The Pullback equation for degenerate forms

DISCRETE AND CONTINUOUS doi:10.3934/dcds.2010.27.657DYNAMICAL SYSTEMSVolume 27, Number 2, June 2010 pp. 657–691

THE PULLBACK EQUATION FOR DEGENERATE FORMS

Saugata Bandyopadhyay

Department of Mathematical SciencesIISER Kolkata

Mohanpur Campus, PO: BCKV Campus Main OfficeMohanpur - 741252, West Bengal, India

Bernard Dacorogna and Olivier Kneuss

Section de MathematiquesStation 8, EPFL

1015 Lausanne, Switzerland

Abstract. We discuss the existence of a diffeomorphism ϕ : Rn → Rn suchthat

ϕ∗ (g) = f,

where f, g : Rn → Λk are closed differential forms and 2 ≤ k ≤ n − 1.

1. Introduction. In this article we discuss the existence of a diffeomorphism ϕ :R

n → Rn such that

ϕ∗ (g) = f (1)

where f, g : Rn → Λk (Rn) , f (x) , g (x) 6= 0 for every x ∈ R

n, are closed differentialforms (i.e. df = dg = 0), 2 ≤ k ≤ n

g =∑

1≤i1<···<ik≤n

gi1···ik(x) dxi1 ∧ · · · ∧ dxik

and similarly for f. The meaning of (1) is that∑

1≤i1<···<ik≤n

gi1···ik(ϕ (x)) dϕi1∧· · ·∧dϕik =

∑

1≤i1<···<ik≤n

fi1···ik(x) dxi1∧· · ·∧dxik .

We will concentrate mainly on local existence, though in some special cases we willobtain global results. A condition which is somewhat necessary (cf. Theorem 2.8)to solve this problem is that the rank of the two forms should be the same (for aprecise definition see below), namely

rank g = rank f.

It will turn out that, when k = 2, k = n−1 or k = n, if the rank of the two forms isequal and constant then, at least locally, there exists a diffeomorphism ϕ satisfying(1).

This problem has been intensively studied in two main cases, that we call non-degenerate.

2000 Mathematics Subject Classification. Primary: 35Fxx; Secondary: 58A10.Key words and phrases. Darboux theorem, Differential forms, Pullback equation.

657

http://dx.doi.org/10.3934/dcds.2010.27.657

658 SAUGATA BANDYOPADHYAY, BERNARD DACOROGNA AND OLIVIER KNEUSS

Case 1: k = 2, n = 2m and rank g = rank f = n, this is the celebrated Darbouxtheorem (cf., for example, Abraham-Marsden-Ratiu [2], McDuff-Salamon [13] orTaylor [22]) which states that if g is the standard symplectic form, namely

g = ωm =

m∑

i=1

dx2i−1 ∧ dx2i

and f : Rn → Λ2 (Rn) with df = 0 and rank g = rank f = n, then there exists a

diffeomorphism ϕ defined locally such that

ϕ∗ (g) = f.

This result was significantly improved in Bandyopadhyay-Dacorogna [3] in two maindirections:

- optimal regularity in Holder spaces was obtained (see Theorem 18 in [3], whichis restated in Theorem 3.1 below),

- more strikingly a global result with Dirichlet boundary data was established(cf. Theorems 13 and 15 in [3]).

Case 2: k = n and rank g = rank f = n (which is equivalent to f (x) , g (x) 6= 0for every x ∈ R

n). This is the fundamental result of Moser in [15], later alsoconsidered by Banyaga [4], Cupini-Dacorogna-Kneuss [7], Dacorogna [8], Reimann[17], Riviere-Ye [18], Tartar [21], Ye [23] and Zehnder [24]. The optimal regularityin Holder spaces has been obtained in Dacorogna-Moser [10].

We here want to concentrate our attention on degenerate cases which are of twotypes.

Case 3: k = 2 and rank g = rank f = 2l with 2 ≤ 2l < n. This case is adegenerate version of Darboux theorem and has also been intensively studied (see,for example, Abraham-Marsden [1]), the result is that if

g = ωl =

l∑

i=1

dx2i−1 ∧ dx2i

and f : Rn → Λ2 (Rn) with df = 0 and rank g = rank f = 2l, then there exists a

diffeomorphism ϕ defined locally such that

ϕ∗ (g) = f.

The usual proof uses Frobenius theorem reducing the dimension to R2l and then

appeals to the non-degenerate case. We here give two proofs. The first one (cf.Theorem 3.2) inspired by the classical one but using the theorem of Bandyopadhyay-Dacorogna (cf. Theorem 3.1) instead of the standard Darboux theorem; it allows usto improve the known regularity. We also give a totally different proof (cf. Theorem3.11) using the flow method. This last proof seems to be more appropriate if onewants to obtain global results.

Case 4: 3 ≤ k ≤ n − 1. In this context we necessarily have (recall that f (x),g (x) 6= 0 for every x ∈ R

n)

rank g = rank f ∈ k, k + 2, · · · , n

and thus when k = n − 1 we only have rank g = rank f = n − 1. The case ofk−forms, with 3 ≤ k ≤ n− 1, has been much less studied than the other ones andit is considerably harder. In particular, except when k = n − 1, the conservationof the rank is not the only necessary condition. The notion of decomposability ofa k−form enters into play (see below). We are able to handle (cf. Theorem 4.1)

PULLBACK EQUATION FOR DEGENERATE FORMS 659

completely the case rank g = rank f = k (and so completely the case k = n − 1)and obtain a non-trivial regularity result, though not optimal. We can also dealwith several other cases (cf. Theorems 4.4, 4.6, 4.8) but the general case is still farfrom complete. For previous work on this subject, we refer, notably for the casek = n− 1, to Bandyopadhyay-Dacorogna in [3].

We should also point out that when k = n, the problem of degeneracy also occursin a different context, for example when g > 0 and f may become 0 or even is allowedto change sign. Of course then one cannot expect to have a diffeomorphism. Thishas been studied in Cupini-Dacorogna-Kneuss [7].

2. Some algebraic preliminaries.

2.1. Notations. We denote a k−form g : Rn → Λk (Rn) by

g =∑

1≤i1<···<ik≤n

gi1···ik(x) dxi1 ∧ · · · ∧ dxik .

Sometimes it will be more convenient to assign meaning to gi1···ikfor any k−index

and we will let

gi1···ik= (sgnσ) giσ(1)···iσ(k)

where σ is a permutation of 1, · · · , k .

Throughout the paper, to avoid burdening the notations, we will identify, when

necessary, k−forms with vectors in R

(nk

). We moreover adopt the following standard

notations.

1) Let 0 ≤ k ≤ n, we recall that the Hodge ∗−operator associates to f ∈ Λk (Rn)the (n− k)−form

∗f ∈ Λn−k (Rn) .

2) For 0 ≤ l, k ≤ n, f ∈ Λl (Rn) and g ∈ Λk (Rn) , the exterior product is denotedby

f ∧ g ∈ Λk+l (Rn) .

3) For f, g ∈ Λk (Rn) we write inner product as

〈f ; g〉 =∑

1≤i1<···<ik≤n

fi1···ikgi1···ik

∈ R.

4) For 0 ≤ l ≤ k ≤ n, f ∈ Λl (Rn) and g ∈ Λk (Rn) , the interior product willthen be defined by

f y g = (−1)n(k−l)

∗ [f ∧ (∗g)] ∈ Λk−l (Rn) .

In particular if f ∈ Λ1 (Rn) and g ∈ Λk (Rn) , we find

f y g =∑

1≤i1<···<ik≤n

gi1···ik

k∑

r=1

(−1)r−1 firdxi1 ∧ · · · ∧ dxir−1 ∧ dxir+1 ∧ · · · ∧ dxik

= (−1)k−1

∑

1≤i1<···<ik−1≤n

n∑

ik=1

gi1···ikfikdxi1 ∧ · · · ∧ dxik−1 (2)

which turns out when k = 2 to give

f y g = −n∑

i=1

n∑

j=1

gijfjdxi .


If l > k, we let

f y g = 0.

Note that if l = k, then

f y g = 〈f ; g〉 .

5) The exterior derivative is denoted by

d : C1(R

n; Λk)→ C0

(R

n; Λk+1).

6) Let ϕ : Rn → R

N

x =(x1, · · · , xN

)= ϕ (y) = ϕ

(y1, · · · , yn

)=(ϕ1 (y) , · · · , ϕN (y)

)

and g : RN → Λk

(R

N), 0 ≤ k ≤ min n,N ,

g =∑

1≤i1<···<ik≤N

gi1···ik(x) dxi1 ∧ · · · ∧ dxik .

We define the pullback of g by ϕ, denoted by f = ϕ∗ (g) : Rn → Λk (Rn) , through

∑

1≤j1<···<jk≤n

fj1···jk(y)dyj1∧· · ·∧dyjk =

∑

1≤i1<···<ik≤N

gi1···ik(ϕ (y)) dϕi1∧· · ·∧dϕik .

In particular ifN = n and k = 2, then the above equation is, in terms of components,a system of

(n2

)first order pdes, namely

∑

1≤p<q≤n

gpq (ϕ (x))

(∂ϕp

∂xi

∂ϕq

∂xj−∂ϕp

∂xj

∂ϕq

∂xi

)= fij , for every 1 ≤ i < j ≤ n.

This explains the reason for which we use, throughout the article, the notationsof differential geometry rather than those with local coordinates, which are morefamiliar to analysts but are unfortunately much heavier.

We have as a direct consequence of the definitions the following.

Proposition 2.1. Let f ∈ Λk (Rn) , g ∈ Λl (Rn) and h ∈ Λ1 (Rn) , then

h y (f ∧ g) = (h y f) ∧ g + (−1)kl

(h y g) ∧ f

and, in particular,

h y (h ∧ g) = (h yh) ∧ g + (−1)k (h y g) ∧ h.

If, moreover, l is even and m is an integer, then

h y gm+1 = (m+ 1) (h y g) ∧ gm

where gm = g ∧ · · · ∧ g︸︷︷︸m−times

.

Finally, if l = k − 1, then

〈f ; g ∧ h〉 = (−1)k−1

〈h y f ; g〉 .


2.2. The rank of a form. The notion of the rank of a 2−form is standard, al-though it is usually presented in a different way. However the corresponding onefor a k−form, 3 ≤ k ≤ n, is less standard but can be found, for example, in [3], [11]or [20]. To define the rank it is more convenient to represent u→ u y g as a matrixoperating on a vector.

Definition 2.2. Let g ∈ Λk (Rn) , 1 ≤ k ≤ n.

(i) The alternate representation of g is the matrix g ∈ R

(n

k−1

)×n

defined through(cf. (2))

u y g = g u, for every u ∈ Λ1 (Rn)

where, as already said and by abuse of notations, in the left hand side the (k − 1)-

form u y g is identified with a vector in R

(n

k−1

)while in the right hand side of the

definition the 1−form u is identified with a vector in Rn. More explicitly, using the

lexicographical order for the columns (index below) and the rows (index above) ofthe matrix g , we have

(g)j1···jk−1

i = gij1···jk−1

for 1 ≤ i ≤ n and 1 ≤ j1 < · · · < jk−1 ≤ n.

