Exterior Differential Calculus and Applications to Economic Theory

Ivar Ekeland

CEREMADE and Institut de Finance University of Paris Dauphine

Notes by

Paolo Guiotto and Tiziano Vargiolu Scuola Normale Superiore of Pisa

Pisa, February – April 1996

Contents

1 Foreword 2

2 From economics to mathematics 3

3 The single consumer model 4 3.1 The indirect approach . . . . . . . . . . . . . . . . . . . . . . . 7

4 Exterior differential calculus 9 4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 4.2 Exterior product . . . . . . . . . . . . . . . . . . . . . . . . . . 10 4.3 Exterior derivative . . . . . . . . . . . . . . . . . . . . . . . . . 11 4.4 Differential ideals . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 The Frobenius theorem 15 5.1 Statement and proof . . . . . . . . . . . . . . . . . . . . . . . . 15 5.2 Application to the one consumer model . . . . . . . . . . . . . 19

6 The household model 21 6.1 Statement of the model and Pareto optimality . . . . . . . . . . 21 6.2 Browning - Chiappori necessary conditions . . . . . . . . . . . 23

7 Pfaff’s theorem 27

8 The Cartan–Kähler theorem 34 8.1 An introductory example . . . . . . . . . . . . . . . . . . . . . 34 8.2 The general problem . . . . . . . . . . . . . . . . . . . . . . . . 35 8.3 Differential ideals . . . . . . . . . . . . . . . . . . . . . . . . . . 36 8.4 A counter example . . . . . . . . . . . . . . . . . . . . . . . . . 37 8.5 Regularity conditions and ordinary points . . . . . . . . . . . . 38 8.6 The main result . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

9 Markets 41

1 Foreword

During the academic year 1995–96, I was invited by the Scuola Normale Superiore to give a series of lectures on my work in progress with Pierre-André Chiappori. MM. Paolo Guiotto and Tiziano Vargiolu attended these lectures and reworked the material to write a set of notes, which I then reviewed and expanded.

These lecture notes are thus the result of a joint effort by MM. Paolo Guiotto and Tiziano Vargiolu and myself. The material is taken from the original paper [7] (see also [6]), and the exposition of exterior differential calculus owes much to the beautiful book [4] by Bryant, Chern, Gardner, Goldschmidt and Griffiths (see also [14]. In the past year, there has been further progress on the subject matter, as recorded in [8], [9], and [10].

The purpose of these notes is to make the underlying economic problems and the mathematical theory of exterior differential systems accessible to a larger number of people. Indeed, the articles are mainly concerned with economic theory, and proceed as quickly as possible to the main results, giving only the barest outline of the underlying mathematical theory, while the book is concerned with exterior differential calculus as a whole, and goes much beyond our immediate needs. It is the purpose of these notes to go over these results at a more leisurely pace, keeping in mind that mathematicians are not familiar with economic theory and that very few people, if any, have read Elie Cartan.

It is not often that one has the privilege of exploring a subject where deep economics and deep mathematics are so intimately related. I thank the Scuola Normale Superiore for giving me the opportunity to lecture on this work, MM. Paolo Guiotto and Tiziano Vargiolu for the very stimulating discussions we had,and their patience in writing these notes, and my very good friend professor Antonio Ambrosetti for his constant help and encouragement. He had the idea of these notes, and we are glad that he did.

Ivar Ekeland

2 From economics to mathematics

Many questions in economic theory boil dawn to one of the following mathematical questions, which we write in order of increasing difficulty:

First question : given a smooth vector field X in ℝ^{n}, and a positive number k < n, is it possible to find real functions λ_{i} and f_{i}, 1 ≤ i ≤ k, such that:

X(x) = ∑_{i=1}^{k} λ_{i}(x)∇f_{i}(x).  (1)

Note that the question is trivial for k ≥ n.

Second question : can we find the λ_{i}, f_{i} in the decomposition (1) such that the λ_{i} are non-negative and the f_{i} are convex ?

Third question : can we ask the λ_{i}, and the f_{i} in the decomposition to satisfy some additional equations of the following type

ϕ_{j}(λ(x), ∇f(x), x) = 0, 1 ≤ j ≤ p.

where the ϕ_{j} are prescribed functions.

In the very particular case when k = 1 and λ ≡ 1, the answer to the first question is given by the well-known:

Theorem 2.1 Given a smooth vector field X in ℝ^{n}, there exists f such that X = ∇f if and only if:

\frac{∂X_{i}}{∂x_{j}} = \frac{∂X_{j}}{∂x_{i}} ∀i, j = 1, …, n

In the general case when λ is a non-constant fuction, the answer to the first question is given by the Frobenius theorem for k = 1 , and by the Pfaff theorem while for k > 1. All solutions are local, that is, they are defined and solve (1) on some neighbourhood of a prescribed point x.

To solve the problem posed in the second and third question, we will need the tools developed at the turn of the century by a score of mathematicians from France and Germany, the most prominent of which is Elie Cartan. In a series of aticles and books extending over fifty years, he systematized these tools and discovered a beautiful mathematical structure, which he called exterior differential calculus, where antisymmetry plays a fundamental role. He has given an overview of the theory in a book [5], which is still in print. This theory has been very useful in understanding the local geometry of submanifolds and the differential structure of Lie groups, but has had few applications to other areas of mathematics. It is no exaggeration to claim taht it has been little known to analysts, let alone economists. And yet, i think that the Cartan-Kähler theorem, for instance, is as basic a result for Partial differential equations as the Cauchy-Lipschitz theorem is for ordinary differential equations.

But before we solve these problems, we have to tell the reader where they come from, and this will require going into basic economic theory. Once the foundation is laid, and some understanding of the basic economic issue is gained, we can proceed with the mathematical analysis. Thus, our hope is to make mathematics and economics walk hand in hand, as so often is the case with physics.

3 The single consumer model

Let us describe the standard framework of microeconomics (see any textbook on the subject,[15] for instance). The basic behavioural assumption is that each each individual makes the best choice among all opportunities which are open to him. Mathematically, this translates into the following kind of problem:

max_{x∈K} U(x),  (2)

K is called the feasible set. Each x ∈ K represents a possible choice (or decision) for the individual. Preferences are given by the real-valued function U, the utility function. If x, y ∈ K, and U(x) > U(y), then we say that the individual prefers x to y; on the other hand if U(x) = U(y), then we say that x and y are indifferent. For a given x, the set of all y ∈ K such that U(x) = U(y) is called the indifference set through x.

In economics, decisions or choices concern goods to be consumed (or produced; there is a well-developed theory of production, but we will not go into it in these notes). Then K is a subset of ℝ^{n}, where n is the number of goods available in the economy. For each x ∈ K, the i-th coordinate x^{i} represents a quantity x^{i} of the i-th good; it is assumed that goods are divisible, so that x^{i} can be any real number, and it is usually assumed also that x^{i} cannot be negative.

