Fluid Mechanics

August 7, 2017

1 Submanifolds and Lie Groups

1.1 Submanifolds

We start with some basic definitions.

Definition 1. Let W,U ⊂ Rn be open subsets. Then f : W → U is called a diffeomor-phism if

• f is C∞ (smooth)

• f is bijective

• f−1 is C∞

Definition 2. A subset M ⊂ Rk is called an n-dimensional smooth submanifold if forevery point p ∈M there is an open neighbourhood W ⊂ Rk, p ∈ W and a diffeomorphism

ϕ : W → U ⊂ Rk = Rk−n × Rn (1)

such that

ϕ(W ∩M) = 0 × Rn ∩ U (2)

Theorem 1. For M ⊂ Rk the following are equivalent

a) M is an n-dimensional submanifold

b) For each p ∈M there is an open neighbourhood W ⊂ Rk, p ∈ W and g : W → Rn−k,such that W ∩M = g−1(0) and g′(0) ∈ Rk−n×k has rank k − n for all q ∈ W .

c) After applying a permutation of coordinates there are open sets U ⊂ Rn, W ⊂ Rk

with p ∈ W and a smooth map f : U → Rk−n such that

W ∩M = (x, f(x)) | x ∈ U

d) For each p ∈ M there is an open neighbourhood W ∈ Rk, an open set U ⊂ Rn anda diffeomorphism ψ : U → W ∩M such that for all x ∈ U the matrix ψ′(x) ∈ Rk×n

has rank n. This means: “locally a submanifold can be parameterized”.

Proof. (Sketch)b) ⇒ c): implicit function theoremc) ⇒ a): simply project it onto Rn i.e

ϕ(x, f(x)) = x (3)

a) ⇒ d) define U := ϕ(W ) ∩ 0 × Rn and set

ψ(x) := ϕ−1(0, x) (4)

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d) ⇒ c) Apply the inverse function theorem to Π ψ where

Π : Rk → Rn

Π(x1, . . . xk) = (xi1 . . . xin)

i1, . . . in ⊂ 1, . . . k (5)

c) ⇒ b) Choose simply

g : W → Rk−n

q 7→ (qn+1, . . . , qk)− (f1(q1, . . . , qn), . . . , fn(q1, . . . , qn)) (6)

Then g′(q) has maximal rank since f is assumed to be smooth.

Definition 3. Let M ⊂ Rk be an n-dimensional submanifold and p ∈M . Then

TpM := γ′(0) | γ : (ε, ε)→M smooth , γ(0) = p (7)

is called tangent space at p.

Theorem 2. TpM is an n-dimensional linear subspace of Rk.

Definition 4. A subset M ⊂ Rn is called a compact domain with smooth boundary, iffor every p ∈ ∂M there is an open neighbourhood W ⊂ Rn and a diffeomorphism

ϕ : W → U ⊂ Rn

such that

ϕ(W ∩ ∂M) = U ∩ (0, x)|x ∈ Rn−1 (8)

Definition 5. Let M ⊂ Rn be a subset with the additional requirement that the interiorM is dense in M . Then f : M → Rk is called smooth if there is an open set U ⊃M anda smooth extention f : U → Rk with f |M = f .

Note that we can define all partial derivatives

∂|α|

∂α1 · · · ∂αkf(p) :=

∂|α|

∂α1 · · · ∂αkf(p)

independent of the choice of f since the partial derivatives of f are continuous.

Definition 6. Let M,M ⊂ Rn be compact domains with smooth boundary, then a mapg : M →M is called an orientation preserving diffeomorphism if

1. g is smooth in the sense of Definition 5

2. g is bijective

3. g−1 : M → M is C∞

4. detg′(p) > 0 for all p ∈ M

3

1.2 Lie Groups

Firstly we recall that GL(n,R) := A ∈ Rn×n | detA 6= 0 is a group with respect tomatrix multiplication and neutral element I.

Definition 7. A subset G ⊂ GL(n,R) ⊂ Rn×n ia called a Lie group if it is a submanifoldof Rn×n and a subgroup of GL(n,R). i.e.

A,B ∈ G⇒ AB ∈ GA ∈ G⇒ A−1 ∈ G (9)

Definition 8. Let G ⊂ Rn×n be a Lie group. Then

g := TIG (10)

is called the Lie algebra of G.

Theorem 3. Let G ⊂ Rn×n be a Lie group and g := TIG its Lie algebra. Then for everyA ∈ G the tangent space TAG is given by

TAG = AX | X ∈ g = XA | X ∈ g (11)

Proof. Take X ∈ g and define the map

B : (ε, ε)→ G with B(0) = I , B′(0) = X (12)

Then the map C : (ε, ε)→ G defined by C := AB satisfies

C(0) = AB(0) = AI = A

C ′(0) = AB′(0) = AX

On the other hand take Y ∈ TAG and consider the curve

C : (ε, ε)→ G with C(0) = A , C ′(0) = Y (13)

Then define B : (ε, ε)→ G by B = A−1C. Obviously B(0) = I and B′(0) ∈ TIG = g.

Definition 9. Let G ∈ Rn×n be a Lie group and g its Lie algebra. Then

exp : g→ Rn×n

X 7→ A(1) (14)

where A : R→ Rn×n solves the initial value problemA(0) = I

A′(t) = A(t)X(15)

is called the exponential map.

Theorem 4. Let G ∈ Rn×n be a Lie group and g its Lie algebra. Then

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a) For all X ∈ g we have exp(X) ∈ G.

b) For X ∈ g the map

R 3 t 7→ exp(tX) =: A(t) (16)

solves A′(t) = A(t)X = XA(t)

c)

exp(X) =∞∑k=0

Xk

k!(17)

d)

exp((t+ s)X) = exp(tX) exp(sX) (18)

Proof.a) We start with an arbitrary X ∈ g and consider exp(tX). Due to the Picard-Lindelöftheorem we find ε such that exp(tX) = A(t) exists uniquely for t ∈ (−ε, ε). Now letB ∈ G be the group element which satisfies B = A( ε

2). Hence A′( ε

2) = BX ∈ TBG. We

now define

A1 : (−ε, ε)→ G

t 7→ A1(t) := A(t)B (19)

then A1 solves the initial value problemA1(0) = B

A′1(t) = A1(t)X = A(t)BX(20)

Indeed A1(t) exists uniquely for t ∈ (−ε, ε), since A(t) ∈ G. We proceed as above andconstruct the iteration

A,A1, A2, A3, . . .

Therefore A(t) ∈ G for all t ∈ R.c) We define

˜exp(X) = I +X +X2

2+X3

3!+ · · ·

A(t) := ˜exp(tX) = I + tX + t2X2

2+ · · ·

Hence A(0) = I and A′(t) = X + tX2

1!+ t2X

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2!+ · · · = XA(t) = A(t)X. By the uniqueness

part of Picard-Lindelöf we have A = A and finally

˜exp(X) = A(1) = A(1) = exp(X)

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b) is proven by c)d) Fix s ∈ R and define B, B : R→ G by

B(t) := exp((t+ s)X)

B(t) := exp(tX) exp(sX)

Then B(0) = B(0) = exp(sX) and

B′(t) = XB(t)

B′(t) = XB(t)

Due to the uniqueness part of Picard-Lindelöf one has B = B.

Definition 10. For X ∈ g the curve

t 7→ exp(tX) ∈ G (21)

is called a one-parameter subgroup of G. (actually a group homomorphism:R→ G)

Example 1. Consider the orthonogal group O(n) := A ∈ Rn×n | ATA = I and thecurve t 7→ A(t) ∈ O(n) where t ∈ (−ε, ε), A(0) = I and X := A′(0). By definition ofO(n) we have

A′(0)T︸ ︷︷ ︸=XT

A(0)︸︷︷︸=I

+A(0)T︸ ︷︷ ︸=I

A′(0)︸ ︷︷ ︸=X

= 0 ⇔ XT +X = 0

This shows that so(n) := g ⊂ X ∈ Rn×n | XT +X = 0On the other hand, if XT +X = 0 and A(t) = exp(tX). Then

(ATA)′ = (XA)TA+ ATXA = ATXTA+ ATXA = −ATXA+ ATXA = 0

Hence ATA = const and by continuity ATA = I for all t ∈ R since ATA(0) = I.

Definition 11. Let G be a Lie group, M a manifold. Then a smooth map

G×M →M

(g, x) 7→ g.x (22)

is called a group action if

a)

I.x = x ∀x ∈M (23)

b) for all A,B ∈ G, x ∈M hold

(AB).x = A.(B.x) (24)

Example 2. Let M := x ∈ Rn | |x| ≤ 1 be the unit ball and G = O(n). Then A.x isjust matrix multiplication.

6

2 Setup of Fluid Mechanics

Definition 12. Let M,M ⊂ Rn be compact domains with smooth boundary. Then wedefine the following useful map

gt : [t0, t1]× M →M (25)

where

1. For each t ∈ [t0, t1]

gt : M →M

is an orientation preserving diffeomorphism.

2. gt itself is smooth in the sense of Definition 5.

Intuition 1. For each fluid molecule p ∈ M its position at time t ∈ [t0, t1] is given bygt(p) ∈M .

Later on we might drop bijectivity (think of waves and air) and very rarely we will dropcompactness, because a huge container (ocean) can be seen as a infinitely long container.

Definition 13. In the space of smooth maps C∞(M,Rn) we define the set

M := g : M →M | g orientation preserving diffeomorphism (26)

We think aboutM as a kind of "submanifold“.

Definition 14. A map

[t0, t1]→M

t 7→ gt (27)

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is called a smooth curve inM if the map

[t0, t1]× M →M

(t, p) 7→ gt(p) (28)

is smooth.

Definition 15. For g ∈M we define the tangent space TgM by

TgM := W : M → Rn smooth | W (p) ∈ Tg(p)∂M if p ∈ ∂M (29)

If t 7→ gt ∈M, t ∈ [t0, t1] is smooth then we define gt ∈ TgtM by

gt(p) :=d

dτ

∣∣∣τ=tgτ (p) (30)

Definition 16. For a smooth family

(−ε, ε) 3 t 7→ ht ∈M (31)

with h0 = g we define

h0 ∈ TgM (32)

by

h0(p) :=∂ht(p)

∂t

∣∣∣t=0

(33)

Theorem 5. For every W ∈ TgM there is a map

R 3 t 7→ ht ∈M (34)

such that

W = h0

Definition 17.

Diff0(M) := h : M →M | h orientation preserving diffeomorphism (35)

This is a group under composition and the neutral element is idM .

diff0(M) := v : M → Rn smooth | q ∈ ∂M ⇒ vq ∈ Tq∂M (36)

Theorem 6. For every g ∈M we have the vector space isomorphism

diff0(M)→ TgMv 7→ v g (37)

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Proof. Since v ∈ diff0(M) the composition v g : M → Rn is smooth. By definitiong(p) ∈ ∂M for p ∈ ∂M and therefore Wp = vg(p) ∈ Tg(p)∂M .In order to construct the inverse linear map we takeW ∈ TgM and define v := W g−1.

At the boundary we have for q ∈ ∂M ⇒ g−1(q) ∈ ∂M by definition and therefore

vq = Wg−1(q) ∈ Tq∂M

Theorem 7. For every v ∈ diff0(M) there is a smooth family

R 3 t 7→ ht ∈ Diff0(M) (38)

such that

1. h0 = idM

2. ht(q) = vht(q) ∀q ∈M, ∀t ∈ R

3. ht+s = ht hs

Proof. Firstly we look at the boundary. For each q ∈ ∂M we choose a coordinate chart(Vq, ϕq) where Vq ⊂ ∂M is an open neighbourhood of q and ϕq : Vq → Uq ⊂ Rn−1 isassumed to be a diffeomorphism, since ∂M is a smooth boundary. Since ∂M is compact,the smooth vector field v|∂M is bounded and therefore Lipschitz continuous. Thus we canapply the Picard-Lindelöf theorem to the vector field

uϕq(x) = dxϕq(vx) x ∈ ∂M (39)

This means there is an open subset Uq ⊂ Uq and εq > 0 such that for any p ∈ Uq there isan integral curve

γ : (−εq, εq)→ Uq (40)

with γp(0) = p

γp(t) = uγp(t)

(41)

Moreover let Vq := ϕ−1q (Uq) be the corresponding neighbourhood on ∂M . By compactness

there are q1, . . . , qk ∈ ∂M such that Vq1 , . . . , Vqk cover ∂M . Hence one can define a globalε := minεq1 , . . . , εqk.We now jump back to the boundary of M . For every q ∈ ∂M there is a solution

ηq : (−ε, ε)→ ∂M (42)

with ηq(0) = q

ηq(t) = vηq(t)(43)

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Since q1 := ηq(ε2) can be used as starting point for

ηq1 : (−ε, ε)→ ∂M

ηq is extendable to

ηq : R→ ∂M

We now define

ht : ∂M → ∂M (44)

by

ht(q) := ηq(t) (45)

for all t ∈ R.