(ii) The rank of the k−form g is then, by definition, the rank of the matrix g,so we can write

rank g = rank g.

When k = 2, the matrix g ∈ Rn×n and it is given by

g = (gij) ∈ Rn×n with gij = −gji .

We immediately have the following elementary result.

Proposition 2.3. Let g ∈ Λk (Rn) , 1 ≤ k ≤ n.

(i) Define the vector space

Λ1g =

a ∈ Λ1 (Rn) : ∃ b ∈ Λk−1 (Rn) with b y g = a

,

then

rank g = dim Λ1g .

(ii) If g 6= 0 and 3 ≤ k ≤ n, then

rank g ∈ k, k + 2, · · · , n

and any of the values in k, k + 2, · · · , n can be achieved by the rank of a k−form.

(iii) If k = 2, then the rank of g, g 6= 0, is even and any even value less than orequal to n can be achieved by the rank of a 2−form. Moreover rank g = 2l if andonly if

gl 6= 0 and gl+1 = 0

where gl = g ∧ · · · ∧ g︸︷︷︸l−times

.

(iv) If f ∈ Λl (Rn) , then

rank (f ∧ g) ≤ rank f + rank g − dim(Λ1

f ∩ Λ1g

).

Moreover,

rank (f ∧ g) = rankf + rank g ⇔ Λ1f ∩ Λ1

g = 0 .


(v) If A ∈ GLn (R) , then

rank (ϕ∗ (g)) = rank g

where ϕ (x) = Ax.

Remark (i) If g ∈ Λ1 (Rn) and g 6= 0, then

rank g = 1.

If g ∈ Λn (Rn) with g 6= 0, then

rank g = n.

(ii) For g ∈ Λk (Rn) , then the rank of g can never be (k + 1) . In particular whenk = n− 1 and g 6= 0, then

rank g = n− 1.

(iii) When k = 2 and rank g = 2l = n, then

det g =1

l!

(gl)2> 0.

(iv) If g ∈ Λ2 (Rn) and f ∈ Λ1 (Rn) with f ∧ g 6= 0, then

rank (f ∧ g) = rank g ± 1

and thus is odd. In particular if rank g = n (and therefore n is even), then

rank (f ∧ g) = rank g − 1 = n− 1.

(v) Let g ∈ Λk (Rn) . Then rank g = k if and only if (cf. Proposition 2.5 below)

g ∧ [f y g] = 0

for every f ∈ Λk−1 (Rn) .

(vi) From (iv) of the proposition we can infer that if g 6= 0, then

rank (∗g) ≥ n− rank g.

When k = 1 then the inequality becomes an equality. In general, however, as soonas 2 ≤ k ≤ n− 2 the inequality can be strict.

2.3. Decomposability of a form. The other important algebraic concept is thenotion of decomposability.

Definition 2.4. Let g ∈ Λk (Rn) , 1 ≤ k ≤ n.

(i) Let 1 ≤ l ≤ k − l. We say that g is l−decomposable if there exist a ∈ Λl (Rn)and b ∈ Λk−l (Rn) such that

g = a ∧ b.

We say that g is indecomposable if it is not l−decomposable for any l.

(ii) We say that g is totally decomposable if there exist a1, · · · , ak ∈ Λ1 (Rn) suchthat

g = a1 ∧ · · · ∧ ak .

We now gather some properties whose proofs are elementary; the last result isknown as Cartan lemma (see, for example, [16]).


Proposition 2.5. Let 1 ≤ k ≤ n and g ∈ Λk (Rn) with g 6= 0.

(i) The form g is 1−decomposable meaning that there exist a ∈ Λ1 (Rn) andb ∈ Λk−1 (Rn) such that

g = a ∧ b

if and only if there exists a ∈ Λ1 (Rn) , a 6= 0, such that

g ∧ a = 0

if and only if there exists f ∈ Λk−1 (Rn) such that f y g 6= 0 and

g ∧ [f y g] = 0.

(ii) The form g is totally decomposable if and only if rank g = k if and only if

g ∧ [f y g] = 0

for every f ∈ Λk−1 (Rn) .

(iii) If k is odd and if rank g = k + 2, then g is 1−decomposable.

(iv) Let 1 ≤ l ≤ k ≤ n and a1, · · · , al ∈ Λ1 (Rn) be such that

a1 ∧ · · · ∧ al 6= 0.

Then there exists f ∈ Λk−l (Rn) verifying

g = f ∧ a1 ∧ · · · ∧ al

if and only if

g ∧ a1 = · · · = g ∧ al = 0.

Remark (i) In the literature the second definition is standard; it goes back toCartan and such a form is, sometimes, also called pure or only decomposable. Thefirst definition seems to be new. Note that we have restricted 1 ≤ l ≤ k − l, since ak−form is l decomposable if and only if it is (k − l) decomposable.

(ii) It is important to stress that, in general, the rank of a k−form and itsdecomposability are two distinct notions. The only exceptions are the cases k =2, n− 1, n, where the rank determines completely the decomposability.

- If k = 2, we have only two possibilities. Either rank g = 2 and then (accordingto (ii) of Proposition 2.5) the form g is totally decomposable, or rank g ≥ 4 and itis indecomposable.

- If k = n−1, n then we are automatically in the framework of (ii) of Proposition2.5 and therefore all forms are totally decomposable.

(iii) We should point out that a form is not uniquely decomposable into inde-composable forms. Indeed consider

g =[dx1 ∧ dx2 + dx3 ∧ dx4

]∧ dx3 = dx1 ∧ dx2 ∧ dx3

and observe that it is a product of one indecomposable 2−form of rank 4 and one(indecomposable) 1−form and, at the same time, a product of three (indecompos-able) 1−forms. However only the second one is an optimal decomposition of g, inthe sense that

g = gk1 ∧ · · · ∧ gks

with k1 + · · · + ks = k, gki∈ Λki (Rn) indecomposable and

rank g =

s∑

i=1

rank gki.


An optimal decomposition of the above type does not always exist as the followingexample shows. Let g ∈ Λ4

(R

6)

with rank g = 6 and given by

g = a ∧ b =[dx1 ∧ dx2 + dx3 ∧ dx4

]∧[dx1 ∧ dx2 + dx5 ∧ dx6

].

Note that a, b ∈ Λ2(R

6)

are indecomposable and ranka = rank b = 4. It can beseen that g cannot be optimally decomposed in the above sense.

(iv) The statement (iii) in the proposition is, in general, false when k is even.Indeed the form g ∈ Λ4 (Rn) given by

g = dx1 ∧ dx2 ∧ dx3 ∧ dx4 + dx1 ∧ dx2 ∧ dx5 ∧ dx6 + dx3 ∧ dx4 ∧ dx5 ∧ dx6

is not 1−decomposable (although it is 2−decomposable) while rank g = k + 2 = 6.

(v) When k = 3, g 6= 0 and rank g is even then g is indecomposable. This easilyfollows from the fact that if g is 1−decomposable, there exists a ∈ Λ1 (Rn) andb ∈ Λ2 (Rn) so that g = a ∧ b and therefore rank g is odd.

2.4. Solutions to some algebraic equations. We here provide an example, thatwill be used later, on how to solve equations involving the interior or the exteriorproduct.

Proposition 2.6. Let g ∈ Λ2 (Rn) with rank g = 2l ≤ n and a ∈ Λ1 (Rn) . Thereexists u ∈ Λ1 (Rn) such that

a = u y g

if and only ifa ∧ gl = 0.

Remark (i) If n = 2l, then a ∧ gl = 0 is always satisfied (since then a ∧ gl ∈Λn+1 (Rn)) and therefore there always exists u ∈ Λ1 (Rn) such that a = u y g.

(ii) More generally if l is even, g ∈ Λk (Rn) with gl+1 = 0 (lk ≤ n) and thereexists u ∈ Λ1 (Rn) such that a = u y g, then necessarily (cf. Proposition 2.1)

a ∧ gl = 0.

However the converse is then, in general, not true.

Before starting our proof we will need the following lemma, whose proof is ele-mentary (see [12]).

Lemma 2.7. Let 2 ≤ 2l ≤ n be integers and g ∈ Λ2 (Rn) . Then, for every 1 ≤m ≤ 2l,

1

l

(gl)i1···i2l

=m−1∑

j=1

(−1)j+m+1 gijim

(gl−1

)i1···ij ···im···i2l

+

2l∑

j=m+1

(−1)j+m+1

gimij

(gl−1

)i1···im···ij ···i2l

for every set of (2l)−indices 1 ≤ i1 , · · · , i2l ≤ n.

Notation. When in an index we write ı, this means that i is omitted. For example

h1···4···k = h1235···k .

We will adopt this notation throughout the present article.

We can now turn our attention to the proof of Proposition 2.6.


Proof. Step 1. We start with the necessary part. Since rank g = 2l, we have gl 6= 0and gl+1 = 0. We hence deduce, using Proposition 2.1, that

0 = u y gl+1 = (l + 1) (u y g) ∧ gl = (l+ 1) a ∧ gl,

which implies that a ∧ gl = 0.

Step 2. We now prove the sufficiency part. Without loss of generality, we canassume that (gl)1···(2l) 6= 0. Since a ∧ gl = 0 we have, for every 2l+ 1 ≤ k ≤ n,

ak(gl)1···(2l) =

2l∑

i=1

(−1)iai(gl)1···i···(2l)k . (3)

Note that (3) is trivially valid for 1 ≤ k ≤ 2l. For every 1 ≤ k ≤ n, it follows, fromLemma 2.7 (with m = 2l, ij = j if j < i, ij = j + 1 if i ≤ j ≤ 2l − 1 and i2l = k),that

1

l(gl)1···i···(2l)k =

i−1∑

j=1

(−1)j+1gjk(gl−1)1···j···i···(2l)

+

2l∑

j=i+1

(−1)jgjk(gl−1)1···i···j···(2l) .

We thus deduce from (3) that, for every 1 ≤ k ≤ n,

ak(gl)1···(2l) = l

2l∑

i=1

(−1)iai

[ ∑i−1j=1(−1)j+1gjk(gl−1)1···j···i···(2l)

+∑2l

j=i+1(−1)jgjk(gl−1)1···i···j···(2l)

]

and thus

ak(gl)1···(2l) = −l

2l∑

j=1

(−1)j+1gjk

[ ∑j−1i=1 (−1)iai(gl−1)1···i···j···(2l)

−∑2l

i=j+1(−1)iai(gl−1)1···j···i···(2l)

].

We now define u ∈ Λ1(Rn) by

uj = 0 if 2l+ 1 ≤ j ≤ n

while, if 1 ≤ j ≤ 2l,

uj =(−1)j+1 l

(gl)1···(2l)

j−1∑

i=1

(−1)iai(gl−1)1···i···j···(2l) −

2l∑

i=j+1

(−1)iai(gl−1)1···j···i···(2l)

It follows from the definition of u that, for every 1 ≤ k ≤ n,

(u y g)k = −2l∑

j=1

gjkuj = ak .