With each good i is associated a price p_{i} per unit of good, so that a bundle x = (x^{1}, …, x^{n}) costs p_{1}x^{1} + … + p_{n}x^{n} - in short, px. The consumer starts out with a wealth which is normalized to 1, and chooses the bundle of goods he (she) prefers among those he (she) can afford. In mathematical terms, this is stated as a classical problem in optimization:

{x ∈ ℝ^{n} : x^{i} ≥ 0, ∀i = 1, …, n, ∑_{i=1}^{n} p_{i}x^{i} ≤ 1} .

So the problem (2) becomes

(P_{p,U}) \{ \begin{matrix}max U(x) \\ x_{i} ≥ 0, \\ p · x ≤ 1 \end{matrix}

To make economic sense of this problem, we need some additional hypotheses on U. For instance, we would like the optimal solution to be unique, which leads us to the idea that U should be strictly concave (there are other, perhaps more compelling, economic reasons for assuming concavity). On the other hand, it is clear that the relevant economic object is the preference relation, not the utility function: if h is any increasing function on the real line, U and V = h ∘ U define the same preference relation (U(x) ≥ U(y) if and only if V (x) ≥ V (y), and yet V need not be concave when U is. So we have to enlarge somewhat the notion of concavity:

Definition 3.1 We say that U is (strictly) quasi–concave if the sets {x : U(x) ≥ a}are (strictly) convex ∀a ∈ ℝ.

Remark 3.1 Is easy to see that in general a quasi–concave function is not concave. More generally if U is quasi-concave, so is h ∘ U for every increasing function h : ℝ → ℝ

We shall henceforth assume that:

Hypotheses 0 (H0)

(i) U is C^{∞}(ℝ^{n}) and strictly quasi-concave;

(ii) if x̄ is optimum for the problem (2), then x_{i} > 0 ∀i;

(iii) \frac{∂U}{∂x_{i}} > 0, ∀i.

If (H0) holds, we have a unique optimal solution x̄. It is characterized by:

U′(x̄) = λp, for some λ ∈ ℝ^{+}, and p · x̄ = 1.

This characterization is a standard result in convex analysis, or in optimization theory. The number λ is called the Lagrange multiplier . Note that in general both λ and x̄ depend on p. Thus in this situation, we have an optimization problem depending on the parameters p = (p_{1}, …, p_{n}) and a map sending p into an unique optimum point x̄(p). The function p ↦ x̄(p) is called demand function of the individual.

It is well-kown that if a function U is concave, then U′′(x) is negative semidefinite, and that if U′′(x) is everywhere negative definite, then U is strictly concave. There are analogous conditions for the quasi-concave case. To state them, we need to introduce a hyperplane E(x), which is in fact the tangent hyperplane to the indifference set hypersurface through x:

E(x) := {y : U′(x) · y = 0};

Consider the restriction U′′(x)_{|E(x)} of the quadratic form U′′(x) to E(x). If U is quasi-concave, then U′′(x)_{|E(x)} is negative semi-definite. Conversely, if U′′(x)_{|E(x)} is negative definite, then U is strictly quasi-concave.

In order to obtain regularity for the demand function we make a further (generic) assumption

Hypothesis 1 (H1) U′′(x)_{|E(x)} is negative semi-definite for all x such that x_{i} > 0 for all i.

Applying the implicit function theorem we obtain the following

Proposition 3.1 If (H0) and (H1) hold, then the demand function is C^{∞}.

Again the proof is left to the reader. We are now ready to set the following problem:

given a C^{∞} function x(p), can it be a demand function?

In other words, is it possible to find a quasi-concave function U such that x is the optimal solution to the problem (P_{p,U}) ?

It may be the case (and in fact, it is) that the answer is no, unless x(p) satisfies additional conditions (in fact, certain systems of nonlinear partial differential equations). Testing these conditions against the data, that is, checking whether the actual demand functions, as observed by econometricians, satisfy these additional conditions, then provides us with a test for the model we started from. In other words, the idea that people maximize their utility over a feasible set, far from flying high up in the air as many people believe, can actually be tested against hard data, just as physicists test their models against experimental data.

So we have to infer the (unobserved) utility function U from the (observed) demand functionx(p). There are two ways to do this

The first one is the so called direct approach. We assume that the function x(p) is invertible (which is the case in general). Then, recalling that

U′(x(p)) = λ(p)p, λ > 0

we have

p(x) = \frac{1}{λ(x)}∇U(x)

Since the left-hand side p(x) is known, and the decompostion on the righthand side is not, we recognize the problem we stated in the first question of the preceding section (with k = 1). This approach has been developed by the Italian mathematician Antonelli []. However, his method has fallen into disuse, mainly because x(p) is a more natural function than p(x), even though they are mathematically equivalent

3.1 The indirect approach

The second method to solve the problem is called the indirect approach. We start by defining the so called indirect utility function or value function:

V (p) := max{U(x) : x ≥ 0, p · x = 1}.

By this definition we have that

V (p) = U(x(p)) = U(x(p)) + λ(p)(1 − p · x(p)),

due to the fact that p · x(p) = 1. Differentiating with respect to p:

\begin{aligned}V′(p) & = (U′(x(p)) − λ(p)p) · x′(p) + λ′(p)(1 − p · x(p)) − λ(p)x(p) = \\ & = −λ(p)x(p). \end{aligned}

So we can write

x(p) = −\frac{1}{λ(p)}∇V (p)

Since the left-hand side is known, and we are looking for the decompostion on the right-ahnd side, we are back to the first question again. If we can solve it, and find λ and V, the following proposition will give us U:

Proposition 3.2 Suppose that (H0) and (H1) holds. Then we have:

a) U(x) = min{V (p) : p · x ≥ 1};

b) U is quasi-concave if and only if V is quasi-convex.

This approach was first developed in Pisa in 1886, by a young researcher in mathematics named Antonelli , who later left academia and went on to a long and distinguished career in engineering. Among other things,in 1923 he promoted Italy’s first international airline, the Genoa-Barcelona. His paper [1] (in fact, a booklet published at his own expense) went unnoticed, and it was rediscovered in 1943, one year before the author’s death, by Herman Wold, who came across it accidentally while browsing through the personal library of Ladislaus von Bortkiewicz, which had been acquired by the university of Uppsala. In the meantime, of course, the result had been rediscovered again, by another Italian named Slutsky (see [21] . The basic characterization is as follows.

Theorem 3.1 Let the map x ∈ C^{∞} satisfy the condition

p · x(p) = 1 ∀p

Then x(p) is a demand function if and only if the matrix Σ = (σ_{ij})_{ij}, defined by:

σ_{ij} := \frac{∂x_{i}}{∂p_{j}} − ∑_{k} p_{k} \frac{∂x_{i}}{∂p_{k}} x_{j}

is symmetric and positive definite.