• By assumption we have a smooth dependency on the initial data, hence ht is smooth.

• Clearly ht+s = ht hs and in particular h−t ht = h0 = idM . Hence the inverseh−1t := h−t exists and ht is bijective.

Putting it all together, we eventually obtain that ht is a diffeomorphism.Now we do the same onM . Let M ⊃M be an open bounded set containingM . Extend

the vector field v to v smoothly on the closure M . Then v is Lipschitz. Consider the opencover

M =⋃q∈M

Vq (46)

Again we send the covering to Rn−1 via ϕq

Uq := ϕq(Vq) (47)

Due to Picard-Lindelöf we find ˜Uq and εq > 0 such that for any p ∈ ˜Uq there is an integralcurve

γ : (−εq, εq)→ Uq (48)

By the above procedure we find a global ε and finitely many qi such that

ηqi : (−ε, ε)→ M (49)

with ηq(0) = q ∈ ˜Vq := ϕq(

˜Uq)

ηq(t) = vηq(t)(50)

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is well defined for all q ∈M .If we start at q ∈ ∂M we stay on ∂M .If we start in the interior at q ∈ M then by the uniqueness part of Picard-Lindelöf wenever hit ∂M . This implies that we never leave M .

As before we construct

ht : M →M ∀t ∈ R (51)

Due to the smooth dependency on the initial data all ht are smooth. On the otherhandwe have ht+s = ht hs which gives us the bijectivity. Therefore all ht are diffeomor-phisms. Since h0 is orientation preserving and by the continuity of the determinant allht ∈ Diff0(M) for all t.

We now say that Diff0(M) is like a Lie group with identity element idM and its Liealgebra ”TidMDiff0(M) is equal to diff0(M) =: g. Moreover for h ∈ M and g ∈ Diff0(M)we define by "g.h := g h“ a "smooth group action” of Diff0(M) onM.

Definition 18. For v ∈ diff0(M) we define the exponential map

exp(v) ∈ Diff0(m) (52)

by

exp(v)(p) = γ(1) (53)

where γ(0) = p

γ(t) = vγ(t)

(54)

Indeed, we have: exp((t+ s)v) = exp(tv) exp(sv) and exp(0) = idM .Suppose now that the volume vol(U) :=

∫U

1 for U ∈ M can be interpreted as the totalmass of the fluid molecules in U .

Suppose t 7→ gt ∈M describes the motion of the fluid in M . Then each fluid moleculep ∈ M has velocity gt(p) ∈ Rn. So we can write the kinetic energy as

1

2m|gt|2 (55)

Moreover the total kinetic energy (up to a constant) is

E(gt) =

∫M

|gt|2 (56)

With this as background we can define a metric on TgtM.

Definition 19. The map

〈〈·, ·〉〉 : TgtM× TgtM→ R

(g, h) 7→ 1

2

∫M

〈g, h〉 (57)

can be viewed as a "Riemannian metric“ onM.

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Definition 20. Suppose [a, b] 3 t 7→ gt is a smooth curve in M. Then a variation of gis a smooth map

α : (−ε, ε)× [a, b]× M →M (58)

such that for all (τ, t) ∈ (−ε, ε)× [a, b] the map

gτ,t : M →M

p 7→ α(τ, t, p) (59)

is inM.Moreover the variational vector field Yt ∈ TgtM is defined as

Yt(p) :=∂α(τ, t, p)

∂τ

∣∣∣τ=0

(60)

Definition 21. The action of a free motion [a, b] 3 t 7→ gt ∈M is defined by

S(g) :=1

2

∫ b

a

〈〈gt, gt〉〉dt (61)

Theorem 8. (First variational formula)Let α be a variation of [a, b] 3 t 7→ gt ∈M with a variational vector field Yt. Then

d

dτ

∣∣∣τ=0

S(gτ,·) = 〈〈Yt, gt〉〉∣∣∣t=bt=a−∫ b

a

〈〈Yt, gt〉〉dt (62)

Before we prove the theorem we introduce the following notation

∂f

∂τ= f (63)

Proof.

d

dτ

∣∣∣τ=0

S(gτ,·) =

∫ b

a

〈〈 ˚go,t︸︷︷︸=:gt

, go,t︸︷︷︸=:gt

〉〉dt

=

∫ b

a

(∂

∂t〈〈gt, gt〉〉 − 〈〈gt, gt〉〉)dt

= 〈〈Yt, gt〉〉∣∣∣t=bt=a−∫ b

a

〈〈Yt, gt〉〉dt

where the last identity is given by the fundamental theorem of calculus.

Theorem 9. Let [a, b] 3 t 7→ Yt ∈ TgtM be smooth as a function on [a, b] × M . Thenthere is a variation α : R× [a, b]× M →M with the variational vector field Y0.

Proof.

gτ,t := exp(τ Yt g−1t︸ ︷︷ ︸

∈diff0(m)

) gt (64)

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Definition 22. The curve

[a, b]→Mt 7→ gt (65)

is called a geodesic in M if for all variations α of (t 7→ gt) with fixed endpoints i.egτ,a =: ga and gτ,b = gb for all τ ∈ (−ε, ε)

d

dτ

∣∣∣τ=0

S(t 7→ gτ,t) = 0 (66)

"principle of least action“

Theorem 10. The curve

[a, b]→Mt 7→ gt

is a geodesic if and only if gt = 0 for all t ∈ [a, b].

Proof. "⇐” Let α be a variation with fixed endpoints. Then the first variational formulagives us d

dτ

∣∣∣τ=0

S(t 7→ gτ,t) = 0 since the endpoints are fixed ga = gb = 0.

"⇒” Suppose t 7→ gt and gt 6= 0. Then there is t ∈ (a, b) and p ∈ ˚M with gt(p) 6= 0.

Choose δ > 0 such that [t − δ, t + δ] ⊂ (a, b) and q ∈ Rn | |q − p| < δ ⊂ ˚M . Then for

s ∈ [a, b] define Ys by Ys := ϕsgs where ϕs(q) := ψ(s)ρ(q) with

ψ : [a, b]→ Rψ(t) > 0

ψ(s) ≥ 0 for s ∈ [t− δ, t+ δ]

ψ(s) = 0 for s /∈ [t− δ, t+ δ]

and

ρ : M → Rρ(p) > 0

ρ(x) ≥ 0 for x ∈ q ∈ Rn | |q − p| < δρ(s) = 0 for x /∈ q ∈ Rn | |q − p| < δ

Now define

α : R× [a, b]× M →M

α(τ, s, p) = exp(τYs) (67)

where Ys = Ys g−1s .

Note that Ys vanishes on ∂M and therefore Ys ∈ diff0(M).Due to the first variational formula one obtains

0geodesic

=d

dτ

∣∣∣τ=0

S(τ(gτ,·)) = −∫ b

a

∫M

〈Ys, gs〉︸ ︷︷ ︸ϕs︸︷︷︸≥0

|gs|2

< 0

This contradicts gt 6= 0.

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Thus, for a geodesic there is a fixed v : M → Rn such that gt = v for all t ∈ [a, b]. Sofor all p ∈ M , t ∈ [a, b] a geodesic containing the point ga(p) is the straight line

gt(p) = ga(p) + (t− a)v(p) (68)

Note that the particles move independently of each other.

Example 3. 1D shock waves.Let M = M = [−2, 2] and g0 = idM .

gt(x) = x+ tv(x) =

x , |x| ≥ 1

x+ t(1− |x|) , |x| < 1

Note that gt is not an orientation preserving diffeomorphism for t > 1.

14

3 Natural Potential Energy for Fluids (Gases)

Before we start investigating the model of interacting particles, we consider a single masspoint under a potetial and derive Newtons law from the principle of least action.

3.1 Mass Point in Rn

Given a mass point in Rn with unit mass and a smooth curve

γ : [a, b]→ Rn

Moreover let V : Rn → R be a potential. Then

• the Kinetic energy at time t is: 12|γ(t)|2

• the action for free motion is: S(γ) =∫ ba

12|γ(t)|2dt

• the action under a conservative force is : S(γ) =∫ ba

12|γ(t)|2 − V (γ(t))dt

Consider the variation of a curve γ with fixed end points (γτ (a) = γ(a), γ − τ(b) = γ(b)).Since for a variational vector field Y we have γ = Y and hence Ya = Yb = 0. Therefore

d

dτ

∣∣∣τ=0

S(γτ ) =

∫ b

a

(〈γ, γ〉 − 〈gradV γ, γ〉)

=

∫ b

a

(〈γ, γ〉 − 〈γ, γ〉 − 〈gradV γ, γ〉)

= 〈Y, γ〉︸ ︷︷ ︸=0

∣∣∣ba−∫ b

a

〈Y, γ + gradV γ〉

Since Y is arbitrary the curve γ is a critical point of S with respect to variations withfixed ends if and only if

γ = − gradV γ (69)

3.2 Barotropic Fluid

Firstly, let us recall the change of volume. Assume that U ⊂ M , V ⊂M and g : M →Mis a diffeomorphism. Then

• vol(g(U)) =∫U

det g′

• vol(U) =∫g(U)

det(g−1)′

• vol(g−1(V )) =∫V

det(g−1)′ =∫V

1det g′

g−1

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Note that for small U one can approximate

vol(V ) = vol(g(U)) =

∫U

det g′ ≈ det g′(p) vol(U)

Thus the density of a fluid at q an be defined by

ρ(q) =vol(U)

vol(V )≈ 1

det g′(70)

This justifies the following definition.

Definition 23. Let g : M →M be an orientation preserving diffeomorphism. Then

•

ρ : M → R

ρ =1

det g′ g−1 (71)

is called Euclidian density.

•

ρ : M → R

ρ =1

det g′(72)

is called Lagrangian density.

More generally we define

1. Lagrangian viewpoint: Following each fluid particle p ∈ M .

2. Eulerian viewpoint: Hover over a fixed point q ∈M .

Given a function W : (0,∞)→ R, then one can define for g ∈M the potential V by

V (g) :=

∫M

W (ρ) =

∫M

W (ρ) g−1︸ ︷︷ ︸W (ρ g−1︸ ︷︷ ︸

ρ

)

det(g−1)′︸ ︷︷ ︸1

det g′ g−1︸ ︷︷ ︸

ρ

=

∫M

ρW (ρ) (73)

If we like to apply the principle of least action an expression for the variation of thedensity is needed.Unfortunately ρ : (−ε, ε) ×M → R lives on (−ε, ε) ×M . Therefore we have to extendour g to

g : (−ε, ε)× M → (−ε, ε)×M(τ, p) 7→ (τ, gτ (p)) (74)

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and obtain

ρ = ρ g : (−ε, ε)× M → R(τ, p) 7→ ρ(g(τ, p)) = ρ(τ, gτ (p)) (75)

For convenience we define the curve

γp : (−ε, ε)→ (−ε, ε)× Mτ 7→ (τ, p) (76)

Then obviously g(γp(τ)) = (τ, gτ (p)).

To prepare for the next theorem we need the following

Proposition 11. Let A : (−ε, ε)→ GL(n,R) be a curve in the invertible n×n matrices,where B := A(0) and B = A′(0). Then

d

dτ

∣∣∣τ=0

detA(τ) = detB tr(B−1B) (77)

Proof. We introduce the following notation B−1Bej =∑n

i=1 cijei.

det(Be1, . . . Ben) = det(Be1, Be2, . . . Ben) + · · ·+ det(Be1, . . . Ben−1, Ben)

= det(BB−1Be1, Be2, . . . Ben) + · · ·+ det(Be1, . . . Ben−1, BB−1Ben)

= detB(det(B−1Be1, Be2, . . . Ben)︸ ︷︷ ︸c11

+ · · ·+ det(Be1, . . . Ben−1, B−1Ben)︸ ︷︷ ︸

cnn

)

= detB tr(B−1B)

Now we are in a position to state and prove the “continuity equation”.

