)?+i−1 (x1, . . . , xn)

So ? + i − 1 = 0 , ? = i − 1 (mod 2).

Final answer is

∂ In =n∑i=1

(−1)i−1(In−1xi=1 − In−1

xi=0

)(“Algebraic Boundary”).

Stokes Formula is∫IndΩn−1 =

n∑i=1

(−1)i(∫

In−1xi=1

Ω −∫In−1xi=0

Ω

)

118

Ω =n∑i=1

Ai(x1, . . . , xn) dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn

Now we calculate Algebraic Boundary for simplex σn = (x0, . . . , xn) ,xj − points in Rn (independent).

It means that(x0x1, . . . , x0xn) = (e1, . . . , en)

form a basis in Rn .

x ∈ σn ↔ x =n∑j=0

αj xj , 0 ≤ αj ≤ 1 ,∑

αj = 1

σ :0

σ :1x0

x1

σ :2

x0x1

x2

σ :3

x0x1

x2

x3

Lemma 1. Orientation of simplex (x0, . . . , xn) changes by sign (−1)κ forthe permutation of vertices

(x0, . . . , xn)κ−→ (xj0 , . . . , xjn)

Proof. Consider

κ =

(x0 x1 . . . xnx1 x0 . . . xn

), x0 ↔ x1

Basis e1 = x0x1 , . . . , en = x0xn should be replaced by basis e′1 =x1x0 , e

′2 = x1x2 , . . . , e

′n = x1xn . We have e′1 = −e1 and

e2 = x0x2 = x0x1 + x1x2 = e1 + e′2

. . .

119

en = x0xn = x0x1 + x1xn = e1 + e′n

So, this basis has opposite orientation because e′1 = −e1 . Iterating thisargument, we obtain all permutations.

Lemma is proved.

Lemma 2. Simplex σn−1i = (x0, . . . , xi, . . . , xn) enters into boundary

∂σn = ∂(x0, . . . , xn) with sign (−1)i−1 .

Proof. Let i = 0 , σn−10 = (x1, . . . , xn) with basis τ ′ = (x1x2, . . . , x1xn) =

(e′2, . . . , e′n) .

We have n ≡ x0x1 = e1 .Basis (n, τ ′) is

(x0x1 = e1, x1x2 = x1x0+x0x2 = −e1+e2 = e′2, . . . , e′n−1 = x1x0+x0xn = −e1+en)

So, we have orientation:

(n, τ ′) = (−1)? (n, τ) ≃ (n, τ ′)

Final result:

∂ σn =n∑i=0

(−1)i σn−1i

(Algebraic Boundary).

Definition.

“Chain” = linear combination of

cubes (cubic chain)simplexes (simplicial chain)

Cn =∑s

λs In(s) (cubic)

Cn =∑s

λs σn(s) (simplicial)

Boundary of chains

120

∂ Cn =∑s

λs ∂In(s) =

∑s

λs

n∑i=1

(−1)i−1(In−1xi=1 − In−1

xi=0

)

∂ Cn =∑s

λs ∂σn(s) =

∑s

λs

n∑i=0

(−1)i−1 σn−1i

Lemma. ∂ ∂ = 0 .Topology uses the so-called

(Singularcubes In(s) : Inψs−→ U , andSingularsimplexes σn(s) : σn

ψs−→ U)

Their algebraic boundary can be defined naturally as well as singularchains as finite linear combination of singular simplices (cubes). The cyclesare chains with zero boundary. Singular homology group is defined as factorof space of cycles by the ”exact ” cycles which are algebraic boundaries ofsingular chains. The space of cycles is very big but homology group is not,it is topologically (homotopy) invariant.

121

Lecture 24. Differential Forms and Homotopy PoincareLemma.

Definition. Homotopy process is a map F (C∞ - map)

U ×R F−→ V

x, t → F (x, t) = y

U , V - manifolds (open domains in euclidean spaces U ⊂ RM , V ⊂ RN

or other)F (x, t = const) = ft : U → V

(deformation or homotopy of map ft).Maps f1 = f and f0 = g are called “homotopic maps” (C∞ - homo-

topy).

Theorem. For every closed differential k - form Ω in Λk(V ) we have

f ∗ Ω − g∗Ω = du , u ∈ Λk−1(U) ,

Λk - space of all C∞ differential forms on any manyfold.

