Frobenious theorem

THE FROBENIUS THEOREM

SOPASSAKIS PANTELIS

Abstract. This text aims to be a self-contained report on the Frobenius

theorem. First, some necessary definitions are given and basic facts about dis-

tributions are stated. The dual objects of distributions - the codistributions -are introduced. We try to describe these notions and engage them to involu-

tiveness and complete integrability, to state and prove the Frobenius Theorem

which is of great importance in nonlinear control theory.

Ferdinand Georg Frobenious (1849-1917) was a German mathe-matician well known for his contributions to differential equations,linear algebra, group theory and differential geometry. He wasthe first man to prove the Cayley-Hamilton theorem. He studiedmathematics at the University of Berlin where he completed hisdoctorinary thesis on the solution of differential equations underWeierstrass.After its completion in 1870, he taught in Berlin for afew years before receiving an appointment at the Polytechnicum inZurich (now ETH Zurich). In 1893 he returned to Berlin, wherehe was elected to the Prussian Academy of Sciences. Group theorywas one of Frobenius’ principal interests in the second half of hiscareer. One of his first notable contributions was the proof of theSylow theorems for abstract groups.

Despite being named for Ferdinand Georg Frobenius, the the-orem we discuss within the following pages was first proven byAlfred Clebsch and F. Deahna. Deahna was the first to establishthe sufficient conditions for the theorem, and Clebsch developedthe necessary conditions. Frobenius is responsible for applying thetheorem to Pfaffian systems, thus paving the way for its usage indifferential topology.

1. Introduction

A Distribution is a central notion in control theory. Hereinafter when using theterm smooth we refer to functions of class C∞. We give the definition straightaway:

Definition 1. A distribution is a mapping D : <n → P (<n) that assigns a subspaceof <n to every x ∈ <n. That is D : <n → P (<n) , x 7→ D (x) ⊂ <n (Here Pstands for the powerset) according to D(x) = span f1(x), f2(x), . . . , fd(x), wheref1, f2, . . . , fd are not necessarily considered to be linearly independent.

1991 Mathematics Subject Classification. Interdisciplinary Program of Post Graduate Stud-

ies in Applied Mathematical Sciences - School of Applied Mathematical and Physical Sciences,National Technical University, Athens, Greece.

Key words and phrases. Distributions, Involutive, Frobenious Theorem.

1

2 SOPASSAKIS PANTELIS

♦

Considering of the distribution as a new mathematical object, we define the bi-nary operations of addition and intersection of two distributions as:

(D1 +D2)(x) = D1(x) +D2(x)(1.1)

(D1 ∩D2)(x) = D1(x) ∩D2(x)(1.2)

Apparently D1(x) + D2(x) and D1(x) ∩ D2(x) are once again distributions. LetD0 = span 0. Then D0 is the unitary element for addition and intersection.LetD(<n) be the space of all distributions D with domain the eucleidian space <n.Then (D,+) and (D,∩) are rendered semigroups with unity the distribution D0.Indeed for every D ∈ D, one has that D +D0 = span f1, . . . , fd+ span 0 = D.We say that D1 ⊆ D2, whenever for every x ∈ <n we have D1(x) ⊆ D2(x). Thedimension of a distribution for a fixed x ∈ <n is the dimension of the subspaceD(x), that is dimD(x) = dim span f1(x), f2(x), . . . , fd(x). We now move on tosome examples:

Example 1. Let D ∈ D(<n) and D(x) = span f1(x), . . . , fd(x) = span F (x),where F (x) ∈Mn×d(<). Thus D = ImF (x). Let now:

F : <3 →M3×3(<), x 7→ F (x) =

x1 0 00 x1 + x2 00 0 x1 + x2 + x3

(1.3)

It is easy to verify that rank(F (x)) = 0 if and only if x = 0, rank(F (x)) = 3 if andonly if x1 6= 0,x2 6= −x1 and x1 +x2 +x3 6= 0. The dimension of D in this exampleis a function of x.

•

Definition 2. A distribution D ∈ D(<n) is said to be of constant degree k if forevery x ∈ <n, dim(D(x)) = k. Such a distribution is sometimes called nonsingular.