There are several proofs of this result (see for instance the textbook by Kreps [] for the most popular one. We will prove it by using the Frobenius theorem, which will give us a first glimpse into exterior differential calculus.

4 Exterior differential calculus

It is now time to point out that the three questions we asked in section 1 are not correctly formulated. Indeed, formula (1) is not invariant by a change of coordinates. If we set y = Φ(x), with Φ a diffeomorphism, the vector field X(x) becomes T Φ(x)X(x), while T Φ(x)∇f(x) is not the gradient of ∇f(Φ^{−1}(y)) at y. So the two sides of formula (1) behave according to different rules, and equality can hold onlying one particular set of coordinates.

This is because the gradient is not the right notion. It requires one to identify a linear form with a vector via a Euclidean structure, which makes changes of coordinates very complicated. We shall forget this identification, and work directly with linear forms, without identifying them with vectors; more generally, we will work directly with differential forms (fields of linear forms) without identifying them with vector fields. Let us first recall some basic facts, referring to the book by V. Arnold [2] for more.

4.1 Preliminaries

Given x ∈ ℝ^{n} let us denote by T_{x}ℝ^{n} the tangent space to ℝ^{n} at the point x. Each element of T_{x}ℝ^{n} is called a tangent vector, and we will pose, as usual,

T ℝ^{n} := {(x, v) : x ∈ ℝ^{n}, v ∈ T_{x}ℝ^{n}} .

A smooth section of T ℝ^{n} is called avector field. In other words a vector field X is a mapping defined on ℝ^{n} such that X(x) ∈ T_{x}ℝ^{n}. For properties of vector fields we refer the reader to any textbook on differential geometry, for instance the previously cited book by Arnold [2].

Let us denote with T_{x}^{∗}ℝ^{n} the cotangent space at x, that is, the dual of T_{x}ℝ^{n}. In other words, and element in T_{x}^{∗}ℝ^{n} is a tangent linear form at x, just as an element in T_{x}ℝ^{n} is a tangent vector at x. Denote by T^{∗}ℝ^{n} the cotangent bundle:

T^{∗}ℝ^{n} = {(x, ξ) : x ∈ ℝ^{n}, ξ ∈ T_{x}^{∗}ℝ^{n}} .

The canonical coordinate system gives us a natural base for the tangent space T_{x}ℝ^{n}: we will denote it by

\frac{∂}{∂x^{1}} , …, \frac{∂}{∂x^{n}} .

This notation follows from our understanding a vector field as a differential operator on functions f defined on ℝ^{n}:

ξ(f) = ∑_{i} \frac{∂f}{∂x^{i}} ξ^{i}

Finally we will denote by dx^{1}, …, dx^{n} the base dual to the previous one in the space T_{x}^{∗}ℝ^{n}.

Definition 4.1 A differential 1-form is a smooth section of T^{∗}ℝ^{n} over ℝ^{n}

In other words, a differential 1-form is a smooth mapping ω defined on ℝ^{n} such that ω(x) ∈ T_{x}^{∗}ℝ^{n} for all x ∈ ℝ^{n}. In canonical coordinates a 1–form ω will be written as w = ∑ w_{i}dx^{i}, so that if ξ = ∑ ξ^{j} \frac{∂}{∂x_{j}} is a vector field, the operation of ω on ξ is given by:

ω(ξ) = ∑ ω_{i}ξ^{i}.

Now we introduce the more general concept of differential k-forms, and we introduce two fundamental operations: the exterior product and the exterior derivative.

4.2 Exterior product

Exterior k−forms can be constructed from 1-forms by a simple algebraic operation, called the exterior product :

Definition 4.2 Let ω_{1}, …, ω_{k} be 1-forms. The exterior product ω_{1} ∧ … ∧ ω_{k} is the k-form defined by :

ω_{1} ∧ … ∧ ω_{k} (π^{1}, …, π^{k}) = ∑_{σ} (−1)^{sign(σ)} ω_{1} (π^{σ(1)}) … ω_{k} (π^{σ(k)})

where the sum ranges over all permutations σ of {1, …, k}

For instance, for k = 2 :

α ∧ β (π, χ) = α(π)β(χ) − α(χ)β(π)
= (∑_{i} α^{i}π_{i}) (∑_{j} β^{j}χ_{j}) − (∑_{j} α^{j}π_{j}) (∑_{i} β^{i}χ_{i})
= ∑_{i< j} (α^{i}β^{j} − α^{j}β^{i}) (π_{i}χ_{j} − π_{j}χ_{i})

Obviously, this is the simplest 2-form related to α and β and satisfying two basic properties, namely bilinearity and antisymmetry.

If M is a n-dimensional manifold, and (p_{1}, …, p_{n}) is a local coordinate system on M, then (dp_{1}, …, dp_{n}) is a basis for the space of differential 1forms, and any differential k−form α on M can be written as :

α(p) = ∑ α_{i_{1},...,i_{k}}(p) dp_{i_{1}} ∧ … ∧ dp_{i_{k}}

where i_{1} < … < i_{k} ranges over all k-subsets of {1, …, n}. It follows that the exterior products immediatly extends to k-forms :

Definition 4.3 Let α be a k-form, and β be a ℓ-form, then α ∧β is a (k +ℓ)- form such that

α ∧ β(p^{1}, …, p^{k+ℓ}) =
= ∑_{σ} \frac{1}{k!ℓ!} (−1)^{sign(σ)}α (p^{σ(1)}, …, p^{σ(k)}) β (p^{σ(k+1)}, …, p^{σ(k+ℓ)})

where the sum ranges over all permutations σ of {1, …, k + ℓ}

Finally, a few properties of the exterior product must be kept in mind :

• The exterior product is associative

• Whenever ω is linear (or of odd order), ω ∧ ω = 0.

• More generally, if ω_{1}, …, ω_{s} are linearly dependent 1-forms, then

ω_{1} ∧ … ∧ ω_{s} = 0

• But : if ω is a 2-form (or a form of even order), ω ∧ ω ≠ 0 in general.

• Finally, for any k-form, (ω)^{s} = ω∧ω∧…∧ω is a (k.s)-form. In particular, (ω)^{s} = 0 as soon as k.s > n.

4.3 Exterior derivative

We now define the exterior derivative of a differential form. It is first defined for 0–forms (that is, smooth functions) and the definition extends easily to k–forms.

Definition 4.4 Let U : ℝ^{n} → ℝ a smooth function. Then dU is the 1–form defined by:

dU = ∑_{i} \frac{∂U}{∂x_{i}} dx^{i}.

More generally if

α = ∑ α_{i_{1},...,i_{k}}dx^{i_{1}} ∧ … ∧ dx^{i_{k}},

is a k-form, then dα is the k + 1–form, defined by:

dα = ∑ dα_{i_{1},...,i_{k}} ∧ dx^{i_{1}} ∧ … ∧ dx^{i_{k}}.