Theorem 12. Continuity Equation - (Euler)Let Y : M ⊂ Rn → Rn be a vector field on M and let ρ : (−ε, ε)×M → R be a density.Then holds

ρ+ 〈grad ρ, Y 〉 = −ρ divY (78)

or equivalently

ρ+ div(ρ Y ) = 0 (79)

17

Proof. Firstly we compute

˚ρ =d

ds

∣∣∣s=τ

ρ g γp(s)

= (ρ, grad ρ) g γ

1 0 . . . 0

g γ g′ γ

10·0

= (ρ, grad ρ) g

1

g

= ρ g + 〈grad g g, g︸︷︷︸

Y g

〉

= (ρ+ 〈grad g, Y 〉) g

On the other hand we have

(det g ′) = (det ρ−1) =

(1

ρ g

)= − (ρ g)

(ρ g)2= −

˚ρ

(ρ g)2

Using Proposition 11 leads to

− ρ+ 〈grad g, Y 〉ρ2

g = det g′︸ ︷︷ ︸1ρg

tr((g′)−1 g′︸︷︷︸(Y ′g)·g′

)

=1

ρ gtr Y ′︸︷︷︸divY

g

⇔ ρ+ 〈grad ρ, Y 〉 = −ρ divY

Theorem 13. Continuity Equation - (Lagrange)Let Y : M ⊂ Rn → Rn be a vector field on M and let ρ : (−ε, ε) × M → R be a densityon M . Then holds

˚ρ = −ρ (divY ) g (80)

Proof.

˚ρ =

(1

det g′

)= − 1

(det g′)2(det g′) = ρ2 det g′ tr(g′−1 g′)

= ρ tr(g′−1(Y g)′g′) = ρ tr(Y ′) g = ρ (divY ) g

18

We now look at a variation of the action of a free motion with a variational vector fieldwith fixed end points i.e. ga = gb = 0. Then

S = −∫ b

a

∫M

〈g, g〉 (81)

Consider a time depending vector field X : (−ε, ε)×M → Rn. Then

(X g)· = (X +∇vX) g

Applying this to X = v yields

g = (v g)· = (v +∇vv) g (82)

Hence

S =

∫ b

a

∫M

−〈Y, v +∇vv〉 g

ρ= 1det g′ g=

∫ b

a

∫M

−ρ〈Y, v +∇vv〉 (83)

• Lagrangian viewpoint: ⇒ g = 0.

• Eulerian viewpoint: watch vt(q) ⇒ v +∇vv = 0.

The next step is to compute the infinitesimal variation of the potential energy.

Theorem 14. Let

V :M→ R

g 7→∫M

W (ρ)

be the potential energy. Where W : (0,∞) → R is a given function and ρ : M → R+ isthe density on M . Then the variation of V is

V =

∫M

ρ 〈Y, grad(W ρ+ ρW ′ ρ)〉 (84)

Proof. Firstly recall the following two identities

• Divergence theorem∫M

divX =

∫∂M

〈X,N〉 = 0 ∀X ∈ diff0(M) (85)

• Let f : M → R be a real valued function, then

div(fY ) = 〈grad f, Y 〉+ fdivY (86)

19

As we have seen before V (g) =∫Mρ W (ρ). Hence

V =

∫M

ρ W (ρ) + ρ W ′(ρ) ρ (87)

Plugging in the continuity equation leads to

V =

∫M

−(ρ divY + 〈grad ρ, Y 〉)(W (ρ) + ρ W ′(ρ))

(86)=

∫M

−div(ρY ) (W (ρ) + ρ W ′(ρ))

Using again (86) by setting f := −(W (ρ) + ρ W ′(ρ)) and (85) completes the proof.

We now consider the variation of S(g) =∫ ba

[12

∫Mρt|vt|2 − V (gt)

]dt

S =

∫ b

a

∫M

ρ〈Y, v +∇vv − grad(W (ρ) + ρW ′(ρ))〉 (88)

Then S = 0 for all variations Y with fixed end points if and only if

v +∇vv = − grad p (89)

where p := −(W (ρ) + ρW ′(ρ)) is called pressure.

Theorem 15. The total energy H := E + V is constant.

Proof. Consider the kinetic energy E = 12

∫Mρ〈v, v〉 on M . From the Eulerian continuity

equation we know

ρ = −div(ρ Y )

Then

E =

∫M

(−1

2div(ρv)〈v, v〉 − ρ〈v, v〉

)(89)=

∫M

(−1

2div(ρv)〈v, v〉 − ρ〈v,∇vv + grad p〉

)(86)=

∫M

(−1

2div(ρv〈v, v〉) +

1

2〈ρv, grad〈v, v〉〉 − 〈ρv,∇vv + grad p〉

)=

∫M

(〈ρv,∇vv〉 − 〈ρv,∇vv + grad p〉)

(89)= −

∫M

ρ〈v, grad p〉

Comparing this with the variation of the potential energy (84) proves the statement.

20

Remark 1. Let λ, µ ∈ R be real constants. Moreover we define W by

W (ρ) := W (ρ) + λ+µ

ρ

Then the corresponding dynamics coincide.We simply compute

p = W (ρ) + ρW ′(ρ) = p+ λ+µ

ρ− ρ µ

ρ2= ρ+ λ

⇒ grad p = grad p

Example 4. Ideal gas W (ρ) = ρ. Then p = ρ+ ρ · 1 = 2ρ

4 Incompressible Fluids

Definition 24. A map g : M →M is called volume preserving diffeomorphism if

1. g is a diffeomorphism

2. vol(U) = vol(g(U)) for all subsets U ⊂ M

Definition 25. The map g : M → M is called incompressible fluid if g is a volumepreserving diffeomorphism.

Intuitively it seems clear that the corresponding vector field is divergence free, becausethe volume of any subset is preserved. But we are going to prove this rigorously.

As above we define

M := g : M →M | det g′ = 1 ⊂ M (90)

and the Lie group

SDiff(M) := h : M →M | deth′ = 1 (91)

Proposition 16. The corresponding Lie algebra is

sdiff(M) := X ∈ Γ(TM) | 〈X,N〉 = 0 on ∂M , divX = 0 (92)

Proof. Set Xt(q) = ddt

∣∣∣t=0ht(q). Since ρt ≡ 1 the Eulerian continuity equation (79) be-

comes

ρ︸︷︷︸=0

+ div(ρX) = 0 ⇔ divX = 0

Conversely take X ∈ sdiff(M) and define for all t ∈ R

ht := exp(tX) ∈ Diff0(M)

21

Due to the Lagrangian continuity equation (80)

˚ρt = −ρt (divXt)︸ ︷︷ ︸=0

ht

one obtains that ρt is constant. Since h0 = id and ρt = 1deth′t

one can conclude thatρt ≡ 1.

Again we define the incompressible motion by the priciple of least action

S(gt) =

∫ b

a

∫M

1

2|gt|2 gt ∈ M

We already know that

S = −∫ b

a

∫M

〈gt, Yt gt〉dt

The only difference now is that S = 0 for all Yt such thatYa = Yb = 0

Yt ∈ sdiff(M)(93)

The same reasoning as before now yields

0 =

∫M

〈gt, Yt gt〉 =

∫M

〈v +∇vv, Yt〉 ∀Yt ∈ sdiff(M)

i.e. v +∇vv ∈ sdiff(M)⊥

In summary can be said

Theorem 17. The action S is critical if and only if

v +∇vv ∈ sdiff(M)⊥ for all t ∈ [a, b]

Before we start to characterize the orthonogal complement of sdiff(M), we introducethe notation of musical isomorphisms.

Definition 26. Let M be a smooth Remanian manifold with metric 〈·, ·〉. Then thepointwise defined operator

[ : TpM → Ω1p(M) = T ∗pM

Xp 7→ 〈Xp, ·〉p (94)

is called flat operator and its inverse] : Ω1

p(M)→ TpM (95)

such that

〈ω]p, ·〉p = ωp (96)

is called sharp operator.

22

Moreover we define Wt ∈ Γ(TM) by

gt = Wt gt (97)

Then the action S is critical if and only if

0 =

∫M

〈Wt gt, Yt gt〉 =

∫M

〈Wt gt, Yt g + t〉 | det g′t|︸ ︷︷ ︸=1

=

∫M

W [t (Yt)

for all t ∈ [a, b] and Yt ∈ sdiff(W ).

Theorem 18.

grad f | f ∈ C∞(M) ⊥ sdiff(M) (98)

Proof. f ∈ C∞(M), Y ∈ sdiff(M). Then∫M

〈grad f, Y 〉 =

∫M

div(fY )−∫M

f div(Y )︸ ︷︷ ︸=0

=

∫∂M

〈fY,N〉 = 0

Theorem 19.

grad f | f ∈ C∞(M) = sdiff(M)⊥ (99)

We will give two proofs for the other inclusion “⊂”.1.Proof. We prove this in several steps.From calculus we know: There exists f ∈ C∞(M) with W = grad f if and only if forevery p, q ∈M the integral ∫

γ

W [ =

∫ b

a

〈W γ, γ′〉

where γ(a) = p, γ(b) = q does not depend on γ.From this follows easily:

Proposition 20. W ∈ grad f | f ∈ C∞(M) if and only if∫γW [ = 0 for all closed

curves γ.

Proof.

• “⇒“∫γ(grad f)[ = f(γ(b))− f(γ(a)).

• ”⇐“ Let γ : [a, b] → M , γ : [c, d] → M be two curves in M with end pointsγ(a) = γ(c) = p and γ(b) = γ(d) = q. Moreover we define

ϕ ∈ C∞([0,1

2], [a, b]) with

ϕ(t) = a for t ∈ [0, ε]

ϕ(t) = b for t ∈ [12− ε, 1

2]

23

and

ϕ ∈ C∞([1

2, 1], [c, d]) with

˜ϕ(t) = d for t ∈ [1

2, 1

2+ ε]

˜ϕ(t) = c for t ∈ [1− ε, 1]

Afterwards we define

η := γ ϕ : [0,1

2]→M

η := γ ϕ : [1

2, 1]→M

and finally

γ : [0, 1]→M

γ(t) =

η(t) for t ∈ [0, 1

2]

η(t) for t ∈ [12, 1]

Then

0 =

∫γ

W [ =

∫η

W [ +

∫η

W [ =

∫γ

W [ −∫γ

W [

Proposition 21. W ∈ grad f | f ∈ C∞(M) if and only if∫γW [ = 0 for all smooth

embeddings

γ : S1 = R/2π →M

The proof needs the following definition.

Definition 27. Let M be a two-dimensional manifold and let γ1 : [a, b] → M , γ2 :[c, d] → M be two curves that intersect at γ1(p) = γ2(q) = r, then the intersection iscalled transversal if

TrM = dγ1(R) + dγ2(R)

Proof.”⇒“ by Proposition 20.”⇐“ We will use two theorems from differential-topology

1. dimM ≥ 2 ⇒ immersions γ : S1 → M are C∞ dense in the space of smooth mapsC∞(S1,M).

2. dimM ≥ 2 ⇒ small C∞ pertubations of γ makes self-intersections transversal.

24

We now proceed as follows:Small C∞ pertubations do not change

∫γW [ much. Hence we can concentrate on im-

mersions γ : S1 → M with transversal self-intersections. Then we use again statement1.

Now we come back to the proof of Theorem19. Due to the above considerations, itsuffices to show that

∫γW [ = 0 for all embeddings γ : S1 = R/L→ M with |γ′| = 1.

Let dimM = n ≥ 2 and let (T,N1, . . . , Nn−1) be an orthonormal frame along γ. Then

ϕ : S1 ×Bε → M

(t, u1, . . . , un−1) 7→ γ(t) +n−1∑i=1

uiNi(t)

is a diffeomorphism for small enough ε.