Proof. We have by definition dΩ = 0 .Remind.

d f ∗ = f ∗ d , d g∗ = g∗ d

Sod (f ∗Ω) = d (g∗Ω) = 0

Take homotopy process

F : U ×R → V

F (x, t) : F |t=0 = g , F |t=1 = f

Consider k - form F ∗Ω in Λk(U ×R) .Every form A in U ×R has the following form

A = a + b ∧ dt (k − form)

wherea =

∑ai1...ik dx

i1 ∧ · · · ∧ dxik = a(x, t)

122

b =∑

bj1...jk−1dxj1 ∧ · · · ∧ dxjk−1 = b(x, t)

a(x, t) = A|t=const

We have:

dA = da + (−1)k a ∧ dt + db ∧ dtU ×R U U

a =∂a

∂t(x, t) , a = A|t=const (put dt = 0)

Define operator

D : Λk(U ×R) → Λk−1(U)

DA =

∫ 1

0

b ∧ dt , b = b(x, t)

Lemma 1:± (dD ± D d) A = A|t=1 − A|t=0

(sign depends on dimension and plays no role here).

Proof of Lemma.

DA =

∫ 1

0

b ∧ dt

D (dA) = D ( da + (−1)k a ∧ dt + db ∧ dt ) =U ×R U U

= (−1)k (a|t=1 − a|t=0) + dU

(∫ 1

0

b(x, t) ∧ dt)

=

= (−1)k (A|t=1 − A|t=0) + dUDA = D dA

Lemma is proved.

Apply lemma to the form A = F ∗(Ω) in U ×R . We have

A|t=1 = f ∗Ω , A|t=0 = g∗ Ω

dA = dF ∗(Ω) = F ∗(dΩ) = 0

because dΩ = 0 (closed form).

123

So we have

f ∗(Ω) − g∗(Ω) = d u , u = DA

Theorem is proved.

Poincare Lemma: Let U = Ball Dn . Every closed k - form in the Ball isexact.

Proof. Mapf : [0, 1]×Dn → Dn

such that f(x) = x is homotopic to the map g , g(x) = 0 .Obviously we have

g∗(Ω) = 0 , 1 ≤ k ≤ n− 1

Sof ∗(Ω) = Ωk , Ωk − 0 = du

Poincare Lemma is proved.

Cohomology:

Hk(U) = Closed forms /Exact forms

Ω ∼ Ω′ iff dΩ = dΩ′ = 0 , Ω − Ω′ = du

Cohomology Ring is given by Product of forms.

Examples:1. U = pointH0(U) = R , Hk(U) = 0, k = 0.

1′. U = l pointsH0(U) = Rl , Hk(U) = 0, k = 0.

2. U = BallH0(U) = R , Hk(U) = 0, k = 0.

3. U = S1 : H0 = R , H1 = R , Hk = 0 , k = 0, 1.

4. U = Sn : H0 = Hn = R , Hk = 0 , k = 0, n (not provedyet).

5. U = connected domain in R2 , H0 = R , H1 = Rp (p =number of holes), H2 = 0 (!) .

124

Lecture 25. Examples of important closed differentialforms.

0 - forms: constant, dc = 0

1 - forms:

1) dφ =x dy − y dx

ρ2, ρ2 = x2 + y2

2) f(z) dz ,∂f

∂z= 0 (Cauchy)

f(z) = zn , f =1

z, f =

P (z)

Q(z)

n - torus Tn: (x1, . . . , xn) , Tn = Rn/Zn(periodic functions)

f(x1, . . . , xi + 1, . . . , xn) = f(x1, . . . , xn)

Forms: 1, dxi, dxi ∧ dxj, . . . , dx1 ∧ · · · ∧ dxn(basis of cohomology)

H∗(Tn) = Λk (exterior algebra)

2 - forms Ω , dΩ = 0 in R3 \ 0 :

x dy ∧ dz − y dx ∧ dz + z dy ∧ dxr3

, r2 = x2 + y2 + z2

Ω = sin θ dθ ∧ dφ on S2 (r = 1) - area form in S2 .