Note that the coefficients α_{i_{1},...,i_{k}} of the k–form α are just smooth functions, so that dα_{i_{1}...i_{k}} is a 1–form given by the first rule, and the definition is complete.

The exterior derivative satisifes the following properties:

(i) d is a linear operator

(ii) d^{2} = 0, that is d(dα) = 0 for every α.

(iii) d(α ∧ β) = dα ∧ β + (−1)^{p}α ∧ dβ, for any p–form α.

Property (ii) is easily checked for α = dU (use formula 4.4), where it amounts to the equality of cross-derivatives : \frac{∂^{2}U}{∂x^{i}∂x^{j}} = \frac{∂^{2}U}{∂x^{j}∂x^{i}}. It is then extended to all k-forms by formula 4.4.

An important feature of property (iii) is that it has a converse, known as Poincaré’s lemma: if a k-form α satisifies dα = 0 is some neighbourhood of some point x, then there is a (possibly smaller) neighbourhood U of x and a (k − 1)-form ω such that α = dω on U.

An important feature of the exterior product and the exterior derivative is that they are natural operations. This means in particular that they do not depend on the particular coordinate system where the computation are made.

As an example of properties that depend on the coordinate system, let us think of the Hessian of a smooth function U at a point x: it is the symmetric matrix U′′(x), with coefficients

\frac{∂^{2}U}{∂x_{i}∂x_{j}}

. If we perform a change of variables x = ϕ(y), the Hessian becomes

∑_{i,j} \frac{∂ϕ_{i}}{∂y_{k}} \frac{∂^{2}U}{∂x_{i}∂x_{j}} \frac{∂ϕ_{j}}{∂y_{l}} + ∑_{i} \frac{∂U}{∂x_{i}} \frac{∂^{2}ϕ_{i}}{∂y_{k}∂y_{l}}

If x is a critical point, that is U′(x) = 0, then the second term vanishes and the first term tells us that U′′(x) behaves like a quadratic form; certain things, like the number of positive can be attached to the Hessian independently of the ccordiante system. In the general case, however, the second term does not vanish, and the only invariant thing that can be said about the Hessian is that it is symmetric. For instance, it may be the case that the Hessian is positive definite in some coordinate system and negative definite in another.

Note that the symmetry of the Hessian is equivalent to the relation d^{2}U = 0, which states that the antisymmetric part is zero. One can think of exterior differential calculus as an antisymmetric calculus, where symmetric parts are systematically discarded and we are left with the antisymmetric ones, which are the only ones to be invariant under changes of coordinates.

To define naturality of operations in a precise way, we need the concept of pullback. It is defined as follows. Let ϕ : ℝ^{n} → ℝ^{m} a differentiable mapping. The mapping ϕ induces a mapping from differential forms on ℝ^{m} to differential forms on ℝ^{n} : if α is a k–form on ℝ^{m} then ϕ^{∗}(α) is a k–form on ℝ^{n}, the action of which is defined by transposition as follows.

If v_{1}, …,v_{k} are vectors of the tangent space T_{x}ℝ^{n}, then we can construct paths starting at the point x with velocity vectors given just by v_{1}, …,v_{k}. Then we transfer the paths on ℝ^{m} by ϕ and we compute their velocities: let us call w_{1}, …,w_{k} these velocities. We set:

w_{i} = T_{x}ϕv_{i}
ϕ^{∗}(α)(v_{1}, …, v_{k}) := α(T_{x}ϕv_{1}, …, T_{x}ϕv_{k}).

To say that the operations ∧ and d are natural means that the following properties hold:

(I) ϕ^{∗}(α ∧ β) = ϕ^{∗}(α) ∧ ϕ^{∗}(β);

(II) d[ϕ^{∗}(α)] = ϕ^{∗}(dα).

4.4 Differential ideals

Having defined a product between differential forms we can introduce the algebraic concept of ideal.

Definition 4.5 A set I of differential forms is called an algebraic ideal if the following conditions holds:

(i) I is a linear subspace;

(ii) if α ∈ I and β are differential forms, then α ∧ β ∈ I.

Moreover we say that the ideal I is generated by α_{1}, …, α_{p} (and we will write I = [α_{1}, …, α_{p}]) if every α ∈ I is of the type

α = ∑_{i} α_{i} ∧ β_{i},

for some differential forms β_{i}.

The following proposition gives conditions for 1–forms to be generators of an ideal.

Proposition 4.1 Let α_{1}, …, α_{p} be linearly independent 1–forms, and I the ideal they generate. Finally let ω be a k–form. Then the following conditions are equivalent:

(i) ω ∈ [α_{1}, …, α_{p}];

(ii) ω ∧ α_{1} ∧ … ∧ α_{p} = 0;

(iii) if ξ_{1}, …, ξ_{k} are vector fields on ℝ^{n} such that

α_{i}(ξ_{j}) = 0, ∀i = 1, …, p, j = 1, …, k,

then

ω(ξ_{1}, …, ξ_{k}) = 0.

Proof — (i)⇒(ii) If ω ∈ I then ω = ∑_{k} α_{i} ∧ β_{i}, hence, by (??) and the definition 4.2,

ω ∧ α_{1} ∧ … ∧ α_{p} = ∑_{i} β_{i} ∧ α_{i} ∧ α_{1} ∧ … ∧ α_{p} = 0.

(ii)⇒(i) Let us complete {α_{1}, …, α_{p}} to a base {α_{1}, …, α_{n}} of T^{∗}ℝ^{n}. So we can rewrite ω as a linear combination ∑_{i_{1}≤...≤i_{k}} a_{i_{1},...,i_{k}}α_{i_{1}} ∧ … ∧ α_{i_{k}}. If (ii) holds then in each term of the linear combination of ω there must be at least one form α_{i} with i ∈ {1, …, p}. This show that ω ∈ [α_{1}, …, α_{p}].

(i)⇒(iii) is obvious.

(iii)⇒(i) is easily proved by choosing the local coordinate system in such a way that ξ_{i} = \frac{∂}{∂x^{i}}. The conditions α_{i}(ξ_{j}) = 0 or ω(ξ_{j}) = 0 then mean that all terms in α_{i} or ω contain dx^{i}, and the result is obvious.

We finish this section introducing the definition of differential ideal, which is an ideal closed with respect to the differentiation.

Definition 4.6 Let I an ideal. We say that is a differential ideal if the following property holds:

ω ∈ I ⟹ dω ∈ I.

It is easy to see that if I = [α_{1}, …, α_{p}] is a differential ideal if and only if there exists forms β_{ij} such that

dα_{i} = ∑_{j} β_{ij} ∧ α_{j}.

An important class of differential ideals is given as follows. Let M be a submanifold of ℝ^{n}, let i_{M} : M → ℝ^{n} be the canonical embedding and i_{M}^{∗} its pullback. Define:

I_{M} := {α : i_{M} ∗ α = 0} .