In other words, we have u y g = a. This is the desired claim.

2.5. Necessary conditions for the pullback equation. In the following theo-rem we gather elementary necessary conditions (statements (ii), (iii) and (iv) beloware straightforward and were already mentioned in [3]).

Theorem 2.8. Let Ω ⊂ Rn be a smooth domain, 1 ≤ k ≤ n, f, g ∈ C1(Ω; Λk) and

ϕ ∈ Diff1(Ω; Rn

)be such that

ϕ∗ (g) = f in Ω.

Then the following conditions hold.


(i) The forms f (x) and g (ϕ (x)) satisfy, for every x ∈ Ω,

rank g (ϕ (x)) = rank f (x) .

(ii) If dg = 0 in Ω, then

df = 0.

(iii) If Ω is bounded, ϕ(Ω)

= Ω and n = mk with m an integer, then∫

Ω

fm =

∫

Ω

gm

where fm = f ∧ · · · ∧ f︸︷︷︸m−times

.

(iv) If ϕ (x) = x for x ∈ ∂Ω, then

f ∧ ν = g ∧ ν on ∂Ω

where ν is the exterior unit normal to Ω.

Remark Note that pullback preserves not only the rank but also the decompos-ability in the sense that if

g = a ∧ b, a ∈ Λl (Rn) , b ∈ Λk−l (Rn)

and ϕ∗ (g) = f, then c = ϕ∗ (a) ∈ Λl (Rn) , d = ϕ∗ (b) ∈ Λk−l (Rn) are such that

f = c ∧ d.

3. New proofs of Darboux theorem in the degenerate case.

3.1. The main result. We start by recalling the classical Darboux theorem. Theversion that we quote below gives the optimal regularity result, which is a partic-ularly delicate point, and has been established by Bandyopadhyay-Dacorogna (seeTheorem 18 in [3]).

Theorem 3.1 (Darboux theorem with optimal regularity). Let r ≥ 0 and n =2m ≥ 4 be integers. Let 0 < α < 1 and x0 ∈ R

n. Let ωm be the standard symplecticform

ωm =

m∑

i=1

dx2i−1 ∧ dx2i.

Let ω be a 2−form. The two following statements are then equivalent.

(i) The 2−form ω is closed, is in Cr,α in a neighbourhood of x0 and verifies

rankω (x0) = n.

(ii) There exist a neighbourhood V of x0 and ϕ ∈ Diffr+1,α(V ; Rn) such that

ϕ∗ (ωm) = ω in V and ϕ (x0) = x0 .

Remark (i) We refer to [3] for global results including Dirichlet data.

(ii) When r = 0, the hypothesis dω = 0 is to be understood in the sense ofdistributions.

(iii) The theorem is still valid when n = 2, but it is then the result of Dacorogna-Moser [10] (cf. Theorem 14.6 in [9]).

(iv) We will use throughout the following notations. Let Ω ⊂ Rn be open, r a

non-negative integer and 0 < α ≤ 1.


- We denote by Cr,α(Ω)

the usual set of Holder functions and by Cr,α(Ω; Λk

)

the set of k−forms

g =∑

1≤i1<···<ik≤n

gi1···ikdxi1 ∧ · · · ∧ dxik

so that gi1···ik∈ Cr,α

(Ω).

- The sets Diffr(Ω; Rn) and Diffr,α(Ω; Rn) denote the sets of diffeomorphisms ϕso that ϕ ∈ Cr

(Ω; Rn

)and ϕ−1 ∈ Cr

(ϕ(Ω); Rn

), Cr,α respectively.

- When α = 0, sometimes, by abuse of notations, we write indistinctly Cr,0 orCr.

The main result of the present section is the following version of Darboux theoremfor degenerate forms.

Theorem 3.2. Let n ≥ 3, r, l ≥ 1 be integers and 0 < α < 1. Let x0 ∈ Rn and ωl

be the standard symplectic form of rank 2l < n, namely

ωl =

l∑

i=1

dx2i−1 ∧ dx2i.

Let ω be a Cr,α closed 2−form such that

rankω = 2l in a neighbourhood of x0 .

Then there exist a neighbourhood V of x0 and ϕ ∈ Diffr,α(V ; Rn) such that

ϕ∗ (ωl) = ω in V and ϕ (x0) = x0 .

Remark (i) The theorem is standard in the C∞ case. In all proofs that we haveseen the regularity that is established is, at best, that if ω ∈ Cr then ϕ ∈ Cr−1.However our result asserts that ω and ϕ have the same regularity in Holder spaces.This is, of course, better but still not optimal as in the non-degenerate case ofTheorem 3.1.

(ii) The standard proof uses Frobenius theorem to reduce the dimension so thatthe forms have maximal rank and then apply the classical Darboux theorem. Wewill follow this path but using the more sophisticated Theorem 3.1. We will alsoprovide a completely different proof in Section 3.4; it will use an argument basedon the flow method. Still a different proof can be found in [3] when n = 2l+ 1.

Proof. Step 1. Without loss of generality, we can assume x0 = 0. We first find,appealing to Theorem 5.2, two neighbourhoods V, Ω of 0 and ψ ∈ Diffr,α(V ; Ω)with ψ (0) = 0 and

ψ∗ (ω)(x1, · · · , xn

)= ω

(x1, · · · , x2l

)=

∑

1≤i<j≤2l

ωij

(x1, · · · , x2l

)dxi ∧ dxj .

Therefore ψ∗ (ω) = ω ∈ Cr−1,α in a neighbourhood of 0 in R2l and rank ω = 2l in

a neighbourhood of 0.

Step 2. We then apply Theorem 3.1 to ω and find a neighbourhood U ⊂ R2l of

0 and χ ∈ Diffr,α(U ; R2l), with χ (0) = 0, such that

χ∗ (ωl) = ω in U.

Setting

χ (y) = χ(y1, · · · , y2l, y2l+1, · · · , yn

)=(χ(y1, · · · , y2l

), y2l+1, · · · , yn

)


and ϕ = χ ψ−1, we have the claim.

3.2. Poincare lemma with constraint. We start with a generalization of Poincarelemma that will play the central role in our second proof (Theorem 3.11) of Darbouxtheorem in the degenerate case.

Theorem 3.3. Let 2 ≤ 2l ≤ n be integers and x0 ∈ Rn. Let f, g be two C∞ closed

2−forms such that, in a neighbourhood of x0 ,

f ∧ gl = 0 and rank g = 2l.

Then there exist a neighbourhood V of x0 and ω ∈ C∞(V ; Λ1) such that

dω = f, ω (x0) = 0 and ω ∧ gl = 0.

Remark We can easily replace C∞ by Cr, but a refined version of our constructionfinds ω only in Cr−1 (see [12]).

Proof. The theorem is easily proved if we can invoke Theorem 3.2. However, sincewe want to use the present theorem to establish Theorem 3.2, we have to find anindependent proof.

Step 1. Without loss of generality, we can assume that x0 = 0. Appealing to theclassical Poincare lemma, we can find a neighbourhood V of 0 and u ∈ C∞(V ; Λ1)with u(0) = 0 and du = f in V. We then set

ω = u− dv.

Our result will follow if we can find v ∈ C∞(V ) verifyingdv ∧ gl = u ∧ gl in V

dv(0) = 0.(4)

Step 2. Since gl 6= 0, we can find a matrix A ∈ GLn(R), so that if ϕ(x) = Ax,then

(ϕ∗(gl))1···(2l)(0) > 0.

The problem (4) is then equivalent to finding w ∈ C∞(V ) such thatdw ∧ ϕ∗(gl) = ϕ∗(u) ∧ ϕ∗(gl) in V

dw(0) = 0.

Indeed it is enough to set v = (ϕ−1)∗(w) to have a solution of (4). So from now onwe will assume, without loss of generality, that

(gl)1···(2l)(0) > 0.

Step 3. We then solve (4) by induction on n, l being fixed. In the case n = 2lnothing is to be proved, just choose v = 0. So we assume that the result has beenproven for n = 2l + j, j ≥ 0, and let us prove it for n = 2l + j + 1. We therefore

assume that we can find a neighbourhood V ⊂ Rn−1 of 0 ∈ R

n−1 and h ∈ C∞(V )with

dh ∧ gl = u ∧ gl

dh(0) = 0

whenever g ∈ C∞(V ; Λ2(Rn−1)) and u ∈ C∞(V ; Λ1(Rn−1)) verify

du ∧ gl = 0, dg = 0, gl+1 = 0 and (gl)1···2l(0) 6= 0


and let us prove that it holds for n. To establish this result we proceed in fivesubsteps.

Step 3.1. It follows from Proposition 3.10 that there exist V and W neighbour-hoods of 0 ∈ R

n and ϕ ∈ Diff∞(W ;V ) with ϕ (0) = 0 and such that

ϕ∗ (g) =l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

rijdxi ∧ dxj

for some rij ∈ C∞(W ). Therefore solving (4) is equivalent to finding a solution v of

dv ∧ (ϕ∗(g))l = ϕ∗(u) ∧ (ϕ∗(g))l

dv(0) = 0.

So from now on we will assume, upon substitution of W, ϕ∗(g) and ϕ∗(u) by V, gand u, that

g =

l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

rijdxi ∧ dxj (5)

and we therefore have to find v ∈ C∞(V ) satisfying (4) only for g as in (5). Notethat for such a g we have

(gl)1···(2l) = l ! 6= 0.

Step 3.2. We then solve, by the method of characteristics, the Cauchy problemfor the first order partial differential equation

(dv ∧ gl)1···(2l)n = (u ∧ gl)1···(2l)n

v(x1, · · · , xn−1, 0) = h(x1, · · · , xn−1)(6)

where h ∈ Cr(V ) is a solution, which exists by hypothesis of induction, ofdh ∧ i∗n(g)l = i∗n(u) ∧ i∗n(g)l

dh(0) = 0.(7)

and

in(x1, · · · , xn−1) = (x1, · · · , xn−1, 0)

Observe that

i∗n(v) = h ⇔ v(x1, · · · , xn−1, 0) = h(x1, · · · , xn−1).

Note that we can apply the hypothesis of induction since

dg = 0, du ∧ gl = 0, gl+1 = 0 and (gl)1···(2l)(0) 6= 0

imply

d(i∗n(g)) = 0, d(i∗n(u)) ∧ i∗n(g)l = 0, i∗n(g)l+1 = 0 and (i∗n(g))1···(2l)(0) 6= 0.

Step 3.3. It now remains to prove that the solution v of (6) is indeed a solutionof (4). We therefore have to show that

dv ∧ gl = u ∧ gl (8)

and

dv(0) = 0. (9)


Using (6), (7) and the fact that u(0) = 0, we easily see that (9) holds. We here onlyestablish (8). Lemma 3.6 implies that to show (8) we only need to prove

(dv ∧ gl

)1···(2l)k

= (u ∧ gl)1···(2l)k , 2l + 1 ≤ k ≤ n. (10)

Step 3.4. We now prove (10). Define, for every 2l+ 1 ≤ k ≤ n,

Lkv = (dv ∧ gl)1···(2l)k and wk = Lkv − (u ∧ gl)1···(2l)k .