25

Furthermore we define

•

r : S1 ×Bε → R

(t, u1, . . . , un−1) 7→

√√√√n−1∑i=1

u2i

•

ψδ : (−ε, ε)→ Rψδ(−x) = ψδ(x)

ψδ(x) ≥ 0

ψδ(x) = 0 for |x| ≥ δ∫ ε0

2πψδ(x)xdx = 1

•

T (ϕ(t, u1, . . . , un−1)) := T (t)

•

f := ψδ r ϕ−1

26

•

Yq ∈ sdiff(M)

Yq :=

f · T for q ∈ ϕ(S1 × Bε)

0 otherwise

Then

0 = limδ→0

∫M

〈Yδ,W 〉 =

∫ L

0

〈γ′,W γ〉 =

∫γ

W [

Finally we have to check: divY = 0.If one define Ni just as T , one can write the divergence as

divY = 〈∇TY, T 〉︸ ︷︷ ︸A

+n−1∑i=1

〈∇NiY, Ni〉︸ ︷︷ ︸B

•

A = 〈(T · f)T + f∇T T , T 〉 = 0

Note that T · f = 0, since

T = dϕ

(∂

∂t

)⇒ df(T ) = d(f ϕ︸ ︷︷ ︸

ψr

)

(∂

∂t

)=∂(ψ r)∂t

and 〈∇T T , T 〉 = 0 since

0 = T 〈T , T 〉 = 2〈∇T T , T 〉

27

•

B = 〈(Ni · f)T + f∇NiT , Ni〉 = 0

∇NiT = 0 since

Ni = dϕ

(∂

∂ui

)is the derivative in direction ui. But T depends on t only.

2.Proof. We start with a definition.

Definition 28. The operator

∆ : C∞(M)→ C∞(M)

f 7→ div grad f (100)

is called Laplace operator.

We will use two theorems from the PDE theory

Theorem 22. (Dirichlet boundary value problem)Let g ∈ C∞(M) and h ∈ C∞(∂M) be smooth functions on M and ∂M respectively. Thenthere exists a unique f ∈ C∞(M) such that

∆f = g

f |∂M = h(101)

Theorem 23. (Neumann boundary value problem)Let g ∈ C∞(M) and h ∈ C∞(∂M) be smooth functions on M and ∂M respectively. Thenthere exists a unique f ∈ C∞(M) such that

∆f = g

〈grad f,N〉 = h(102)

if and only if∫Mg =

∫∂M

h. In this case f is unique up to an additive constant.

Proof. Fortunately we are in the situation that M is a subset of Rn. So we can use theresults of the classical PDE theory. For the sake of completeness we prove one directionof the Neumann boundary value problem. ”⇒“ f exists. Thus∫

M

g =

∫M

div grad f =

∫∂M

〈grad f,N〉 =

∫∂M

h

Now we have all ingedients to prove the

28

Theorem 24. (Helmhotz decomposition)

Γ(TM) = grad f | f ∈ C∞(M)⊕

sdiff(M) (103)

Proof.

Γ(TM) = grad f | f ∈ C∞(M) ∩ sdiff(M) = 0follows from Theorem 18.Conversely let X ∈ Γ(TM) then∫

M

divX︸ ︷︷ ︸=:g

=

∫∂M

〈X,N〉︸ ︷︷ ︸=:h

Due to Theorem 23 there exists f ∈ C∞ with∆f = g

〈grad f,N〉 = 〈X,N〉

Then Y := X − grad f satisfies divY = 0, 〈Y,N〉 = 0. Hence Y ∈ sdiff(M) and thereforeX = grad f + Y

We now finish the 2. Proof of Theorem 19.Let X ∈ sdiff(M)⊥, then Helmhotz tells X = grad f + Y with Y ∈ sdiff(M). Thus

0 =

∫M

〈X, Y 〉 =

∫M

〈grad f, Y 〉︸ ︷︷ ︸=0

+

∫M

〈Y, Y 〉

Hence Y = 0 and X = grad f .With Theorem 19 established we know: t 7→ gt ∈ M is critical for the action if and

only if ∀t there is pt ∈ C∞(M) such that v +∇vv = − grad pt.How to determine pt for given vt?

Since vt ∈ sdiff(M) one obtains0 = divv = −div(∇vv + grad pt)

0 = 〈vt, N〉 = −〈∇vv + grad pt, N〉⇔

∆pt = −div(∇vv)

〈grad pt, N〉 = −〈∇vv,N〉(104)

Roughly speaking, all numerical methods proceed as follows

1. Given vt ∈ sdiff(M)

2. Compute vt+δ as vt+δ = vt − δ∇vtvt − grad q where q is determined by0 = div(∇vtvt + grad q)

0 = 〈δ∇vtvt + grad q,N〉⇔

∆q = −δdiv(∇vtvt)

〈grad q,N〉 = −δ〈∇vtvt, N〉(105)

Solving this Neumann problem for q is called pressure projection of vt+δ∇vtvt to sdiff(M)

Definition 29. The equations v +∇vv = − grad p

div v = 0, 〈v, N〉 = 0(106)

are called incompressible Euler equations.

29

4.1 Circulation

Definition 30. Let v ∈ diff0(TM) be a vector field tangent to the boundary and letγ : S1 →M be a closed curve, then∫

γ

v[ =

∫ 2π

0

〈v(γ(t)), γ′(t)〉dt (107)

is called the circulation of v around γ.

We already know: There exists a function f ∈ C∞(M) such that grad f = v (df = v[)if and only if the circulation of v along every closed curve γ vanishes.

Example 5. Let M = D2 = (x, y) ∈ R2 | x2 + y2 ≤ 1 de the unit disc. Moreover thevector field is given by

v(x, y) :=

(−yx

)then the circulation of any closed curve γ is∫

γ

v[ =

∫ 2π

0

〈(−γ2

γ1

),

(γ′1γ′2

)〉dt =

∫ 2π

0

det(γ, γ′)dtStokes

= 2 area(γ)

More generally, let f : D2 → M be a surface in M and let γ := f |∂D2 be its boundary.Then ∫

γ

v[Stokes

=

∫f

〈curl v,N〉 (108)

Figuratively speaking this means: If γ bounds a piece of a surface f then the circulationof v along γ equals the flux of curl v through f . Thus, the circulation of v around smallloops determines curl v.

Example 6. Let M = D2 = (x, y) ∈ R2 | r ≤ x2 + y2 ≤ R be an anulus and

v(x, y) :=1

x2 + y2

(−yx

)Furthermore we define the argument function arg : C \ 0 such that arg is smooth awayfrom the negative real axis. Then the circulation of v around any loop γ is zero if thenumber of transversal intersections with the negative real axis is even. The counterexamplefor odd intersections is as follows

r < ρ < R γ(t) =

(ρ cos tρ sin t

)∫γ

v[ =

∫ 2π

0

〈 1

ρ2

(−ρ sin tρ cos t

), ρ

(− sin tcos t

)〉dt =

∫ 2π

0

1 = 2π

30

So, a circulation around γ’s comtain mor information then curl v.

Theorem 25. Let v, v ∈ sdiff(M) be two divergence free vector fields tangent to theboundary such that

∫γv[ =

∫γv[ for all loops γ ∈M then v and v coincide.

Proof. Since∫γv[ − v[ = 0 for all γ we have v − v︸ ︷︷ ︸

∈sdiff(M)

= grad f︸ ︷︷ ︸∈sdiff(M)⊥

. Due to the Helmhotz

decomposition one has sdiff(M) ∩ sdiff(M)⊥ = 0 and therefore v − v = 0.

Thus a 1-form η encodes all circulations around γ’s independent of the metric (dependson the manifold structure of M only).

We now consider 1-forms under the pullback. Let gt : M →M then we define

ηt : = g∗t (v[t)

ηt(X) = 〈vt gt, dgt(X)〉 ∀X ∈ Γ(TM) (109)

Recall

gt = vt gtgt = (vt +∇vtvt) gt

Euler= − grad pt gt

Now the time derivative reads˙ηt(X) = 〈grad pt gt, dgt(X)〉︸ ︷︷ ︸

−d(ptgt)(X)

+ 〈vt gt, (dgt(x))·︸ ︷︷ ︸d(vtgt)(X)

〉

︸ ︷︷ ︸12d(〈vt gt, vt gt〉︸ ︷︷ ︸

〈vt,vt〉gt

)(X)

= d((1

2|vt|2 − pt︸ ︷︷ ︸

=:−βt

) gt

︸ ︷︷ ︸−βt

)

31

Thus˙ηt = −dβt, βt ∈ C∞(M) (110)

and ∫γt

ηt =

∫ 2π

0

ηt( γ′t︸︷︷︸dgt(γ′)

) =

∫ 2π

0

ηt(γ′) =

∫γ

ηt (111)

Theorem 26. (Kelvin - Helmhotz circulation theorem)Let t 7→ vt ∈ diff0(M) corresponding to the notation of a barotropic fluid(i.e vt + ∇vtvt = − grad pt with pt ∈ C∞(M)) then for every loop γ ∈ M there is aconstant c ∈ R such that for γt = gt γ and ηt = v[t we have∫

γt

ηt = c

Proof. (∫γt

ηt

)·(111)=

∫γ

˙η = −∫γ

dα = 0

4.2 Lie Derivative

Let g : M →M be a diffeomorphism. Then every object on M can be transported via gto M

• f ∈ C∞(M)

Tgf ∈ C∞(M)

(Tgf)(q) : = f(g−1(q))

Tgf : = f g−1 = (g−1)∗f (112)

32

• X ∈ Γ(TM)

TgX ∈ Γ(TM)

(TgX)q = dg(Xg−1(q))

TgX = g∗X (113)

• ω ∈ Ωk(M) and X1, . . . Xk ∈ TqM

Tgω ∈ Ωk(M)

Tgω(X1, . . . Xk) = ωg−1(dg−1(X1), . . . , dg−1(Xk))

Tgω = (g−1)∗ω (114)

Tg plays together nicely with all natural operations.

• Tg(dω) = d(Tgω), Tg([X, Y ]) = [TgX,TgY ]

• T (X · f) = (TgX) · (Tgf)

• Tg(ω(X1, . . . , Xk)) = Tgω(TgX1, . . . , TgXk)

Definition 31. Let X ∈ Γ(TM) a smooth vector field on a compact manifold M withboundary and let gt : M → M be a diffeomorphism for t ∈ R satisfying the initial valueproblem

ddtgt(p) = Xgt(p)

g0 = id(115)

then the Fisherman’s derivative F is given by

• for ω ∈ Ωk(M)

(FXω)q :=d

dt

∣∣∣t=0

(Tgtω)q (116)

and similarly

• for Y ∈ Γ(TM)

(FXY )q :=d

dt

∣∣∣t=0

(TgtY )q (117)

Properties

• FX(dω) = d(FXω)

• FX([Y, Z]) = [FXY, Z] + [Y,FXZ]

• FX(ω(Y1, . . . , Yk)) = (FXω)(Y1, . . . , Yk) + ω(FXY1, . . . , Yk) + · · ·+ ω(Y1, . . . ,FXYk)

• (FXf)q = df(−Xq) = −(Xf)q

33

Definition 32. Let F be the Fisherman’s derivative, then the Lie derivative L is givenby

LXω := −FXω (118)

LXY := −FXY (119)

and

LXf = Xf (120)

Analogously to the Fisherman’s derivative one obtains the following properties

• LX(dω) = d(LXω)

• LX(ω(Y1, . . . , Yk)) = (LXω)(Y1, . . . , Yk) + ω(LXY1, . . . , Yk) + · · ·+ ω(Y1, . . . ,LXYk)

Lemma 1.

LXY = [X, Y ] (121)

Proof.

X(Y f) = LX(Y f) = (LXY ) · f + Y · LXf︸ ︷︷ ︸Y (Xf)

Definition 33. Let ω ∈ Ωk(M) be a k-form on M , then for X ∈ Γ(TM) the map

ιX : Ωk(M)→ Ωk−1(M)

ιXω(Y2, . . . , Yk) : = ω(X, Y2, . . . , Yk) (122)

is called interior derivative

Lemma 2. Let ω ∈ Ω1(M) be a 1-form on M , then for X ∈ Γ(TM) holds

LXω = ιXdω + dιXω (123)

Proof. Take Y ∈ Γ(TM), then

Xω(Y ) = LX(ω(Y )) = (LXω)(Y ) + ω([X, Y ])

and

dω(X, Y ) = Xω(Y )− Y ω(X)− ω([X, Y ])

hence

(LXω)(Y ) = dω(X, Y ) + Y ω(X) = ιXdω(Y ) + d(ιXω)(Y )

34

Theorem 27. (Cartans magic formula)Let ω ∈ Ωk(M) be a k-form on M and let X ∈ Γ(TM), then

LXω = ιXdω + dιXω (124)

Proof. by induction over k. Use Lemma 2 as initial step.