It turns out by naturality that I_{M} is a differential ideal. Indeed if α ∈ I_{M} and β is any differential form, we have

i_{M}^{∗} (β ∧ α) = (i_{M}^{∗}β) ∧ (i_{M}^{∗}α) = 0.

5 The Frobenius theorem

5.1 Statement and proof

Let ω be a differential form on ℝ^{n} of class C^{∞}. We can now try to solve our first problem, which is to find functions λ and V on ℝ^{n} such that ω = λdV. Suppose for the moment that the problem is solved, and let us investigate the consequences. The first one is that the manifolds {V = constant} are interesting. Indeed if we pose

M(h) := {x : V (x) = h} ,

then we have i_{M(h)}^{∗}dV = 0, hence i_{M(h)}^{∗}ω = 0 so i_{M(h)}^{∗}dω = 0. In other words if E := T_{p}M(h) then i_{E}ω = 0 and i_{E}dω = 0. M(h) is n − 1 dimensional so that by (iii) of proposition 4.1 there exists some 1–form α such that dω = α ∧ ω.

Another way to see this fact is to take the differential of ω = λdV. In this case we obtain

dω = \frac{dλ}{λ} ∧ ω,

so that α = \frac{dλ}{λ}. The condition

dω = α ∧ ω

is called the Frobenius condition. It is satisfied if and only if ω generates a differential ideal. It can also be written:

ω ∧ dω = 0

The relevance of this condition is explained by the following

Theorem 5.1 (Frobenius) Let ω be a 1–form which satisfies the Frobenius condition. Then there exists two functions λ and V such that

ω = λdV.

We will deduce the proof of the Frobenius theorem to a more general fact, which we now state as a

Lemma 5.1 Let ω_{r+1}, …, ω_{n} be n − r linearly independent 1–forms. Assume that [ω_{r+1}, …, ω_{n}] is a differential ideal or, in other words, that ω_{r+1}, …, ω_{n} satisfies the generalized Frobenius condition:

∀i > r ∃α_{ij}, s.t. dω_{i} = ∑_{j} α_{ij} ∧ ω_{j}.  (3)

Then it is possible to find a system of local coordinates (y_{1}, …, y_{n}) in such a way that the dy_{i} and the ω_{i}, for i > r, generate the same differential ideal:

[ω_{r+1}, …, ω_{n}] = [dy_{r+1}, …, dy_{n}].

Proof — Let us consider first the case r = 1. This means that we have n − 1 forms ω_{2}, …, ω_{n} on ℝ^{n}. Incidentally we note that in this case the generalized Frobenius condition (3) is always satisfied. Indeed, completing ω_{2}, …, ω_{n} to a base by adding an ω_{1} we can decompose dω_{i} according to that base:

dω_{i} = ∑_{i,j=1}^{n} a_{ij}ω_{i} ∧ ω_{j},

for some a_{ij} with a_{ii} = 0. Clearly, for i > 1, every term on the right contains some ω_{j} with j > 1.

To construct the local coordinate system we ask for, we fix a point P_{0} and we associate with every P in a neighbourhood of P_{0} the kernel of the ω_{i}:

E(P) := {ξ ∈ T_{P} ℝ^{n} : ω_{i}(ξ) = 0, i = 2, …, n} ,

It follows frome the assumptions that E(P) is a one dimensional subspace, spanned by a tangent vector ξ(P). So we can define a C^{∞} vector field ξ(P) in a neighbourhood of P_{0}, to which is associated a C^{∞} flow. It is well known that there is a local coordinate system (y_{1}, …, y_{n}) such that ξ = \frac{∂}{∂y_{1}} . This is done by solving the the Cauchy problem

\{ \begin{matrix}\frac{d}{dt}γ(t) = ξ(γ(t)), \\ γ(0) = P_{0}. \end{matrix}

and by posing

y_{1}(γ(t)) = t, y_{i}(γ(t)) = 0, i = 2, …, n.

By construction, the families ω_{i}(P) and dy_{i}(P), for i > 1, have the same kernel ξ(P). This means that they generate the same (n − 1)-dimensional subspace, and hence the same differential ideal (remember that the generalized Frobenius condition is automatically satisfied). Hence the result for r = 1.

Now we give the argument for r = 2. In this case I = [ω_{3}, …, ω_{n}]. Let us fix a local coordinate system x_{1}, …, x_{n} in such a way that dx_{1} and dx_{2} are linearly independent of ω_{3}, …, ω_{n} and let J be the subspace spanned by (dx_{2}, ω_{3}, …, ω_{n}). Again, since J is (n − 1)-dimensional, it is also a differential ideal, and we write J = [dx_{2}, ω_{3}, …, ω_{n}].

By the first part (induction hypothesis) there exists a coordinate system y_{1}, …, y_{n} such that J = [dy_{2}, …, dy_{n}]. Since the dy_{k} generate J, there will be some β_{2}, …, β_{n} such that

dx_{2} = ∑_{k=2}^{n} β_{k}dy_{k}.

At least one of the β_{k} is not zero. We can suppose that it is β_{2}. From this it follows that dx_{2}, dy_{3}, …, dy_{n} are linearly independent and that

[dy_{2}, …, dy_{n}] = [dx_{2}, dy_{3}, …, dy_{n}],

Hence we can find, for each k ≥ 3, some p_{k}(y_{1}, …, y_{n}) and some ω′_{k} ∈ I such that

dy_{k} = p_{k}dx_{2} + ω′_{k},  (4)

We claim that I = [ω′_{3}, …, ω′_{n}]. Clearly [ω′_{3}, …, ω′_{n}] ⊂ I. If we show that ω′_{3}, …, ω′_{n} are linearly independent the equality follows. Suppose that there exists q_{3}, …, q_{n} such that

∑_{k=3}^{n} q_{k}ω′_{k} = 0.

Then

∑_{k=3}^{n} q_{k}dy_{k} − (∑_{k=3}^{n} q_{k}p_{k}) dx_{2} = 0.

But the forms dx_{2}, dy_{3}, …, dy_{n} are linearly independent (by the preceding discussion) so that q_{k} = 0 for all k ≥ 3, as desired.

Now differentiating (4) and recalling that I is a differential ideal we find dp_{k} ∧ dx_{2} = −dω′_{k} ∈ I, or explicitly

∑_{i} \frac{∂p_{k}}{∂y_{i}} dy_{i} ∧ dx_{2} ∈ I.

Hence \frac{∂p_{k}}{∂y_{1}} dy_{1} ∧dx_{2} ∈ I and, by our choice of dx_{2} and dy_{1}, this is possible only if

\frac{∂p_{k}}{∂y_{1}} = 0, k ≥ 3.