♦

Example 2. Consider of the distribution D(x) = ImF (x) ∈ D(<4), with:

F (x) =

x24 + 1 x22 + 2 x24 + 1 x1

0 x21 + x1 + 2 1 2x1 + x2 0 −x43 − 1 0

1 1 1 1

(1.4)

Then dim(D(x)) = 3 for every x ∈ <4 and thus D is said to be a distribution ofconstant degree 3.

•

Definition 3. A distribution D is said to be smooth if there exist smooth vectorsf1, . . . , fd : <n → < such that span f1, . . . , fd = D

♦

NONLINEAR CONTROL THEORY 3

A smooth distribution D can be considered as vector space over the field ofsmooth functions. That is, every vector field f ∈ D = span f1, . . . , fd is expressedas a (finite) linear combination:

f(x) =

d∑j=1

aj(x)fj(x)(1.5)

where f, fj , a(x) ∈ C∞(U).

Proposition 1. Let D ∈ D(U) be a nonsingular smooth distribution of constantdegree d and let Y : U → <n be a smooth vector field with Y (x) ∈ D(U) for allx ∈ U . Then there exist k real smooth functions m1,m2, . . . ,mk : S → <, suchthat:

Y (x) = m1(x)X1(x) +m2(x)X2(x) + . . .+mk(x)Xk(x)(1.6)

for x ∈ U and where X1, X2, . . . , Xk are vector fields of D(U).

Proof.

We suppose that there exist X1, X2, . . . , Xk and m1,m2, . . . ,mk : S → < such thatY (x) = m1(x)X1(x) +m2(x)X2(x) + . . .+mk(x)Xk(x) holds for all x ∈ U and weshow that m1, . . . ,mk are smooth functions. We have:

Y (x) = m1(x)X1(x) +m2(x)X2(x) + . . .+mk(x)Xk(x)

⇔

Y1(x)Y1(x)

...Yk(x)

Yk + 1(x)...

Yn(x)

= m1(x) ·

X1,1(x)X1,2(x)

...X1,k(x)X1,k+1(x)

...X1,n(x)

+ . . .+mk(x) ·

Xk,1(x)Xk,2(x)

...Xk,k(x)Xk,k+1(x)

...Xk,n(x)

Now, introducing the notation:

y =

Y1(x)...

Yk(x)

, ξ1(x) =

X1,1(x)X1,2(x)

...X1,k(x)

, . . . , ξk(x) =

Xk,1(x)Xk,2(x)

...Xk,k(x)

We suppose without loss of generality that for every x the corresponding ξi = ξi(x),for i = 1, 2, . . . , k are linearly independent. Furthermore we have that y = m1ξ1 +. . . + mkξk = [ξ1 ξ2 . . . ξk] · [m1 m2 . . . mk]T . The matrix Ξ = [ξ1 ξ2 . . . ξk] isk × k and invertible; therefore we have:

m1

m2

...mk

= Ξ−1︸︷︷︸smooth

· y(x)︸︷︷︸smooth

(1.7)

Consequently, m1, . . . ,mk are (linearly independent) smooth functions.


Proposition 2. Let D1, D2 be two smooth distributions, then:

(1) D1 +D2 is a smooth distribution(2) D1 ∩D2 is not necessarily smooth.

We skip the proof of 1 as it is beyond the purpose of this essay. For 2, we givea counter example. So let:

D1 = span

[11

](1.8)

and

D2 = span

[1 + x1

1

](1.9)

D1 and D2 are smooth distributions and their intersection is given by:

(D1 ∩D2)(x) = span

[11

]= D1, for x1 = 0(1.10)

(D1 ∩D2)(x) = 0 , for x1 6= 0(1.11)

Definition 4. Let D ∈ D(U) be a distribution. We call x0 ∈ U a regular point ofD if there exists an open neighborhood U0 of x0 such that D is nonsingular in U0

♦

The following propositions states that every distribution, either if it is nonsin-gular of not, has ”‘many”’ regular points.

Proposition 3. The set RD = x0 ∈ U : x0 is a regular point of D is an opendense subset of U

In fact the (Lebesgue) measure of nonregular points is 0. µL(RCD) = 0

Proposition 4. Let D1 and D2 be two smooth distributions in U , D2 is nonsingularand D1 ⊂ D2 in a dense subset of U . Then D1 ⊂ D2 in U .