We already know: gt is volume preserving (i.e Tgt det = det) if and only if divX = 0. Thenext theorem is a reformulation in terms of Lie derivative.

Theorem 28. Let gt : M →M be a diffeomorphism, then gt is volume preserving if andonly if LX det = 0.

Proof. Firstly we prove

d(ιX det) = (divX) det (125)

Let p ∈ M , choose an orthonormal frame field Y1, . . . Yn near p with (∇Yj)p = 0 andexpress X as

X :=n∑i=1

uiYi

then

d(ιX det)(Y1, . . . , Yk)p =n∑j=1

(−1)j+1Yj(ιX det)(Y1, . . . , Yj, . . . , Yn)

=n∑j=1

Yj(det(Y1, . . . , X↑j

, . . . , Yn))

and

divXp = tr(∇X)p =n∑j=1

〈∇YjX, Yj〉

=n∑j=1

〈∇Yj

n∑i=1

uiYi, Yj〉 =n∑j=1

〈n∑i=1

(Yjui)Yi, Yj〉

Finally we use Cartan’s magic formula

LX det = ιX d det︸︷︷︸(n+1) form=0)

+dιX det

35

This theorem can be seen as a non Riemannian definition of divergence.We now derive the Euler equations in terms of the Lie derivative of the velocity 1-formη = v[.

(Lη]η)(X) = Lη](η(X))− η(Lη]X)

= v · 〈v,X〉 − 〈v, [v,X]〉= 〈∇vv,X〉+ v,∇vX〉 − 〈v,∇vX −∇Xv〉

= 〈∇vv,X〉+1

2X · 〈v, v〉

= 〈∇vv +1

2grad |v|2, X〉

hence

(Lη]η)] = ∇vv +1

2grad |v|2 ⇔ ∇vv = (Lη]η)] − 1

2grad |v|2

plugging in the equation of motion v +∇vv = − grad p yields

v + (Lη]η)] = − grad(p− 1

2|v|2︸ ︷︷ ︸

=:β

)

Applying the [ operator and using (grad f)[ = df gives us the Euler equation

η + Lη]η = −dβ (126)

4.3 Vorticity

Let v ∈ Γ(TM) be a vector field. At each p ∈M decompose

∇v : X 7→ ∇Xv (127)

into its symmetric and skew parts

∇v =1

2(∇v + (∇v)∗)︸ ︷︷ ︸

=:A

+1

2(∇v − (∇v)∗)︸ ︷︷ ︸

=:B

(128)

Consider the taylor expansion of the vector field around p

36

Theorem 29. Let 〈·, ·〉 := g(·, ·) denote the metric on M then

〈∇Xv, Y 〉 =1

2(Lvg + dη)(X, Y ) (129)

Proof.

(Lvg)(X, Y ) = Lv(g(X, Y ))− g(LvX, Y )− g(X,LvY )

= v · 〈X, Y 〉︸ ︷︷ ︸〈∇vX,Y 〉+〈X,∇vY 〉

−〈∇vX −∇Xv, Y 〉 − 〈X,∇vY −∇Y v〉

= 〈 ∇Xv︸︷︷︸(∇v)(X)

, Y 〉+ 〈X, ∇Y v︸︷︷︸(∇v)(Y )

〉

︸ ︷︷ ︸〈(∇v)∗X,Y 〉

= 〈(∇v + (∇v)∗)X, Y 〉 = 〈2AX, Y 〉

In addition, this shows that the Lie derivative of a metric is still a symmetric tensor.We now look at the exterior derivative of the velocity 1-form η = v[.

dη(X, Y ) = X〈v, Y 〉 − Y 〈v,X〉 − 〈v,∇XY −∇YX〉= 〈∇Xv, Y 〉 − 〈∇Y v,X〉= 〈(∇v − (∇v)

∗)X, Y 〉 = 〈2BX, Y 〉

At this point we investigate the special case dimM = 3.Fore each ω ∈ Ω2(M) there is a unique w ∈ Γ(TM) such that

ω = ιw det

ω(X, Y ) : = det(w, X, Y ) =: 〈w, X × Y 〉 (130)

In particular for ω = dη

∇Xv =1

2(ιXLvg)] +

1

2w×X (131)

Proposition 30. For dimM = 3 and dη = ιw det

v ×w =1

2grad |v|2 −∇vv (132)

Proof.

〈 v ×w, Y 〉 = 〈w, Y × v 〉 = dη(Y × v)

= Y η(v)︸︷︷︸〈 v,v 〉

− vη(Y )︸ ︷︷ ︸v 〈 v, Y 〉︸ ︷︷ ︸

〈∇vv,Y 〉+ 〈 v,∇vY 〉

− η([Y, v])︸ ︷︷ ︸〈 v,∇Y v−∇vY 〉

= Y 〈 v, v 〉− 〈∇vv, Y 〉− 〈 v,∇Y v 〉︸ ︷︷ ︸12Y 〈 v,v 〉

= 〈 1

2grad |v|2 −∇vv, Y 〉

37

Let’s come back to the equations ot motion v +∇vv = − grad p. Plugging in equation(132) leads to

v + w× v = − grad(p+1

2|v|2︸ ︷︷ ︸

=:α

) (133)

Definition 34. α is called the Bernoulli funtion of v.

We have three kinds of equations of motion

v +

w× v∇vv

(Lvη)]

= − grad

p+ 12|v|2

pp− 1

2|v|2

(134)

4.4 Stationary Flows

Before we proceed we racall some facts from calculus

Definition 35. Let U ⊂ Rn be an open set and f : U → Rk be a smooth function, thenq ∈ Rk is called a critical point of f if there is p ∈ U with f(p) = q and rank(f ′(g)) < k( regular value otherwise).

Lemma 3. (Lemma of Sarde)The set of critical values of f has measure zero.

Note q /∈ f(U) then q is a regular value.

Example 7. k = 1: Then q ∈ R is a regular value if there is no p ∈ U such that f(p) = qand f ′(p) = 0.

The preimage of a regular value q ∈ Rk is either empty or a smooth submanifold of U .

Definition 36. A flow is called stationary if v = 0.

Then the equation of motion becomes

v ×w = gradα

Case 1: α is not a constant function.Let q ∈ R be a regular value of α then Σ := α−1q 6= ∅ and hence Σ is a smooth,orientable, compact surface possibly with boundary ∂Σ = Σ ∩ ∂M .

• v|Σ, w|Σ are tangent to Σ since gradα points in the normal direction of the levelset.

• v|Σ, w|Σ are linearly independent since q was assumed to be a regular value.

38

One can say for v ∈ sdiff(M) stationary: vt = v is a solution of equations of motion ifand only if there exists α ∈ C∞(M) such that v × curl v = gradα.

Example 8.

M := (x, y, z) ∈ R3 | x2 + y2 + z2 ≤ 1 v(x, y, z) =

−yx0

then

curl v(x, y, z) =

00

1 + 1

=

002

and

(v ×w)(x, y, z) =

2x2y0

= grad(α(x, y, z))

where α(x, y, z) = x2 + y2.

Theorem 31. (proof later)Let Σ be a compact oriented surface with boundary which admits a nowhere vanishingvector field v ∈ diff(M). Then every connected component of Σ is diffeomorph to [0, 1]×S1

or S1 × S1.

What is about the motion on Σ?Applying the exterior derivative to the vorticity equation η + Lvη = −dβ and settingω = dη leads to

ω + Lvω = 0 (135)

39

Using the chain rule yields

(Lvω)(X, Y ) = Lv(ω(X, Y ))− ω(LvX, Y )− ω(X,LvY )

Since ω is defined as ω = ιw det and Lv det = 0 because it is divergence free, one obtains

Lv(ω(X, Y )) = Lv(det(w, X, Y ))

= (Lv det)︸ ︷︷ ︸=0

+ det([v,w], X, Y ) + det(w, [v,X], Y ) + det(w, X, [v, Y ])

Hence

ω = (ιw det)· = −ι[v,w] det

Since the determinant depends not on time

ιw det = −ι[v,w] det

Therefore the vorticity equation becomes

w = [v,w] (136)

Theorem 32. If t 7→ vt ∈ sdiff(M) satisfying the Euler equations, then the vorticityvector field t 7→ w is frozen into the fluid in the sense that there is a time independentvector field w ∈ Γ(TM) such that

wt = gt∗w (137)

Proof. Since incompressible fluids are volume preserving we have Tgt det = det and there-fore

Tgtιwt det = ιwt det

Using the definitions of ω and η one obtains

ιwt det = dηt

Recall identity (110)

˙ηt = −dβt

Hence

d ˙ηt = 0 ⇒ ˙wt = 0

40

Figuratively speaking, at time 0 draw a curved arrow along a vortex line with blue ink,wait a time t, look what became of the blue arrow.

Then then the new evolved arrow is a vortex line at time t.

Example 9. Construct w to be non-zere only inside a certain knotted tube. Moreover letv ∈ sdiff(M) with curl v = w. Then after evolving v the vorticity wt is concentrated inthe knotted tube for all time.

Note that the vorticity equation for a stationary flow is

[v,w] = 0

i.e. v and w commute.

Theorem 33. Let M be a compact manifold with boundary and X, Y ∈ sdiff(M) with[X, Y ] = 0. Moreover let gt, ht ∈ Diff0(M) be solutions of the initial value problems

∂gt∂t

= X gtg0 = idM

∂hs∂s

= Y hsh0 = idM

then for all s, t ∈ R

hs gt = gt hs

Proof. Fix t ∈ R and define hs ∈ Diff0(M) piontwise by

hs := gt hs(p) g−1t (p)

41

Then the corresponding vector field is Y = TgtY = gt∗Y . The fact that

Y = Y ⇔ LXY = [X, Y ] = 0

proves the statement.

42

5 Cohomology of Compact Manifolds with Boundary

LetM be a compact oriented n-dimensional manifold with boundary, M = M1∪· · ·∪Mm

where M1, . . . ,Mm are the connected components of M .

Ω−1(M)d−1−→ Ω0(M)

d0−→ Ω1(M)d1−→ · · · dn−1−→ Ωn(M)

dn−→ Ωn+1(M)‖ ‖ ‖0 C∞(M) 0

Definition 37. Let ω ∈ Ωk(M) be a k-form. Then ω

• is called closed if ω ∈ Ker dk

• is called exact if ω ∈ Im dk−1

Since dk dk−1 = 0 ⇒ Im dk−1 ⊂ Ker dk.

Definition 38.

Hk(M) := Ker dk/

Im dk−1(138)

is called the k-th cohomology.

Theorem 34. Hk = 0 for k < 0 and k > n. H0(M), . . . , Hn(M) are all finitedimensional.

Definition 39. βk(M) := dimHk(M) is called the k-th Betti number of M .

43

Example 10. Consider Ker d0 = f ∈ C∞(M) | df = 0 then

f ∈ Ker d0 ⇔ f |Mj= const ⇒ H0(M) ' Rm and β0(M) = m

since Im d−1 = 0.Ker dn = Ωn(M) because all n-forms are closed.

Suppose ∂M = ∅. Due to Stokes theorem∫M

dη = 0 ∀η ∈ Ωn−1(M)

So, if ω ∈ Im dd−1 ⇒∫Mω = 0. On the other hand, if ω is a volume form

∫Mω =

vol(ω) > 0. Hence the volume form ω is not exaxt.

Theorem 35. (Ditichlet Problem, manifold version)Let M be a compact connected Riemannian manifold with boundary and ∂M 6= ∅, H ∈C∞(M), g ∈ C∞(M). Then there is a unique f ∈ C∞(M) with

∆f = g

f |∂M = h

Theorem 36. Let M be a compact connected oriented n-dim manifold with ∂M 6= ∅.Then Hn(M) = 0. “All n-forms are exact”.