Gauss 2 - form ( for calculation of Linking Number)

x(t) = x(t+ 1) , y(τ) = y(τ + 1) in R3

Ω =(dx× dy , x− y)

|x− y|3− 2− form in R6 : (x1, x2, x3, y1, y2, y3)

minus diagonal ∆: xi = yi, i = 1, 2, 3.

dΩ = 0 (calculation)

125

x(t)

y( )τ

Property (Gauss): if x(t) does not crossy(τ) then

1

4π

∫ 1

0

∫ 1

0

dt dτ( ˙x× y′ , x− y)|x− y|3

= integer

represents a topological invariant (linkingnumber).

Degree of map:Take any n-form in n-sphere Ω such the∫Sn Ω = 0, for

example, volume form in Sn : (Ω = dφ for n = 1 , Ω = sin θ dθ ∧ dφ forn = 2) ∫

Sn

Ω = 2π

Definition. For ft : Sn → Sn

1

2π

∫Sn

f ∗t (Ω)

divided by∫Sn Ω, is homotopy invariant dI/dt = 0

It is called Degree of Map.It is integer.

Proof for n = 1 :

f : S1 → S1 , Ω = dφψ φ

f (ψ + 2π) = f(ψ) + 2πm

Partial cases: m = degree of map φ = f(ψ)

1) m =1

2π

∫ 2π

0

f ∗ (dφ) =1

2π

∫ 2π

0

dφ

dψdψ = m

2) Let n be any and f is orientation preserving diffeomorphism (one-to-one), so ∫

Sn

f ∗ (Ω) =

∫Sn

Ω

, so degree =1.

126

Gauss Map

First case: Curve M1 ⊂ R2 , (x(s), y(s))

Gaussmap:

τn

Μ1

M1 G−→ S1 , s → τ(s)

τ = (dx/ds, dy/ds) , |τ | = 1

Definition: G∗(dφ) = k ds , s is length, k - curvature.

Corollary:1

2π

∮M1

G∗ dφ = m ∈ Z

(number of rotations).

Second case: M2 ⊂ R3 is a boundary of some convex body

n

Μ 2

(x,y)

G : M2 → S2 , (x, y) → n(x, y)

Ω = area form ,

∫∫S2

Ω = 2π

G∗ Ω = K d2σ

d2σ - area element in M2 .

Corollary: ∫∫M2

K d2σ = 4π

G is one-to one: degree of map = 1 .

Kelvin Integral

X − 1 - form in S3 (“vector field” X =∑ai dx

i )X ∧ dX − 3 - form in S3 (“vorticity” dX = Ω = “curl X” − closed

2 - form)

“Hopf invariant”

S3 f−→ S2

127

(Whitehead, 1950’s). Let

dX = f ∗Ω ,

∫S2

Ω = 0

Claim: Kelvin - Whitehead Integral is homotopy invariant.

I(t) ≡∫S3

X ∧ dX , dX = f ∗t (Ω) , X(t) , dX(t)

Proof:dI

dt= 0 ?

Consider Homotopy Process

F = ft : S3 × R → S2 , F ∗Ω = dX

“Kelvin Integral” in S3 × R :

3 - form: X ∧ dX , dX = F ∗(Ω) .

t

t

1

2

t

It is closed 3 - form in S3 × R . Why?

0 = F ∗ (Ω ∧ Ω) = dX ∧ dX = d (X ∧ dX)

So ∫t=const

X ∧ dX = I(t)

is t - independent.OK.

128

Lecture 26.

Riemannian Metric in U : x1, . . . , xn :

ds2 =∑i,j

gij(x) dxi dxj , gij = gji , dxi dxj = dxj dxi

Riemannian: ds2 > 0 .

Pseudo-Riemannian: det gij = 0 (indefinite).

Lorentzian Case: Signature (p, q) = (1, n− 1) .

Inner Product in every point x :

⟨ei , ej⟩ = gij , ei ↔∂

∂xi− basis

n = 2 : ds2 = F du2 + 2Gdu dv + H dv2

g11 = F (u, v) , g12 = G(u, v) , g22 = H(u, v)

(1st quadratic form).

Euclidean Metric

ds2 =n∑i=1

(dxi)2

Pseudoeuclidean Metric

ds2 =(dx0)2 − n∑

i=1

(dxi)2

(Lorentzian).

Curve in Rn : (x1(t), . . . , xn(t)) . What is restriction of metric to curve?

ds2|γ = gij (x(t))dxi

dt

dxj

dt= ⟨x(t) , x(t)⟩ = |x|2

129

“Length” (arc length) is:

l(γ) =

∫ b

a

|x| dt t=a t=b

It is NOT a differential 1-form.