Proposition 5. Let D1 and D2 be two smooth distributions in U , D1 is nonsingularin U and D1 ⊂ D2 in U and D1 = D2 in a dense subset of U . Then D1 = D2 in U

The following result is very important as it postulates a sufficient condition forthe intersection of two distributions to be locally smooth.

Proposition 6. Let x0 be a regular point of D1, D2 and D1 ∩D2 of D(U). Thenthere exists a neighborhood U0 of x0 such as the restriction of D1 ∩D2 in U0 is asmooth distribution.


A common regular point for D1, D2 and D1 ∩D2 does exist because the set RDof regular points is dense in U . The proof of this proposition is realized by showingthat D1 ∩D2 is spanned by smooth vector fields, but this is easy to show since x0is a regular point of all D1, D2 and D1 ∩D2.Before proceeding to the definition of the notion of involutive distributions, wedefine the Lie bracket of two vector fields. In general a Lie bracket is a binaryoperation which satisfies certain properties.

Definition 5. Let L be a vector field over the field F . A Lie algebra over F is anF-vector space with a bilinear map, the Lie bracket:

[·, ·] : L× L→ L, (x, y) 7→ [x, y](1.12)

satisfying the following properties:

[x, x] = 0, for all x ∈ L(1.13)

[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0, for all x, y, z ∈ L(1.14)

The second property is known as the Jacobi identity. The Lie Bracket is sometimesreferred to as commutator.

♦

When studying smooth (in the sense of at least C1) vector fields, we define theLie bracket to be the mapping:

C1(<n;<m)× C1(<n;<m)→ C1(<n;<m), (x(t), y(t)) 7→ [x, y](t) =∂y

∂tx(t)− ∂x

∂ty(t)

(1.15)

We define the derived algebra of a Lie algebra L.

Definition 6. Let (L, [·, ·]) be a Lie algebra. Then the set:

L1 = [L,L] = z : ∃x, y ∈ L : [x, y] = z(1.16)

with the restriction of the Lie bracket of L, on L1 is called the derived algebra or L.

♦

The following important proposition guarantees that the restriction of the Liebracket of L, on L1 has a meaning since L1 is a subset of L.

Proposition 7. L1 is an ideal of L, hence L1 ⊆ L

To a fixed x0 ∈ U , we assign a vector space D(x0) ∈ D(U) which becomes a Liealgebra with the aforementioned Lie bracket. By means of the the above proposi-tion (7), one has that [D(x0), D(x0)] ⊆ D(x0). However this does not imply thatfor every x0, y0 ∈ U there exists a z0 ∈ U such that [D(x0), D(y0)] ⊆ D(z0). Ifa distribution has this property we say that is involutive. Let us define [D,D] =[D(x), D(y)], x, y ∈ U. Then we may give a formal definition for involutive dis-tributions.

Definition 7. We say that a distribution D is involutive if [D,D] ⊆ D, that is forevery f, g ∈ D it follows that [f, g] ∈ D. An involutive distribution is a []-invariantdistribution.

♦


Let D = span f1, . . . , fd be a nonsingular distribution. Then the smooth func-tions f, g can be expressed as a finite linear combination of f1, . . . , fd. So wehave:

f(x) =

d∑j=1

aj(x)fj(x)(1.17)

g(x) =

d∑j=1

bj(x)fj(x)(1.18)

Thus:

[f, g](x) = [

d∑j=1

aj(x)fj(x),

d∑j=1

bj(x)fj(x)](1.19)

=

d∑i=1

d∑j=1

ai(x)bj(x)[fi(x), fj(x)](1.20)

Hence for a distribution to be involutive, it is sufficient (and necessary) that

[fi, fj ] ∈ D, for 1 ≤ i, j ≤ d(1.21)

Note that for i = j, from the definition of the Lie bracket, [fi, fj ] = 0, and fromthe skew-commutativity property for i 6= j we have [fi, fj ] = −[fj , fi]. If D =span f1, f2 then it follows that for every f, g ∈ D, we have:

[f, g] =

d∑i=1,j=1,i6=j

det

(ai ajbi bj

)[fi, fj ](1.22)