Proof. Choose a Riemannian metric on M . Rewrite ω ∈ Ωn as

ω = g · det with g ∈ C∞(M)

where det denotes the volume form. Due to the Dirichlet theorem there exists f ∈ C∞(M)with ∆f = g, f |∂M = 0. Then

div grad f = g

⇔ div grad f · det = g · det

(125)⇔ d(ιgrad f det) = g · det = ω

Theorem 37. (Neuman Problem, manifold version)Let M be a compact connected Riemannian manifold with boundary. Moreover let g ∈C∞(M) and h ∈ C∞(∂M) be smooth functions on M and ∂M respectively satisfying∫Mg =

∫∂M

h. Then there is a f ∈ C∞(M) (unique up to an additive constant) such that∆f = g

df(N) = h

where N is the outward pointing unit normal along ∂M .

44

Remark 2. If∫Mg 6=

∫∂M

h then ∆f = g

df(N) = h

has no solution f .

Proof. Using the Divergence theorem yields∫M

∆f =

∫M

div grad f =

∫∂M

〈 grad f,N 〉 =

∫∂M

df(N)

Theorem 38. Let M be a compact connected oriented manifold without boundary. ThendimHn(M) = 1.

Proof. By Stokes∫Mη = 0 for all η ∈ Im dn−1. So if ω ∈ Ωn(M) then

∫Mω + η =

∫Mω.

Hence∫Mω depends on the cohomology class of ω only

[ω] ∈ Hn(M) = Ωn(M)/

Im dn−1(139)

So∫M

[ω] :=∫Mω is well defined.

Finally we prove that the linear map∫M

: Hn(M)→ R (140)

is bijective.

• surjective: set ω = c · det with c ∈ R

• injective: Choose a Riemannian metric, [ω] ∈ Ker∫M

i.e. [ω] ∈ Hn(M) with∫Mω = 0. Set ω = g · det hence

∫Mω = 0. By Neumann ∃f ∈ C∞(M) with

∆f = g. Therefore

div(grad f) = g

(125)⇔ d(ιgrad f det) = g · det = ω

⇒ ω ∈ Im dn−1 ⇒ [ω] = 0

Since (140) is linear this proves the injectivity.

Easy to see: M1, . . . ,Mm compact oriented manifolds with boundary andM = M1

∐· · ·∐Mm

is the disjoint union. Then Hk(M) = Hk(M1)⊕· · ·⊕

Hk(Mm).

Corollary 1. Let M be a compact oriented manifold with boundary, M1, . . . ,Mm con-nected components of M . Then

dimHn(M) = #j | ∂M = ∅ (141)

45

Corollary 2. Let M be a compact oriented manifold without boundary. Then

dimH0(M) = dimHn(M) (142)

This is a special case of the Poincaré duality theorem.In order to state the theorem in its full generality we need a few definitions.

Definition 40. Let V,W be vector spaces. Then 〈 ·, · 〉 : V ×W → R bilinear yields alinear map

f : V → W ∗

(f(v))(w) : = 〈 v, w 〉 (143)

〈 ·, · 〉 is called non-degenerate if

• 〈 v, w 〉 = 0 for all w ⇒ v = 0

• 〈 v, w 〉 = 0 for all v ⇒ w = 0

Note that, 〈 ·, · 〉 is non-degenerate if and only if f is an isomorphism.

Theorem 39. (Poincaré duality theorem)Let M be a compact oriented manifold without boundary. Then

Hk(M)×Hn−k(M)→ R

(ω, η) 7→ 〈ω, η 〉 =

∫M

ω ∧ η (144)

is well defined and it is a non-degenerate pairing between Hk(M) and Hn−k(M). So itestablishes an isomorphism Hk ↔ Hn−k(M)∗. In particular dimHk = dimHn−k.

Remark 3. Let β ∈ Ωk−1(M) then∫M

(ω + dβ) ∧ η =

∫M

ω ∧ η∫M

dβ ∧ η

dη=0=

∫M

ω ∧ η∫M

d(β ∧ η)

=

∫M

ω ∧ η

Similarly for α ∈ Ωn−k−1(M) ∫M

ω ∧ (η + dα) =

∫M

ω ∧ η

Definition 41.

Hk(M) := Hk(M)∗ (145)

is called the k’th homology vector space of M .

46

Let Σ be a compact oriented manifold, dim Σ = k and ∂Σ = ∅. Moreover let f : Σ→Mbe a smooth map. Then define f ∈ Hk(M) by

f([ω]) :=

∫Σ

f ∗ω (146)

where ω ∈ Ωk(M), dω = 0. This is well defined because∫Σ

f ∗(ω + dη) =

∫Σ

f ∗ω +

∫Σ

f ∗dη︸︷︷︸df∗η

=

∫Σ

f ∗ω

Example 11. Σ = S1

ω ∈ Ω1(M)

γ[ω] =

∫S1γ∗ω =

∫γ

ω

Definition 42. f, f : Σ→M are called homologous if f = ˆf i.e.∫Σ

f ∗ω =

∫Σ

f ∗ω ∀ω ∈ Ωk(M) with dω = 0

If you compare with books, They would call our cohomology "de Rham-cohomology".Only one book does everything.Bott & Tu "Differetial Forms in Algebraic Topology"See alsoMilnor "Topology from the Differentiable Viewpoint"

Example 12. LetM := (x, y) ∈ R2 | 1 ≤ x2+y2 ≤ 2 be an anulus. Then H1(M) 6= 0because H1(M) 6= 0. To see this choose

ω :=−ydx+ xdy

x2 + y2

47

then

dω = −2xdx+ 2ydy

(x2 + y2)2∧ (−ydx+ xdy) +

−dy ∧ dx+ dx ∧ dyx2 + y2

=

[2−x2 − y2

(x2 + y2)2+

2

x2 + y2

]dx ∧ dy = 0

We now integrate over γ(t) :=

(cos tsin t

)∫γ

ω =

∫ 2π

0

sin2 t+ cos2 t dt = 2π

Thus γ is not homologous to zero, i.e. γ 6= 0 because γ([ω]) 6= 0.

To prepare for the next theorem we recall some basic facts from linear algebra.

• Let V be a vector space and U ⊂ V be a linear subspace. Then

U := f ∈ V ∗ | f(u) = 0 ∀u ∈ U (147)

is called the annihilator of U .

• In finite dimensional vector spaces the dimension of U is given by

dimU = dimV − dimU (148)

• Thus we can conclude that U = V if U = 0.

Theorem 40. H1(M) is spanned by γ | γ : S1 →M smooth.

Proof. Due to the above considerations, the claim is equivalent to:If γ([ω]) = 0 for all γ : S1 →M then [ω] = 0.So let [ω] ∈ H1(M). Suppose

∫γω = 0. Then ω is exact, i.e. [ω] = 0.

Definition 43. f, f : Σ → M are called homotopic, if there is a smooth map F :[0, 1]× Σ→M such that

F (0, p) = f(p)

F (1, p) = f(p)(149)

Theorem 41. Let f, f : Σ→M are homotopic then f, f are homologous.

Proof. Furnish the disjoint boundary ∂([0, 1]×Σ) = 0×Σ∐1×Σ with the induced

orientation of [0, 1]× Σ.

48

Choose ω ∈ Ωk(M) with dω = 0. Then using the usual Stokes argument leads to

0 =

∫[0,1]×Σ

F ∗dω =

∫[0,1]×Σ

d(F ∗ω) =

∫Σ

f ∗ω −∫

Σ

f ∗ω

Hence ˆf = f .

Definition 44. M is called simply connected if every γ : S1 → M is homotopic to aconstant map.

Theorem 42. If M is simply connected then H1(M) = 0.

Proof. For a constant loop η : S1 →M we have η = 0. Due to the last theorem γ = 0 forall γ : S1 →M .

Example 13.

• Dn := x ∈ Rn | |x| ≤ 1 has H1(Dn) = 0 because it is simply connected.

• By Sarde’s lemma Sn, n ≥ 2 is simply connected and therefore H1(Sn) = 0.

Classification of 3-dimensional manifolds up to diffeomorphisms is difficult.

Example 14. Poincaré conjecture Every connected and simply connected 3-dimensionalmanifold without boundary is diffeomorphis to S3.

Surfaces

• Only S2 is simply connected without boundary.

• Only D2 is simply connected with ∂M 6= ∅.

49

• The torus

T2 = S1 × S1 = R2/Z2 (150)

is not simply connected.

γ1, γ1 form a homology basis, so H1(T2) = 2.Constant 1-forms ω = adx + bdy are pullbacks of closed forms on T2 under theprojection

Π : R2 → R2/Z2 = T2

Hence H1(T2) = adx+ bdy | a, b ∈ R.Let M, M be connected n-dimensional oriented manifolds.

Cut out a hole from M diffeomorphic to D2. Similar cut a hole in M . Glue holeboundaries to form another n-dimensional manifold M#M (connected sum of M, M).

Remark 4. The diffeomorphism type of M#M is independent of how we cut holes andhow we glue.

Theorem 43.

Hk(M#M) = Hk(M)⊕

Hk(M) (151)

Theorem 44. Every connected oriented compact 2-manifold without boundary is diffeo-morphic to one of these S2, T2, T2# · · ·#T2.

So dimH1(M) = 2g for some g ∈ N and g is called the genus of M .

50

5.1 Vorticity from the Riemannian View Point

In this section we investigate vorticity expressed in the language of cohomology. For thispurpose we recall the essential facts from section 4.3.Let M,M be compact n-dimensional oriented manifolds with boundary. M comes withdet ∈ Ω(M) and M comes with a Riemannian metric 〈 ·, · 〉. Moreover let gt : M →M bea diffeomorphism with g∗t det = det and gt = vt gt for some time dependent divergencefree vector field vt ∈ Γ(TM), i.e. divvt = 0 ⇔ Lvt det = 0.Then the equation of motion in terms of η = v[ reads

ηt + Lvtηt = −dpt

for some family of functions t 7→ pt ∈ C∞(M). Furthermore we derived

˙ηt = −dpt

where η = g∗t ηt and pt = pt gt. This means “ η modulo exact 1-forms is fixed in time"which implies

˙ω = d ˙η = 0

is fixed in time. For ωt = g−1∗t ω one can say it is

• "just advected"

• "flows with the fluid"

• "frozen in the fluis"

For each time t the euqivalence class

η + exaxt 1− forms = Ω1(M)/

Im d0(152)

determines v completely.

Theorem 45. Given u ∈ Γ(TM) then there is q ∈ C∞(M) (unique up to an additiveconstant) such that

u+ grad q ∈ sdiff(M) (153)

Proof.

u+ grad q ∈ sdiff(M)⇔

divu+ ∆q = 0

〈u+ grad q,N 〉 = 0⇔

∆q = −divu

〈 grad q,N 〉 = −〈u,N 〉

The Neumann problem for q is solvable because∫M−div u =

∫∂M−〈u,N 〉

51

Taking the vorticity

η + Imd0 7→ ω = dη

does not capture all information about η + Imd0.dη = dη only means d(η − η) = 0. In general it does not imlpy that

η + Imd0 = η + Imd0

⇔ η − η ∈ Imd0

So we want to understand how v, v ∈ sdiff(M) can be different even if ω = ω.

Definition 45. v ∈ sdiff(M) is called a harmonic vector field if dv[ = 0.

Example 15.

Theorem 46. For each cohomology class [α] ∈ H1(M) there is a unique harmonic vectorfield v ∈ sdiff(M) such that v[ = [α].

Proof. Given α ∈ Ω1(M) with dα = 0. Then there is q ∈ C∞(M) (unique up to anadditive constant) such that v = (α + dq)] ∈ sdiff(M). This v is harmonic since dv[ =0.

Definition 46. v ∈ sdiff(M) is called a potetial flow if there is q ∈ C∞(M) such thatv = grad q.

Theorem 47. There is no potential flow:

v potential flow ⇒ v = 0

Proof.

v = grad q ⇒ q solves

div grad q = 0

〈 grad q,N 〉 = 0

Then q ≡ 0. The solution is unique up to a constant. Hence v ≡ 0.

52

Remark 5. Locally every flow with no vorticity (dv[ = 0) is a potential flow, if you forgetabout tangancy to the boundary. (If the boundary is at infinity then every closed form isexact)

The question is: What are the conditions for ω ∈ Ω2(M), dω = 0 such that there existsv ∈ sdiff(M) with ω = dv[?

• v exists ⇔ ω is exact.

But, if v exists, how many different v’s are there?

• Difference of any two v’s with the same vorticity ω, is a harmonic vector field. i.e.