Since we already have from (6) that wn = 0, our claim (10) reduces to proving that

wk = 0 for every 2l+ 1 ≤ k ≤ n− 1. (11)

Since

0 = f ∧ gl = du ∧ gl

and (5) holds, we can apply Lemma 3.7 and Lemma 3.8 to obtain

Lnwk = LnLkv − Ln

((u ∧ gl)1···(2l)k

)

= LkLnv − Lk

((u ∧ gl)1···(2l)n

)= Lkwn = 0.

Assume, cf. Step 3.5, that we can prove that

wk(x1, · · · , xn−1, 0) = 0 (12)

we will then have, by uniqueness of the solutions of the Cauchy problem, that theonly solution of

Lnwk = 0

wk = 0 on xn = 0.

is wk = 0. This is exactly our claim (11).

Step 3.5. Finally we show (12), which is equivalent to proving that i∗n(wk) = 0.We have that, recalling that i∗n(v) = h,

i∗n(wk) = i∗n((dv ∧ gl − u ∧ gl)1···(2l)k

)=(i∗n(dv ∧ gl − u ∧ gl)

)1···(2l)k

=(d(i∗n(v)) ∧ i∗n(gl)) − i∗n(u) ∧ i∗n(gl)

)1···(2l)k

and thus, appealing to (7),

i∗n(wk) =(dh ∧ i∗n(gl) − i∗n(u) ∧ i∗n(gl)

)1···(2l)k

= 0.

This concludes the proof of the theorem.

With substantially the same proof we can get a global result (see Kneuss [12]).

Theorem 3.4. Let 2 ≤ 2l < n and g, f ∈ C∞(Rn; Λ2) be closed. Assume that g isof the form

g =

l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

gij dxi ∧ dxj

for some gij ∈ C∞(Rn) and, for every x ∈ Rn,

gl+1 (x) = 0 and |g (x)| ≤ a |x| + b

where a, b > 0 are constants. Then there exists w ∈ C∞(Rn; Λ1) so that the follow-ing equations are satisfied on the whole of R

n

dw = f and w ∧ gl = 0.


3.3. Some technical algebraic lemmas. In this subsection we gather all alge-braic lemmas that we have used in the proof of Theorem 3.3. In the sequel we willadopt Notation 2.4.

Lemma 3.5. Let 2 ≤ 2l < n be integers and g ∈ Λ2(Rn) with rank g = 2l and ofthe form

g =l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

gij dxi ∧ dxj .

Then, for every 1 ≤ i, j ≤ 2l < k ≤ n, the following holds

(gl)1···i···(2l)n (gl)1···j···(2l)k − (gl)1···i···(2l)k (gl)1···j···(2l)n

=

0 if i = j

l! (gl)1···i···j···(2l)knif i < j

−l! (gl)1···j···i···(2l)kn if i > j

Remark When l = 1 then g2 = 0 and the conclusion of the lemma is immediateand reads as

g2ng1k − g2kg1n = gkn = g12gkn .

Proof. We begin by noting that, when i = j, the proof is trivial. We prove theresult for i < j, which in turn implies the case i > j. Since rank g = 2l, we have

gl+1 = 0. (13)

Moreover, the special structure of g gives

g12 = g34 = · · · = g(2l−1)(2l) = 1 (14)

gij = 0, for every 1 ≤ i < j ≤ 2l with i, j 6= 2s− 1, 2s, 1 ≤ s ≤ l. (15)

We define ǫ : N → −1, 1 as

ǫ(i) =

−1 if i is even

1 if i is odd

Note that i+ ǫ(i), i = 2s− 1, 2s for some s ∈ N.

Step 1. We first show that, for every 1 ≤ i ≤ 2l < k ≤ n, we have

(gl)1···i···(2l)k = l! g(i+ǫ(i))k . (16)

We assume that i is even (the case i odd, can be handled exactly in same way), inwhich case i+ ǫ(i) = i− 1. It follows, from Lemma 2.7, that

(gl)1···i···(2l)k = l

i+ǫ(i)−1∑

j=1

(−1)j+i+ǫ(i)+1gj(i+ǫ(i)) (gl−1)1···j···i+ǫ(i)i···(2l)k

+ l

2l∑

j=i+1

(−1)j+i+ǫ(i)g(i+ǫ(i))j (gl−1)1···i+ǫ(i)i···j···(2l)k

+ l g(i+ǫ(i))k

(gl−1

)1···i+ǫ(i)i······(2l)

. (17)

Using (15), it follows that gj(i+ǫ(i)) = 0, for every 1 ≤ j ≤ i+ǫ(i)−1 and g(i+ǫ(i))j =0, for every i+ 1 ≤ j ≤ 2l. We now appeal to (14) and (15) to deduce that

gl−1

1···i+ǫ(i)i······(2l)= (l − 1)! . (18)


Finally, we combine (17) and (18) to arrive at (16).

Step 2. Thanks to Step 1, to prove the lemma, it is enough to show that

l! g(i+ǫ(i))n g(j+ǫ(j))k − l! g(i+ǫ(i))k g(j+ǫ(j))n = (gl)1···i···j···(2l)kn. (19)

Recall that we discuss only the case i < j. We will consider two cases to establish(19).

Case 1: i, j 6= 2s− 1, 2s, for every 1 ≤ s ≤ l. We only deal with the case ieven (the case i odd is handled similarly). This implies that i+ ǫ(i) = i− 1. FromLemma 2.7 it follows that

(gl)1···i···j···(2l)kn = l

i+ǫ(i)−1∑

t=1

(−1)t+i+ǫ(i)+1gt(i+ǫ(i)) (gl−1)1···t···i+ǫ(i)i···j···(2l)kn

+ l

j−1∑

t=i+1

(−1)t+i+ǫ(i)g(i+ǫ(i))t (gl−1)1···i+ǫ(i)i···t···j···(2l)kn

+ l

2l∑

t=j+1

(−1)t+i+ǫ(i)+1g(i+ǫ(i))t (gl−1)1···i+ǫ(i)i···j···t···(2l)kn

− l g(i+ǫ(i))k (gl−1)1···i+ǫ(i)i···j···(2l)n

+ l g(i+ǫ(i))n (gl−1)1···i+ǫ(i)i···j···(2l)k

.

Proceeding exactly as in Step 1, we can show that

(gl−1)1···i+ǫ(i)i···j···(2l)k

= (l − 1)! g(j+ǫ(j))k . (20)

Using (15) and (20), it follows that

(gl)1···i···j···(2l)kn = −l! g(i+ǫ(i))k g(j+ǫ(j))n + l! g(i+ǫ(i))n g(j+ǫ(j))k

which is nothing else than (19).

Case 2: i, j = 2s − 1, 2s, for some 1 ≤ s ≤ l. Since (gl+1)1···(2l)kn = 0, weget, using once more Lemma 2.7, that

0 = (gl+1)1···(2l)kn = (l + 1)2s−2∑

t=1

(−1)t+2sgt(2s−1) (gl)1···t···2s−1···(2l)kn

+ (l + 1)g(2s−1)(2s) (gl)1···2s−12s···(2l)kn

+ (l + 1)

2l∑

t=2s+1

(−1)t+2lg(2s−1)t(gl)1···2s−1···t···(2l)kn

− (l + 1)g(2s−1)k (gl)1···2s−1···(2l)n

+ (l + 1)g(2s−1)n (gl)1···2s−1···(2l)k

.

We next invoke Step 1, (14) and (15), to deduce that

0 = (gl)1···2s−12s···(2l)kn

− l! g(2s−1)k g(2s)n + l! g(2s−1)n g(2s)k . (21)

Note that we then have i = 2s−1, j = 2s, i+ ǫ(i) = 2s and j+ ǫ(j) = 2s−1. Hence(21) implies that

l! g(i+ǫ(i))n g(j+ǫ(j))k − l! g(i+ǫ(i))k g(j+ǫ(j))n = (gl)1···i···j···(2l)kn

which is exactly (19). This finishes the proof.


The next lemma has been used in Step 3.3 of the proof of Theorem 3.3.

Lemma 3.6. Let 2 ≤ 2l ≤ n be integers, g ∈ Λ2(Rn) and ω ∈ Λ1(Rn) with

gl+1 = 0, (gl)12······(2l−1)(2l) 6= 0

(ω ∧ gl)1···(2l)k = 0, for every 2l+ 1 ≤ k ≤ n. (22)

Then

ω ∧ gl = 0.

Proof. Step 1. Without loss of generality, we can assume that

(gl)12······(2l−1)(2l) = 1.

It follows from (22) that, for every 2l+ 1 ≤ k ≤ n,

ω1(gl)2···(2l)k − ω2(g

l)13···(2l)k + · · · − ω2l(gl)1···(2l−1)k + ωk(gl)1···(2l) = 0

and therefore

ωk =

2l∑

s=1

(−1)sωs (gl)1···s···(2l)k . (23)

Note that (23) is trivially valid for 1 ≤ k ≤ 2l.

Step 2. We have to prove that

(ω ∧ gl)i1···i2l+1= 0, for every 1 ≤ i1 < · · · < i2l+1 ≤ n.

Using (23), it follows that

(ω ∧ gl)i1···i2l+1=

2l+1∑

j=1

(−1)j+1ωij(gl)

i1···ij ···i2l+1

=

2l+1∑

j=1

(−1)j+1

[2l∑

s=1

(−1)sωs(gl)1···s···(2l)ij

](gl)

i1···ij ···i2l+1

=

2l∑

s=1

(−1)sωs

2l+1∑

j=1

(−1)j+1(gl)1···s···(2l)ij(gl)

i1···ij ···i2l+1

.

Step 3. Let

As =

2l+1∑

j=1

(−1)j+1(gl)1···s···(2l)ij(gl)

i1···ij ···i2l+1.

If we show that As = 0, for every 1 ≤ s ≤ 2l, the lemma will be proved. UsingLemma 2.7, we have

(gl)1···s···(2l)ij= l

s−1∑

r=1

(−1)r+1grij(gl−1)1···r···s···(2l)

+ l

2l∑

r=s+1

(−1)rgrij(gl−1)1···s···r···(2l).

We therefore find

As = l2l+1∑

j=1

(−1)j+1(gl)i1···ij ···i2l+1

[ ∑s−1r=1(−1)r+1grij

(gl−1)1···r···s···(2l)

+∑2l

r=s+1(−1)rgrij(gl−1)1···s···r···(2l)

]


and hence

As = l

s−1∑

r=1

(−1)r+1(gl−1)1···r···s···(2l)

[∑2l+1

j=1(−1)j+1grij

(gl)i1···ij ···i2l+1

]

+ l2l∑

r=s+1

(−1)r(gl−1)1···s···r···(2l)

[∑2l+1

j=1(−1)j+1grij

(gl)i1···ij ···i2l+1

].