For example, if d = 3, then:

[f, g] = det

(a1 a2b1 b2

)[f1, f2] + det

(a2 a3b2 b3

)[f2, f3] + det

(a1 a3b1 b3

)[f1, f3]

Employing a well known result from linear algebra, we state the following proposi-tion:

Proposition 8. A nonsingular distribution D ∈ D(U) is involutive if and only if

rank f1(x), . . . , fd(x) = rank f1(x), . . . , fd(x), [fi, fj ](x)(1.23)

for every x ∈ U and 1 ≤ i 6= j ≤ d

Example 3. Consider of the distribution D = span f1, f2, defined on a subset Uof <3, namely U = <∗ ×<∗ ×< and suppose that f1, f2 are defined as:

f1(x) =

x1x1 + x2

1

(1.24)

f1(x) =

x2x23 + 1

1

(1.25)


then the rank of F (x) = [f1(x) f2(x)] is 3 for all x ∈ U and the distribution Dis of constant rank k = 2 (nonsingular). Employing the abovementioned criterion(Proposition 8), we first compute the Lie bracket of f1 with f2.

[f1, f2] =df2(x)

dxf1(x)− df1(x)

dxf2(x)(1.26)

=

1 0 01 1 00 0 0

x1x1 + x2

1

−0 1 0

0 0 2x30 0 0

x2x23 + 1

1

(1.27)

=

x1 + x23 + 12x3(x23 + 1)

0

(1.28)

We now have:

[f1, f2, [f1, f2]](x) =

x1 x2 x23 + 1x1 + x2 x23 + 1 2x3(x23 + 1)

1 1 0

(1.29)

This matrix has constant rank 3 for all x ∈ U , therefore this distribution is notinvolutive.

•It is now clear that is a distribution defined on U with dim(U) = d, is nonsingu-

lar with constant rank k = d, is also involutive. In the special case that D is onedimensional, we have D = span f1. Then [D,D] is the zero-dimensional subspaceof U and by the above criterion (Proposition 8), every one dimensional distribu-tion is involutive. The following proposition, easily follows from the definition ofinvolutiveness.

Proposition 9. Suppose that D1, D2 are two involutive distributive, the same holdsfor D1 ∩D2, however D1 +D2 might not be involutive.

We state a counter-example to prove the second statement. As mentioned be-fore, every one-dimensional distribution is involutive. Furthermore, every two-dimensional distribution can be expressed as a sum of two one-dimensional distri-butions. That is D = span f1, f2 = span f + 1+ span f2 = D1 +D2 and notevery two-dimensional distribution is involutive. However every distribution canbe decomposed in a sum of involutive ones. We now give the definition of covectorfields and codistributions - a couple of crucial and important notions in controltheory.

Definition 8. A covector field in U ⊆ <n is a map w : <n → (<n)∗, where (<n)∗

is the dual space of <n, i.e. the space of linear continuous functionals φ : <n → <.A codistribution is a map Ω : <n → P((<n)∗), x 7→ Vx such that Vx is a subspaceof (<n)∗.

♦Hereinafter we will denote the set of all codistributions on U by D∗(U). The

notions of smoothness and involutiveness are defined analogously for codistribu-tions. The binary operations of addition and intersection and the binary relation of


inclusion are also defined the same way. Given a distribution D we may correlateto that a codistribution D⊥ called the annihilator of D and defined to be:

D⊥(x) = w∗ ∈ (<n)∗| 〈w∗, u〉 = 0,∀u ∈ D(x)(1.30)

Here 〈·, ·〉, stands for the eucleidian inner product in <n. For every x ∈ U ⊆ <n,D⊥(x) is a subspace of (<n)∗. The annihilator of a distribution is a codistribu-tion and since the second dual of <n is isomorphic to <n, the annihilator of acodistribution is again a distribution. In that sense we define the annihilator of acodistribution to be:

Ω⊥(x) = u ∈ <n| 〈w∗, u〉 = 0,∀w∗ ∈ Ω(x)(1.31)

Remark 1. Let D ∈ D(U) be a smooth distribution and D⊥ ∈ D∗(U) its annihi-lator. Then D⊥ is not necessarily smooth. Indeed, if we consider of the followingone-dimensional distribution D ∈ D(<n)

D(x) = span x(1.32)

then we have that:

D⊥(x) =

0 if x 6= 0

(<)∗ if x = 0(1.33)

Observe that D is smooth while D⊥ is not.