H1(M)↔ harmonic vector field

Theorem 48. Let Σ1, . . . ,Σk be the connected components of the boundary ∂M = Σ1 q· · · q Σk. Moreover define

α : Rk → R(x1, . . . , xk) 7→ x1 + · · ·+ xk (154)

Then the map

σ : Hk−1(M)→ Kerα

[ω] 7→

∫

Σ1ω

...∫Σkω

(155)

is well defined and surjetiv.

Proof.

• well defined: Since the connected components Σi of ∂M themself have no boundary,adding an exact η ∈ Ωk−1(M) to the representive ω does not change the integral i.e.∫

Σi

ω =

∫Σi

(ω + η)

• surjective: For each i, j ∈ 1, . . . , k choose a regular embedded curve

γij : [0, 1]→M

such that

γij(0) ∈Σi γ′ij(0) ⊥ Tγij(0)Σi

γij(1) ∈Σj γ′ij(1) ⊥ Tγij(0)Σj

53

Construct a divergence free vector field Xij supported in a neighbouhood of γij,such that

∫Σl

〈Xij, N 〉 =

−1 l = i

1 l = j

0 otherwise

For ωij := ιXij det one obtains

∫Σl

ωij =

−1 l = i

1 l = j

0 otherwise

Hence

ρ([ωij]) = eij eij =

0...0−1i0...01j0...0

Now use spaneij | i 6= j = Kerα

54

Theorem 49. If M is a domain in Rn then ρ is injective.

Corollary 3. Let M be a compact domain in Rn with smooth boundary. Then ω ∈Ωk−1(M) is exact if and only if ∫

Σ1

ω = · · · =∫

Σk

ω = 0

Example 16. dimM = 3

• Think of the earth with atmosphere.

Corollary 3. tells us, there can not be a single tornado that tuches both, the earthand the heaven.

• A spoon in a coffee cup. Vortexes arise at surfaces only.

Example 17. dimM = 2

55

Let M be compact oriented manifold with boundary Σ1, . . . ,Σk. Glue a disc to each Σi.This gives us M ' Mg. So every compact oriented surfaces with boundary arised from asurface M without boundary by deleting k disc-shaped holes. M 'Mg,k

Theorem 50. dimH1(Mg,k) = 2g + k − 1In particular, if M is a domain in R2 then dimH1(M) = k − 1.

6 Fluid Dynamics of Surfaces

Definition 47.

• A bilinear form σ : V ×W → R is called non-degenerate ifσ(X, Y ) = 0 ∀Y ∈ W ⇒ X = 0

σ(X, Y ) = 0 ∀X ∈ V ⇒ Y = 0(156)

• Let M be a manifold. Then σ ∈ Ω2(M) is called symplectic form if σp is non-degenerate for each p ∈M and dσ = 0.

Remark 6. If σ is a non-degenerated 2-form then dimM = 2k and σ ∧ · · · ∧ σ︸ ︷︷ ︸k

6= 0 is a

volume form.

Definition 48. Let σ be a symplectic form.

• A vector field X ∈ sp(M) := Y ∈ diff0(M) | LXσ = 0 is called symplectic.

• A map g : M →M is called symplectic if

g∗σ = σ

So: X is a symplectic vector field on a compact oriented manifold M if and only ifexp(tX) is symplectic for all t ∈ R.

Theorem 51.

X ∈ Γ(TM) symplectic ⇔ d(ιXσ = 0) (157)

56

Proof.

X symplectic ⇔ 0 = LXσCartan

= d(ιXσ) + ιX dσ︸︷︷︸=0

Definition 49. A vector field X ∈ sp(M) is called Hamiltonian if ιXσ = −dH for someH ∈ C∞(M).

Definition 50. Let M be a symplectic manifold, then the symplectic gradient

sgradf ∈ Γ(TM) (158)

is defined by

ιsgradfσ = −df (159)

Intuition 2. Analogoulsly one can define the ordinary gradient as

Y = grad f

the unique vector field such that

〈X, Y 〉 = df(X) ∀X ∈ Γ(TM)

So now we define

Y = sgradf

the unique vector field such that

σ(X, Y ) = df(X) ∀X ∈ Γ(TM)

Remark 7. If H1(M) = 0 (every closed 1-form is exact, like Bn or Sn) then everysymplectic vector field is Hamiltonian.

Example 18. Let M = T2 = R2/2πZ2 be a torus and σ = dx ∧ dy, X =

(ab

), a, b ∈ R

its symplectic form and vector field respectively.

H ∈ C∞(T2) H : R2 → R2

H(x+ 2π, y) = H(x, y)

H(x, y + 2π) = H(x, y)

then

−(∂H

∂xdx+

∂H

∂ydy

)(vw

)︸ ︷︷ ︸

− ∂H∂xv− ∂H

∂yw

= σ

((ab

);

(vw

))just det

= aw − bv

⇒ ∂H

∂y= −a ∂H

∂x= b

57

Thus (ab

)= −J

(∂H∂x∂H∂y

)= −JsgradH

and

0 =

∫γ2

∂H

∂y= −2πa 0 =

∫γ2

∂H

∂x= 2πb X = 0

So X is symplectic but not Hamiltonian.

Theorem 52. Let X, Y ∈ sp(M) be symplectic vector fields. Then [X, Y ] ∈ sp(M)and [X, Y ] = sgradσ(X, Y ). "The Lie bracket of two symplectic vector fields is alwaysHamiltonian".

Proof.

1.

0 = dσ(X, Y, Z) =Xσ(Y, Z) + Y σ(Z,X) + Zσ(X, Y )

− σ([X, Y ], Z)− σ([Y, Z], X)− σ([Z,X], Y )

2.

0 = (LXσ)(Y, Z) = LX(σ(Y, Z))− σ(LXY, Z)− σ(Y,LXZ)

= Xσ(X, Y )− σ([X, Y ], Z)− σ(Y, [X,Z])

3.

0 = (LY σ)(Z,X) = Y σ(Z,X)− σ([Y, Z], X)− σ(Z, [Y,X])

1.− 2.− 3. = 0 = Zσ(X, Y ) + σ(Z, [Y,X]) = (dσ(X, Y ))(Z) + σ([X, Y ], Z)︸ ︷︷ ︸ι[X,Y ]σ(Z)

ι[X,Y ]σ = −dσ(X, Y )) sgradσ(X, Y ) = [X, Y ]

58

Theorem 53. Let H, H ∈ C∞ be smooth functions and X := sgradH, X := sgradH.then the following are equivalent

1. LXH = 0

2. LXH = 0

3. σ(X, X) = 0

Proof.

0 = LXH = dH(sgradH) = −σ(sgradH, sgrad H) = · · · = −LXH

In addition we have σ(X, X) = 0 [XX] = sgrad σ(X, X) = 0.The main example in physics of a symplectic manifold is M = T ∗Q, where Q is anarbitrary manifold.

Let π : T ∗Q→ Q be the projection map and α ∈ Ω1(T ∗Q) given by

αω(X) := ω(dπ(X)︸ ︷︷ ︸∈Tπ(ω)Q

) (160)

where X ∈ TωM and ω ∈ T ∗qQ. Or in another notation

TωM 3 Xα7→ ω(dπ(X)) ∈ R (161)

Locally we have coordinates q1...qn

: U → Rn on Q

Define q1 := q1 π, . . . , qn := qn π ∈ C∞(π−1(U)). Then ω ∈ T ∗qQ

ω = p1dq1|TqQ + · · ·+ pndqn|TqQ p1, . . . , pn : π−1(U)→ R

So p1, . . . , pn, q1, . . . qn are really coordinates on π−1(Q)

α = p1dq1, . . . , pndqn

59

Theorem 54. σ := dα is a cannonical symplectic form on T ∗Q.

Proof. In the above coordinates σ = dp1 ∧ dq1, . . . , dpn ∧ dqn is non-degenerate.

Main example for H is 〈 ·, · 〉 Riemannian metric on Q (kinetic energy) and V ∈ C∞(Q)(potential energy).

Identify T ∗Q with TQ via 〈 , 〉.Theorem 55. γ : [a, b] → M = TQ is an integral curve of − sgradH if and only ifγ = π γ : [a, b]→ Q satisfies

γ′′(t) = − gradγ(t) V (162)

Proof. Let X ∈ TQ be an element of the tangent bundle. Near q choose a vector field Yon Q with

Yq = X ∇ZY = 0 ∀Z ∈ TqQ

and define

HX := dY (Z) | Z ∈ TqM

Then we split the tangent space into

TXM = TX(TqQ)︸ ︷︷ ︸vertical space

⊕HX︸︷︷︸

horizontal space

Let γ : [−ε, ε]→M be a curve in M with γ(0) = X. Then

γ′(0)︸︷︷︸∈TXM

=Dγ

dt(0)︸ ︷︷ ︸

∈TX(TqQ)

+ γH︸︷︷︸∈HX

60

We can identify TX(TqM) ' TqQ via the vector space isomorphism

dXπ|HX : HX → TqM

Define a Riemannian metric on M by demanding that VX ⊥ HX . VX ' TqM already hasa metric, HX gets a metric by asking that TX(TqM) ' TqQ is an isometrie.In other words: Both V and H as rank n vector bundles over M can be identified withE := π∗(TQ) i.e. TM = E

⊕E. So they come with the pullback of the Levi-Civita

connection.Now let ω be a tautological 1-form on Q, ω(Y ) = Y d∇ω = 0 since ∇ is torsion free.Rename π to q ”dq” ∈ Ω1(E), dq(Z) = π(Z) for Z ∈ TXM and q(X) = q sinceω(X) = X. With this notation α becomes α = 〈 p, dq 〉. Therefore σ = 〈 dp ∧ dq 〉 where

〈 η ∧ ν 〉(X, Y ) = 〈 η(X), ν(Y ) 〉− 〈 η(Y ), ν(X) 〉 (163)

We would like to determine sgradX H =

(YVYH

)So

〈X,ZV 〉+ 〈 gradq V, ZH 〉 = dXH

(ZVZH

)= −(ιYV

YH

σ)

(ZVZH

)= −σ

((YVYH

),

(ZVZH

))

= σ

((ZVZH

),

(YVYH

))= 〈ZV , YH 〉− 〈ZH , YV 〉

YH = X

YV = − gradq V

We now apply the result to γ := π γ then

γ′ = Y γ′′ = − gradγ(t) V

If dimM = 2 then the volume form det =: σ is a symplectic form and the velocityfield v ∈ sdiff(M) is symplectic. Moreover if M is a toplogical S2 or M ⊂ R2 then V isautomatically Hamiltonian.

X ∈ T∂M ω(X) = σ(v,X) = 0 ω|T∂M

Therfore we have∫

Σkω = 0 for all boundary components. Hence ω = dH and v =

− sgradH.H ∈ C∞(M) is a constant of motion.

LvH = v ·H = dH(v) = σ(v, v) = 0

61

The flow of a divergence free vector field v on a 2-dim surface is orderly. At least on S2

or planar domains

v = sgradφ

where φ is the "steam function". If Lvφ = 0 then φ is a constant of motion and the flowlines are level lines of φ almost all flow loines are closed.

A divergence free vector field v on a 3-dim manifold M can have a chaotic flow. Let

62

M = Σ× S1 and ut ∈ sdiff(Σ) for all t ∈ S1. Then define

v(q,t) :=

(ut(q)

1

)Let σ be the volume form on Σ then det = σ ∧ dt is the volume form on M .Start with

(q0

)∈M , q ∈ Σ then define

g2π

(q0

)=

(F (q)

2π(≡ 0)

)F : Σ→ Σ is called the Poincaré map of v. F is a symplectic map, since

(F ∗σ)(X, Y ) = σ(dF (X), dF (Y )) = σ(X, Y )

We are now concerned with some numericel methods.Verlet methodThe Verlet method works for equations of the form

u = f(u) (164)

In order to approximate f(nh) by un we use the mean value theorem twice

u(t) ≈u(t+ h

2)− u(t− h

2)

h≈

u(t+h)−u(t)h

− u(t)−u(t−h)h

h

=u(t+ h)− 2u(t) + u(t− h)

h2

This gives us the following recursion formula for the un:

f(un) =un−1 − 2un + un+1

h2

⇔ un+1 = 2un − un−1 − h2f(un)

If we consider that as a two dimensional system, we have(un−1

un

)F7−→(

un2un − un−1h

2f(un)

)In particular

F ′(x, y) =

(0 1−1 2− h2f ′(y)

) detF ′ = 1

Thus F is volume preserving.