Going back to (4) we can say that we have found a local coordinate system such that the forms {ω′_{k}}_{3≤k≤n} depend only on the coordinates y_{2}, …, y_{n}. So we are reduced to the first case, and this finishes the proof.

5.2 Application to the one consumer model

As an application of the Frobenius theorem, we shall prove the so-called Slutsky conditions, i.e. theorem 3.1. Recall from the one-consumer model that we want to determine when a given function p ↦ x(p) from ℝ^{n} into itself, satisfying x(p) · p = 1 ∀p is a demand function. This can formally written as follows. Introduce the 1–form ω given by:

ω(p) := ∑_{i} x_{i}(p)dp_{i}

and find functions λ(p) and V (p) satisfying

ω = λdV

In addition, we should require that λ is positive and V quasi-convex, but we rae not gioing into that problem for the moment. The Frobenius condition is then necessary and sufficient; let us find out what it looks like.

To this purpose we introduce the vector field

P := ∑_{i} p_{i} \frac{∂}{∂p_{i}} .

Then the first condition in (??) becomes ω(P) = 1, and the Frobenius condition gives

dω(P, ξ) = α(P)ω(ξ) − α(ξ)ω(P) = α(P)ω(ξ) − α(ξ),

for aby vector field ξ . From this we deduce a formula for α:

α(ξ) = −dω(P, ξ) + α(P)ω(ξ).

In the product α ∧ ω the last term will disappear. For this reason we can choose

α_{0}(ξ) = −dω(P, ξ).

Now we have to check that

dω(ξ, η) = α_{0} ∧ ω(ξ, η),  (5)

for any vector fields ξ, η. It is already the case when ξ = P and η is arbitrary, by our definition of α_{0}. For arbitrary ξ we have:

α_{0} ∧ ω(ξ, η) = −dω(P, ξ)ω(η) + dω(P, η)ω(ξ).

Note that

dω = ∑_{i,j} \frac{∂x_{i}}{∂p_{j}} dp_{j} ∧ dp_{i}.

Then

dω(P, ξ) = ∑_{i,j} \frac{∂x_{i}}{∂p_{j}} dp_{j} ∧ dp_{i}(P, ξ) = ∑_{i,j} \frac{∂x_{i}}{∂p_{j}} (p_{j}ξ^{i} − ξ^{j}p_{i}).

Rearranging indexes we can write:

α_{0} ∧ ω(ξ, η) = ∑ (\frac{∂x_{i}}{∂p_{k}} p_{k}x_{j} + \frac{∂x_{k}}{∂p_{j}} p_{k}x_{i} − \frac{∂x_{j}}{∂p_{k}} p_{k}x_{i} − \frac{∂x_{k}}{∂p_{i}} p_{k}x_{j}) ξ^{j}η^{i}.

By the first condition in (??) we have that

\frac{∂x_{k}}{∂p_{j}} p_{k} = \frac{∂}{∂p_{j}} (x_{k}p_{k}) − x_{k} \frac{∂p_{k}}{∂p_{j}} = −x_{k}δ_{j}^{k} = −x_{j},

so that finally we arrive to

α_{0} ∧ ω(ξ, η) = (\frac{∂x_{i}}{∂p_{k}} p_{k}x_{j} − \frac{∂x_{j}}{∂p_{k}} p_{k}x_{i}) ξ^{j}η^{i}.

On the other hand

dω(ξ, η) = \frac{∂x_{i}}{∂p_{j}} dp_{j} ∧ dp_{i}(ξ, η) =

= \frac{∂x_{i}}{∂p_{j}} (dp_{j}(ξ)dp_{i}(η) − dp_{j}(η)dp_{i}(ξ)) =

= \frac{∂x_{i}}{∂p_{j}} (ξ^{j}η^{i} − η^{j}ξ^{i}) =

= (\frac{∂x_{i}}{∂p_{j}} − \frac{∂x_{j}}{∂p_{i}} ) ξ^{j}η^{i}.

Equality (5) is true if and only if

\frac{∂x_{i}}{∂p_{j}} − \frac{∂x_{i}}{∂p_{k}} p_{k}x_{j} = \frac{∂x_{j}}{∂p_{i}} − \frac{∂x_{j}}{∂p_{k}} p_{k}x_{i},

or, equivalently, if and only if the matrix (σ_{ij})_{i,j} is symmetric, where

σ_{ij} = \frac{∂x_{i}}{∂p_{j}} − \frac{∂x_{i}}{∂p_{k}} p_{k}x_{j}.

This is just the Slutsky conditions, as stated in section 2.

6 The household model

6.1 Statement of the model and Pareto optimality

Up to now we have considered the simplest model, namely a situation where there is only one consumer. We have deduced the Slutsky conditions for this case which are necessary and sufficient to rebuild the utility functions. In this section we discuss what happens if we try to study more complex systems.

Let us suppose that there are n + N goods and two consumers (one can easily imagine that the situation can be more complicated, nevertheless what we consider here contains all the obstructions which appear in more general situations). Our way to denote the number of goods is due to the fact that we denote by n the number of private goods and by N the number of public goods within the household. Food and clothes are private goods, because they are consumed singly; lodging and TV are public goods, because they can be enjoyed simultaneously by both partners.

Each consumer is characterized by his own individual utility function:

U_{i} : ℝ^{n} × ℝ^{n} × ℝ^{N} → ℝ.

for i = 1, 2. We will denote by x_{1} ∈ ℝ^{n} the private consumption of the first consumer, by x_{2} ∈ ℝ^{n} the private consumption of the second one and by X ∈ ℝ^{n} the common consumption of public goods. Note that the utility function of a consumer depends not only on his own consumption, but also on the private consumption of the other consumer.If one partner smokes, it may inconvenience the other; on the other hand, if one partner plays an instrument, the other one may enjoy the sound. These are called externalities.

Of course various restrictive assumptions can be made: no externalities (in this case U_{i} = U_{i}(x_{i}, X)), no public goods (U_{i} = U_{i}(x_{1}, x_{2})), no private goods (U_{i} = U_{i}(X)). This problem is called the household problem, because it is typical of two people living in the same house, and pooling their resources, which they then have to share between their private and public consumptions.

This problem differs from the preceding one in the fact that there are now two criterions to be optimized simultaneously. Multicriterion optimization has been well developed for the purpose of economics, and the appropriate definition is that of Pareto optimality, to which we now proceed.

Denote by p the price of private goods, and by P the price of public goods. For the sake of simplicity, set the wealth of the household equal to 1.

Definition 6.1 We say that an allocation (x_{1}, x_{2}, X) is feasible iff:

p · (x_{1} + x_{2}) + P · X ≤ 1

We say that it is Pareto optimal if it is feasible and if there do not exist other feasible allocations which can strictly increase at least one individual’s utility U_{i} without decreasing the utilities of the other: p · (y_{1} + y_{2}) + P · Y ≤ 1 implies U_{1}(y_{1}, y_{2}, Y) < U_{1}(x_{1}, x_{2}, X) or U_{2}(y_{1}, y_{2}, Y) < U_{2}(x_{1}, x_{2}, X).