The annihilator of a distribution D, is a complementary object since the followingdimension identity holds:

dim(D) + dim(D⊥) = dim(U)(1.34)

Additionaly, some fine properties of annihilators are presented:

D1 ⊆ D2 ⇔ D⊥1 ⊇ D⊥2(1.35)

[D1 ∩D2]⊥ = D⊥1 +D⊥2(1.36)

The above properties can be easily checked. If we consider of a smooth distributionas a matrix F (with smooth entries), in terms of D is spanned by the columns ofF , then D⊥F can be realized as a ”‘distribution”’ spanned by some rows w∗. In thatsense, we can write the annihilator of DF as:

D⊥F (x) = w∗ ∈M1×d(<)|w∗ · F (x) = 0, x ∈ U(1.37)

Indeed (<n)∗ is isomorphic to M1×d(<) therefore every k-subspace of (<n)∗ is inone-to-one correspondence with a matrix in Mk×d(<). Vice versa, supposing thata smooth codistribution ΩG is spanned by the rows of a matrix G(x) (with smoothentries), the annihilator of ΩG is:

Ω⊥G(x) = u ∈Md×1(<)|G(x) · u = 0x ∈ U(1.38)

That is

Ω⊥G(x) = ker(G(x))(1.39)

At this point we have obtained a better realization of codistributions. We state thefollowing important proposition:

Proposition 10. Let x0 ∈ U be a regular point of a smooth distribution D ∈ D(U).Then x0 is a regular point of D⊥ and there exists a neighborhood S 3 x0 such thatD⊥|S is a smooth codistribution.


Let D be a smooth distribution defined on U ⊆ <n and nonsingular with constantdegree d. In a neighborhood U0 3 x0 of every x0 ∈ U , by definition there exist dsmooth functions f1, . . . , fd that span D.

D = span(f1, . . . , fd)(1.40)

According to the last proposition, x0 are regular points for D⊥ = Ω and Ω|U0 is asmooth nonsingular codistribution. Therefore by the dimension theorem dim(Ω) =n − d in U0 and consequently there are n − d covector fields w1, . . . , wd that spanΩ.

Ω = span(w1, . . . , wd)(1.41)

By definition 〈wj , fi〉 = 0 for 1 ≤ j ≤ n − d and 1 ≤ j ≤ d. Set F (x) =[f1(x) f2(x) . . . fd(x)]. Then wj(x)·F (x) = 0 for every x ∈ U , with rank(F (x)) = dfor all x ∈ U . We seek solutions to wj(x) · F (x) = 0 with 1 ≤ j ≤ n− d subject tospanwj = D. In particular we seek for solutions in the form:

wj =∂λj∂x

(1.42)

Then, it is easy to verify that:

wj(x) · F (x) = 0⇔ ∂λj∂x· F (x) = 0⇔ 〈dλj(x), fi(x)〉 = 0(1.43)

We want to solve the above system of partial differential equations for λj and findn− d linearly independent solutions. The Frobenius theorem provides some handy(sufficient and necessary) conditions for these functions to exist. We have first togive the following definition:

Definition 9. A nonsingular d-dimensional distribution D, defined on U ⊆ (<n)is said to be completely integrable if for every x0 ∈ U there exist a neighborhoodU0 3 x0 and n − d real smooth functions λ1, λ2, . . . , λn−d defined on U0, such thatspan(dλ1, dλ2, . . . , dλn−d) = D⊥

♦

It is now time for us to state and prove the Frobenious Theorem:

Theorem 1. [Frobenius] A nonsingular distribution is completely integrable if andonly if it is involutive.