Example 19. (Symplectic map F : T2 → T2)Let U : R → S1 and f : S1 → R satisfying u′′ = −f(u). For f(u) = sinu this is thependulum equation.Start with a discetisation u0, u1, . . . given by un = u(n · h) for some time step h > 0.Compare the following methods for some initial data.

• Use a Runge-Kutta method and draw 1000 points of the form(un−1

un

).

• For p :=

(u0

u1

)draw p, F (p), F 2(p), . . . , F 10000(p).

63

6.1 Sationary Flows in 3D

For stationary flows the equation of motion becomev ×w = gradα

[v,w] = 0

We distinguish between the following cases

• α is non constantRecall that the Lie derivative of a function is just the directional derivative. There-fore

Lvα = 〈 gradα, v 〉 = 〈 v ×w 〉 = 0

Thus the Bernoulli pressure α is a constant of motion.

• α is constantThen v ×w = 0, hence w = λv locally and 0 = [v, λv] = (Lvλ)v.

– λ is not constant everything is orderly

– the only possily chaotic case is:

curlv = λv with λ ∈ R

Then the vector field v is called Beltrami field.Example 20. On T3 = S1 × S1 × S1 the Beltrami field with curlv = v, divv = 0

v(x, y, z) :=

A sin z + C cos yB sinx+ A cos zC sin y +B cosx

(165)

is called the ABC-flow (Arnold-Beltrami-Childress flow).

7 Numerical Methods

1.

η + Lvη = −dp

Devide a box M into cubical voxels of side length h. Model v as the "flux 2-form"ιv det. For each oriented cube face ϕ assign a number vϕ · h supposed to model∫

ϕ

ιv det =

∫ϕ

〈 v,N 〉

ϕi,j, (i, j) ∈ Z3 differ by 1 in just 1 coordinate (neighbouring voxels) vij = −vji. Thedivergence free condition becomes

∑j vij = 0. Draw an edge eij from the center of

voxel i to the center of each neighbouring voxel j and assign a number vij · h = ηij

64

supposed to model∫eijη.

Velocity is modeled as a staggered vector field :component vx, vy, vz given on face centers facing in x, y, z-direction respectively andvϕ = 0 on boundary faces.The velocity is linearly interpolated to all points of M component wise.• Time evolution of η:Given ηt at time t. Use "time splitting" to compute ηt+δa) First solve η + Lvη = 0 over the time interval [t, tδ] with v fixed to vt.

Use the Kelvin circulation theorem. Flow edge eij backward in time over thetime step leads to the curve γij.

∫eijηt+δ =

∫γijηt.

b) Add −dp to η to reestablish divv = 0.

2.

v +∇vv = − grad p

Model v as vi ∈ R3 at each voxel center. Define the flux through face ϕij ash2 〈 vi,N 〉+ 〈 vj ,N 〉

2. In the advection step advect v as an R3-valued function.

Given a voxelcenter i, find that point p ∈M that flows to i within time δ, based onthe current vt. Then assign to i the value vt(p).

7.1 Schrödinger Flow

In order to prepare ourselves for the Schrödinger flow we derive another version of varia-tional formula.Let M be a Riemannian manifold and let Y ∈ Γ(γ∗TM) with γ = Y be a variationalvector field.

65

Then the first variational formula of the Dirichlet energy E(γ) = 12

∫ ba|γ′|2 (compare

Theorem 8) is given by

E = 〈Y, γ′ 〉 |ba −∫ b

a

〈Y, γ′ 〉

We now consider f : M → M whereM is a compact Riemannian manifold with boundary.

Let V,W vector spaces od finite dimension and A : V → W a linear map. Then wedefine

|A|2 := tr(A∗A) (166)

If v1, . . . , vn is a orthnonormal basis of V and w1, . . . , wm is a orthnonormal basis of Wthen Avj =

∑Mi=1 aijwj and therefore

|A|2 =n∑j=1

〈A∗Avj, vj 〉 =n∑j=1

〈Avj, Avj 〉 =n∑j=1

|Avj|2 =∑i,j

a2ij

On a Riemannian manifoldM , locally there exists a orthnonormal vector field X2, . . . , Xn

66

an we can write v ∈ Γ(TM) as v =∑n

j=1 vjXj. Hence

divv =∑j

〈∇Xjv,Xj 〉 =∑i,j

〈∇Xj(vjXi), Xj 〉

=∑i,j

(Xjvj) 〈Xi, Xj 〉︸ ︷︷ ︸=δij

+∑i,j

vi 〈∇XjXi, Xi 〉︸ ︷︷ ︸=divXi

=∑j

Xjvj +∑i

vi divXi

If we apply this to the gradient of a function f ∈ C∞(M) then

∆f = div grad f =∑i

XiXif +∑i

(Xif)divXi

Locally at a fixed p ∈M one can assume that (∇Xj)p = 0 and hence divXi = 0. So

∆f =∑i

XiXif

More genarally let E qa vector bundle over M with Levi Civita connection ∇. Then wedefine for Ψ ∈ Γ(E)

∆Ψ =∑i

∇Xi∇XiΨ +∑i

(divXi)∇XiΨ (167)

This is independent of the the choice of Xi, . . . , Xn.If f : M → M is defined as above and ∇ is the Levi Civita connection on M . Then

∆f ∈ Γ(f ∗TM)

∆f =∑

i ∇Xidf(Xi) +∑

i(divXi)df(Xi)(168)

where ∇ = f ∗∇. ∆f is called the tension field of f .

67

We now consider a variation f = Y . Thend

dt

∣∣∣t=0E(ft) =

d

dt

∣∣∣t=0

1

2

∫M

∑i

〈 df(Xi), df(Xi) 〉

=∑i

∫M

〈 ∂∂tXif, df(Xi) 〉

=∑i

∫M

〈 ∇XiY, df(Xi) 〉

=∑i

∫M

(Xi 〈Y, df(Xi) 〉− 〈Y, ∇Xidf(Xi) 〉

)=∑i

∫M

Xi 〈 df ∗(Y ), Xi 〉−∫M

〈Y,∆f 〉

=

∫M

div(df ∗(Y ))−∫M

〈Y,∆f 〉

div Thm=

∫∂M

〈(df ∗(Y ), N 〉−∫M

〈Y,∆f 〉

=

∫∂M

〈Y, df ∗(N) 〉−∫M

〈Y,∆f 〉

Definition 51. f : M → M is called a harmonic map if f is a critical point of theDirichlet energy with respect to a variation supported in M , i.e. ∆f = 0.Example 21. Let M be a manifold of dimension n. Then an immersion f : M → Rn+1

has constan mean curvature if and only if the Gauss-map N : M → Sn is harmonic.Definition 52. A family of smooth maps ft : M → M , t ∈ [a, b] solves the heat equationif t 7→ ft is an integral curve of − gradE.Example 22. In case of M = S1 use the heat equation to prove: If M is compact withsectional curvature K ≤ 0 then each homotopy class of loops in M contains a a uniqueclosed geodesic.

68

A complex vector space V cqan be viewed as a real vector space V together withJ ∈ End(V ) with J2 = −id.Definition 53. An almost complex manifold is a manifold M together with J ∈ End(V )with J2 = −id.Definition 54. An almost complex manifold which is also Riemannian is called almostKähler id J∗ = −JRemark 8. In an almost Kähler manifold holds

〈 JX, JY 〉 = 〈X, J∗JY 〉 = 〈X, Y 〉

Definition 55. An almost Kähler manifold is called Kähler if ∇J = 0, i.e. ∇X(JY ) =J∇XY for all X, Y ∈ Γ(TM).

Theorem 56. M Kähler ⇒ M complex.

Example 23. Every oriented 2-manifold is Káhler.

On every Kähler manifold there is σ ∈ Ω2(M) defined by

σ(X, Y ) = 〈 JX, Y 〉 (169)

This σ is symplectic, since it is obviouly skew symmetric and non-degenerate. Moreoverfor parallel vector fields (∇X)p = (∇Y )p = (∇Z)p = 0 we have

(dσ(X, Y, Z))p = (Xσ(Y, Z) + Y σ(Z,X) + Zσ(x, y))p

= (X 〈 JY, Z 〉+Y 〈 JZ,X 〉+Z 〈 JX, Y 〉)p= (〈∇X(JY ), Z 〉+ 〈 JY,∇XZ 〉︸ ︷︷ ︸

=0

+ 〈∇Y (JZ), X 〉+ 〈 JZ,∇XZ 〉︸ ︷︷ ︸=0

+ · · · )p

= (〈 J ∇XY︸ ︷︷ ︸=0

, Z 〉+(〈 J ∇YZ︸ ︷︷ ︸=0

, X 〉+(〈 J ∇ZX︸ ︷︷ ︸=0

, Y 〉)p = 0

if M is Kähler then also M = C∞(M, M) is Kähler. For Y ∈ TfM = Γ(f ∗TM) JY isalready defined and also for Y, Z ∈ TfM

σ(Y, Z) =

∫M

σ(Y, Z) (170)

With some work σ is closed in a appropriate sense.The symplectic gradient flow of the Dirichlet energy E

f = J∆f

iψ = −∆ψ(171)

gives us the Schrödinger equation.

Example 24.

• M = Cm ordinary Schrödinger equation.

• M = S2 Then for Y ∈ TSS2 we have −J = S × Y and we obtain

S = S ×∆S (172)

the Landau-Lifschitz equation.

69

7.2 Incompressible Schrödinger Flow

Let M be a Riemannian manifold and let ρ : M → R be a density. Moreover we defineψ : M → Cn satisfying the Schrödinger equation ψ = i

2∆ψ and ρ = |ψ|2. Via the

Madelung transform one can define the velocity

η =〈 dψ, iψ 〉|ψ|2

(173)

where v = η] and 〈 ·, · 〉 : Cn × Cn → R is the ordinary scalar product.

Theorem 57.

(ρ det)· + Lv(ρ det) = 0

⇔ ρ det + 〈 grad ρ, v 〉 det +ρ(divv) det = 0

⇔ ρ+ div(ρv) = 0 (174)

Proof.

ρ = 2 〈 ψ, ψ 〉 = 〈 i∆ψ, ψ 〉 = −〈∆ψ, iψ 〉

div(ρv) = div(〈 dψ, iψ 〉]) =∑j

Xj 〈 dψ(Xj)︸ ︷︷ ︸Xjψ

, iψ 〉

=∑j

〈XjXjψ, iψ 〉+ 〈Xjψ, iXjψ 〉︸ ︷︷ ︸=0

= 〈∆ψ, iψ 〉

ρ+ div(ρv) = 0

We want incompressible fluids. So we impose the constraint |ψ| = 1 ( divv = 0).Suppose at time t = 0 we have ψ with |ψ| = 1 and 0 = divv = 〈∆ψ, iψ 〉. Then ψ = i

2∆ψ

automatically implies ρ = 0.Evolve ψ over the time intervall [t, t + δ] according to ψ = i

2∆ψ. Afterwards repair

constraints: ψ ψ|ψ| .

To repair the constraint divv = 0 we replace ψ by eiqψ with q ∈ C∞(M) and define thenew velocity by

η = 〈 d(eiqψ), ieiqψ 〉 = 〈 idqψ + dψ, iψ 〉= η + dq

therefore

v = v + grad q and div v = div v + ∆q

70

So div v = 0 if ∆q = −div v.In case of ∂M = ∅ this determines q up to a constant.So the final ψ is determined up to ψ 7→ eiαψ, α ∈ R. Indeed this is exactly pressureprojection.

In praxis we use ψ : M → S3 ⊂ C2 and define the Hopf map

π : S3 → S2

ψ 7→

ψtσ1ψψtσ2ψψtσ3ψ

=

ψ1ψ2 + ψ2ψ11iψ1ψ2 − ψ2

|ψ1|2 + |ψ2|2

=: s ∈ R3 (175)

where

σ1 :=

(0 11 0

)σ2 :=

(0 −ii 0

)σ3 :=

(1 00 −1

)(176)

are the Pauli matrices.Then the vorticity 2-form ω = ιw det where w = curl v or simply ω = dη is given by

ω =1

2s∗dAS2 (177)

71