MATH 8230 LECTURE NOTES, SPRING 2015 · 2018. 9. 11. · MATH 8230 LECTURE NOTES, SPRING 2015 3...

MATH 8230 LECTURE NOTES, SPRING 2015

MIKE USHER

1. AREAS AND THE CHANGE OF VARIABLES FORMULA

The start of this course will be about area-preserving diffeomorphisms of R2 and of more generaltwo-manifolds; this will later be our entry point into symplectic geometry. The first question is, then,given a diffeomorphism φ : R2→ R2 (i.e., a smooth1 map which is a bijection and whose inverse isalso smooth), how do we tell what effect this map has on areas? First we will work this out in therather special case when the map φ is linear; this is useful for the more general case since to saythat φ is differentiable is essentially to say that at small scales it behaves, to good approximation,as though it were linear.

To be perfectly precise I should say what the “area” of a subset of R2 is. Naturally enough wewill use two-dimensional Lebesgue measure (or maybe one should say “outer measure” to havesomething defined on all subsets, though in practice we will not have any reason to take measuresof sets which are not Borel)—in other words we first define the area of a rectangle R= (a, b)×(c, d)for a ≤ b and c ≤ d to be (b− a)(d − c) and then define Area(S) to be the infimum over countablecovers S ⊂ ∪∞i=1Ri by rectangles of the sum

∑∞i=1 Area(Ri). It is not hard to see that standard

formulas from Euclidean geometry about the areas of polygons are consistent with this definition.Here is a simple fact about vectors with which you may already be familiar.

Proposition 1.1. Given two-dimensional vectors ~v = (v1, v2) and ~w = (w1, w2) define the parallelo-gram spanned by ~v and ~w to be

Π(~v, ~w) = {s~v + t ~w|0≤ s ≤ 1, 0≤ t ≤ 1}

Then

Area(Π(~v, ~w)) = |v1w2 − v2w1|

Proof. Where θ is the angle between ~v and ~w we have Area(Π(~v, ~w)) = ‖~v‖‖~w‖ sinθ , since theparallelogram has base ‖~v‖ and altitude ‖~w‖ sinθ . Meanwhile recall the standard dot productformula ~v · ~w= ‖~v‖‖~w‖ cosθ . So we obtain

Area(Π(~v, ~w))2 = ‖~v‖2‖~w‖2(1− cos2 θ ) = (v21 + v22 )(w

21 +w

22)− (~v · ~w)

2

= (v21 + v22 )(w

21 +w

22)− (v1w1 + v2w2)

2

= v21 w22 + v

22 w

21 − 2v1w1v2w2 = (v1w2 − v2w1)

2,

whereupon taking the square root of both sides implies the result. �

Corollary 1.2. Let Π = Π(~v, ~w) be any parallelogram and let A: R2 → R2 be a linear map. ThenArea(A(Π)) = |det A|Area(Π).

1As is customary, “smooth” means the same thing as C∞; for the most part the assumption of infinite differentiabilitywill just be a convenience and C∞ could be replaced by Ck for some low value of k like 1 or 2, but usually not by C0

1

2 MIKE USHER

Proof. Since A is linear we have

A(Π) = {sA~v + tA~w|0≤ s ≤ 1,0≤ t ≤ 1}= Π(A~v, A~w)

Now where P =�

~v ~w�

(i.e. P is the 2 × 2 matrix whose first column is ~v and whose secondcolumn is ~w), we have AP =

�

A~v A~w�

. Now the preceding proposition can be rephrased assaying that Area(Π) = |det P| and likewise that Area(A(Π)) = |det(AP)|, so by the multiplicativityof the determinant we indeed have

Area(A(Π)) = |det(A)det(P)|= |det A|Area(Π)

�

Corollary 1.3. If S ⊂ R2 is any subset and A is a linear map then Area(A(S)) = |det A|Area(S).

Proof. If det A= 0 then the image of A is contained in a line, which is easily seen to have measurezero, so the corollary certainly holds in this case. So assume that det A 6= 0.

For one inequality, given ε > 0 let {Ri}∞i=1 be a sequence of rectangles with S ⊂ ∪iRi and∑

i Area(Ri) ≤ Area(S) +ε

2|det A| . For each i, A(Ri) is a parallelogram of area |det A|Area(Ri), sothere is a sequence of rectangles {Ci j}∞j=1 with ARi ⊂ ∪ jCi j and

∑

j Area(Ci j) ≤ Area(ARi) +ε

2i+1 .Then the entire countable family {Ci j}∞i, j=1 collectively covers A(S) and so

Area(A(S))≤∑

i, j

Area(Ci j)≤∑

i

�

Area(ARi) +ε

2i+1

�

=ε

2+ |det A|

∑

i

Area(Ri)≤ε

2+ε

2+ |det A|Area(S).

So since ε > 0 was arbitrary this shows that Area(A(S))≤ |det A|Area(S). For the reverse inequality,just note that since we are assuming det A 6= 0, A is invertible, and what we have already done showsthat

Area(S) = Area(A−1(A(S)))≤ |det A−1|Area(A(S)) =1

|det A|Area(A(S))

�

So a linear map affects areas by multiplying them by the absolute value of its determinant, andin particular a linear map will be area-preserving if and only if its determinant is ±1.

We now consider a general smooth map φ : R2 → R2. Let ~x0 = (x0, y0) ∈ R2. There is then aderivative dφ~x0 : R

2 → R2; by definition this means that dφ~x0 is a linear map having the propertythat

lim~v→0

f (~x0 + ~v)− f (~x0)− dφ~x0 ~v

‖~v‖= 0

The linear map dφ~x0 is represented by the matrix

∂ φ1∂ x

∂ φ1∂ y

∂ φ2∂ x

∂ φ2∂ y

�

�

�

�

�

~x0

where we write φ in coordinates as φ(x , y) = (φ1(x , y),φ2(x , y)).Recall also that given two smooth mapsψ,φ : R2→ R2 the chain rule is most concisely expressed

by the formulad(ψ ◦φ)~x0 = (dψφ(x0)) ◦ (dφx0).

Theorem 1.4. Let ε > 0. Then there is δ > 0 such that if ∆x ,∆y ∈ [0,δ] then�

�Area(φ([x0 −∆x/2, x0 +∆x/2]× [y0 −∆y/2, y0 +∆y/2]))− |det dφx0 |∆x∆y�

�< ε‖(∆x ,∆y)‖2

MATH 8230 LECTURE NOTES, SPRING 2015 3

Proof. Step 1: We prove the theorem in the special case that dφ~x0 is the identity.It then follows from the definition of dφ~x0 that there is δ > 0 such that if h, k ∈ [−δ/2,δ/2] then(1)|φ1(x0 + h, y0 + k)− (φ1(x0, y0) + h)| ≤ ε‖(h, k)‖ |φ2(x0 + h, y0 + k)− (φ2(x0, y0) + k)| ≤ ε‖(h, k)‖

So if ∆x ,∆y ∈ [0,δ] and |h| ≤∆x/2 while |k| ≤∆y/2, we in particular obtain that

φ1(x0 + h, y0 + k) ∈ [φ1(x0, y0)−∆x/2− ε‖(∆x ,∆y)‖/2,φ1(x0, y0) +∆x/2+ ε‖(∆x ,∆y)‖/2]

and likewise

φ2(x0 + h, y0 + k) ∈ [φ2(x0, y0)−∆y/2− ε‖(∆x ,∆y)‖/2,φ2(x0, y0) +∆y/2+ ε‖(∆x ,∆y)‖/2]

Thus

φ([x0 −∆x/2, x0 +∆x/2]×[y0 −∆y/2, y0 +∆y/2]) ⊂[φ1(x0, y0)−∆x/2− ε‖(∆x ,∆y)‖/2,φ1(x0, y0) +∆x/2+ ε‖(∆x ,∆y)‖/2]× [φ2(x0, y0)−∆y/2− ε‖(∆x ,∆y)‖/2,φ2(x0, y0) +∆y/2+ ε‖(∆x ,∆y)‖/2](2)

and since areas are monotone this implies that

Area(φ([x0 −∆x/2, x0 +∆x/2]× [y0 −∆y/2, y0 +∆y/2]))≤ (∆x + ε‖(∆x ,∆y)‖)(∆y + ε‖(∆x ,∆y)‖)

=∆x∆y + ε(∆x +∆y)‖(∆x ,∆y)‖+ ε2‖(∆x ,∆y)‖2

Modulo replacing ε by a suitable multiple of itself this proves the upper bound forArea(φ([x0 −∆x/2, x0 +∆x/2]× [y0 −∆y/2, y0 +∆y/2])) required in the theorem; it remainsto prove the lower bound. For this we claim that, dually to (2), we have

φ([x0 −∆x/2, x0 +∆x/2]×[y0 −∆y/2, y0 +∆y/2]) ⊃[φ1(x0, y0)−∆x/2+ ε‖(∆x ,∆y)‖/2,φ1(x0, y0) +∆x/2− ε‖(∆x ,∆y)‖/2]× [φ2(x0, y0)−∆x/2+ ε‖(∆x ,∆y)‖/2,φ2(x0, y0) +∆y/2− ε‖(∆x ,∆y)‖/2](3)

Since the area of the rectangle on the right hand side differs from ∆x∆y by at most an amountproportional to ε‖(∆x ,∆y)‖2 this would complete Step 1.

To do this, observe that, for |h| ≤∆x/2 and |k| ≤∆y/2, we have, using (1),

φ2(x0 + h, y0 −∆y/2)≤ φ2(x0, y0)−∆y/2+ ε‖(∆x ,∆y)‖/2φ1(x0 +∆x/2, y0 + k)≥ φ1(x0, y0) +∆x/2− ε‖(∆x ,∆y)‖/2φ2(x0 + h, y0 +∆y/2)≥ φ2(x0, y0) +∆y/2− ε‖(∆x ,∆y)‖/2φ1(x0 −∆x/2, y0 + k)≤ φ1(x0, y0)−∆x/2+ ε‖(∆x ,∆y)‖/2

Let γ: S1 → R2 be the a loop which goes once counterclockwise around the boundary of [x0 −∆x/2, x0 + ∆x/2] × [y0 − ∆y/2, y0 + ∆y/2], beginning at (x0 − ∆x/2, y0 − ∆y/2). Considerany point (x1, y1) such that φ1(x0, y0) − ∆x/2 + ε‖(∆x ,∆y)‖/2 ≤ x1 ≤ φ1(x0, y0) + ∆x/2 −ε‖(∆x ,∆y)‖/2 andφ2(x0, y0)−∆y/2+ε‖(∆x ,∆y)‖/2≤ y1 ≤ φ2(x0, y0)+∆y/2−ε‖(∆x ,∆y)‖/2.The above four inequalities then imply that the loop φ ◦ γ has winding number 1 around thepoint (x1, y1).

2 I now claim that this implies that (x1, y1) lies in the image φ([x0 −∆x/2, x0 +∆x/2] × [y0 − ∆y/2, y0 + ∆y/2]). Indeed if this were not the case then φ would give a map[x0 −∆x/2, x0 +∆x/2] × [y0 −∆y/2, y0 +∆y/2] → R2 \ {(x1, y1)}, and the winding number

2In other words, if the map t 7→ φ(γ(t))− (x1, y1) is represented in polar coordinates as t 7→ (r(t) cosθ (t), r(t) sinθ (t))for continuous functions r,θ , then θ (2π)− θ (0) = 2π. The point is that the loop in question moves in the lower half-planefrom the third to the fourth quadrant, then in the right half-plane from the fourth to the first quadrant, then in the upperhalf-plane from the first to the second quadrant, and finally in the left half-plane from the second back to the third quadrant.This implies that it makes one net counterclockwise rotation around the origin.