In other words, if (x_{1}, x_{2}, X) is not Pareto optimal, then the total wealth of the household can be spent is such a way that both partners are better off. Take the elementary case where n = 1 and U_{1}(x_{1}) = x_{1}, U_{2}(x_{2}) = x_{2} (no externalities nor public goods). In that case, the problem is simply to split 1 dollar. Then (X_{1}, x_{2}) is Pareto optimal iff x_{1} + x_{2} = 1, that is, there is no money left. This gives us the intuitive meaning of Pareto optimality: no resource is wasted.

It can be proved, using the standard separation theorem for convex sets, that (x_{1}, x_{2}, X) is Pareto optimal iff there exists some constant μ > 0 such that (x_{1}, x_{2}, X) maximizes U_{1} + μU_{2} over all feasible allocations. It follows that there are infinitelu many Pareto optima, one for every choice of μ. In the household problem, the parameter μ ∈ ℝ_{+} indicates the relative power of the two consumers in the allocation process. It will be taken to be a smooth function of the price system (p, P).

The household problem can now be stated as an optimization problem:

(P_{p,P,μ,U_{1},U_{2}}) \{ \begin{matrix}max \{U_{1}(x_{1}, x_{2}, X) + μ(p, P)U_{2}(x_{1}, x_{2}, X)\}), \\ x_{1}, x_{2} ∈ ℝ_{+}^{n}, X ∈ ℝ_{+}^{N}, \\ p · (x_{1} + x_{2}) + P · X ≤ 1. \end{matrix}

As in the case of the single consumer we introduce the first order necessary condition (in the follow (FONC)) for optimality.

Definition 6.2 Let Θ an open subset of ℝ^{n} × ℝ^{N}, (x̂_{1}, x̂_{2}, X̂) : Θ → ℝ_{+}^{n} × ℝ_{+}^{n} × ℝ_{+}^{N} a smooth mapping. We shall say that

(x̂_{1}, x̂_{2}, X̂) ∈ FONC(U_{1}, U_{2}, μ, Θ),

if there exists a smooth function λ : Θ → ℝ such that

(i) p · (x̂_{1}(p, P) + x̂_{2}(p, P)) + P · X̂(p, P) = 1

(ii)

\frac{∂}{∂x_{i}} [U_{1} + μ(p, P)U_{2}]_{|(x̂_{1},x̂_{2},X̂)} = λ(p, P)p_{i}, i = 1, …, n

\frac{∂}{∂X_{j}} [U_{1} + μ(p, P)U_{2}]_{|(x̂_{1},x̂_{2},X̂)} = λ(p, P)P_{j}, j = 1, …, N

(iv) x̂_{k}^{i}(p, P) > 0, X̂^{j}(p, P) > 0, λ(p, P) ≠ 0

It is well known that if (x̄_{1}, x̄_{2}, X̄) solves the problem (P_{p,P,μ,U_{1},U_{2}}) for the particular value (p̄, P̄) ∈ ℝ_{+}^{n} × ℝ_{+}^{N}, then there will usually exists some neighbourhood Θ of (p̄, P̄) and a function (x̂_{1}, x̂_{2}, X̂) ∈ FONC(U_{1}, U_{2}, μ, Θ) with x̂_{i}(p̄, P̄) = x̄_{i} and X̂(p̄, P̄) = X̄. ”Usually” means here that some non– degeneracy conditions, involving first and second derivatives of U_{1} and U_{2} at the point (x̄_{1}, x̄_{2}, X̄) have to hold: we shall not go into the details. It is also the case that FONC are not sufficient, unless U_{1} and U_{2} are concave (and μ positive).

6.2 Browning - Chiappori necessary conditions

As in the single consumer case, we want to characterize demand functions which arise from the problem (P_{p,P,μ,U_{1},U_{2}}). But this time we cannot observe the individual consumptions of private goods x̂_{1} and x̂_{2}. Only the aggregate demand function of the household

x̂(p, P) := x̂_{1}(p, P) + x̂_{2}(p, P),

can be observed together with X̂(p, P), the demand for public goods. Browning and Chiappori [3] have found a necessary condition for (x̂, X̂) to be an household demand function. They have proved the following

Proposition 6.1 If there exists U_{1}, U_{2}, μ, Θ and x̂_{1}, x̂_{2} in such a way that (x̂_{1}, x̂_{2}, X̂) ∈ FONC(U_{1}, U_{2}, μ, Θ) and x̂ = x̂_{1} + x̂_{2}, then the Slutsky matrix associated to (x̂, X̂) is a sum of a symmetric matrix and a matrix with rank one.

To write down explicitly the conclusion of the proposition 6.1 let us denote by π := (p, P) ∈ ℝ^{n+N} and ξ := (x̂, X̂). Recall that the Slutsky matrix is given by

σ^{ij} = \frac{∂ξ^{i}}{∂π_{j}} − ∑_{k} p_{k} \frac{∂ξ^{i}}{∂p_{k}} ξ^{j}.

So the Browning–Chiappori result states that

σ^{ij} = s^{ij} + a^{i}b^{j}, with s^{ij} = s^{ji}.

Browning and Chiappori also raise the question whether this condition is sufficient. We will give a positive answer by using Cartan calculus (and simultaneously give a direct proof that the above condition is necessary).

In order to approach the problem we have presented in the previous section we follow the same procedure as in the single consumer case, namely we translate the problem into the language of differential forms. The first step is given by the following

Lemma 6.1 Let ξ : Θ → ℝ_{+}^{n} × ℝ_{+}^{N} be given. If there are smooth functions (x̂_{1}, x̂_{2}) and (U_{1}, U_{2}, μ) such that

ξ = (x̂_{1} + x̂_{2}, X̂), (x̂_{1}, x̂_{2}, X̂) ∈ FONC(U_{1}, U_{2}, μ, Θ),  (6)

then there exist smooth functions φ, ψ, f and g on Θ such that

∑_{i=1}^{n+N} ξ^{i}(π)dπ_{i} = φdf + ψdg.  (7)

Conversely, if there exists an open subset U ⊂ ℝ_{+}^{l} such that ξ : Θ → U is smoothly invertible, if (7) holds for some smooth functions (φ, ψ, f, g) and if π · ξ(π) = 1 on Θ, then, setting n = 0, N = l and X̂ = ξ (no private goods), smooth functions (U_{1}, U_{2}) can be found in such a way that (6) holds.