Surfaces: xi(u, v) , i = 1, . . . , n (M2) .

Restriction of metric:

ds2|M2 = gij (x(u, v)) (dxi)|M2 (dxj)|M2

dxi|M2 =∂xi

∂udu +

∂xi

∂vdv

Euclidean Case: M2 ⊂ Rn :

ds2|M2 =n∑i=1

(dxi)2|M2 =n∑i=1

(∂xi

∂udu +

∂xi

∂vdv

)2

=

= F du2 + 2Gdu dv + H dv2

F =n∑i=1

(∂xi

∂u

)2

, G =n∑i=1

(∂xi

∂u

)(∂xi

∂v

), H =

n∑i=1

(∂xi

∂v

)2

n = 3 : (x, y, z) :

Let surface is given as

z = f(x, y) , x = u , y = v

ds2 = dx2 + dy2 + dz2 = dx2(1 + f 2

x

)+ dy2

(1 + f 2

y

)+ 2dxdy (fxfy)

fx =∂f

∂x, fy =

∂f

∂y

130

Letfx|(x,y)=(0,0) = fy|(x,y)=(0,0) = 0

x

z y

x

z y

z = a x2 + 2b xy + c y2 + o (x2 + y2)

Definition: “Curvature form” at the point (0, 0) is equal to 2nd differentiald2z if

z = f(x, y) and fx|(x,y)=(0,0) = fy|(x,y)=(0,0) = 0

Riemannian Metric gij at the surface is equal to

g11 = 1 + z2z , g12 = zx zy , g22 = 1 + z2y

if z = f(x, y) .

It is unit matrix at the point (0, 0) if zx = zy = 0 at this point.

Curves: γ ⊂ R2 , x(t) , y(t) .

ds|γ =√(dx)2 + (dy)2 =

√x2 + y2 dt

“Natural Parameter” = arc length.

x(s) , y(s) , τ(s) =

(dx

ds,dy

ds

)

Lemma 1. |τ(s)| = 1 .

131

Proof: (dx

ds

)2

+

(dy

ds

)2

= 1 by definition.

Lemma 2. dτ/ds ⊥ τ .

Proof.d

ds⟨τ , τ⟩ = 0 = 2 ⟨dτ

ds, τ⟩

OK.

Conclusion.

dτ

ds= k n

where n is normal vector.

τn

Lemma 3. dn/ds = − k τ(s) .

Proof.A(s) = τ(s) , n(s) ∈ SO2

⟨A(s)η , A(s)ζ⟩ = ⟨η , ζ⟩

A(s = 0) = I

d

ds⟨A(s)η , A(s)ζ⟩ = 0 = ⟨Aη , ζ⟩ + ⟨η , Aζ⟩

Conclusion: At = −A

d

ds(τ , n) =

(0 k−k 0

)(τ , n)

Statement: (Frenet Formulas are partial cases of the following general prop-erty:)

A(t) ∈ On ⇒ (dA/dt)A−1 − skew symmetric

as well as A−1(dA/dt).

132

Lecture 27. Curvature.

γ : x(t) , y(t) in R2 , x = (x, y)

Let t be a “Natural parameter”

t = s : |(x, y)|2 = x2 + y2 = 1

Arc Length

l =

∫ b

a

| ˙x| dt = b − a , t = s

˙x = τ(s) , |τ | = 1

Lemmas.Curvature.

1.dτ

ds= k n , n ⊥ τ

Proof.

τn

x(t),y(t)γ

⟨τ, τ⟩ = 1 ,d⟨τ, τ⟩ds

= 0 = ⟨τ, dτds⟩

dτ

ds= k n

OK.

2. A ∈ On , A = A(t) , A(0) = 1

We have

Bt = −B , B =dA

dt

∣∣∣∣t=0

Proof.

⟨Aη,Aζ⟩ = ⟨η, ζ⟩ , A = A(t) , A(t) = 1 + B t + O(t2)

133

d

dt⟨Aη,Aζ⟩ = 0 =

d

dt

[⟨η, ζ⟩ + t ⟨Bη, ζ⟩ + t ⟨η,Bζ⟩ + O(t2)

] ∣∣t=0

Conclusion:⟨Bη, ζ⟩ = −⟨η,Bζ⟩

OK.