4 MIKE USHER

computation implies that the induced map on π1 would send γ to an element representing a gen-erator for π1(R2 \ {(x1, y1)}). But of course γ represents the trivial element of π1([x0−∆x/2, x0+∆x/2]× [y0 −∆y/2, y0 +∆y/2]) and so this is a contradiction. So our arbitrary element (x1, y1)belongs to φ([x0−∆x/2, x0+∆x/2]×[y0−∆y/2, y0+∆y/2]), verifying the claim (3) and hencecompleting the proof of Step 1.

Step 2: We deduce the general case of the theorem from Step 1.Since we are assuming that φ is a diffeomorphism there is ψ: R2 → R2 such that ψ ◦ φ is theidentity. Let A= dψφ(x0) (so A is a linear map), and define φ̃ = A ◦φ. Then the chain rule showsthat dφ̃~x0 is the identity, and so Step 1 applies to prove that, given ε > 0, there is δ > 0 such thatif ∆x ,∆y ∈ [0,δ] then(4)�

�Area(φ̃([x0 −∆x/2, x0 +∆x/2]× [y0 −∆y/2, y0 +∆y/2]))−∆x∆y�

�< ε|det A|‖(∆x ,∆y)‖2

By Corollary 1.3, since φ̃([x0 − ∆x/2, x0 + ∆x/2] × [y0 − ∆y/2, y0 + ∆y/2]) = A(φ([x0 −∆x/2, x0 +∆x/2]× [y0 −∆y/2, y0 +∆y/2]) we have

Area(φ([x0 −∆x/2, x0 +∆x/2]×[y0 −∆y/2, y0 +∆y/2]))

=1

|det A|Area(φ̃([x0 −∆x/2, x0 +∆x/2]× [y0 −∆y/2, y0 +∆y/2])),

so multiplying both sides of (4) by 1|det A| (which by the chain rule is equal to |det dφ~x0 |) immediatelyyields the result. �

Remark 1.5. Inspection of the proof shows that the maximal allowable value for the parameter δ inTheorem 1.4 can be bounded from below in terms of the magnitudes of |det dφ~x0 | and of the matrixof second derivatives of φ on the rectangle under consideration. In particular for a given compactregion K one can find a single parameter δ obeying the conclusion of Theorem 1.4 for all ~x0 ∈ K .

We can now determine precisely how a smooth map affects areas of rectangles.

Theorem 1.6. Let R = [a, b]× [c, d] ⊂ R2 be a compact rectangle and let φ : R2 → R2 be a diffeo-morphism. Then

Area(φ(R)) =

∫∫

R

|det dφ(x ,y)|d xd y

More generally, for any continuous function f : R2→ R,∫∫

φ(R)f (u, v)dudv =

∫∫

R

f (φ(x , y))|det dφ(x ,y)|d xd y

Proof. Let ε > 0. Since φ(R) is compact, there is M ∈ R with | f (u, v)| ≤ M for all (u, v) ∈ φ(R), andthere is δ > 0 such that | f (u1, v1)− f (u2, v2)| ≤

ε4Area(φ(R)) whenever ‖(u1−u2, v1− v2)‖< δ. Since

R is compact, φ and ( f ◦φ)|det dφ| are both uniformly continuous, and so there is η > 0 such thatboth ‖φ(x1, y1)−φ(x2, y2)‖ < δ and

�

�( f (φ(x1, y1))|det dφ(x1,y1)| − f (φ(x2, y2))|det dφ(x2,y2)|�

� <ε

4Area(R) whenever ‖(x1 − x2, y1 − y2)‖< η.Divide the rectangle R into a grid of subrectangles {Ri}Ni=1 (intersecting each other only along

their boundaries) such that for each i the rectangle Ri has the following properties:

(i) Any two points in Ri lie a distance at most η apart.(ii) The length ∆x i and width ∆yi of Ri obey

12 ≤

∆yi∆x i≤ 2; note that this implies that

‖(∆x i ,∆yi)‖2 ≤ 4∆x∆y .


(iii) Where pi is the point at the center of Ri ,�

�Area(φ(Ri))− |det(dφ)pi |Area(Ri)�

�<ε

2MArea(R)Area(Ri)

The fact that the last property can be arranged follows readily from Theorem 1.4, the observationat the end of (ii) above, and Remark 1.5.

We then have, using (i) and the bound on the rate of change of ( f ◦φ)|det dφ|,(5)�

�

�

�

�

∫∫

R

f (φ(x , y))|det dφ(x ,y)|d xd y −N∑

i=1

f (φ(pi))|det dφpi |Area(Ri)

�

�

�

�

�

≤N∑

i=1

ε

4Area(R)Area(Ri) =

ε

4

while (using that (i) and the choice of η imply that f varies by at most ε4Area(φ(R)) over each φ(Ri))

(6)

�

�

�

�

�

∫∫

φ(R)f (u, v)d xd y −

N∑

i=1

f (φ(pi))Area(φ(Ri))

�

�

�

�

�

≤ε

4

Furthermore, by (iii),

(7)

�

�

�

�

�

N∑

i=1

f (φ(pi))|det dφpi |Area(Ri)−N∑

i=1

f (φ(pi))Area(φ(Ri))

�

�

�

�

�

< Mε

2MArea(R)Area(R) =

ε

2.

Together (5),(6), and (7) show that�

�

�

∫∫

φ(R) f (u, v)dudv =∫∫

R f (φ(x , y))|det dφ(x ,y)|d xd y�

�

� <

ε. Since ε > 0 was arbitrary the theorem follows. �

Corollary 1.7 (Change of variables formula). For a diffeomorphism φ : R2 → R2, a compact set K,and a continuous function f : R2→ R2 we have

∫∫

φ(K)f (u, v)dudv =

∫∫

K

f (φ(x , y))|det dφ(x ,y)|d xd y

Proof. Approximate K to within a subset of arbitrarily small measure by the union of a family ofrectangles {Ri}Ni=1 with disjoint interiors, apply Theorem 1.6 to each Ri , and add. (Details are leftto the reader.) �

This gives a complete answer to the question of when a diffeomorphism φ : R2 → R2 is area-preserving (i.e., has the property that Area(φ(K)) = Area(K) for every compact K): φ is area-preserving if and only if the function x 7→ |det dφ~x | is identically equal to 1. (Indeed, the backwardimplication is a special case of Corollary 1.7; for the forward implication, if there is some ~x where|det dφ~x | 6= 1 then if we take K to be a sufficiently small closed disk centered at ~x the continuity ofdφ together with Corollary 1.7 show that Area(φ(K)) 6= Area(K).) Evidently a “randomly-chosen”map φ is unlikely to have the property that |det dφ| = 1 everywhere; however we will see laterhow to use flows of appropriate vector fields to produce a very large family of examples of such φ.

Remark 1.8. While I haven’t explicitly incorporated this in order to avoid cluttering the statements,it should be clear from the proofs that Corollary 1.7 applies equally well if φ : U → V is a diffeo-morphism between two open subsets of R2 where K ⊂ U , and if f is just defined on a neighborhoodof K .

Example 1.9. Define φ : (0,∞) × (−π,π) → R2 \ {(u, 0)|u ≤ 0} by φ(r,θ ) = (r cosθ , r sinθ ).This is a diffeomorphism which transforms the polar coordinates of a point to its Cartesian coordinates

6 MIKE USHER

(we delete the negative horizontal axis to make the map one-to-one). We see that, for any (r,θ ) ∈(0,∞)× (−π,π),

dφ(r,θ ) =

cosθ −r sinθsinθ r cosθ

and so det dφ(r,θ ) = r(cos2 θ + sin2 θ ) = r. So the change of variables formula recovers the familiar

formula∫∫

φ(R)f (u, v)dudv =

∫∫

R

f (r cosθ , r sinθ )rdrdθ

for integration in polar coordinates.

Remark 1.10. Although we’re focusing on dimension two, Corollary 1.7 is also true (after obviousnotational changes) for diffeomorphisms φ : Rn → Rn for any n. Almost all of the above proofextends in a straightforward way to prove this; the only real exception is the proof of (3), whichused a winding number argument to show that the image of a rectangle under a diffeomorphismhad to contain certain points. In higher dimensions one could try to make a similar argument usingdegree theory, though this seems more difficult. Probably a more efficient approach (which I’ll leaveto interested readers to work out the details of for themselves) would be based on the followingexercise (which is closely related to the inverse function theorem; indeed the exercise is also usefulin one proof of that theorem).

Exercise 1.11. Let φ : Rn→ Rn be a smooth map such that φ(~0) = ~0 and dφ~0 is the identity. Provethat there is δ > 0 such that if ‖~y‖ < δ then the sequence {~xn}∞n=0 defined recursively by ~x0 = ~0 and~xn+1 = ~xn+ ~y −φ(~xn) converges to some ~x with φ(~x) = y. (Hints: You should probably take δ smallenough that dφ remains close to the identity throughout the ball of radius (say) 3δ, as is possible sincedφ is continuous. Show that ~xn is a Cauchy sequence using an approach similar to that in the proof ofthe Contraction Mapping Principle—or perhaps just apply the Contraction Mapping Principle directlyif you can arrange to satisfy its hypotheses.)

2. INTEGRATION OF DIFFERENTIAL FORMS

The following is a key notion in differential geometry and topology, and in particular in thiscourse.

Definition 2.1. Let M be a smooth manifold and k ≥ 0. A differential k-form ω on M is a rulewhich assigns to each p ∈ M a map ωp : (Tp M)k → R which is:

• k-linear, i.e. for v1, . . . , vk, w ∈ Tp M and c ∈ R we have, for any 1≤ i ≤ k,

ωp(v1, . . . , vi−1, cvi +w, vi+1, . . . , vk) = cωp(v1, . . . , vi , . . . , vk) +ωp(v1, . . . , w, . . . , vk)

• alternating, i.e. for 1≤ i < j ≤ k,

ωp(v1, . . . , vi , . . . , v j , . . . , vk) = −ωp(v1, . . . , v j , . . . , vi , . . . , vk)

(so switching the ith and jth entries without altering the others reverses the sign).• smooth in the sense that if X1, . . . , Xk are (smooth) vector fields on M , the function p 7→ωp(X1(p), . . . , Xk(p)) is a C∞ function on M .

We will write Ωk(M) for the set of differential k-forms on M .

(When k = 0 the above should just be interpreted as saying that a 0-form is a C∞ function onM .)

The most important special case for this course will be k = 2 (in particular, a symplectic manifoldis a smooth manifold equipped with a particular kind of differential 2-form). For the moment let


us just consider 2-forms on R2. The standard area form on R2 is by definition the 2-form d x ∧ d ydefined by

(d x ∧ d y)p((a, b), (c, d)) = ad − bc(we always use the tautological identification of TpR2 with R2 for all p ∈ R2. The reason forthe notation d x ∧ d y will be indicated later.) It is easy to see that this satisfies the axioms for adifferential 2-form. It is also easy to see that, quite generally, if ω is a differential k-form on asmooth manifold M , and if f ∈ C∞(M), then fω is also a differential k-form (where fω is definedby ( fω)p = f (p)ωp). So we have a family of differential 2-forms on R2 defined by f d x ∧ d y forany f ∈ C∞(R2). In fact any differential 2-form on R2 can be obtained by this procedure:

Proposition 2.2. If U ⊂ R2 is an open subset and ω ∈ Ω2(U) then there is a function f ∈ C∞(R2)such that ω= f d x ∧ d y.