Before giving the proof of the Frobenius theorem we have to recall some prop-erties of the Lie bracket and directional derivative. The derivative of g along f isdefined as:

Lfg(x) = fT (x)[∂gT

∂x]T + g(x)

∂f

∂x(1.44)

where g is a covector field and f is a vector field. The result of this operationis again a covector field. An alternative definition is adopted by some authors,according to which:

(Lfg)(x) = 〈dg(x), f(x)〉 = g(f)|x(1.45)

where g(f)|x is a convenient symbolic representation of the directional derivativeand should not be confused with the synthesis (g f)(x) which is something com-pletely different. Employing this notation, we have the following properties:


Lemma 1. Let A,B,C be vector fields of proper smoothness, φ a real valued smoothfunction and l1, l2 ∈ <. Then:

(1) [A,B](φ) = A(B(φ))−B(A(φ))(2) [A,B] = −[B,A](3) [A, l1B + l2C] = l1[A,B] + l2[A,C](4) [A, φ ·B] = B(φ)A+ φ · [A,B](5) A(〈B,C〉) = 〈A(B), C〉+ 〈B,A(C)〉(6) A(dφ) = dA(φ)

We now come to the proof of the Frobenius Theorem:

Proof.

We first prove the that complete integrability ⇒ involutiveness. So suppose thata nonsingular distribution D of constant degree d is completely integrable. Bydefinition of complete integrability, there are n − d covector fields λ1, . . . , λn− ddefined on neighborhoods U0 3 x0 such that span(dλ1, dλ2, . . . , dλn−d) = D⊥, orequivalently:

∂λj∂x

fi(x) = 0(1.46)

⇔ 〈dλj(x), fi(x)〉 = 0(1.47)

⇔ Lfiλj(x) = 0(1.48)

⇔ λj(fi)|x = 0(1.49)

For x ∈ U0 and i = 1, 2, . . . , d. Now differentiating λj along [fi, fk] we have:

[fi, fk](λj)|x = fi(fk(λj))|x − fk(fi(λj))|x = 0(1.50)

Applying this operation for j = 1, 2, . . . , n− d we obtain:[fi, fk](λ1)|x[fi, fk](λ2)|x

...[fi, fk](λn−d)|x

=

dλ1dλ2

...dλn−d

[fi, fk]|x = 0(1.51)

For x ∈ U0. By assumption the covector fields dλ1, . . . , dλn−d span D⊥. There-fore [fi, fk]|x is in D for every index i, k = 1, 2, . . . , d and we conclude that D isinvloutive. We now prove that Involutiveness ⇒ Complete Integrability. So wesuppose that D is a nonsingular involutive distribution of constant degree d definedon some subset U of <n. From a proposition stated before (Proposition 1), we knowthat in a neighborhood U0 of every point x0 ∈ U there exist d smooth vector fieldsdefined on U0 such that they span D(x). That is:

D(x) = span(f1(x), . . . , fd(x)), x ∈ U0(1.52)

Now let fd+1, . . . , fnx be a complementary set of vector fields defined on U0 suchthat:

span(f1(x), . . . , fd(x))⊕ span(fd+1(x), . . . , fn(x)) = <n, x ∈ U0(1.53)

Let Φft (x) stand for the flow of the vector field f , that is the function Φft (x0) assignsto every initial point x0 for the differential equation x = f(x), a function φ(t;x0)


that is a solution to the IVP (initial value problem) x = f(x), with x(0) = x0 =

φ(0;x0). So φ(t;x0) = Φft (x0). By definition we have:

∂

∂tΦft (x0) = (f Φft )|x0(1.54)

Φf0 (x0) = x0(1.55)

Formally, we say that Φft is a mapping:

Φft : S → C1(I), x0 7→ Φft (x0) = φ(t, x0)(1.56)

for x0 ∈ S and t ∈ I, and f is Lipschitz with respect to x in S and the solution of

the related IVP exists for every t ∈ I. Φft (x) is a local diffeomorphism. One caneasily verify that Φ has the semigroup property, that is:

Φft+T |x0 = Φft ΦfT |x0(1.57)

To every function f1, f2, . . . , fd, fd+1, . . . , fn we assign a flow Φfit |x for i = 1, 2, . . . , nand for all x ∈ U0. We define the mapping:

F : Bε(<n)→ <n, z = (z1, . . . , zn) 7→ Φf1z1 · · ·Φf1z1 |x0(1.58)

Claim 1. For ε sufficiently small, F has the following properties:

(1) Is defined for all z ∈ Bε and is a diffeomorphism(2) For all z ∈ Bε, the first d columns of the matrix ∂F/∂x are linearly inde-

pendent vectors in D(F (z)).