Conclusion. Let

A(s) = [τ(s) , n(s)] ,dA

ds

∣∣∣∣s=s0

= A(s0) B(s0) , Bt = −B

dτ

ds= k n ,

dn

ds= − k τ

It is true because

A(s) = A(s0)A(s0)−1A(s) ,

A(s0)−1A(s) = 1 + B(s0) (s − s0) + O

((s − s0)

2)∈ O2 ,

Bt(s0) = −B(s0)

Another definition of curvature:

x(s) , y(s) , x(s0) = 0 , y(s0) = 0

γs=s0

n

yx

τ y = f(x) (curve γ) ,dy

dx

∣∣∣∣x=0

= 0

y = y0 + a x2 + O(x2)/2

Definition: k = a

τ = ex |x=0 , n = ey |x=0

We have

s = s(x) ,ds

dx

∣∣∣∣x=0

= 1

134

Calculate dn/dx |x=0 = ?

For the curve F (x, y) = 0 we have

n =(Fx, Fy)√F 2x + F 2

y

, F (x, y) = y − f(x) , n =(−fx, 1)√1 + f 2

x

dn

dx

∣∣∣∣x=0

= − (fxx, 0) =dn

ds

ds

dx

∣∣∣∣x=0

=dn

ds

∣∣∣∣s=0

Sodn

ds= − k τ = − fxx

(Frenet) .

Conclusion: k = fxx .

OK.

One more definition of Curvature (differential 1 - form)

γ = M1 ⊂ R2 , x(s) , y(s) , |(x(s), y(s))| = 1

(τ(s) = 0 for all s).

Gauss map

γ = M1 G−→ S1

n τ

τ

γ=M1

G S1

τ

Another form

(x, y) → n(x, y) , (x, y) ∈ γ = M1

135

Definition. k ds = G∗ (dφ) (pull back) .

We have (x(s), y(s)) − periodic functions

x (s+ T ) = x(s) , y (s+ T ) = y(s)

s − local parameter on γ ⊂ R2

φ − local parameter on S1 ⊂ R2

G(s+ T ) = G(s) + m · (2π) . Why?

BecauseG(s) = φ , φ + 2πm ≃ φ

Claim. m ∈ Z .

m = “degree of map” , m =1

2π

∮k ds (!)

136

Lecture 28. Curvature.

Curves:γ = M1 ⊂ R2 , x(t) , y(t)

Arc length

ds =√x1 + y2 dt , l(γ) =

∫ b

a

ds

Curvature: ± k = |dτ(s)/ds| .

τn |τ(s)| = 1 , dτ/ds = k n

Oriented frame: (τ, n) .

Frene:dτ

ds= k n ,

dn

ds= − k τ

A(s) = (τ(s) n(s)) , B(s) = A−1(s)dA

ds=

(0 k−k 0

)Curvature: y = f(x) .

yx dy

dx

∣∣∣∣x=0

= 0 , k |x=0 =d2y

dx2

(last lecture).

ds

dx

∣∣∣∣x=0

= 1 , ds2∣∣x=0

= dx2 , ds2 = dx2

(1 +

(dy

dx

)2)

137

Gauss Map

G : M1 → S1 , G(x) = n(x) (or G(x) = τ(x))

F (x, y) = 0 :

n =(Fx, Fy)√F 2x + F 2

y

, F = y − f(x) : n(x) =(−fx, 1)√1 + f 2

x

τ(x) : rotate n by 90o

Definition:

G∗(dφ)/2π = k ds ,

∮S1

dφ/2π = 1 (?)

Claim: For closed curve M1 = c :∮c

k ds = m ∈ Z

What is “degree of map” ?

G : M1 → S1

(s) (φ)

closed curve x (s+ T ) = x(s) , y (s+ T ) = y(s) :

G (s+ T ) = G(s) + 2πm , m ∈ Z

Proof: Points G(s) and G(s+T ) are the same on S1 , φ + 2πm ≃ φ .OK.

Proof that G∗(dφ)/2π = k ds (later).