This leads (according to Lemma 2.7) to

As =l

l+ 1

s−1∑

r=1

(−1)r+1(gl−1)1···r···s···(2l)(gl+1)ri1···i2l+1

+l

l + 1

2l∑

r=s+1

(−1)r(gl−1)1···s···r···(2l)(gl+1)ri1···i2l+1

= 0

where we have used the fact that gl+1 = 0. This finishes the proof.

The following two lemmas have been used in Step 3.4 of Theorem 3.3.

Lemma 3.7. Let 2 ≤ 2l < n be integers, Ω ⊂ Rn an open set. Let g ∈ C∞(Ω; Λ2)

be closed with rank g = 2l in Ω and of the form

g (x) =

l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

gij (x) dxi ∧ dxj

where gij ∈ C∞(Ω). Then, for every 2l+ 1 ≤ k ≤ n,

LnLk = LkLn (24)

where Lk : C∞(Ω) → C∞(Ω), 2l+ 1 ≤ k ≤ n, is defined by

Lk(z) = (dz ∧ gl)1···(2l)k .

Proof. We begin by the noting that the structure of g implies that

(gl)1···(2l) = l! in Ω. (25)

For z ∈ C∞(Ω), we have

Lk(z) = (dz ∧ gl)1···(2l)k = zxk l! +

2l∑

i=1

(−1)i+1zxi(gl)1···i···(2l)k

where we have denoted partial differentiation of z by xk with zxk . We therefore find

LnLk(z) = Ln

(zxk l! +

2l∑

i=1

(−1)i+1zxi(gl)1···i···(2l)k

)

=

(zxk l! +

2l∑

i=1

(−1)i+1zxi(gl)1···i···(2l)k

)

xn

l!

+

2l∑

j=1

(−1)j+1

(zxk l! +

2l∑

i=1

(−1)i+1zxi(gl)1···i···(2l)k

)

xj

(gl)1···j···(2l)n .


Setting

A1 = l!2zxkxn , A2 = l!

2l∑

i=1

(−1)i+1zxixn(gl)1···i···(2l)k

A3 = l!

2l∑

i=1

(−1)i+1zxi

((gl)1···i···(2l)k

)

xn

A4 = l!

2l∑

j=1

(−1)j+1zxkxj (gl)1···j···(2l)n

A5 =2l∑

i,j=1

(−1)i+1(−1)j+1zxixj (gl)1···i···(2l)k(gl)1···j···(2l)n

A6 =

2l∑

i,j=1

(−1)i+1(−1)j+1zxi

((gl)1···i···(2l)k

)

xj(gl)1···j···(2l)n

we find that

LnLk(z) = A1 +A2 +A3 +A4 +A5 +A6 .

Note that A1 , A2 + A4 and A5 are symmetric in k and n. Therefore, for provingthat LkLn(z) = LnLk(z), it is enough to show that A3 +A6 is symmetric in k andn, which is equivalent to

2l∑

i=1

(−1)i+1zxi

l!((gl)1···i···(2l)k

)

xn

+∑2l

j=1(−1)j+1((gl)1···i···(2l)k

)

xj(gl)1···j···(2l)n

=2l∑

i=1

(−1)i+1zxi

l!((gl)1···i···(2l)n

)

xk

+∑2l

j=1(−1)j+1((gl)1···i···(2l)n

)

xj(gl)1···j···(2l)k

. (26)

To prove this, note first that, for every 2l+ 1 ≤ k ≤ n

2l∑

j=1

(−1)j+1((gl)1···j···(2l)k

)

xj= 0 (27)

since (dgl)1···(2l)k = 0, g being closed, and((gl)1···(2l)

)xk = 0 according to (25).

Hence, it follows from (27) that (26) is equivalent to

2l∑

i=1

(−1)i+1zxiCi = 0

where

Ci = l!((gl)1···i···(2l)k

)

xn− l!

((gl)1···i···(2l)n

)

xk

+

2l∑

j=1

(−1)j+1((gl)1···i···(2l)k(gl)1···j···(2l)n

)

xj

−

2l∑

j=1

(−1)j+1((gl)1···i···(2l)n(gl)1···j···(2l)k

)

xj.


To finish the proof, it is enough to prove that Ci = 0, for every 1 ≤ i ≤ 2l. Indeed,using Lemma 3.5, we deduce that

Ci = l!((gl)1···i···(2l)k

)

xn− l!

((gl)1···i···(2l)n

)

xk

+ l!

i−1∑

j=1

(−1)j+1((gl)1···j···i···(2l)kn

)

xj+ l!

2l∑

j=i+1

(−1)j((gl)1···i···j···(2l)kn

)

xj

= l! (dgl)1···i···(2l)kn= 0

since g is closed. This finishes the proof of the lemma.

Lemma 3.8. Let 2 ≤ 2l < n be integers, Ω ⊂ Rn an open set. Let g ∈ C∞(Ω; Λ2)

be closed with rank g = 2l in Ω and of the form

g (x) =l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

gij (x) dxi ∧ dxj

where gij ∈ C∞(Ω). Let u ∈ C∞(Ω) be such that

du ∧ gl = 0 in Ω.

Then, for every integer 2l + 1 ≤ k ≤ n, the following holds

Ln((u ∧ gl)1···(2l)k) = Lk((u ∧ gl)1···(2l)n)

where Lk : C∞(Ω) → C∞(Ω), 2l+ 1 ≤ k ≤ n, is given by

Lk(z) = (dz ∧ gl)1···(2l)k .

Proof. We divide the proof of the lemma into three steps.

Step 1. We have, since (gl)1···(2l) = l!,

Lk((u ∧ gl)1···(2l)n) = (d[(u ∧ gl)1···(2l)n] ∧ gl)1···(2l)k

= l!((u ∧ gl)1···(2l)n

)xk

+

2l∑

i=1

(−1)i+1((u ∧ gl)1···(2l)n

)xi (gl)1···i···(2l)k .

Since (dgl)1···(2l)k = 0, g being closed, and ((gl)1···(2l))xk = 0, it follows that

2l∑

i=1

(−1)i+1((gl)1···i···(2l)k

)

xi= 0

and therefore

Lk((u ∧ gl)1···(2l)n) = l!((u ∧ gl)1···(2l)n

)xk

+2l∑

i=1

(−1)i+1((u ∧ gl)1···(2l)n(gl)1···i···(2l)k

)

xi.

Similarly, we have

Ln((u ∧ gl)1···(2l)k) = l!((u ∧ gl)1···(2l)k

)xn

+

2l∑

i=1

(−1)i+1((u ∧ gl)1···(2l)k(gl)1···i···(2l)n

)

xi.


We then set, for 1 ≤ i ≤ 2l,

Ai = (u ∧ gl)1···(2l)n(gl)1···i···(2l)k − (u ∧ gl)1···(2l)k(gl)1···i···(2l)n .

In order to prove the lemma, we therefore have to show the following

l!((u ∧ gl)1···(2l)n

)xk − l!

((u ∧ gl)1···(2l)k

)xn +

2l∑

i=1

(−1)i+1 (Ai)xi = 0. (28)

Step 2. In this step, we prove that, for every 1 ≤ i ≤ 2l,

Ai = l! (u ∧ gl)1···i···(2l)kn . (29)

To show this, we note that

Ai =

2l∑

j=1

(−1)j+1uj(gl)1···j···(2l)n + l!un

(gl)1···i···(2l)k

−

2l∑

j=1

(−1)j+1uj(gl)1···j···(2l)k + l!uk

(gl)1···i···(2l)n.

We therefore get

Ai = l!un(gl)1···i···(2l)k − l!uk(gl)1···i···(2l)n

+

2l∑

j=1

(−1)j+1uj[(gl)1···j···(2l)n(gl)1···i···(2l)k − (gl)1···j···(2l)k(gl)1···i···(2l)n

].

Invoking Lemma 3.5 at this point, it follows that

Ai = l!un(gl)1···i···(2l)k − l!uk(gl)1···i···(2l)n

+

i−1∑

j=1

(−1)j+1l!uj(gl)1···j···i···(2l)kn +

2l∑

j=i+1

(−1)jl!uj(gl)1···i···j···(2l)kn

= l! (u ∧ gl)1···i···(2l)kn.

Step 3. We finally use (29) in (28) to deduce that

l!((u ∧ gl)1···(2l)n

)xk − l!

((u ∧ gl)1···(2l)k

)xn +

l!

2l∑

i=1

(−1)i+1((u ∧ gl)1···i···(2l)kn

)

xi=

l!(d(u ∧ gl)

)1···(2l)kn

= 0,

since dg = 0 and du ∧ gl = 0. The proof is finished.

The following lemma is standard, cf. [12], [19].

Lemma 3.9. Let 2 ≤ 2l ≤ n and g ∈ Λ2(Rn) with gl 6= 0 and gl+1 = 0. Then thereexists A ∈ GLn(R), letting ϕ(x) = Ax, such that

ϕ∗(g) = ωl =

l∑

i=1

dx2i−1 ∧ dx2i.

Moreover, if n = 2l, then, in addition to the above conclusion, ϕ can be chosen sothat ϕ1 (x) = x1.


The final proposition has been used in Step 3.1 of Theorem 3.3.

Proposition 3.10. Let 2 ≤ 2l < n be integers and x0 ∈ Rn. Let g be a C∞ closed

2−form with

gl 6= 0 and gl+1 = 0 in a neighbourhood of x0 . (30)

Then there exist U and V two neighbourhoods of x0 , rij ∈ C∞(U) and ϕ ∈Diff∞(U ;V ) such that ϕ (x0) = x0 and

ϕ∗ (g) =

l∑

i=1

dx2i−1 ∧ dx2i +∑

1≤i<j≤n2l<j

rij dxi ∧ dxj .

Proof. Step 1. Without loss of generality, we can assume that x0 = 0. In addition,using Lemma 3.9, we can take (as in Step 2 of Theorem 3.3)

g(0) = ωl =

l∑

i=1

dx2i−1 ∧ dx2i.

We next introduce some notations. We let

x = (y, z) =(x1, · · · , x2l, x2l+1, · · · , xn

)∈ R

2l × Rn−2l

and we define, for every z ∈ Rn−2l, the map iz : R

2l → Rn through

iz (y) = (y, z) = x.

Step 2. We define, for every x = (y, z) ∈ R2l ×R

n−2l with |x| small, the 2−form

gz,t : R2l → Λ2

(R

2l)

by gz,t (y) = i∗z [tg + (1 − t)ωl] (y) .

Note that

gz,0 (y) = i∗z [ωl] = ωl and gz,1 (y) =∑

1≤i<j≤2l

gij (x) dxi ∧ dxj .

Our assumption in Step 1 leads to gz,t (0) = ωl and therefore, in a sufficiently smallcube C centred at 0 ∈ R

2l × Rn−2l, we can ensure that

(gz,t)l (y) > 0 for every (y, z) ∈ C and t ∈ [0, 1]. (31)

Furthermore, gz,t has the property that

dygz,t = 0 in C, for every t ∈ [0, 1], (32)

where dy is understood as the exterior differential operator involving only the vari-able y =

(x1, · · · , x2l

), namely dygz,t = 0 is equivalent to

∂ (gz,t)ij

∂xk−∂ (gz,t)ik

∂xj+∂ (gz,t)jk

∂xi= 0 for every i, j, k = 1, · · · , 2l.