We prove the first claim: From the Picard-Lindelof theorem, we have that for all

x ∈ <n and sufficiently small |t|, the flow Φft (x) of a (Locally Lipschitz) vector fieldf is well defined and therefore F is well defined for all z ∈ Bε(<n) for sufficientlysmall ε. In order to show that F is a local diffeomorphism, it is sufficient to showthat rank(F ) = n for all z ∈ Bε(<n). To this end, we define for convenience:

M? =∂M

∂x(1.59)

We have that:

∂F

∂zi= (Φf1z1)? · · · (Φfi−1

zi−1)?

∂

∂zi(Φfizi · · · Φfnzn)|x0(1.60)

= (Φf1z1)? · · · (Φfi−1zi−1

)? · fi Φfizi · · · Φfnzn |x0(1.61)

= (Φf1z1)? · · · (Φfi−1zi−1

)? · fi Φfi−1

−zi−1 · · · Φfi−z1 |F (z)(1.62)

For z = 0 we have F (0) = x0 and

∂F

∂zi|0 = fi|x0(1.63)

The vector fields f1(x0), · · · , fn(x0)are linearly independent and tanget to F , hencethe n columns of (F )? are linearly independent at z = 0 and due to the fact thatF is continuous, the martix F is of full rank at 0 and in a neighborhood of 0.

We now prove the second claim: We have already shown that for all z ∈ Bε(<n),


(Φf1z1)? · · · (Φfi−1zi−1

)? · fi Φfi−1

−zi−1 · · · Φfi−z1 |x=F (z) =

∂F

∂zi(1.64)

Instead of proving the claim (1.2) we formulate and prove an other one. If thefollowing claim holds, so does the claim (1.2).

Claim 2. For all x ∈ U0, where U0 is a neighborhood of x0, for |t| sufficientlysmall and for any two vector fields τ, ϑ in D the following holds:

(Φϑt )?τ (Φϑ−t)|x ∈ D(x)(1.65)

This claim imlies that (Φϑt )?τ (Φϑ−t)|x is a locally defined vector space in D and(1.2) follows. So we prove the claim (2). Let ϑ be a vector space in D and let usdefine:

Vi(t) = (Φϑt )?fi (Φϑ−t)|x(1.66)

for i = 1, 2, . . . , d. Observe that:

I = (Φϑ−t)?(Φϑt )?

⇒ 0 =d

dt(Φϑ−t)?(Φ

ϑt )?

⇒ 0 =d

dt(Φϑ−t)? (Φϑt )|x + (Φϑ−t)star

∂ϑ

∂x Φϑt |x

⇒ d

dt(Φϑ−t)? (Φϑt )|x = −(Φϑ−t)star

∂ϑ

∂x Φϑt |x

Furthermore:d

dt(fi Φϑt |x) =

∂fi∂x

ϑ Φϑt |x(1.67)

The functions Vi(t) defined earlier satisfy:

dVidt

= (Φϑ−t)?[ϑ, fi] Φϑt(1.68)

By assumtion D is involutive and both ϑ and fi belong to D, thus their Lie bracketbelongs to D so there exist functions λij defined locally in U0 such that:

[ϑ, fi] =

d∑j=1

λijfj ∈ D(1.69)

Hence we can express dVi/dt as:

dVidt

= (Φϑ−t)?[ϑ, fi] Φϑt = (Φϑ−t)?(

d∑j=1

λijfj) Φϑt(1.70)

=

d∑j=1

λij(Φϑt |x)Vj(t)(1.71)

So the functions Vi(t) are the solutions of a system of linear diffenential equations,and for that reason we may set:

[V1(t) · · · Vd(t)] = [V1(0) · · · Vd(0)] ·X(t)(1.72)

Where X(t) ∈ Md×d(<) in (1.72) is the fundamental matrix of solutions. Wemultiply (1.72) on the left both sides by (Φϑt )? and we get

[f1 Φϑt |x · · · fd Φϑt |x] = [((Φϑt )?)f1|x · · · ((Φϑt )?)fd|x]X(t)(1.73)


and we replace x by Φϑ−t|x to get

[f1(x) f2(x) . . . fd(x)] = [(Φϑt )?f1|x Φϑ−t|x · · · (Φϑt )?fd|x Φϑ−t|x](1.74)