Curvature: n = 2 :

M2 ⊂ R3 : x (u, v) , y (u, v) , z (u, v)

138

Riemannian Metric: dx2 + dy2 + dz2 = ds2

g11 = F = x2u + y2u + z2u , g12 = G = xu xv + yu yv + zu zv ,

g22 = H = x2v + y2v + z2v

Let

Μ 2y

x(0,0)

z=n u = x , v = y , z = f(x, y)

and

zx|(0,0) = zy|(0,0) = 0

z = a +1

2

(fxx x

2 + 2 fxy x y + fyy y2)

+ o (x2 + y2) (near(0, 0))

Definition.

kij =∂2f

∂xi∂xj

∣∣∣∣(0,0)

, x1 = x , x2 = y

k11 + k22 = kmean

k11 k22 − k212 = K (Gaussian Curvature)

Area Form =√det gij dx

1 ∧ dx2 = d2σ (= dx ∧ dy at (0, 0))

Theorem.K d2σ = G∗ (Ω)

139

− pull-back Ω is an area 2 - form on S2 ⊂ R3

G : (x, y, z) → n (x, y, z)

M2 → S2

n

Ω = sin θ dθ ∧ dφ (spherical coordinates)

Ω =(x dy ∧ dz − y dx ∧ dz + z dx ∧ dy

)∣∣∣r2=1

, r2 = x2 + y2 + z2

Describe form Ω in special coordinates:

z

N

z′ =√1− x′2 − y′2 near pole N (in S2)

z′x′ |(0,0) = z′y′ |(0,0) = 0

Sphere S2

(gij = δij) |(0,0) , g′11 = 1 , g′12 = 0 , g′22 = 1 for x′ = 0 , y′ = 0

d2σ = Φ(x′, y′) dx′∧dy′ , Φ (0, 0) = 1 in S2 (x′ = 0 , y′ = 0)

Define Guass Map in such coordinates:

zn M2 ⊂ R3 , z = f (x, y) ,

z ⊥ M2 for x′ = 0 , y′ = 0

G (x, y) : (x, y) → n , n =(−fx, −fy, 1)√1 + f 2

x + f 2y

Gauss

G : n = n (x, y) , n (0, 0) = (0, 0, 1) (x′, y′, z′)

140

(x, y) − coordinates on S2 :

x′ = − fx√1 + f 2

x + f 2y

, y′ = − fy√1 + f 2

x + f 2y

, z′ =1√

1 + f 2x + f 2

y

dx′ → ∂x′

∂xdx +

∂x′

∂ydy , dy′ → ∂y′

∂xdx +

∂y′

∂ydy

At the point x = 0 , y = 0 we have

dx′G∗−→

(∂x′

∂xdx +

∂x′

∂ydy

)∣∣∣∣(0,0)

= − fxx dx − fxy dy

dy′G∗−→

(∂y′

∂xdx +

∂y′

∂ydy

)∣∣∣∣(0,0)

= − fxy dx − fyy dy

Ω = dx′ ∧ dy′ (x = 0, y = 0)

ΩG∗−→ G∗ dx′ ∧ G∗ dy′ = (−fxx dx − fxy dy) ∧ (−fxy dx − fyy dy) =

=(fxx fyy − f 2

xy

)dx ∧ dy = K dx ∧ dy (x = 0, y = 0)

Theorem is proved.

Corollary: Let M2 ⊂ R3 is a closed oriented surface.

Then ∫∫M2

K d2σ

Μ 2t

remains unchanged under smooth deformation M2t of closed surface M2 ⊂

R3 .Proof: ∫∫

M2t

K d2σ =

∫∫M2

t

G∗Ω , dΩ = 0 in S2

141

So K d2σ changes by exact form. By Stokes Formula we have

d

dt

∫∫M2

K d2σ = 0

OK.

142

Lecture 29.

Gauss Map:G : M2 → S2 ⊂ R3

CurvatureG∗ (Ω) = K d2σ

Ω = sin θ dθ ∧ dφ (area in S2)

Ω = dx′ ∧ dy′ |(0,0) , z′ =√

1− x′2 − y′2

Surface

M2 : z = f(x, y) fx|(0,0) = fy|(0,0) = 0

Μ 2y

x

zd2σ |(0,0) = dx ∧ dy

ds2 |(0,0) = dx2 + dy2

Gauss Map : n =(−fx, −fy, 1)√

1 + f 2x + f 2

y

, (x, y)G−→ n (x, y)

x′ = − fx√1 + f 2

x + f 2y

, y′ = − fy√1 + f 2

x + f 2y

, z′ =1√

1 + f 2x + f 2

y

dx′|(0,0) = − fxx dx − fxy dy , dy′|(0,0) = − fyx dx − fyy dy

dx′ ∧ dy′ = Ω |(0,0) → K dx ∧ dy = G∗ (Ω)

143

K |(0,0) = fxx fyy − f 2yx

Remark. ∫∫S2

Ω = 4π

Normalization∫∫S2

Ω

2π= 2 (Euler characteristics).