Step 3. Using (31), (32) and Poincare lemma, we find a C∞ vector field uz,t :R

2l → R2l such that

dy (uz,t y gz,t) = −d

dtgz,t = ωl − i∗z [g] in C, for every t ∈ [0, 1].

We now consider the initial value problem, for every x = (y, z) ∈ C,

d

dtϕz,t = uz,t(ϕz,t) and ϕz,0(y) = y.

Using Moser flow method, we deduce that, up to restricting the set C,

ϕ∗z,1 (gz,1) = gz,0 = ωl


which means that∑

1≤i<j≤2l

(gz,1)ij(ϕz,1 (y)) dyϕ

iz,1 ∧ dyϕ

jz,1 = ωl (33)

where for u : Rn → R we set

dyu =2l∑

i=1

∂u

∂xidxi and dzu =

n∑

i=2l+1

∂u

∂xidxi.

We finally let, for x = (y, z) ∈ C,

ϕ (x) = (ϕz,1 (y) , z)

and we claim that this is the diffeomorphism we are looking for. Indeed first observethat

∑

1≤i<j≤2l

(gz,1)ij(ϕz,1 (y)) dyϕ

iz,1 ∧ dyϕ

jz,1 =

∑

1≤i<j≤2l

gij (ϕ (y)) dyϕi ∧ dyϕ

j . (34)

We moreover have

ϕ∗ (g) =∑

1≤i<j≤n

gij (ϕ (y)) dϕi ∧ dϕj

=∑

1≤i<j≤2l

gij (ϕ (y)) dϕi ∧ dϕj +∑

1≤i<j≤n2l<j

gij (ϕ (y)) dϕi ∧ dϕj

=∑

1≤i<j≤2l

gij (ϕ (y))(dyϕ

i + dzϕi)∧(dyϕ

j + dzϕj)

+∑

1≤i<j≤n2l<j

gij (ϕ (y)) dϕi ∧ dxj

and thus

ϕ∗ (g) =∑

1≤i<j≤2l

gij (ϕ (y)) dyϕi ∧ dyϕ

j

+∑

1≤i<j≤2l

gij (ϕ (y)) dyϕi ∧ dzϕ

j +∑

1≤i<j≤2l

gij (ϕ (y)) dzϕi ∧ dyϕ

j

+∑

1≤i<j≤2l

gij (ϕ (y)) dzϕi ∧ dzϕ

j +∑

1≤i<j≤n2l<j

gij (ϕ (y)) dϕi ∧ dxj . (35)

Appealing to (33), (34) and (35), we get

ϕ∗ (g) = ωl +∑

1≤i<j≤n2l<j

rij dxi ∧ dxj

for appropriate rij . This finishes the proof.

3.4. An alternative proof of Darboux theorem in the degenerate case.The second proof that we provide in this section (with the help of the previoussections) may seem to be much longer than the first one given in Theorem 3.2.However this is misleading because to really compare the length of the two proofs,one should take into account the full proof of Frobenius theorem and the one ofthe non-degenerate Darboux theorem (Theorem 3.1). It also seems that the presentproof is more appropriate if one wants to look for global results.


Theorem 3.11. Let n ≥ 3 and x0 ∈ Rn. Let ωl be the standard symplectic form of

rank 2l < n, namely

ωl =l∑

i=1

dx2i−1 ∧ dx2i.

Let ω be a C∞ closed 2−form such that

rankω = 2l in a neighbourhood of x0 .

Then there exist a neighbourhood V of x0 and ϕ ∈ Diff∞(V ; Rn) such that

ϕ∗ (ωl) = ω in V and ϕ (x0) = x0 .

Proof. Step 1 . It is easy to see (cf. [12]) that we can construct a homotopy hwith the following properties. There exist a neighbourhood V1 of x0 and h ∈C∞(V1 × [0, 1]; Λ2), h(x, t) = ht(x), such that, for every (x, t) ∈ V1 × [0, 1],

dht = 0, hlt 6= 0 and hl+1

t = 0 (36)

whileh0 = ω (x0) and h1 = ω.

Step 2. Since (36) holds and

hlt ∧

∂ht

∂t=

1

l + 1

∂hl+1t

∂t= 0

we can apply Theorem 3.3. We therefore can find a neighbourhood V2 ⊂ V1 of x0

and w ∈ C∞(V2× [0, 1]; Rn), w(x, t) = wt(x), satisfying, for every (x, t) ∈ V2× [0, 1],

dwt = −∂ht

∂t, wt ∧ h

lt = 0 and wt(x0) = 0.

We then apply Proposition 2.6 to find u ∈ C∞(V2 × [0, 1]; Rn), u(x, t) = ut(x), with

ut y ht = wt and ut(x0) = 0.

Step 3. We next find the flow, associated to the vector field ut ,ddtϕt(x) = ut(ϕt(x)) t ∈ [0, 1]

ϕ0 = id .

Classical results (cf. for example Theorem 10 in [3]) show the existence of a neigh-bourhood V3 ⊂ V2 of x0 such that

ϕ∗1(h1) = h0 in V3 and ϕ1(x0) = x0 .

Step 4. Since h0 is constant, we can use Lemma 3.9 to find a linear diffeomorphismψ so that

ψ∗(h0) = ωl =

l∑

i=1

dx2i−1 ∧ dx2i.

Letting ϕ = ϕ1 ψ we have the claim.


4. Simultaneous resolutions and the case of k-forms when 3 ≤ k ≤ n− 1.

4.1. A general theorem for forms of rank k. Our first result concerns k−formsof minimal non-zero rank.

Theorem 4.1. Let 2 ≤ k ≤ n, r ≥ 1 be integers, 0 < α < 1 and x0 ∈ Rn. Let f, g

be two Cr,α closed k−forms satisfying

rank f = rank g = k in a neighbourhood of x0 .

Then there exists a diffeomorphism ϕ of class Cr,α such that, in a neighbourhood ofx0 ,

ϕ∗ (g) = f and ϕ (x0) = x0 .

Moreover if k = n, the diffeomorphism ϕ can be chosen of class Cr+1,α.

Remark The theorem solves completely the case k = n− 1. Moreover in this casethe condition on the rank is equivalent to

f (x0) 6= 0 and g (x0) 6= 0.

Proof. The statement with k = n is the result of Dacorogna-Moser [10] (cf. Theorem14.6 in [9]) and will be used below after reduction of dimension. Note that, it isenough to find a ϕ ∈ Diffr,α(Ω; Rn) such that

ϕ∗(dx1 ∧ · · · dxk) = f in Ω,

where Ω is a neighbourhood of x0 and we take, without loss of generality, x0 = 0.Using Theorem 5.2, we find two neighbourhood V,Ω of 0 in R

n and ψ ∈ Diffr,α(V ; Ω)satisfying ψ(0) = 0 and

ψ∗(f)(x1, · · · , xn) = a(x1, · · · , xk) dx1 ∧ · · · ∧ dxk,

where a ∈ Cr−1,α(W ) and W is a sufficiently small open ball in Rk containing 0.

Note that a 6= 0 in W, since rankf = k. We now apply the result of Dacorogna-Moser [10] to find χ ∈ Diffr,α(W ;W ) such that

χ∗(c dx1 ∧ · · · ∧ dxk) = a dx1 ∧ · · · ∧ dxk in W,

where

c =1

measW

∫

W

a.

At this point, we can perform a linear change of variables which allows us to assumethat c = 1. Next, we construct a map χ ∈ Cr,α(W × R

n−k; Rn) by

χ(y) = (χ(y1, · · · , yk), yk+1, · · · , yn), for every y = (y1, · · · , yn) ∈W × Rn−k.

Finally, we define ϕ : Ω → Rn as

ϕ = χ ψ−1.

Hence we have found ϕ ∈ Diffr,α(Ω; Rn) satisfying

ϕ∗(dx1 ∧ · · ·dxk) = f in Ω

as wished.

We have as an immediate corollary the following.


Corollary 4.2. Let 2 ≤ k ≤ n, r ≥ 1 be integers, 0 < α < 1 and x0 ∈ Rn. Let f be

a Cr,α closed k−form such that

rank f = k in a neighbourhood of x0 .


f = ∇ϕ1 ∧ · · · ∧ ∇ϕk and ϕ (x0) = x0 .

Proof. It is enough to choose

g = dx1 ∧ · · · ∧ dxk

in the theorem.

The corollary reads in a more analytical way when k = n− 1 (cf. also [5]), sincethe exterior derivative of an (n− 1)−form is then essentially the classical divergenceoperator.

Corollary 4.3. Let r ≥ 1 be an integer and 0 < α < 1 and x0 ∈ Rn. Let f be a

Cr,α vector field satisfying

f (x0) 6= 0 and div f = 0 in a neighbourhood of x0 .


f = ∗(∇ϕ1 ∧ · · · ∧ ∇ϕn−1

)and ϕ (x0) = x0 .

4.2. Simultaneous resolutions and applications. We start with a simultaneousresolution of closed 1−forms. We will also obtain a global result with Dirichlet data.We follow here the idea explained in [3].

Theorem 4.4. Let r ≥ 0, 1 ≤ m ≤ n be integers and x0 ∈ Rn. Let b1, · · · , bm,

a1, · · · , am be Cr closed 1−forms satisfying, in a neighbourhood of x0 ,

b1 ∧ · · · ∧ bm 6= 0 and a1 ∧ · · · ∧ am 6= 0.

Then there exists a Cr+1 diffeomorphism such that ϕ (x0) = x0 and, in a neigh-bourhood of x0 ,

ϕ∗ (bi) = ai , for every i = 1, · · · ,m.

Remark (i) When r = 0, the fact that the forms are closed has to be understoodin the sense of distributions.

(ii) It is interesting to compare the above theorem and Theorem 4.1. In viewof Proposition 2.5, we know that any m−form g with rank g = m, is a product of1−forms b1, · · · , bm so that

g = b1 ∧ · · · ∧ bm ;

however we do not know, a priori, that b1, · · · , bm are closed if g is closed (and eventhat b1, · · · , bm ∈ Cr if g ∈ Cr); we know it only a posteriori from the conclusion ofTheorem 4.1, but we have lost one degree of regularity, namely b1, · · · , bm ∈ Cr−1,α.Therefore if we assume that b1, · · · , bm are closed, then the above theorem is betterfrom the point of view of regularity than Theorem 4.1.

(iii) When m = n and g ∈ C0 it is, in general, impossible (according to [6] and[14]) to find closed 1−forms b1, · · · , bn ∈ C0 so that

g = b1 ∧ · · · ∧ bn ;


although, in view of Theorem 4.1, we can do so if g ∈ C0,α, finding even thatb1, · · · , bn ∈ C0,α.