X(t) is nonsingular for all t with |t| sufficiently small, therefore for i = 1, 2, · · · , dwe have:

(Φϑt )?fi|x Φϑt |x ∈ span f1|x, · · · , fd|x(1.75)

Or equivalently,

(Φϑt )?fi|x Φϑ−t|x ∈ D(x)(1.76)

Employing the fact that the arbitary τ ∈ D is written as:

τ =

d∑j=1

ajfj(1.77)

(where aj are functions from the field of C∞ functions), the proof of the claim (2)is complete.

We now use the results from the claims we have proven so far to complete the proofof the Frobenius’ theorem. So let U be the image of the mapping F . Then U isan open neighborhood of x0 for F (0) = x0. Since F is a (local) diffeomorphism(We know that from claim 1), it is one-to-one, and therefore it has an inverseF−1 : U→ Bε(<n) and is smooth. Set:

φ1(x)φ2(x)

...φn(x)

= F−1(x)(1.78)

where φ1(x), . . . , φn(x) are real-valued smooth functions fi(x) : U→ <n. We claimthat the last n− d of these functions defined in (1.78) are independent solutions of(1.43). By definition of F in (1.58), we have:[

∂F−1

∂x

]x=F (z)

[∂F

∂z

]= I(1.79)

for all z ∈ Bε(<n) that is for all x ∈ U. By the claim (1.2), the first d columnsof the second factor on the left-hand side of (1.79) form a basis of D at any pointx ∈ U. Hence the vectors dφd+1, . . . , dφn are annihilated by the vectors of D ateach x ∈ U and therefore form a basis for D⊥. The proof of the Frobenius’ Theoremis now complete.

The proof of the sufficiency of involutiveness for a nonsingular distribution to becompletely integrable is constructive and offers us a frame for the construction offunctions such that their gradients form a basis for the annihilator of a given distri-bution. We give an example straihtforward to explain the constructive procedure.


Example 4. Consider of a distribution D = span(f1, f2) defined on U =x ∈ <3 : x21 + x23 6= 0

with:

f1 =

2x3−10

, f1 =

−x1−2x2x3

(1.80)

We define:

G(x) = [f1(x) f2(x)] =

2x3 −x1−1 −2x20 x3

,(1.81)

Since rankG(x) = 2, for all x ∈ U , D is nonsingular with constant degree d = 2.The Lie Bracket of f1 with f2 is:

[f1, f2] = (f2)?f1 − (f1)?f2(1.82)

=

−1 0 00 −2 00 0 1

2x3−10

−0 0 2

0 0 00 0 0

−x1−2x2x3

(1.83)

=

−4x320

(1.84)

the matrix

[f1 f2 [f1, f2]] =

2x3 −x1 −4x3−1 −2x2 20 x3 0

(1.85)

has rank 2 for all x ∈ U and the distribution is involutive (According to proposition8). Set f3(x) = e3. We calculate the flows of f1, f2 and f3.

Φf1z1(x) =

2z1x3 + x1−z1 + x2

x3

,Φf2z2(x) =

exp(−z2)x1exp(−2z2)x2

exp(z2)x3

,Φf3z3(x) =

z3 + x1x2x3

.(1.86)

The mapping F is defined to be:

F (z)|x=x0 = Φf1z1 Φf2z2 Φf3z3 |x0(1.87)

=

2z1 exp(z2)x03 + exp(−z2)(z3 + x01)−z1 + exp(−2z2)x02

exp(z2)x03

(1.88)

The partial differential equation:

∂λ

∂xG(x) = [0 0](1.89)

has the solution:

λ(x) = z3(x) = (x1 + 2x2x3)x3(1.90)

•


References

[1] Isidori A., Nonlinear Control Systems - An introduction, Springer Verlag Editions Interna-tional, 2nd edition, 1989.

[2] Erdmann K., Wildon J., K., Introduction to Lie Algebras, Springer Verlag Editions Interna-

tional, 2006.[3] Wikipedia, the free encyclopedia, available online: http://www.wikipedia.com

E-mail address, Sopassakis P.: [email protected]

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