Proof. For p ∈ U define f (p) = ωp((1,0), (0,1)). By the smoothness condition for ω, f ∈ C∞(U).Moreover the bilinearity condition on ω implies that, for p ∈ U and (a, b), (c, d) ∈ TpU ∼= R2,

ωp((a, b), (c, d)) =ωp((a, 0), (c, d)) +ωp((0, b), (c, d))

=ωp((a, 0), (c, 0)) +ωp((a, 0), (0, d)) +ωp((0, b), (c, 0)) +ωp((0, b), (0, d))

= acωp((1,0), (1, 0)) + adωp((1, 0), (0,1)) + bcωp((0,1), (1, 0)) + bdωp((0, 1), (0,1))

Now our definition was that ωp((1,0), (0,1)) = f (p), and then the alternating condition showsωp((0,1), (1,0)) = − f (p). Meanwhile for any v ∈ TpR2 the alternating condition shows thatωp(v, v) = −ωp(v, v) and hence that ωp(v, v) = 0. So the other terms above vanish and we have

ωp((a, b), (c, d)) = (ad − bc) f (p) = f (p)(d x ∧ d y)p((a, b), (c, d))

Since p, (a, b), (c, d) were arbitrary this confirms that ω= f d x ∧ d y . �

Definition 2.3. Let M and N be smooth manifolds, ω ∈ Ωk(N), and φ ∈ C∞(M , N) a smooth mapfrom M to N . The pullback of ω by φ is the differential form φ∗ω ∈ Ωk(M) defined by, for p ∈ M ,

(φ∗ω)p(v1, . . . , vk) =ωφ(p)(dφp v1, . . . , dφp vk)

for p ∈ M and v1, . . . , vk ∈ Tp M , where dφp : Tp M → T f (p)N is the derivative of φ at p.

It is easy to see from the chain rule d(ψ ◦φ)p = dψφ(p) ◦ dφp that pullback of differential formssatisfies the contravariance property

(ψ ◦φ)∗ω= φ∗ψ∗ω

when φ : M → N and ψ: N → P are smooth maps and ω ∈ Ωk(P).

Proposition 2.4. Let U , V ⊂ R2 be open subsets and φ : U → V a smooth map. Then for f ∈ C∞(V )we have

φ∗( f d x ∧ d y) = ( f ◦φ)(det dφ)d x ∧ d y

Proof. Write φ(p) = (φ1(p),φ2(p)) for p ∈ U . We have

(φ∗( f d x ∧ d y))p ((1,0), (0,1)) = ( f d x ∧ d y)φ(p)�

dφp(1, 0), dφp(0,1)�

= f (φ(p))(d x ∧ d y)φ(p)

∂ φ1∂ x

,∂ φ2∂ x

,

∂ φ1∂ y

,∂ φ2∂ y

= f (φ(p))

∂ φ1∂ x

∂ φ2∂ y−∂ φ1∂ y

∂ φ2∂ x

= ( f ◦φ)(det dφ)(p)

So evaluating at ((1,0), (0, 1)) and using Proposition 2.2 shows thatφ∗( f d x∧d y) = ( f ◦φ)det dφd x∧d y . �

8 MIKE USHER

Proposition 2.4 should be reminiscient of the change of variables formula (Corollary 1.7); thereis however the matter of the absolute value that appears in Corollary 1.7 but not in Proposition 2.4.This partly motivates the following definition:

Definition 2.5. Let U and V be open subsets of Rn, φ : U → V a smooth map, and p ∈ U .• φ is called orientation-preserving at p if det dφp > 0.• φ is called orientation-reversing at p if det dφp < 0• φ is called singular at p if det dφp = 0.

φ is simply called orientation-preserving (resp. orientation-reversing) if it is orientation preservingat p for all p ∈ U (resp. orientation-reversing at p for all p ∈ U).

Here is a simple observation:

Proposition 2.6. If φ : U → V is a diffeomorphism between two open subsets of Rn and if U isconnected then either φ is orientation-preserving or φ is orientation-reversing.

Proof. By the chain rule, for each p ∈ U d(φ−1)φ(p) is an inverse to dφp and so det dφ never takesthe value zero on U . So p 7→ det dφp is a continuous map from the connected set U to R \ {0} andthus is either everywhere positive or everywhere negative. �

Definition 2.7. If U ⊂ R2 is open, K ⊂ U is compact, and ω ∈ Ω2(U), we define∫

K

ω=

∫∫

K

f (x , y)d xd y

where the function f : U → R2 is chosen such that ω= f d x ∧ d y .

This definition causes the change of variables formula to take a very elegant form, at least fororientation-preserving diffeomorphisms.

Corollary 2.8 (Change of variables formula, differential forms version). If U , V ⊂ R2 are open,φ : U → V is a diffeomorphism, ω ∈ Ω2(V ), and K ⊂ U is compact, then

∫

K

φ∗ω=

¨ ∫

φ(K)ω if φ is orientation-preserving−∫

φ(K)ω if φ is orientation-reversing

Proof. This follows directly from Corollary 1.7, Proposition 2.4, and the definitions. �

Remark 2.9. Analogues of the above results hold for all dimensions n, not just 2. There is a standardn-form d x1∧· · ·∧ d xn whose value on an n-tuple of tangent vectors v1, . . . , vn is the determinant ofthe n× n matrix whose columns are the vi (in order). In linear algebra you may have learned thatall alternating, n-linear forms on an n-dimensional vector space are multiples of the determinant;this leads to the conclusion that every differential n-formω on Rn can be written as f d x1∧· · ·∧d xnwhere f ∈ C∞(Rn) can be obtained by evaluating ω on the standard basis of vector fields. Giventhis, the proofs of Proposition 2.4 and Corollary 2.8 extend to pullbacks of n-forms on Rn, wherewe use the obvious definition that the integral of an n-form f d x1 ∧ · · · ∧ d xn over a compact set Kis just the Lebesgue integral of f .

In addition to yielding an elegant formulation for the change of variables formula, the languageof differential forms is well-suited for adapting the notion of integration to general oriented smoothmanifolds. First we define this notion:

Definition 2.10. An oriented smooth manifold of dimension n is a second-countable Hausdorff spaceM equipped with an oriented atlas, i.e., a collection {φα : Uα→ Rn} where:

• The sets Uα form an open cover of M .


• The maps φα : Uα→ Rn are homeomorphisms to their images φα(Uα), where each φα(Uα)is an open subset of Rn.

• The transition functions φβ ◦φ−1α : φα(Uα∩Uβ )→ φβ (Uα∩Uβ ) are orientation-preservingdiffeomorphisms.

Definition 2.11. If M is an oriented smooth manifold of dimension n, U ⊂ M is open, an orientedchart for M is a map φ : U → Rn such that φ(U) is open, φ : U → φ(U) is a homeomorphism, andfor each element φα : Uα→ Rn of the oriented atlas for M , the transition function φ ◦φ−1α : φα(U ∩Uα)→ φ(U ∩ Uα) is an orientation-preserving diffeomorphism.

As you should recall, the definition of a smooth manifold (as opposed to an oriented smoothmanifold) is just the same as the above except that the orientation-preserving condition on thetransition functions is dropped. Some authors use the convention that the oriented atlas should bemaximal with respect to inclusion; given a not-necessarily-maximal oriented atlas one can obtainthe unique maximal oriented atlas containing it by simply adding in all of the oriented charts.

Exercise 2.12. Construct, with proof, oriented atlases for the sphere S2 and for the torus S1 × S1.

Definition 2.13. A smooth manifold M is orientable if it has an oriented atlas consisting of chartswhich are compatible with the charts in its atlas.3

Proposition 2.14. A smooth n-manifold M is orientable if and only if there is ω ∈ Ωn(M) such thatωp 6= 0 for all p ∈ M.

Proof. We first prove the backward implication. Given such an ω, for any smooth chart φ : U →φ(U) ⊂ Rn where U is connected we obtain a pullback (φ−1)∗ω ∈ Ωn(φ(U)). We can write(φ−1)∗ω = fφd x1 ∧ · · · ∧ d xn where fφ ∈ C∞(φ(U)) is nowhere-vanishing (since ωp 6= 0 forall p), and so since φ(U) is connected we have either fφ > 0 everywhere or fφ < 0 everywhere.Note that if we define I(x1, . . . , xn) = (x1, . . . , xn−1,−xn), then φ̄ := I ◦φ is also a smooth chart andobeys fφ̄ ◦ I = − fφ; thus for any smooth chart φ either fφ > 0 everywhere or fφ̄ > 0 everywhere.

Take the oriented atlas to consist of those φ having fφ > 0 everywhere. If φ : U → Rn andψ: V → Rn are two such charts then since (on ψ(U ∩ V ))

fψd x1 ∧ · · · ∧ d xn = (φ ◦ψ−1)∗( fφd x1 ∧ d xn)

it follows from Proposition 2.4 (and its higher-dimensional analogue) that det d(φ◦ψ−1) is positivewherever it is defined. Moreover the domains of such charts φ cover M : indeded since we beganwith a smooth atlas, if p ∈ M there is a smooth chart φ : U → Rn defined on a connected neighbor-hood U of p, and then either φ or φ̄ will belong to our proposed atlas. Thus we have constructedan oriented atlas for M consisting of charts in the initial atlas.

For the forward implication, suppose that {φα : Uα→ Rn} is an oriented atlas for M , let {χα} bea partition of unity subordinate to {Uα} and define

ω=∑

α

χαφ∗α(d x1 ∧ · · · ∧ d xn)

If p ∈ Uα0 ⊂ M the fact that each φα ◦ φ−1α0

is orientation-preserving is easily seen (again usingProposition 2.4) to imply that (φ−1α0 )

∗ω is a positive multiple of d x1 ∧ · · · ∧ d xn at φα0(p), so thatωp 6= 0. �

We can now define the notion of the integral of a differential form over compact subsets of anoriented manifold:

3If one requires atlases in the definition of a smooth manifold to be maximal this is the same as saying that it admits anoriented subatlas.

10 MIKE USHER

Definition 2.15. Let M be a smooth oriented n-dimensional manifold, K ⊂ M a compact subset, andω ∈ Ωn(M). Choose oriented charts φα : Uα→ Rn where M = ∪αUα and let {χα} be a partition ofunity subordinate to {Uα}. Then we define

∫

K

ω=∑

α

∫

φα(K∩Uα)(φ−1α )

∗(χαω)

Exercise 2.16. Prove that a different choice of oriented charts ψβ : Vβ → Rn and partition of unity{ζβ} subordinate to {Vβ} in the above definition will yield the same result for

∫

Kω. (Suggestion: Thinkabout the partition of unity {χαζβ} subordinate to the cover {Uα ∩ Vβ}.)

Exercise 2.17. The standard area form ω on S2 is defined as follows. Recall that, viewing S2 asembedded in the unit sphere in R3, at any p ∈ S2 the tangent space TpS2 is naturally identified as theset of vectors in R3 that are orthogonal to p. Then for v, w ∈ TpS2 define

ωp(v, w) = p · (v ×w)

Compute, with a complete proof, the integral∫

S2ω. (The answer should not surprise you. You’llpresumably use your atlas from Exercise 2.12.)

The change of variables formula extends in a straightforward way to our definition of integrationfor general oriented manifolds. Recall that, by definition, if M and N are smooth manifolds, a mapf : M → N is smooth provided that, for each x ∈ M , there is a coordinate chartφ : U → Rm from theatlas for M with x ∈ U , and a coordinate chartψ: V → Rn from the atlas for N with f (x) ∈ V , suchthat the compositionψ◦ f ◦φ−1 is smooth (as a map from the open subset φ(U∩ f −1ψ−1(V )) ⊂ Rmto the open subset ψ( f (φ−1(U)) ∩ V ) ⊂ Rn). (It is easy to check that this is independent of thechoices of coordinate charts around x and f (x), using that the transition functions of the atlasesare smooth.) If f : M → N is a diffeomorphism (as before this means smooth, bijective, and suchthat f −1 is also smooth—of course this forces m = n) between oriented smooth manifolds, thenwe say that it is orientation-preserving provided that, for each x ∈ M , if φ and ψ as above aretaken to be oriented charts then ψ ◦ f ◦φ−1 is an orientation-preserving diffeomorphism betweenopen subsets of Rn. Again this is independent of the choice of oriented chart due to the transitionfunctions being orientation-preserving.