Theorem.a) For every closed oriented surface M2 ⊂ R3 integral∫∫

M2

G ∗ Ω =

∫∫M2

K d2σ

does not change under deformation of the surface M2t .

b) For a convex body ∂D = M2 we have∫∫M2

K d2σ = 4π

Proof.a) We have homotopy process Gt : M2

t → S2 . By theorem,

G∗t1Ω − G∗

t2Ω = du

Sod

dt

∫∫M2

G∗t Ω = 0

b) For M2 = S2 we have∫∫S2

Ω = 4π , G ≡ 1 : S2 → S2

Boundary of the convex body can be obtained from S2 by deformationM2

t .

144

Theorem is proved.

Remark:

Μ2

g holes

1

2π

∫∫M2

K d2σ = 2 − 2g

Euler Characteristics, does NOT depend on embedding.

Reminder: Theorem (homotopy)

F : M × R → U(x, t)

- smooth homotopy.

ft = F |t : M → U

Claim For the closed form Ω we have

f ∗1 (Ω) − f ∗

0 (Ω) = du

Proof:

Dω =

∫ 1

0

ω1 dt , ω = ω0 + ω1 dt

D ω ∈ Λk−1(M) , ω ∈ Λk (M × R)

d(M×R) ω = dMω ± ω0 dt + dMω1 ∧ dt

D d(M×R) ω =

∫ 1

0

ω0 dt + dM

∫ 1

0

ω1 dt = ω0 |t=1 − ω0 |t=0 + dM Dω

Now let dω = 0 . We have

Dd(M×R) ω = 0 ⇒ ω0 |t=1 − ω0 |t=0 = − dM Dω

145

Apply to ω = F ∗(Ω) ⊂ Λk(M × R)Theorem follows.

146

Lecture 30.

Differential forms and homotopy:

Degree of map Snf−→ Sn :

Take n - form Ω : ∫Sn

Ω = 1

Degree of Map ∫Sn

f ∗ (Ω)

Claim: For the closed form Ω and homotopy

F : Sn × R → Sn

(x, t)

we havef ∗1 (Ω) − f ∗

0 (Ω) = du

dΩ = d (f ∗Ω) = 0

So ∫Sn

f ∗1 (Ω) =

∫Sn

f ∗0 (Ω)

Remark: deg f ∈ Z (proof for n = 1 was given):

f(x+ T ) = f(x) + m · 2π

S1 f−→ S1 mod 2π , x + T ∼ x mod 2π

f(x) + 2πm ∼ f(x) in S1 mod 2π

Hopf invariant

f : S3 → S2

147

Take Ω in S2 , ∫S2

Ω = 1

Find ω in S3 such that

dω = f ∗Ω

Calculate “Kelvin-Whitehead Integral”∫S3

ω ∧ dω = I(f)

Theorem. For

F : S3 × R → S2 , ft = F |t=const

(x, t)

we havedI

dt= 0

Proof.

Take F ∗(Ω) in Λ2(S3 × R) .

Find ω ∈ Λ1(S3 × R) such that dω = F ∗(Ω) in S3 × R .

Calculate ∫S3: t=const

ω ∧ dω = I(t)

We have Ω ∧ Ω = 0 in Λ4 (S2) ≡ 0 .

So F ∗ (Ω) ∧ F ∗ (Ω) = 0 in Λ4(S3 × R

)d (ω ∧ dω) = dω ∧ dω = F ∗ (Ω) ∧ F ∗ (Ω) = 0

By Stokes we have dI/dt = 0 .

OK.

148

Contentssnovikov/math437.pdf · Diﬀerential 2-forms in Rn. Change of coordinates and restriction of 2-forms to 2-surfaces in Rn. Case n= 2. Integration. Case of any k-forms in Rn.

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