Proof. We assume that m = n, otherwise we choose 1 ≤ k1 < · · · < kn−m ≤ n and1 ≤ l1 < · · · < ln−m ≤ n such that

b1 ∧ · · · ∧ bm ∧ dxk1 ∧ · · · ∧ dxkn−m 6= 0,

a1 ∧ · · · ∧ am ∧ dxl1 ∧ · · · ∧ dxln−m 6= 0

and we prove the theorem forb1, · · · , bm, dx

k1 , · · · , dxkn−m

anda1, · · · , am, dx

l1 , · · · , dxln−m.

As usual, we may assume that

ai = dxi, i = 1, · · · , n.

Since (b1 ∧ · · · ∧ bn)(x0) 6= 0, there exists j1 ∈ 1, · · · , n satisfying bj11 (x0) 6= 0which, in turn, implies that

bj11 (x) 6= 0, for every x ∈ U1 ,

where U1 is some ball around x0 . We now use Proposition 19 of [3] to find ϕ1 ∈Diffr+1(U1; R

n) satisfying ϕ1 (x0) = x0 and

ϕ∗1(b1) = dxj1 , in U1

as well as

ϕp1(x) = xp, for every p ∈ 1, · · · , n, p 6= j1 .

Having found ϕ1 , we note that ϕ∗1(b2)

j2 (x0) 6= 0, for some j2 ∈ 1, · · · , n withj2 6= j1 . This follows from the fact that

ϕ∗1(b1 ∧ · · · ∧ bn)(x0) = dxj1 ∧ ϕ∗

1(b2)(x0) · · · ∧ ϕ∗1(bn)(x0) 6= 0.

We hence deduce that there exists a ball U2 ⊂ U1 centred at x0 such that ϕ∗1(b2)

j2 (x)6= 0, for every x ∈ U2 . Invoking Proposition 19 of [3] again, we findϕ2 ∈ Diffr+1(U2; R

n) such that ϕ2 (x0) = x0 ,

ϕ∗2 (ϕ∗

1(b1)) = dxj2 , in U2 ,

and

ϕp2(x) = xp, for every p ∈ 1, · · · , n, p 6= j2 .

Continuing this process for n steps, we find a finite sequence Un ⊂ · · · ⊂ U1 ofballs centred at x0 and ϕq ∈ Diffr+1(Uq; R

n), for every q = 1, · · · , n, such thatϕq (x0) = x0 and

ϕ∗q

(ϕ∗

q−1 · · · ϕ∗1(bq)

)= dxjq , for every q = 1, · · · , n,

where

jq /∈ j1, · · · , jq−1, for every q = 2, · · · , n.

Furthermore, we have, for every q = 1, · · · , n,

ϕpq(x) = xp, for every p ∈ 1, · · · , n, p 6= jq.

Finally, we set V = Un and we define ϕ ∈ Diffr+1(V ; Rn) by

ϕ = ϕ1 · · · ϕn .

By construction we have

ϕ∗(bi) = dxji , for every i = 1, · · · , n.


Finally, composing ϕ with a linear map leads us to the map we have been lookingfor. This proves the theorem.

It is interesting to see that the above theorem can also be global.

Theorem 4.5. Let Ω ⊂ Rn be a bounded, smooth and simply connected domain.

Let r ≥ 0 and 1 ≤ m ≤ n be integers. Let b1, · · · , bm ∈ Cr(Ω; Λ1

)be closed forms.

For all i = 1, · · · ,m− 1 and j, k = 1, · · · ,m, define B(i)jk ∈ Cr(Ω) by

B(1)jk = bkj and B

(i+1)jk = B

(i)ii B

(i)jk −B

(i)ji B

(i)ik .

Assume that, for every i = 1, · · · ,m,

B(i)ii > 0 in Ω and bi ∧ ν = dxi ∧ ν on ∂Ω

where ν is the exterior unit normal to Ω. Then there exists ϕ ∈ Diffr+1(Ω; Ω

)

satisfyingϕ∗ (bi) = dxi in Ω, for every i = 1, · · · ,m

ϕ∗(dxi)

= dxi, for every i = m+ 1, · · · , n.

ϕ (x) = x on ∂Ω.

Proof. We prove the theorem only for m = 2 and we write b1 = b and b2 = a. Thegeneral case is proved similarly (cf. [12] for details). Note that we have

B(1)11 = b1 > 0 and B

(2)22 = (b ∧ a)12 > 0.

Since b1 > 0, using Proposition 19 of [3], we find a diffeomorphism

ψ (x) =(ψ1 (x) , x2, · · · , xn

)

such thatψ∗ (b) = dx1 in Ω and ψ (x) = x on ∂Ω.

Moreover, it follows, since ψ∗ (b) = dx1, that, if j 6= 1,

b1 (ψ)∂ψ1

∂xj+ bj (ψ) = 0

and thus∂ψ1

∂xj= −

bj (ψ)

b1 (ψ).

We therefore deduce that

ψ∗ (a) =n∑

j=1

aj (ψ) dψj = a1 (ψ) dψ1 +n∑

j=2

aj (ψ) dxj

= a1 (ψ)

n∑

j=1

∂ψ1

∂xjdxj +

n∑

j=2

aj (ψ) dxj

= a1 (ψ)∂ψ1

∂x1dx1 +

n∑

j=2

(aj (ψ) + a1 (ψ)

∂ψ1

∂xj

)dxj

= a1 (ψ)∂ψ1

∂x1dx1 +

n∑

j=2

(aj (ψ) − a1 (ψ)

bj (ψ)

b1 (ψ)

)dxj .

Since (b ∧ a)12 > 0, invoking Proposition 19 of [3] again, we find a diffeomorphismχ, where

χ (x) =(x1, χ2 (x) , x3, · · · , xn

),


such that

χ∗ (ψ∗ (a)) = dx2 in Ω and χ (x) = x on ∂Ω.

Setting

ϕ = ψ χ

we observe that, in Ω,

ϕ∗ (b) = dx1 and ϕ∗ (a) = dx2

while ϕ (x) = x on ∂Ω. This finishes the proof.

We next generalize Theorem 4.4 by mixing 1−forms and 2−forms.

Theorem 4.6. Let m, l ≥ 0 be integers and x0 ∈ Rn. Let b1, · · · , bm, a1, · · · , am be

C∞ closed 1−forms satisfying, in a neighbourhood of x0 ,

b1 ∧ · · · ∧ bm 6= 0 and a1 ∧ · · · ∧ am 6= 0.

Let g1, · · · , gl, f1, · · · , fl be C∞ closed 2−forms verifying, in a neighbourhood of x0 ,

rank gi = rank fi = 2si , i = 1, · · · , l

rank [g1 ∧ · · · ∧ gl ∧ b1 ∧ · · · ∧ bm] = rank [f1 ∧ · · · ∧ fl ∧ a1 ∧ · · · ∧ am]

= 2 (s1 + · · · + sl) +m ≤ n.

Then there exists a C∞ diffeomorphism ϕ such that ϕ (x0) = x0 and, in a neigh-bourhood of x0 ,

ϕ∗ (gi) = fi for every i = 1, · · · , l

ϕ∗ (bi) = ai for every i = 1, · · · ,m.

Remark (i) When m = 0, respectively l = 0, the theorem is to be understood asa statement only on 2−forms, respectively only on 1−forms (in this last case seeTheorem 4.4).

(ii) When gi, fi ∈ Cr,α and bj, aj ∈ Cr,α, we have ϕ ∈ Diffr−l+1,α .

(iii) Of course the theorem applies to k−forms, k = 2l +m, of the type

G = g1 ∧ · · · ∧ gl ∧ b1 ∧ · · · ∧ bm and F = f1 ∧ · · · ∧ fl ∧ a1 ∧ · · · ∧ am .

We therefore obtain that there exists a diffeomorphism ϕ such that

ϕ∗ (G) = F

generalizing the result obtained in [3].

It is interesting to contrast the algebraic result of Proposition 2.5 (iii) with theanalytical result of the above theorem, where it is essential to require that the1−forms and the 2−forms be closed.

Example 4.7. Although every constant 3−form of rank = 5 is a linear pullback(according to Proposition 2.5 (iii)) of

G = g ∧ b

where

g = dx1 ∧ dx2 + dx3 ∧ dx4 and b = dx5,

there does exist a (non-constant) closed form F ∈ Λ3 (Rn) with rankF = 5 whichcannot be pulled back to G. One such example is

F = f ∧ a


where

f =1

(x3)4+ 1

dx1 ∧ dx5 + dx3 ∧ dx4 and a = ((x3)2

+ 1)dx1 + ((x3)4

+ 1)dx2.

The result follows from the following observations.

1) Any 1−divisor c of F must be of the form c = λa where λ takes value in R.Indeed if this is not the case we have that the 1−form c is linearly independent ofa. We therefore have

F ∧ a = F ∧ c = 0 and c ∧ a 6= 0.

Appealing to Proposition 2.5 (iv), we deduce that rankF = 3, a contradiction.

2) Note that λa is not closed in any open set unless λ is identically zero.

3) Therefore, if there exists a local diffeomorphism ϕ satisfying

F = ϕ∗(G) = ϕ∗(g) ∧ ϕ∗(b),

it follows from our aforementioned observations that

ϕ∗(b) = ϕ∗(dx5) = λa.

But this leads to contradiction because the form on the left hand side is closed andnon-zero whereas the form on the right hand side is not closed.

We now turn our attention to the proof of Theorem 4.6.

Proof. It is enough to prove the theorem when

fj =

s1+···+sj∑

i=(s1+···+sj−1)+1

dx2i−1 ∧ dx2i, j = 1, · · · , l

and when

ai = dx2(s1+···+sl)+i for every i = 1, · · · ,m.

We establish the result by induction on l. When l = 0, we are in the situation ofTheorem 4.4 which has already been proved. Let us suppose that the theorem istrue for l = k. It remains to prove the result when l = k+1. We find, using Theorem3.2, a neighbourhood V1 of x0 and ϕ1 ∈ Diff∞(V1; R

n) such that ϕ1(x0) = x0 and

ϕ∗1(g1) = f1 in V1 .

Since

rank (ϕ∗1(g1) ∧ · · · ∧ ϕ∗

1(gl) ∧ ϕ∗1(b1) ∧ · · · ∧ ϕ∗

1(bm))

= rank (ϕ∗1(g1 ∧ · · · ∧ gl ∧ b1 ∧ · · · ∧ bm))

= rank(g1 ∧ · · · ∧ gl ∧ b1 ∧ · · · ∧ bm) = 2(s1 + · · · + sl) +m

it follows from Proposition 2.3 (iv) that

rank(dx1 ∧ · · · ∧ dx2s1 ∧ ϕ∗

1(g2) ∧ · · · ∧ ϕ∗1(gl) ∧ ϕ

∗1(b1) ∧ · · · ∧ ϕ∗

1(bm))

= 2(s1 + · · · + sl) +m.