Corollary 2.18. Let φ : M → N be an orientation-preserving diffeomorphism between oriented man-ifolds, ω ∈ Ωn(N), and K ⊂ M is compact, then

∫

K

f ∗ω=

∫

f (K)ω

Proof. If {ψα : Uα → Rn} is an oriented atlas for N , then {ψα ◦ f : f −1(Uα) → Rn} is likewisean oriented atlas for N . (Indeed the f −1(Uα) cover M since the Uα cover N ; the fact that f is ahomeomorphism implies that each ψα ◦ f is a homeomorphism from f −1(Uα) to ψα(Uα), and ifφ : U → Rn is any other oriented chart for M then the statement that f is orientation-preservingmeans that (ψα ◦ f ) ◦ φ−1 is an orientation-preserving diffeomorphism between open subsets ofRn.) Where {χα} is a partition of unity subordinate to {Uα}, observe that {χα ◦ f } is a partition ofunity subordinate to { f −1(Uα)}; moreover from the definition of pullback it is easy to see that, forany α, f ∗(χαω) = (χα ◦ f ) f ∗ω. Then we have∫

K

f ∗ω=∑

α

∫

ψα(Uα)∩(ψα◦ f )(K)((ψ−1α ◦ f )

−1)∗ ((χα ◦ f ) f ∗ω) =∑

α

∫

ψα(Uα∩ f (K))( f −1 ◦ψ−1α )

∗ ( f ∗(χαω))

=∑

α

∫

ψα(Uα∩ f (K))(ψ−1α )

∗( f −1)∗ f ∗(χαω) =∑

α

∫

ψα(Uα∩ f (K))(ψ−1α )

∗(χαω) =

∫

f (K)ω


�

The manifolds that we deal with in this course will typically be oriented (in particular any sym-plectic manifold carrries a natural orientation); non-orientable manifolds do however exist, with thesimplest being the (open) Möbius strip. We now define this carefully as a smooth manifold, afterwhich we will prove that it is indeed not orientable. The standard description of the Möbius strip asa topolgical space calls for beginning with a rectangle [0, 1]×(−1, 1) and then identifying the edgesby gluing each point (0, y) on the left edge to the point (1,−y) on the right edge. Equivalently(from the standpoint of topological spaces) one can choose some small ε > 0 (what follows willwork precisely as written as long as ε < 1/4), start with the open strip S̃ = (−ε, 1+ε)×(−1, 1), andlet S denote the quotient of S̃ by the equivalence relation that identifies (x , y)with (1+x ,−y) for allx ∈ (−ε,ε)×(−1, 1). Let S0 = (−ε, 1/2+ε)×(−1, 1) and S1 = (1/2−ε, 1+ε)×(−1,1). So S̃ = S0∪S1,and the quotient map π: S̃ → S has the property that both π|S0 and π|S1 are homeomorphisms totheir respective images. (This follows easily from the definition of the quotient topology and thefact that the π|Si are injective.) So where, for i = 0, 1, we write Ui = π(Si) and φi = π

−1i : Ui → Si ,

the Ui give an open cover of S and theφi are homeomorphisms to open subsets of R2. We claim that{U0, U1} is a smooth atlas; to see this it is enough to check thatφ1◦φ−10 : φ0(U0∩U1)→ φ1(U0∩U1)is a diffeomorphism.

Now U0 ∩ U1 consists of two components: one of these is the image under both π0 and π1 ofS̃0∩S̃1 = (1/2−ε, 1/2+ε)×(−1,1) and the other isπ0((−ε,ε)×(−1,1)) = π1((1−ε, 1+ε)×(−1,1)).So the transition function φ1 ◦φ−10 : φ0(U0 ∩ U1)→ φ1(U0 ∩ U1) is the map

((−ε,ε)∪ (1/2− ε, 1/2+ ε))× (−1, 1)→ ((1/2− ε, 1/2+ ε)∪ (1− ε, 1+ ε))× (−1, 1)

defined by

(8) (φ1 ◦φ−10 )(x , y) =§

(1+ x ,−y) x ∈ (−ε,ε)(x , y) x ∈ (1/2− ε, 1/2+ ε)

This transition function is indeed a diffeomorphism, so our atlas is a smooth atlas and we havemade S into a smooth manifold. Since the transition function is not orientation preserving (on

(−ε,ε) × (−1,1) the derivative is

1 00 −1

, which has negative determinant), our atlas is not

an oriented atlas. This does not by itself prove that S is not orientable, since one could imagine ithaving another atlas that is oriented; however we can use Proposition 2.14 to rule this out:

Proposition 2.19. For any ω ∈ Ω2(S) there is p ∈ S such that ωp = 0. Hence S is not orientable.

Proof. Suppose for contradiction that ω ∈ Ω2(S) is nonzero at every point. Then where π0 = φ−10andπ1 = φ−11 are as above,π

∗0ω ∈ Ω

2 ((−ε, 1/2+ ε)× (−1, 1)) andπ∗1ω ∈ Ω2 ((1/2− ε, 1+ ε)× (−1, 1))

would both be nowhere-zero two-forms on open subsets of R2, so we could write π∗0ω= f0d x ∧d yand π∗1ω= f1d x ∧ d y for some smooth, nowhere-zero functions f0 : S0→ R and f1 : S1→ R.

By Proposition 2.4, we see that for (x , y) ∈ S0,

f0(x , y) = π∗0ω(x ,y) ((1, 0), (0, 1)) = (π

∗0(π

−11 )∗π∗1ω)(x ,y) ((1,0), (0,1))

= (π−11 ◦π0)∗(π∗1ω)(x ,y) ((1, 0), (0, 1)) = det d(π

−11 ◦π0)(x ,y) f1(π

−11 ◦π0(x , y))

So using the formula (8) for φ1 ◦φ−10 = π−11 ◦π0 we see that

(9) f0(0, 0) = − f1(1,0) f0(1/2,0) = f1(1/2, 0)

But the functions x 7→ f0(x , 0) and x 7→ f1(x , 0) are both continuous, nowhere-vanishing func-tions on, respectively, the intervals (−ε, 1/2 + ε) and (1/2 − ε, 1). So by the intermediate valuetheorem f0(1/2,0) has the same sign as f0(0, 0), while f1(1/2, 0) has the same sign as f1(1, 0).

12 MIKE USHER

This contradicts (9), proving one or both of f0, f1 vanishes somewhere, and hence that ω vanishessomewhere. �

It follows directly from Proposition 2.19 that any smooth two-manifold containing a Möbiusstrip as an open subset is non-orientable: indeed, if there were a nonvanishing two-form on such amanifold then we could restrict that two-form to the Möbius strip, contradicting Proposition 2.19.Thus connected sums of RP2 (which may be familiar to you from the classification of surfacesas comprising all of the “non-orientable” surfaces) are indeed non-orientable in our sense, sincethe Möbius strip is the complement of a disk in RP2. The remaining compact surfaces from theclassification (namely S2 and connected sums of tori) are orientable: to see this, we constructnonvanishing two-forms ω on these surfaces which generalize the area form from Exercise 2.17 asfollows. Consider the standard embedding of such a surface S ⊂ R3 as the boundary of a handlebody,for each x ∈ S let Nx denote the unit vector in R3 that points outward from the handlebody andperpendicular to S at x , and for v, w ∈ TxS defineωx(v, w) as the scalar triple product of the vectorsNx , v, w (i.e. the determinant of the matrix with these three vectors as columns, where we use theembedding of S in R3 to think of v and w as vectors in R3).

Proposition 2.14 and its proof show that, very generally, a nowhere-vanishing n-form ω on asmooth n-dimensional manifold M induces an orientation on M , which is uniquely specified by theproperty that, when this orientation is used to define integration of n-forms, one has

∫

Kω ≥ 0for every compact set K (with a strict inequality when the interior of K is nonempty). The mapK 7→

∫

Kω extends to a measure on the Borel σ-algebra of M , which we will denote volω. Anowhere-vanishing n-form on an n-manifold is often called a volume form (or “area form” if n= 2),since as just explained it induces a notion of volume on the manifold.

By working in charts and using Propositions 2.2 and 2.4, it is not hard to see that if ω is anyvolume form on M then all other volume forms on M can be written as hω for some nowhere-zerofunction h ∈ C∞(M), and that if f : M → M is a diffeomorphism then f is orientation-preservingif and only if, when we write f ∗ω= hω, we have h> 0 everywhere.

Proposition 2.20. Let f : M → M be an orientation-preserving diffeomorphism of a smooth manifoldM with a volume form ω. Then f is volume-preserving (i.e., volω( f (U)) = volω(U) for all open setsU) if and only if f ∗ω=ω.

Proof. Note that, since f is orientation-preserving, the change of variables formula shows that, forany open set U ,

volω( f (U)) =

∫

f (U)ω=

∫

U

f ∗ω

If f ∗ω = ω then the above is equal to∫

U ω = volω(U), proving that f is volume-preserving. Thisproves the forward implication; we leave the backward implication as the following exercise. �

Exercise 2.21. Prove that a volume-preserving diffeomorphism f : M → M of a manifold with avolume form ω must obey the equation f ∗ω=ω. (Hint: Working in a coordinate chart, compare theareas of small balls around any points where f ∗ω and ω are not equal.)

A symplectic 2-manifold is a pair (Σ,ω) where Σ is a smooth 2-manifold and ω ∈ Ω2(Σ) is anarea form. (We will later discuss symplectic manifolds of other even dimensions; the general defi-nition is a bit more complicated but reduces to this one in the two-dimensional case.) A symplecto-morphism of (Σ,ω) is by definition a diffeomorphism f : Σ→ Σ with the property that f ∗ω = ω.So according to what we have just said, a symplectomorphism is (in the two-dimensional case)precisely a diffeomorphism which preserves both orientation and area. We will soon see how toconstruct many symplectomorphisms of symplectic two-manifolds, beginning with the case of R2

with its standard area form d x ∧ d y .


3. FLOWS OF VECTOR FIELDS AND AREA PRESERVATION

A key method for constructing diffeomorphisms in general and area-preserving diffeomorphismsin particular rests on the following two results from the theory of ordinary differential equations.(We omit the proofs; they can be found in textbooks on ODE theory like [9].)

Theorem 3.1 (Existence and Uniqueness for ODE’s). Let

X : R×Rn→ Rn

(t, p) 7→ X t(p)

be continuous and uniformly Lipschitz in its second argument (i.e. there is a constant C such that‖X t(p) − X t(q)‖ ≤ C‖p − q‖ for all t ∈ R, p, q ∈ Rn). Then for all (t0, p0) ∈ R × Rn there is aunique γt0 p0 : R → R

n such that γt0 p0(t0) = p0 and, for all t ∈ R, γ′t0 p0(t) = X t(γt0 p0(t)). In fact,

for any ε > 0 and any η: (t0 − ε, t0 + ε)→ Rn such that η(t0) = p0 and η′(t) = X t(η(t)) we haveη= γt0 p0 |(t0−ε,t0+ε).

Thus specifying a velocity vector field (which is allowed to depend on time) and an initial con-dition uniquely determines the future and past trajectories of an object. Note that if X is smoothand compactly supported then it satisfies the hypotheses (with the Lipschitz constant C determinedby the maximum norm of the derivative of X ). If X is just smooth without satisfying a Lipschitzproperty then the theorem will generally not hold because of the possibility of solutions escapingto infinity in finite time.

Theorem 3.2 (Smooth dependence on initial conditions). Under the hypotheses of Theorem 3.1, ifX is smooth then the function ΨX : R× Rn → Rn defined by ΨX (t, p) = γ0p(t) is smooth. Hence forany T ∈ R the map ψTX : R

n→ Rn defined by ψTX (p) = ΨX (T, p) = γ0p(T ) is also smooth.

The map ψTX is called the time-T flow of the (time-dependent) vector field X .While writing down an arbitrary, complicated formula for a map from Rn to Rn will not generally

give a diffeomorphism since (for instance) it may not be bijective, if one instead writes down anarbitrary time-dependent vector field X and takes its flow then one will obtain a diffeomorphism:

Proposition 3.3. For X as in Theorem 3.2, the map ψTX is a diffeomorphism.

Proof. Theorem 3.2 says that ψTX is smooth, so we just need to find its inverse and show that thisinverse is smooth. Now ψTX is characterized by the property that if γ: [0, T] → R

n solves theequation γ′(t) = X t(γ(t)) then ψTX (γ(0)) = γ(T ). Given such a map γ, define γ̄: [0, T] → R

n byγ̄(t) = γ(T − t). Thus γ̄(0) = γ(T ) and γ̄(T ) = γ(0). We see that

γ̄′(t) = −γ′(T − t) = −XT−t(γ(T − t)) = −XT−t(γ̄(t))

So defining Y : R × Rn → Rn by Yt(p) = −XT−t(p), it follows that ψTY (γ̄(0)) = γ̄(T ), i.e. thatγ(0) = ψTY (γ(T )) = ψ

TY (ψ

TX (γ(0))). Since γ(0) can be chosen to take any value in R

n this provesthat ψTY ◦ψ

TX = 1Rn . Now we obtained Y by applying the operation Vt 7→ −VT−t to X . If we instead

apply this operation to Y we obtain X back, so the same argument shows that ψTX ◦ψTY = 1Rn . Thus

ψTX has inverse given by ψTY . Theorem 3.2 (applied to Y ) shows that this inverse is smooth. �

Exercise 3.4. Let X , Y : R×Rn → Rn be time-dependent vector fields and φ : Rn → Rn an arbitrarydiffeomorphism. Give, with proof, explicit formulas (in terms of X , Y,φ) for:

(a) a time-dependent vector field Z with ψtZ = (ψtX )−1 for all t.

(b) a time-dependent vector field Z with ψtZ = φ ◦ψtX ◦φ

−1 for all t.(c) a time-dependent vector field Z with ψtZ =ψ

tX ◦ψ

tY for all t.

14 MIKE USHER

Now that we have a very general and flexible method for producing diffeomorphisms we addressthe question of when the resulting diffeomorphisms are area-preserving, beginning with the case ofR2. Initially we will use the notion of area given by Lebesgue measure (i.e. using the standard areaform d x ∧ d y), though the result will quickly generalize to other area forms.

So let X be a time-dependent vector field4 on R as above (with the the vector field at time tdenoted by X t = X (t, ·) just as before), inducing a smooth map ΨX : R×R2→ R2 whose restrictionto {t}×R2 defines a diffeomorphismψtX : R

2→ R2. Note thatψ0X is the identity (reread the relevanttheorems and definitions if this isn’t obvious to you).

Below we will assume that S is the closed ball {q ∈ R2 : ‖q − p‖ ≤ r} of some radius r > 0around some point p ∈ R2. So S has a boundary ∂ S = {q ∈ R2|‖q − p‖ = r} (more generally theargument that we will give would work for S equal to any codimension-zero compact submanifoldwith boundary in R2). Note that, by an argument similar to that used in the proof of Corollary 1.3,a diffeomorphism ψ will be area-preserving if and only if one has Area(ψ(S)) = Area(S) for allclosed balls S.

Theorem 3.5. Let S be a closed ball in R2 and let X be a time-dependent vector field. Write X t(x , y) =〈Pt(x , y),Q t(x , y)〉 for all t ∈ R and (x , y) ∈ R2. If

(10)∂ Pt∂ x+∂Q t∂ y

= 0

then we will have Area(ψTX (S)) = Area(S) for all T ≥ 0.

Exercise 3.6. Prove that, conversely, if there is some t ≥ 0 and some point at which ∂ Pt∂ x +∂Q t∂ y 6= 0,

then there is a closed ball s and a time T ≥ 0 such that Area(ψTX (S)) 6= Area(S) (you’ll probably wantto use (11) below).

(Actually both statements hold for all T ∈ R, not just T ≥ 0, but to make the argument clearerwe will restrict to positive time.)

Proof. The plan is to, for any given T ≥ 0, apply the divergence theorem to the following subset ofR3, sometimes called the time-T trace of S under X :

ŜT = {(ψtX (p), t)|p ∈ S, t ∈ [0, T]}

where the notation reflects the obvious identification of R3 with R2 ×R. Aside from a region in R3the divergence theorem requires a (3-dimensional, time-independent) vector field; for this we willuse

~F(x , y, z) = (Xz(x , y), 1) = (Pz(x , y),Qz(x , y), 1)The divergence theorem then says that, where ∂ ŜT is the boundary of the 3-dimensional region ŜT ,and we use ~n to denote the unit normal and dσ to denote the standard area measure on a surfaceembedded in R3,

∫∫∫

ŜT

div ~Fd xd ydz =

∫∫

∂ ŜT

~F · ~ndσ

Now ∂ ŜT consists of three pieces:(i) The part of ST where the third coordinate is equal to T , namely ψ

TX (S) × {T}. Here the

outward normal ~n is the unit vector that points straight up, and so ~F · ~n is just the thirdcomponent of ~F , namely 1. So the flux

∫∫

~F · ~ndσ through this part of the boundary isArea(ψTX (S)).

4Here and below we’ll always implicitly assume that our vector fields obey the conditions of Theorems 3.1 and 3.2, aswill be the case for instance if they are smooth and compactly supported or Lipschitz.


(ii) The part of ST where the third coordinate is equal to 0, namely S × {0}. Here the unitnormal ~n points straight down (i.e. ~n= (0, 0,−1)), so ~F · ~n= −1 and the flux through thispart of the boundary is −Area(S).

(iii) The trace Õ(∂ S)T of the boundary ∂ S. This consists of points of the form (ψtX (p), t) forany p ∈ ∂ S and t ∈ [0, T]. In other words, where for any p ∈ ∂ S we define the mapΓp : [0, T]→ R3 by Γp(t) = (ψtX (p), t), we have Õ(∂ S)T = {Γp(t)|p ∈ ∂ S, t ∈ [0, T]}. Nowat any point Γp(t) ∈Õ(∂ S)T , since the image of Γp is contained in Õ(∂ S)T it follows that thevector Γ ′p(t) is tangent to

Õ(∂ S)T . But we see from the formulas that Γ ′p(t) = ~F(Γp(t)). In

other words the vector field ~F is tangent to Õ(∂ S)T , and therefore is perpendicular to theunit normal ~n at every point of Õ(∂ S)T . So the flux

∫∫

Õ(∂ S)T~F · ~ndσ is zero.

Adding up the contributions of (i), (ii), and (iii) and using the divergence theorem shows that, forany time-dependent vector field X on R2, leading to the three-dimensional vector field ~F(x , y, z) =(Xz(x , y), z) we have (regardless of whether (10) holds)

(11)

∫∫∫

S

div ~Fd xd ydz = Area(ψTX (S))− Area(S)

But we compute

div ~F(x , y, z) =∂ Pz∂ x(x , y) +

∂Qz∂ y(x , y) + 0= 0

by the assumption 10. So the left hand side of (11) is zero and indeed Area(ψTX (S)) = Area(S). �

So our criterion for a time-dependent vector field X on R2 to have area-preserving flow is thatthe divergence of X t should be zero at all times t. This is likely somewhat consistent with (thoughprobably not quite identical to) intuition that you may have about the geometric meaning of thedivergence from multivariable calculus, and so what we have done gives some rigorous justificationto this intuition.

The next goal is to find an analogous criterion for a time-dependent vector field on a symplectictwo-manifold (M ,ω) to have area-preserving flow; we have just solved this problem in the specialcase that (M ,ω) = (R2, d x ∧ d y). First we will work this out when M = R2 but ω is allowed tobe an arbitrary area form inducing the standard orientation of R2; thus ω = gd x ∧ d y for somesmooth function g : R2 → R with g > 0 everywhere. This area form determines the area of anymeasurable subset of R2 via the formula Areag(S) =

∫∫

S gd xd y . In this case we have:

Theorem 3.7. A time-dependent vector field X = {X t}t∈R onR2 has the property that Areag(ψtX (S)) =Areag(S) for all closed balls S if and only if, for all t ∈ R, div(gX t) = 0 (i.e., if X t(x , y) = (Pt(x , y),Q t(x , y)),∂ (gPt )∂ x +

∂ (gQ t )∂ y = 0) everywhere on R

2.

Proof. This is proven very similarly to Theorem 3.5 and Exercise 3.6, so we will just indicate the nec-essary modification to the proof. Writing X t(x , y) = (Pt(x , y),Q t(x , y)), apply the divergence theo-rem to the trace ŜT ⊂ R3 and the the vector field ~G(x , y, z) = (g(x , y)Pz(x , y), g(x , y)Qz(x , y), g(x , y)).In comparison to the proof of (11), the integrand in the flux integral just gets multiplied by g, andso we obtain

(12)

∫∫∫

ŜT

div ~Gd xd ydz = Areag(ψTX (S))− Areag(S).

Meanwhile the (3-dimensional) divergence of ~G at (x , y, z) is just the same as the (2-dimensional)divergence of gXz , so the integrand on the left of (12) vanishes identically for all T and all balls S

16 MIKE USHER

if and only if div(gX t) = 0 everywhere on R2. Once this is noted the rest of the proofs of Theorem3.5 and Exercise 3.6 proceed unchanged. �

Since any symplectic 2-manifold (M ,ω) is covered by oriented coordinate charts φα : Uα → R2under which ω appears as (i.e., (φ−1α )

∗ω is equal to) an area form gαd x ∧ d y for some smooth,everywhere positive function gα : φα(Uα)→ R, Theorem 3.7 indicates a way of checking whethera time-dependent vector field on (M ,ω) induces an area-preserving flow: push the vector fieldforward by the various charts φα to obtain a vector field X

α and see whether div(gαX αt ) = 0 for allt and α. It should be intuitively plausible (and indeed can be checked directly) that this criterion isindependent of the coordinate charts used.

In general in differential geometry and topology it is desirable to express ideas (such as that ofan area-preserving vector field) in explicitly coordinate-free language rather than writing formulasin coordinates—even if one can do a calculation with transition functions to show that a coordinate-based formula yields results that are independent of the coordinate chart, a coordinate-free interpre-tation is closer to giving a real explanation of this independence (and moreover saves the trouble ofdoing the calculation). The criterion in Theorem 3.7 appears to be rather coordinate-bound in thatthe formula for the divergence of a vector field obviously makes reference to the standard Cartesiancoordinates on R2. The language of differential forms, however, will let us recast the criterion in acoordinate-independent way.

Definition 3.8. Let M be a smooth manifold, let X be a vector field on M , and letω ∈ Ωk(M) be anydifferential k-form. The interior product of X and ω is the (k − 1)-form ιXω ∈ Ωk−1(M) definedby, for p ∈ M and v1, . . . , vk−1 ∈ Tp M ,

(ιXω)p(v1, . . . , vk−1) =ωp(X (p), v1, . . . , vk−1)

(So ιXω is given by simply “inserting X into the first slot” of the k-linear form ω to obtain a(k− 1)-linear form.)