Hence, using induction hypothesis, there exists V2 ⊂ V1 of x0 and ϕ2 ∈ Diff∞(V2; Rn)

such that ϕ2(x0) = x0 and, for every i = 2, · · · , l, j = 1, · · · , 2s1 and k = 1, · · · ,m,

ϕ∗2(ϕ

∗1(gi)) = fi , ϕ∗

2(dxj) = dxj and ϕ∗

2(bk) = dx2(s1+···+sl)+k,


Note, in particular, that ϕ∗2(ϕ

∗1(g1)) = ϕ∗

2(f1) = f1 . Setting, choosing if necessarya smaller V2 ,

ϕ = ϕ1 ϕ2

we have ϕ ∈ Diff∞(V2; Rn) with the claimed properties.

In the same spirit we have the following result, a particular case of which wasproved in [3].

Theorem 4.8. Let n = 2m be even and x0 ∈ Rn. Let a, b be C∞ closed 1−forms

witha (x0) 6= 0 and b (x0) 6= 0.

Let f, g be C∞ closed 2−forms such that

rank f (x0) = rank g (x0) = n = 2m.

Then there exists a C∞ diffeomorphism ϕ such that, in a neighbourhood of x0 ,

ϕ∗ (b) = a and ϕ∗ (g) = f.

Proof. It is enough to prove the theorem when

f = ωm =m∑

i=1

dx2i−1 ∧ dx2i and a = dx1.

Using Theorem 4.4, it follows that there exists a C∞ diffeomorphism ϕ defined ina small ball U centred at x0 such that

ϕ∗1(b) = dx1 in U and ϕ1(x0) = x0 .

We now define a homotopy gt through

gt(x) = ϕ∗1(g)(tx + (1 − t)x0) for every x ∈ U and t ∈ [0, 1].

Note that, gt is closed and non-degenerate (choosing U smaller if necessary), forevery t ∈ [0, 1]. We now use the example in Subsection 6.3 in [3] to find ϕ2 suchthat

ϕ∗2(g1) = g0 and ϕ∗

2(dx1) = dx1.

Finally, we find, according to Lemma 3.9, A ∈ GLn(R) satisfying

ϕ∗3(g0) = ωm and ϕ∗

3(dx1) = dx1

where ϕ3 (x) = Ax. It is now easy to check that the map

ϕ = ϕ1 ϕ2 ϕ3

is the one that we are looking for. This finishes the proof.

5. Appendix: reduction of dimension. We turn our attention to a very usefulresult, which is well known in the case of 2−forms. But it can be extended in astraightforward way to the case of k−forms; it seems however that this extensionhas never been noticed. We will provide two proofs of the theorem. The first oneis based on Frobenius theorem, while the second one is much more elementary andself contained. Both versions lead to the same result when k = n−1, while the firstone is better from the point of view of regularity when 2 ≤ k ≤ n− 2.

We begin by recalling few notions and results related to the theory of differentialforms. For details, see [2], [22].

Notation (Lie derivative and involutive family). Let U ⊂ Rn be open, let a, b ∈

C1(U ; Rn) and let ω ∈ C1(U ; Λk).


1. Laω stands for the Lie derivative of ω with respect to a. Recall that, for avector field b, Lab is also known as the Lie bracket of a with b and is denotedby [a, b].

2. Cartan formula states that

Laω = a y dω + d(a yω) (37)

and moreover

[a, b] yω = La(b yω)− b y (Laω). (38)

3. For a1, · · · , am ∈ C1(U ; Rn), we say that a1, · · · , am is an involutive familyif, for every 1 ≤ i, j ≤ m, there exist cpij ∈ C0(U), 1 ≤ p ≤ m, satisfying

[ai, aj ](x) =

m∑

p=1

cpij(x)am(x), for every x ∈ U.

We now recall Frobenius theorem.

Theorem 5.1 (Frobenius theorem). Let Ω ⊂ Rn be open, r ≥ 1 and 1 ≤ m < n be

integers. Let 0 ≤ α ≤ 1 and x0 ∈ Ω. Let a1, · · · , am ∈ Cr,α(Ω; Rn) be an involutivefamily satisfying, for every x ∈ Ω,

a1(x), · · · , am(x) are linearly independent.

Then, there exist two neighbourhoods U, V ⊂ Ω of x0 and ϕ ∈ Diffr,α(V ;U) suchthat

ϕ(x0) = x0

and, for every x ∈ V and 1 ≤ i ≤ m,

∂ϕ

∂xi(x) ∈ span (a1 ϕ)(x), · · · , (am ϕ)(x) .

The main result on dimension reduction is the following.

Theorem 5.2 (Reduction of dimension). Let r ≥ 1, 1 ≤ k ≤ l ≤ n− 1 be integers,0 ≤ α ≤ 1 and x0 ∈ R

n. Let g be a Cr,α closed k−form verifying

rank g = l in a neighbourhood of x0 .

Then, there exist two neighbourhoods U, V of x0 and ϕ ∈ Diffr,α(V ;U) with ϕ (x0) =x0 and such that, for every x =

(x1, · · · , xn

)∈ V,

ϕ∗ (g)(x1, · · · , xn

)= f

(x1, · · · , xl

)

=∑

1≤i1<···<ik≤l

fi1···ik

(x1, · · · , xl

)dxi1 ∧ · · · ∧ dxik .

Thus f = ϕ∗ (g) can be seen as a k−form with maximal rank (i.e. rank f = l) onR

l.

Before starting with the two proofs of the theorem, we need the following simplelemma, the proof of which is straightforward.

Lemma 5.3. Let U, V ⊂ Rn be open, g ∈ C0

(U ; Λk

)and a ∈ C0 (U ; Rn) such that

a(x) y g(x) = 0 for every x ∈ U.

Let ϕ ∈ Diff1(V ;U) satisfy

∂ϕ

∂xn= a ϕ in V .


Then, for every x ∈ V,

(ϕ∗g)i1···ik−1n(x) = 0 for every 1 ≤ i1 < · · · < ik−1 ≤ n− 1.

We now turn our attention to the first proof of our theorem.

First proof of Theorem 5.2. We divide the proof into four steps.

Step 1. Since rank g = l ≤ n− 1, it is easy to find a neighbourhood U1 of x0 andai ∈ Cr,α(U1; R

n) for every l+ 1 ≤ i ≤ n such that, for every x ∈ U1 ,

al+1(x), · · · , an(x) are linearly independent (39)

and

span al+1(x), · · · , an(x) = ker g(x); (40)

so, in particular, for every l + 1 ≤ i ≤ n and every x ∈ U1 ,

ai(x) y g(x) = 0.

Step 2. We now show that the family al+1, · · · , an of vector fields is involutive,i.e. for every l + 1 ≤ i, j ≤ m, there exist cpij ∈ C0(U1), l + 1 ≤ p ≤ m, satisfying

[ai, aj ](x) =n∑

p=l+1

cpij(x)ap(x) for every x ∈ U1 .

It follows, from the fact that g is closed, (37), (38) and (40), that

[ai, aj ] y g = Lai(aj y g) − aj y (Lai

g) = −aj y (Laig)

= −aj y (ai y dg + d(ai y g)) = 0 in U1 .

Hence, we have [ai, aj](x) ∈ ker g(x), for every x ∈ U1 , from where, using (39) and(40), the existence of unique coefficients cpij(x), for every x ∈ U1 , follows. It is easy

to check that cpij ∈ C0(U1).

Step 3. Therefore, using Theorem 5.1, we find two neighbourhoods U, V of x0 ,U, V ⊂ U1 and ϕ ∈ Diffr,α(V ;U) such that ϕ(x0) = x0 and

∂ϕ

∂xi(x) ∈ ker g(ϕ(x)) for every x ∈ V and l + 1 ≤ i ≤ n. (41)

It remains to show that satisfy

ϕ∗(g)i1···ik= 0 in V,

for every 1 ≤ i1 < · · · < ik ≤ n with ik ≥ l + 1. This follows at once from Lemma5.3 and (41).

Step 4. Finally, since dg = 0, we have dϕ∗(g) = 0. Hence, on writing

ϕ∗(g) =∑

1≤i1<···<ik≤l

ri1···ikdxi1 ∧ · · · ∧ dxik ,

we obtain, for every s ≥ l + 1,

∂ri1···ik

∂xs= (dϕ∗(g))i1···iks = 0.

We thus have the result.


We finish this article with the second proof of the theorem. As already said it ismuch more elementary but it gives a less sharp regularity; indeed we will only beable to establish that

ϕ ∈ Diffr+l−n+1,α(V ;U).

Note that when k = n− 1 (and thus l = n− 1) both proofs give the same result.

Second proof of Theorem 5.2. Without loss of generality, we assume that x0 = 0.In the following proof, U, V stand for generic neighbourhoods of 0.

Step 1. Since rank g = l ≤ n− 1, there exists a ∈ Cr,α(U ; Rn) satisfying

a(x) 6= 0 and a(x) y g(x) = 0 for every x ∈ U. (42)

Step 2. We first find ψ = ψ (x, t) ∈ Rn, for |x| and |t| small, such that

∂

∂tψ (x, t) = a (ψ (x, t)) and ψ (x, 0) = x.

Since a(0) 6= 0, there exist b1, · · · , bn−1 ∈ Rn so that

b1, · · · , bn−1, a(0)

are linearly independent.

Let B ∈ Rn×(n−1) be the matrix whose ith column is bi. Finally define

ϕn(x) = ϕn(x1, · · · , xn) = ψ(B(x1, · · · , xn−1), xn

)

and observe that

ϕn(0) = 0, det∇ϕn(0) 6= 0 and∂ϕn

∂xn= a ϕn in V .

We therefore have found a small neighbourhood V of 0 and ϕn ∈ Diffr,α(V ;U) suchthat

ϕn(0) = 0 and∂ϕn

∂xn= a ϕn in V . (43)

Step 3. From (42), (43) and Lemma 5.3, it follows that, for every 1 ≤ i1 < · · · <ik−1 ≤ n− 1,

ϕ∗n(g)i1···ik−1n = 0 in V,

and therefore

ϕ∗n(g) ∈ Cr−1,α(V ; Λk(Rn−1)).

If l = n − 1, the proof is finished. Henceforth, we will assume that 1 ≤ l < n − 1.Since rank(ϕ∗

n(g)) = l, repeating the argument aforementioned, we find ϕn−1 ∈Diffr−1,α(V, U) satisfying, for every 1 ≤ i1 < · · · < ik−1 ≤ n− 2,

ϕ∗n−1(ϕ

∗n(g))i1···ik−1(n−1) = 0 in V.

After repeating the same argument n− l times, we set

ϕ = ϕn · · · ϕl .

It is clear that ϕ ∈ Diffr+l−n+1,α(V ;U) and

ϕ∗(g) ∈ Cr+l−n,α(V ; Λk(Rl)).

Step 4. Furthermore, since dg = 0, it follows, exactly as in Step 4 of the firstproof of Theorem 5.2, that the coefficients of ϕ∗(g) do not depend on xl+1, · · · , xn.This finishes the proof.


Acknowledgments. We have benefitted of interesting discussions with M. Troy-anov. Part of the present research was achieved while the first author was visitingEPFL. The research of the third author has been, in part, subsidized by a grant ofthe Fonds National Suisse de la Recherche Scientifique.

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Received October 2009; revised February 2010.

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The Pullback equation for degenerate forms

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