In particular if X is a vector field and ω is a 2-form on R2 we will get a 1-form ιXω on R2. Todescribe this 1-form we introduce general notation for 1-forms on R2. The standard basis of 1-formson R2 consists of the two elements d x and d y , given by, at each point p ∈ R2, (d x)p((a, b)) = a and(d y)p((a, b)) = b, where as usual we use the canonical identification of TpR2 with R2. Clearly anyθ ∈ Ω1(R2) can be written as θ = f d x+ gd y for some f , g ∈ C∞(M): simply put f (p) = θp((1, 0))and g(p) = θp((0,1)) for all p ∈ R2 and apply the linearity axiom for 1-forms to see that θ andf d x + gd y agree on all tangent vectors.5

Proposition 3.9. If X is a vector field onR2 given by X (x , y) = (P(x , y),Q(x , y)) and ifω= gd x∧d ythen

ιXω= (−gQ)d x + (gP)d yThus if we write ιXω= Fd x + Gd y, we have div(gX ) = Gx − Fy .

Proof. Indeed

ιXω((1,0)) =ω((P,Q), (1, 0)) = gd x ∧ d y((P,Q), (1,0)) = −gQ

and similarlyιXω((0, 1)) = gd x ∧ d y((P,Q), (0, 1)) = gP

5Brief digression: I can now partially justify the notation d x∧d y for the standard volume form by mentioning that thereis a “wedge product” operation ∧: Ωk(M)×Ωl (M)→ Ωk+l (M) on differential forms, which has a complicated formula (tobe discussed later) in general but in the case k = l = 1 is given by (α∧β)(v, w) = α(v)β(w)−α(w)β(v). It’s easy to see thatwhat we have been calling d x ∧ d y is indeed equal to the wedge product of d x and d y . Shortly we will see why the namesd x and d y for these one-forms are appropriate.


so the first sentence follows by the remarks before the proposition. Representing ιXω= Fd x+Gd ythen means taking F = −gQ and G = gP, so indeed Gx − Fy = (gP)x + (gQ)y = div(gX ). �

The above is a modest step toward the goal of coordinate-independence: the one-form ιXω isdefined in a coordinate-independent way (in the sense that if instead ω was a 2-from and X was avector field on a general manifold one wouldn’t need to work in coordinate charts to say preciselywhat ιXω is), and so instead of needing coordinate representations for two things (X and ω) inorder to apply Theorem 3.7 we only need a coordinate representation for the single object ιXω:the theorem says that if we write ιXω as Fd x + Gd y then the flow of X will be ω-preserving iffGx − Fy = 0 (and one has a similar criterion if X = (X t) is time-dependent). By itself this may notseem like real progress, but results from multivariable calculus imply that (at least if everything isdefined on all of R2, or more generally on a simply connected open subset) then the requirementthat Gx − Fy = 0 can be expressed in a coordinate-independent way. To see this requires anotherdifferential forms definition:

Definition 3.10. Let H : M → R be a smooth function on a smooth manifold M (in other words,H ∈ Ω0(M)). We then define dH ∈ Ω1(M) to be the one-form given by setting, for p ∈ M andv ∈ Tp M ,

(dH)p(v) =dd t

H(γ(t))

�

�

�

�

t=0

where γ: R→ M is any smooth path with γ(0) = p and γ′(0) = v.

In other words, (dH)p(v) is the directional derivative of H in the direction v. Or if you arefamiliar with the more abstract definition of tangent vectors as derivations on the space of germs ofsmooth functions, one could just say that (dH)p(v) = v(H). It is easy to see that the dH as describedabove is well-defined (i.e. independent of the choice of curve γ, and linear in v so that dH really isa one-form): just work in local coordinates and apply the multivariable chain rule to H ◦ γ. Whilethis proof depends on local coordinates, once one knows that dH is well-defined its definition isclearly independent of any choice of coordinates. So we have a map d : Ω0(M) → Ω1(M); moregenerally we will see later how to generalize this to a map d : Ωk(M)→ Ωk+1(M) for any k.

Returning to the case of R2, it should be clear from the definitions and from the remarks beforeProposition 3.9 that one has

dH =∂ H∂ x

d x +∂ H∂ y

d y

for any H ∈ Ω0(M). (Just apply dH to (1,0) and (0, 1).)6 Now if F = ∂ H∂ x and G =∂ H∂ y then we

have Gx − Fy = 0 by equality of mixed partials. Together with Theorem 3.7 and Proposition 3.9 thisyields the following corollary:

Corollary 3.11. Let H : R×R2 → R be a smooth function, ω = gd x ∧ d y a 2-form on R2 inducingthe standard orientation (so g > 0) and suppose that we have time-dependent vector field (X t) obeying,for all t, ιX tω= d(H(t, ·)). Then the time-t flow ψ

tX of (X t) is ω-area-preserving for all t.

The criterion for area-preservation above is now coordinate-independent and so can be gen-eralized to any symplectic (2-)manifold: if (X t) is a time-dependent vector field on a symplectic2-manifold (M ,ω) it makes sense to ask whether there exists a smooth function H : R × M → Rsuch that ιX t = d(H(t, ·)) for all t. In this case one can show (for by working in local coordinatecharts and applying the above corollary) that the flow of (X t) preserves area as defined by ω. Foreach t, X t is then called the Hamiltonian vector field of H(t, ·). One can show quite generally that,

6Note that this means that the one-form that we have been calling d x really does result from applying the operator d tothe function x , and similarly for d y .

18 MIKE USHER

given H, the Hamiltonian vector field exists and is unique: in any local coordinate chart, if the formis given by gd x ∧ d y then the unique such vector field in this chart is X =

−g−1 ∂ H∂ y , g−1 ∂ H∂ x

, andsince this is the unique solution it’s easy to see that the vector fields defined chart-wise in this fash-ion are consistent with each other on the overlaps of the charts and so give a well-defined vectorfield on all of M .

Now the criterion in Corollary 3.11 is a sufficient criterion to get an area-preserving flow; next wewill discuss whether it is also necessary—i.e. whether any vector field whose flow is area-preservingis a Hamiltonian vector field. The subtleties in the answer again extend something that you mayrecall from multivariable calculus.

Proposition 3.12. Let α= Fd x+Gd y be a 1-form on an open, simply connected set U ⊂ R2 with theproperty that ∂ G∂ x −

∂ F∂ y = 0 and let γ0,γ1 : [0,1]→ U be two smooth curves with the same endpoints,

i.e. γ0(0) = γ1(0) and γ0(1) = γ1(1). Then∫

γ0α=

∫

γ1α

(Here∫

γα can be understood in terms of our earlier notation as

∫

[0,1] γ∗α. We use this notation

because it is consistent with multivariable calculus.)

Proof. Write γ̄1(t) = γ1(1 − t). In simple cases the argument can proceed by observing that theconcatenation γ0 ∗ γ̄1 is the boundary (oriented either clockwise or counterclockwise) of a regionR ⊂ U , and then applying Green’s theorem to see that

∫

γ0

α−∫

γ1

α=

∫

γ0∗γ̄1

α= ±∫∫

R

∂ G∂ x−∂ F∂ y

d xd y = 0

In general it may not be true that γ0 ∗ γ̄1 is the oriented boundary of a region; however the simple-connectivity assumption implies that there is a map Γ : [0,1]×[0, 1]→ U with Γ (·, 0) = γ0, Γ (·, 1) =γ1, and Γ (0, ·) and Γ (1, ·) both constant. The change of variables formula then implies that, where∂ [0,1]2 is the counterclockwise-oriented boundary of [0, 1]2, we have

(13)

∫

∂ [0,1]2Γ ∗α=

∫

γ0

α−∫

γ1

α

Where s and t are the standard coordinates on [0, 1]2, write Γ (s, t) = (x(s, t), y(s, t)). Then

Γ ∗α((1, 0)) = α(dΓ (1,0)) = α

∂ x∂ s

,∂ y∂ s

= F(Γ (s, t))∂ x∂ s+ G(Γ (s, t))

∂ y∂ s

and likewise

Γ ∗α((0,1)) = F(Γ (s, t))∂ x∂ t+ G(Γ (s, t)

∂ y∂ t

ThusΓ ∗α= F̃ ds+ G̃d t

whereF̃ = (F ◦ Γ )xs + (G ◦ Γ )ys G̃ = (F ◦ Γ )x t + (G ◦ Γ )yt

A chain rule computation shows that

G̃s − F̃t = (F ◦ Γ )s x t + (G ◦ Γ )s yt − (F ◦ Γ )t xs − (G ◦ Γ )t ys= Fx xs x t + Fy ys x t + Gx xs yt + Gy ys yt − Fx x t xs − Fy yt xs − Gx x t ys − Gy yt ys= Fy ys x t + Gx xs yt − Fy yt xs − Gx x t ys = (Gx − Fy)(xs yt − x t ys) = 0

by the assumption that Gx − Fy = 0. So Green’s theorem gives∫∫

∂ [0,1]2Γ ∗α=

∫∫

[0,1]2(G̃s − F̃t)dsd t = 0


which together with (13) implies the result. �

Theorem 3.13. Let U be a simply connected open set in R2 and let α = Fd x + Gd y ∈ Ω1(U) be aone-form with ∂ G∂ x −

∂ F∂ y = 0. Then there is H ∈ C

∞(U) such that dH = α.

Proof. Fix p0 ∈ U and define H : U → R by H(p) =∫

γpα where γp : [0, 1] → U is an arbitrary

smooth curve with γp(0) = p0 and γp(1) = p. By Proposition 3.12 the value of Hp is independentof the choice of γp used to compute it. Where p = (x , y), we may compute H(x + t, y) for smallt by using a curve which is the concatenation of γp with a horizontal line segment; this yields

H(x + t, y)− H(x , y) =∫ x+t

x F(x′, y)d x ′ and hence ∂ H∂ x = F . Similarly using the concatenation of

γp with vertical line segments shows that∂ H∂ y = G. So dH = Fd x + Gd y = α. �

So for time-dependent vector fields (X t) on all of R2 our condition that the flow of (X t) preserveω is equivalent to the statement that, for all t, there should be a smooth function Ht : R2 → Rsuch that ιX tω = dHt (in other words ιX tω should be what is called an exact one-form). Moregenerally if the vector field is defined on some simply connected open subset 7 U then the exactnessof ιX tω is again equivalent to the condition of preserving ω. If we no longer assume U to be simplyconnected the most that we can say from the above is that the exactness of ιX tω is sufficient for theψtX to preserve ω, while conversely if the ψ

tX preserve ω then for all t and for all simply connected

open subsets V ⊂ U there is a function Ht,V : V → R such that ιX tω|V = dHt,V . So ιX tω should be“locally exact” (or equivalently “closed,” though the conventional definition of “closed” is phraseddifferently from this, despite being logically equivalent). However a priori the various functionsHt,V for different simply connected open subsets V ⊂ U may not piece together to give a functionon all of U .

On first exposure it may seem like this issue could be overcome, but in fact it cannot; on non-simply-connected sets closed one-forms are typically not exact, and so area-preserving vector fieldsdo not necessarily have ιXω equal to an exact one-form. The following is an example of this. Workon U = R2 \{(0,0)} with the standard symplectic form d x∧d y , and consider the time-independentvector field

X =

� xx2+y2

yx2+y2

�

so that X repels points from the origin, but with a strength that decreases as one goes further fromthe origin. Let us find the flow of X directly. Since X points radially it is reasonable to assumethat the integral curves of X will likewise point radially. So we look for integral curves of theform γ(t) = r(t)(x0, y0) with initial condition γ(0) = ~x0 = (x0, y0), thus leading to the initialcondition r(t) = 1. The integral curve equation γ′(t) = X (γ(t)) then reduces to r ′(t) = 1‖~x0‖2 r(t) ,

and multiplying both sides of this by 2r(t) gives dd t�

r(t)2�

= 2‖~x0‖2 . which together with the initial

condition r(t) = 1 gives r(t) =Ç

1+ 2t‖~x0‖2 . This shows that the time t flow on R2 \ {0} is well-

defined at least for t ≥ 0 by the formula

ψtX (~x) =

√

√

1+2t‖~x‖2

~x

One can check directly that these maps are area-preserving: to argue geometrically, they mapthe region given in polar coordinates as {r0 ≤ r ≤ r1, θ0 ≤ θ ≤ θ1} to the region

¦q

r20 + 2t ≤ r ≤q

r21 + 2t, θ0 ≤ θ ≤ θ1©

7If the vector field is only defined on a subset of R2 there arises the question of whether the flowψtX exists and preservesthis subset; here we will implicitly assume that it does.

20 MIKE USHER

and both the initial region and its image have area 12 (r21 − r

20 )(θ1−θ0). A simple covering argument

shows that the fact that ψtX preserves areas of these sectors of annuli is enough to imply that itpreserves areas of all measurable sets.

Alternatively we could show that ψtX is area-preserving using our criterion. We see that

ιXω((1,0)) = −y

x2 + y2ιXω((0,1)) =

xx2 + y2

so ιXω= Fd x+Gd y where F(x , y) =−y

x2+y2 and G(x , y) =x

x2+y2 . A simple computation shows that

Gx = Fy =y2−x2(x2+y2)2 , so ιXω is closed. However if it were true that ιXω= dH for some smooth func-

tion H then for any smooth γ: [0,1]→ R2 \ {(0,0)}, written in coordinates as γ(t) = (x(t), y(t)),we would have

∫

γ

ιXω=

∫

γ

dH =

∫ 1

0

∂ H∂ x

x ′(t) +∂ H∂ y

y ′(t)

d t

=

∫ 1

0

dd t

H(γ(t))d t = H(γ(1))−H(γ(0))

In the special case that γ(1) = γ(0) this implies that∫

γιXω = 0. But ιXω = Fd x + Gd y where

the vector field (F, G) points counterclockwise everywhere, so if γ(t) = (cos(2πt), sin(2πt)) is acounterclockwise parametrization of the unit circle then we will have

∫

γιXω> 0 (in fact the integral

is equal to 2π). Thus there can be no H such that ιXω = dH, and ιXω is a closed but not exactone-form, i.e. H is not a Hamiltonian vector field, even though its flow is area-preserving.

3.1. Flows and one-forms on general smooth manifolds. We now adapt some of the above ar-guments to work on manifolds more general than R2.

Definition 3.14. Let M be a smooth manifold and α ∈ Ω1(M). α is said to be exact if there ifH ∈ Ω0(M) such that α = dH. α is said to be closed if it is locally exact, i.e., if for all x ∈ M thereis an open neighborhood U of M and a smooth function HU : U → M such that dHU = α|U .Remark 3.15. By “α|U” I mean the one-form on the open subset U ⊂ M whose corresponding linearfunctional TpU → R at each p ∈ U ⊂ M is αp (using the identification of TpU with Tp M comingfrom U being an open subset of M). Equivalently, where iU : U → M is the inclusion, α|U = i∗Uα.The case could be made that this would better be called α|T U .

The notions of exactness and closedness behave nicely with respect to smooth maps:

Proposition 3.16. Let f : N → M be a smooth map between two smooth manifolds and let H ∈ Ω0(M).Then f ∗dH = d( f ∗H). Consequently if α ∈ Ω1(M) is exact then f ∗α ∈ Ω1(N) is also exact, and if α isclosed then f ∗α is closed.

Proof. Let p ∈ N and v ∈ TpN ; we are to show that (d( f ∗H))p (v) = ( f ∗dH)p(v). Note that bydefinition the zero-form (i.e. smooth function) f ∗H is simply H ◦ f .

The quantity (d( f ∗H))p (v) may be computed by choosing an arbitrary smooth γ: R→ N withγ(0) = p and γ′(0) = v; we will then have

(d( f ∗H))p (v) =dd t(( f ∗H)(γ(t)))

�

�

�

�

t=0=

dd t((H ◦ f ) ◦ γ) (t)

�

�

�

�

t=0

Meanwhile by the definition of pullback ( f ∗dH)p(v) = dH f (p)(d fp v), and where γ: R → N is asabove the map f ◦ γ: R→ M obeys ( f ◦ γ)(0) = f (p) and ( f ◦ γ)′(0) = d fp v. Hence

( f ∗dH)p(v) =dd t

H(( f ◦ γ)(t))�

�

�

�

t=0


which is indeed equal to (d( f ∗H))p (v).The first part of the last sentence now follows immediately: to say that α is exact is to say that

α= dH for some H, and then f ∗α= d( f ∗H) so f ∗α is exact. As for the second part, the closednessof αmeans that there is an open cover {Ui} of M such that each α|Ui is exact. Applying what we haveproven to the map f | f −1(Ui) : f

−1(Ui)→ Ui shows that each ( f ∗α)| f −1(Ui) is exact; since { f−1(Ui)} is

an open cover of N this proves that f ∗α is closed. �

Corollary 3.17. Let α ∈ Ω1(M) be any exact one-form. If two smooth paths γ0 : [0, T0] → M andγ1 : [0, T1]→ M have the same endpoints (i.e. γ0(0) = γ1(0) and γ0(T0) = γ1(T1)) then

∫

[0,T0]γ∗0α=

∫

[0,T1]γ∗1α. In particular if γ: [0, T]→ M is a closed path (i.e. γ(0) = γ(T )) then

∫

[0,T] γ∗α= 0.

Proof. If α = dH and γ: [0, T] → M is a smooth path then by Proposition 3.16 we have γ∗α =d(H ◦γ). Now in general for a smooth function F : [0, T]→ R the derivative dF is simply given byF ′(t)d t. So

∫

[0,T]γ∗α=

∫ T

0

(H ◦ γ)′(t)d t = H(γ(T ))−H(γ(0))

where we have used the fundamental theorem of calculus. In particular∫

[0,T] γ∗α depends only on

the endpoints of γ, and vanishes when these endpoints are the same. �

Thus a way of showing that a one-form is not exact is by finding a closed curve over which itsintegral is not zero; indeed this is what we did in the example of −yd x+xd yx2+y2 on R

2 \ {(0,0)} at theend of the last subsection.

Conversely we have the following generalization of an argument from the previous subsection:

Proposition 3.18. Let α ∈ Ω1(M) be a one-form with the property that whenever two smooth pathsγ0 : [0, T0]→ M and γ1 : [0, T1]→ M have the same endpoints it holds that

∫

[0,T0]γ∗0α =

∫

[0,T1]γ∗1α.

Then α is exact.

Proof. Let S be a subset of M consisting of one point from each of the path components of M .Define H : M → R by, for p ∈ M , letting p0 be the unique element of S belonging to the same pathcomponent as p, and putting H(p) =

∫

[0,T] γ∗pα where γp : [0, T]→ M is smooth with γp(0) = p0

and γp(T ) = p. By hypothesis this is independent of the choice of γp subject to the requirements.Let p ∈ M and v ∈ Tp M ; we shall show that dHp(v) = αp(v). Let γ: R→ M have the property

γ(1) = p, and γ′(1) = v, and γ(0) = p0 where p0 is the element of S belonging to the same pathcomponent as p; by definition dHp(v) =

dds H(γ(s))

�

�

s=1. For any s > 0 we can use the path γ|[0,s] tocompute H(γ(s)). Writing γ∗α= F(t)d t, we have on the one hand

F(1) = (γ∗α)1(1) = αp(γ′(1)) = αp(v)

while on the other hand

H(γ(s)) =

∫

[0,s]γ∗α=

∫ s

0

F(t)d t

So by the fundamental theorem of calculus we indeed have

dHp(v) =dds

H(γ(s))

�

�

�

�

s=1=

dds

�∫ s

0

F(t)d t

��

�

�

�

s=1

= F(1) = αp(v)

Thus α= dH is an exact one-form. �

22 MIKE USHER

Remark 3.19. To check the condition in Corollary 3.18 it suffices to use paths γ0,γ1 : [0, 1] → Mdefined on the unit interval. Indeed if γ: [0, T]→ M is any path then where σT : [0, 1]→ [0, T]is the rescaling map t 7→ T t (which is an orientation-preserving diffeomorphism), the map γ ◦σT : [0, 1]→ M has the same endpoints as γ and

∫

[0,1](γ ◦σT )∗α=

∫

[0,1]σ∗Tγ

∗α=

∫

[0,T]γ∗α

where we have used the change of variables formula. So if we just assume that the integral of αis the same on all unit-interval parametrized paths with the same endpoints, then for paths γ0,γ1as above α will have the same integrals along γ0 ◦σT0 and γ1 ◦σT1 , and hence will have the sameintegrals along γ0 and γ1.

Since we have defined a closed one-form to be a locally exact one-form, it is clear that any exactform is closed. In general the converse is false: this is quantified by what is called the first de Rham

cohomology of M , which by definition is the vector space quotient H1dR(M) ={closed 1-forms}{exact 1-forms} .

However if M is simply-connected the notions are equivalent:

Theorem 3.20. Let M be a simply connected smooth manifold. Then every closed one-form on M isexact.

Proof. Let α ∈ Ω1(M) be a closed one-form. By Proposition 3.18 and Remark 3.19, to show that αis exact it suffices to show that, if γ0,γ1 : [0,1]→ M are two smooth paths with γ0(0) = γ1(0) andγ0(1) = γ1(1) then

∫

[0,1] γ∗0α =

∫

[0,1] γ∗1α. Now because M is simply-connected, there is a smooth

8

map Γ : [0,1] × [0,1] → M such that Γ (·, 0) = γ0, Γ (·, 1) = γ1, and Γ (0, ·) and Γ (1, ·) are bothconstant. Just as in the proof of Theorem 3.13 we have

∫

[0,1]γ∗0α−

∫

[0,1]γ∗1α=

∫

∂ [0,1]2Γ ∗α

By Proposition 3.16, the fact that α is closed implies that Γ ∗α ∈ Ω1([0, 1]2) is closed, i.e., locallyexact. So writing Γ ∗α = Fd x + Gd y it follows (by equality of mixed partials, just as earlier) that∂ G∂ x =

∂ F∂ y . So by Green’s theorem we obtain

∫

∂ [0,1]2Γ ∗α=

∫∫

[0,1]2

∂ G∂ x−∂ F∂ y

d xd y = 0

�

Exercise 3.21. If M is an arbitrary smooth manifold and p0 ∈ M, prove that a closed one-form α ∈Ω1(M) gives rise to a well-defined homomorphism Iα : π1(M , p0)→ R by setting Iα([γ]) =

∫

[0,1] γ∗α

for any smooth γ: [0,1] → M with γ(0) = γ(1) = p0. Thus you need to show that this definitionis independent of the choice of γ within its homotopy class, and that it satisfies the homomorphismproperty (using the standard group operation on π1 and addition on R). (You may assume withoutproof that any loop based at p0 is based-homotopic to a smooth loop, and that any two based-homotopicsmooth loops are homotopic by a based homotopy which is smooth.)

The above exercise gives a map {closed 1-forms} → Hom(π1(M , p0),R) (sending α to Iα); boththe domain and codomain of this map are vector spaces in an obvious fashion and the map is clearly

8The fact that Γ can be taken to be smooth rather than just continuous follows by a standard argument with mollifiers.


linear. Moreover if α is exact then Corollary 3.17 says that Iα = 0, and so we get an induced mapon the quotient

H1dR(M) ={closed 1-forms}{exact 1-forms}

→ Hom(π1(M , p0),R)

If M is connected this map is in fact an isomorphism.

4. SYMPLECTIC AND HAMILTONIAN FLOWS

Definition 4.1. An almost symplectic manifold is a pair (M ,ω) where M is a smooth manifold andω ∈ Ω2(M) is a nondegenerate 2-form (i.e. if p ∈ M and 0 6= v ∈ Tp M then there is w ∈ Tp M suchthat ωp(v, w) 6= 0).

We will later define a symplectic manifold to be an almost symplectic manifold (M ,ω) such thatthe 2-form ω is also closed—however we have not yet said what it means for a 2-form to be closed.However we do already have a definition of a two-dimensional symplectic manifold (namely as apair (M ,ω) where M is a smooth two-manifold and ω ∈ Ω2(M) has ωp 6= 0 for all p), so we shouldcheck that the following is true:

Proposition 4.2. A symplectic two-manifold (M ,ω) is also an almost symplectic manifold.

Proof. Let (M ,ω) be a symplectic two-manifold, p ∈ M , and 0 6= v ∈ Tp M . Since Tp M is a two-dimensional vector space there is w ∈ Tp M such that {v, w} is a basis for Tp M . It suffices to show thatωp(v, w) 6= 0, so suppose for contradiction that ωp(v, w) = 0. Then the alternating condition for adifferential form yields that also ωp(w, v) = −ωp(v, w) = 0, and also that ωp(v, v) =ωp(w, w) = 0.But then any x , y ∈ Tp M can be written x = av+ bw and y = cv+ dw for some a, b, c, d ∈ Tp M , soby multilinearity

ωp(x , y) = acωp(v, v) + adωp(v, w) + bcωp(w, v) + bdωp(w, w) = 0.

Thus ωp 6= 0, contradicting the fact that (M ,ω) is a symplectic 2-manifold. �

Now let (M ,ω) be any almost symplectic manifold. The two-form ω then gives rise, for eachp ∈ M , to a linear map Tp M → T ∗p M defined by v 7→ ωp(v, ·). The nondegeneracy conditionprecisely says that this map is nonsingular, and so since dim Tp M = dim T ∗p M the map is bijective.

Thus if α ∈ Ω1(M) is an arbitrary one-form, there is a unique vector field α# characterized bythe property that ια#ω= α.

Now if (M ,ω) is a symplectic two-manifold a time-dependent vector field (X t) induces (at leastif each X t is compactly supported or satisfies an appropriate growth condition) a flow {ψtX }t∈Rcharacterized by the properties that ψ0X is the identity and, for all p ∈ M ,

dd tψ

tX (p) = X t(ψ

tX (p)).

The discussion after Theorem 3.13 shows that ψtX is area-preserving (i.e. (ψtX )∗ω = ω) for all t

if and only if each ιX tω is a closed one-form.9 This gives us a quite flexible way of constructing

area-preserving diffeomorphisms: choose a one-parameter family of closed one-forms αt all havingsupport in some fixed compact set, let α#t be the one-parameter family of vector fields characterizedby ια#t ω= αt , and take the flow {ψ

t(α#t )} of this time-dependent vector field (which exists because of

the compact support assumption). In fact, if one puts an appropriate (C∞) topology on the groupS ympc(M ,ω) of diffeomorphisms φ : M → M which are the identity outside a compact set andobey φ∗ω = ω, then any element of the identity component of S ympc(M ,ω) can be obtained asthe time-one map of such a flow.

A simple way of choosing a one-parameter family of closed one-forms αt for use in the aboveconstruction is to choose a one-parameter family of smooth functions Ht and then let αt = dHt .

9We will see later that this statement—as well as the discussion that follows—is also true on symplectic manifolds of anydimension, though not generally on almost symplectic manifolds

24 MIKE USHER

Definition 4.3. Let (M ,ω) be an almost symplectic manifold and H : M → R a smooth function.The Hamiltonian vector field of H is the unique vector field XH on M that satisfies the equation

(14) ιXHω= dH

Definition 4.4. Let (M ,ω) be an almost symplectic manifold. Let H : [0,1]× M → R be a smoothfunction and write H(t, ·) = Ht . The Hamiltonian isotopy generated by H is the family of diffeo-morphisms10 {φ tH}t∈[0,1] given by φ

tH = ψ

t(XHt )

, i.e. characterized by the conditions that φ0H = 1Mand, for each p ∈ M , dd tφ

tH(p) = XHt (φ

tH(p)). We let

Ham(M ,ω) =�

φ1H |H : [0,1]×M → R is smooth and compactly supported

Since an exact one-form is closed, it follows from the definition that, on a symplectic 2-manifold(and indeed on any symplectic manifold), Ham(M ,ω) is a subset of S ympc(M ,ω). Elements ofHam(M ,ω) are called Hamiltonian diffeomorphisms.

Exercise 4.5. Prove that Ham(M ,ω) is a group under composition. (So in particular you need toshow that the inverse of a Hamiltonian diffeomorphism is a Hamiltonian diffeomorphism, and thatthe composition of two Hamiltonian diffeomorphisms is a Hamiltonian diffeomorphism; both of theseshould require some effort.)

Let us consider Hamiltonian diffeomorphisms on the simplest symplectic 2-manifold, namely(R2, d x ∧ d y). Our examples will use Hamiltonians H which are independent of t (these are some-times called “autonomous”), and for ease of exposition we will include examples which are notcompactly supported (rather than multiplying them by a cutoff function to make them compactlysupported), though their flows will exist for all time because the resulting Hamiltonian vector fieldis Lipschitz.

Example 4.6. For a very simple example, suppose H(x , y) = c for some constant c. Then dH = 0, sothe Hamiltonian vector field XH—i.e., the unique vector field obeying ιXHω = dH, is the zero vector

field. So the diffeomorphisms φ tH are characterized by φ0H being the identity and

dφ tH (p)d t = 0 for all t;

thus φ tH is the identity for all t.

Example 4.7. Perhaps the second simplest example is given by taking H(x , y) = x. Then dH = d x,so the vector field XH is characterized by ιXH (d x ∧ d y) = d x, i.e., for (a, b) ∈ T(x ,y)R

2,

d x ∧ d y (XH , (a, b)) = d x(a, b) = a

Writing XH(x , y) = (P(x , y),Q(x , y)), the left hand side above is P(x , y)b −Q(x , y)a so we deducethat P(x , y) = 0 and Q(x , y) = −1. So XH(x , y) = (0,−1) for all (x , y), i.e. the vector field (x , y)points downward with unit length. So for (x0, y0) ∈ R2, we can write φ tH(x0, y0) = (x(t), y(t)) wherex(0) = x0, y(0) = y0, and (x ′(t), y ′(t)) = (0,−1). Thus φ tH(x0, y0) = (x0, y0 − t), i.e. φ

tH is given

by downward translation by t.

Example 4.8. Let H(x , y) = 12 (x2 + y2). Then dH = xd x + yd y, i.e. dH(x ,y)(a, b) = xa + y b

for (a, b) ∈ T(x ,y)R2. To find XH , write XH(x , y) = (P(x , y),Q(x , y)), so just as in the previousexample (ιXHω)(x ,y)(a, b) = P(x , y)b−Q(x , y)a. Setting this equal to dH(x ,y)(a, b) for all (a, b) givesP(x , y) = y and Q(x , y) = −x, i.e. XH(x , y) = (y,−x).

It is convenient to express this in terms of complex numbers, writing z = x + i y; then XH(z) =y− i x = −iz. So φ tH(z0) = z(t) where z(t) solves the initial value problem z(0) = z0 and z

′(t) = −iz.So z(t) = e−i tz0 and φ tH is the diffeomorphism of R

2 ∼= C given by clockwise rotation around the originby angle t.

10assuming that the ODE to follow has solutions, as will be the case if H is compactly supported


A similar analysis to that in the previous two examples shows that if H : R2 → R is any smoothfunction then its Hamiltonian vector field is given by XH(x , y) =

∂ H∂ y ,−

∂ H∂ x

. We could more gener-

ally consider R2 with the symplectic form ω= gd x ∧ d y for a nowhere-vanishing smooth functiong; note then that in the equation ιXHω= dH, if we leave H unchanged and multiplyω by g then wemust divide XH by g. So the Hamiltonian vector field of H with respect to the new form gd x ∧ d yis

1g∂ H∂ y ,−

1g∂ H∂ x

.

Exercise 4.9. Consider the sphere S2 with the standard area form from Example 2.17. Show that ifH : S2→ R is the height function given by H(x , y, z) = z for (x , y, z) ∈ S2, then the flow φ tH is given bya rotation around the z axis (and find the direction and speed of this rotation). Moreover find a formula(expressed in either spherical or Cartesian coordinates) for the Hamiltonian flow of K(x , y, z) = z2 onS2.

The above examples above all share the following property: the Hamiltonian flow φ tH of thetime-independent function H has the property that for any given p ∈ M the point φ tH(p) lies onthe same level curve of H for all time (these level curves were vertical lines when H(x , y) = x andcircles when H(x , y) = 12 (x

2 + y2). This is a general phenomenon with a simple explanation: wehave, using the chain rule and the definition of H

ddt

H(φ tH(p)) = dHφ tH (p)

�

dφ tH(p)

d t

�

= dHφ tH (p)(XH) =ωφ tH (p)(XH , XH) = 0.

In classical mechanics (for a particle moving in one dimension with position x and momentum pin our current R2 case), the function H represents the total (kinetic plus potential) energy of theparticle, which evolves according to the flow φ tH . Thus the above calculation is a derivation of theconservation of energy in classical mechanics.

The next few lectures discussed the wedge product and exterior derivative on differential formsand proved Stokes’ theorem; since this is covered in many introductory texts on smooth manifolds(for instance [6, Chapters 9-10], [11, Chapters 2 and 4]) I did not write notes for it.

4.1. Flux. Let (M ,ω) be a symplectic 2-manifold (actually the assumption that dim M = 2 will notreally be necessary). A compactly-supported, time-dependent vector field (X t)t∈[0,1] induces a flow{ψtX }t∈[0,1] characterized by the equations ψ

0X = 1M and

dd tψ

tX = X t ◦ψ

tX . We write

S ymp0(M ,ω) = {ψ1X |ιX tω is closed for all t}

andHam(M ,ω) = {ψ1X |ιX tω is exact for all t}

So obviously Ham(M ,ω) ≤ S ymp0(M ,ω) (both are groups; for Ham you proved this in Ex-ercise 4.5). In dimension two, it follows from what we have already done that every elementφ ∈ S ymp0(M ,ω) has φ∗ω = ω. (This is true, though we haven’t proven it yet, for symplec-tic manifolds of all dimensions; the proof relies on the fact that ω is closed. See Proposition 4.23below.) We will call a vector field X symplectic if ιXω is closed, and Hamiltonian if ιXω is exact. Sim-ilarly a time-dependent vector field (X t)will be called symplectic (resp. Hamiltonian) provided thatX t is symplectic (resp. Hamiltonian) for all t, and likewise for families of vector fields parametrizedby sets other than [0, 1].

Exercise 4.10. Let φ ∈ S ymp(M ,ω) (i.e. φ is a diffeomorphism with φ∗ω = ω; for instance φcould belong to S ymp0(M ,ω)), and let X be any vector field on M. Prove that φ∗X is a symplecticvector field if and only if X is symplectic, and that φ∗X is Hamiltonian if and only if X is Hamiltonian.

26 MIKE USHER

One can ask the question: given φ ∈ S ymp0(M ,ω), does φ in fact belong to Ham(M ,ω)?Evidently the answer is always yes if all closed one-forms are exact (e.g. if M is simply connected).But if there are closed forms that are not exact then one would probably expect to get a non-Hamiltonian element of S ymp0(M ,ω) by taki

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