Proof — Suppose that (6) holds. Let us introduce the indirect utility functions:

Û_{i}(p, P) := U_{i}(x̂_{1}(p, P), x̂_{2}(p, P), X̂(p, P)), i = 1, 2

and

Û(p, P) = Û_{1}(p, P) + μ(p, P)Û_{2}(p, P) =

= max_{x_{1},x_{2},X} \{ \begin{matrix}U_{1}(x_{1}, x_{2}, X) + μ(p, P)U_{2}(x_{1}, x_{2}, X)+ \\ +λ(p, P) [1 − p · (x_{1} + x_{2}) − P · X] \end{matrix} \} .

By the envelope theorem

\frac{∂Û}{∂π_{i}} = \frac{∂μ}{∂π_{i}} Û_{2} − λξ^{i},

which we rewrite as

ξ^{i} = \frac{Û_{2}}{λ} \frac{∂μ}{∂π_{i}} − \frac{1}{λ} \frac{∂Û}{∂π_{i}} .

Hence we deduce

∑_{i=1}^{n+N} ξ^{i}dπ_{i} = \frac{Û_{2}}{λ} dμ − \frac{1}{λ}dÛ,

as announced. Note that in particular we have shown that ψ = −λ^{−1}, g = Û, −φ = Û_{2}ψ and f = μ as a solution (not unique).

For the converse let us suppose that (7) holds for some functions (φ, ψ, f, g) and π · ξ(π) = 1 on Θ. This last condition implies that ξ cannot vanish, hence we can assume that ψ does not vanish on Θ (eventually restricting Θ). Now we define

λ := −\frac{1}{ψ} , Û := g, Û_{2} := −\frac{φ}{ψ}, μ := f, Û_{1} := Û − μÛ_{2},

and

U_{1}(X) := Û_{1}(ξ^{−1}(X)), U_{2}(X) := Û_{2}(ξ^{−1}(X)).

Finally we can pass to verify conditions (6). The first one, in our setting, is trivial. Let us check only the second one, which means to check conditions (i)–(iv) stated in the definition 6.2. In our setting (i) and (iv) are immediate. It remains only the (ii),(iii). Note that by π · ξ(π) = 1 and our definitions,

\begin{aligned}Û(π) & = U_{1}(ξ(π)) + μ(π)U_{2}(ξ(π)) + λ(π)(1 − π · ξ(π)) = \\ & = Û_{1}(π) + μ(π)Û_{2}(π) + λ(π)(1 − π · ξ(π)). \end{aligned}

Differentiating we get

dÛ = dÛ_{1} + μdÛ_{2} + Û_{2}dμ − λξ^{i}dπ_{i} − λπ_{i} \frac{∂ξ^{i}}{∂π_{j}} dπ_{j} + (1 − π · ξ(π))dλ. (8)

But the last term in the previous equality is equal to zero due to the condition π · ξ(π) = 1. On the other hand writing down (7) in our setting we deduce

dÛ = dg = \frac{1}{ψ} (ξ^{i}dπ_{i} − φdf) = −λξ^{i}dπ_{i} + Û_{2}dμ.  (9)

Comparing (8) with (9) we deduce the following equation

dÛ_{1} + μdÛ_{2} − λ\frac{∂ξ^{i}}{∂π_{j}} π_{i}dπ_{j} = 0.  (10)

Inverting the (10) it is easy to find conditions (ii) and (iii) of the definition 6.2.

Remark 6.1 Lemma 6.1 states that if the Browning–Chiappori condition is satisfied, then the problem has a solution with public goods only. If we know in advance that there are really private goods, then the Browning-Chiappori condition is no longer sufficient, and additional conditions must be satisfied. They will be described in the next section.

We now have to solve the following problem: given a one form ω find functions (φ, ψ, f, g) in such a way that ω = φdf + ψdg.

This is a generalization of the problem that we have posed previously and that we have solved via the Frobenius theorem. Let us first look for necessary condition on the form ω. Differentiating the equation satisfied by ω we obtain dω = dφ ∧ df + dψ ∧ dg. Hence

dω ∧ dω = 2dφ ∧ df ∧ dψ ∧ dg.

Multiplying once again by ω we find the following necessary condition:

ω ∧ dω ∧ dω = 0.  (11)

It is possible to give other conditions, as in the following

Proposition 6.2 Let ω a differential form. Each of the following conditions is equivalent to the (11):

(a) there exists a 3–form α such that dω ∧ dω = α ∧ ω;

(b) there exists a 2–form β and a 1–form γ such that dω = β + γ ∧ ω and β ∧ β = 0.

Proof — Let us construct a base of 1–forms, let say ω_{1}, …, ω_{n} with ω_{1} = ω. Then

dω = ∑_{ij} a_{ij}ω_{i} ∧ ω_{j} = ω_{1} ∧ γ + β.

where β contains no term of the form ω_{i} ∧ ω_{j}. From this the equivalence between (a) and (b) follows immediately. Indeed if (a) holds we have

α ∧ ω = dω ∧ dω = β ∧ β + 2ω_{1} ∧ γ ∧ β,

and by construction β does not contain ω = ω_{1} so β ∧ β has to vanish. On the other hand if (b) holds (a) follows immediately.

It is also clear that (a) implies (11). Conversely,if (11) is satisfied, we may again use the decompostion dω = ω_{1} ∧ γ + β; writing this in (11) with ω_{1} = ω, we get ω ∧ β ∧ β = 0, and since β does not contain ω, this leads to β ∧ β = 0.

7 Pfaff’s theorem

In this section we will finally prove that the Browning - Chiappori condition is necessary and sufficient (in the case of public goods). This is done by applying the Pfaff theorem, which solves (among others) the following problem: given a 1–form ω, when can we find functions (φ, ψ, f, g) satisfying ω = φdf + ψdg.

Theorem 7.1 Pfaff Let ω be a differential form. Suppose that there exists an open set U ⊂ ℝ^{n} such that

ω ∧ dω ≠ 0 in U.  (12)
ω ∧ dω ∧ dω = 0 in U.  (13)

Then there exists an open subset V ⊂ U, functions φ, ψ, f, g ∈ C^{∞}(V; ℝ) such that

ω = φdf + ψdg.

Remark 7.1 Note that the Pfaff and the Frobenius theorems have conditions on the form ω which exclude the case of a form ω that satisfies both of them. The Pfaff theorem tells us when there exists a two-terms decompostion ω = φdf + ψdg, while the Frobenius theorem gives us a single one ω = φdf. This is interesting in applications, because we could have to test a demand function of unknown type: we might not know if we are dealing with a single consumer or a household). If the Frobenius condition does not hold but the Pfaff condition does, then we have in hand the demand function of a.

Proof — First step: Suppose for a moment that ω = φdf + ψdg. Then dω = dφ ∧ df + dψ ∧ dg. So, the first step will be to search a decomposition for dω of the type

dω = ω ∧ ω′ + γ ∧ γ′.

By proposition 6.2 we know that dω has the form

dω = ω ∧ ω′ + σ,

and we have the information that σ ∧σ = 0. We need from this a factorization for σ. This purpose comes from the following

Lemma 7.1 Let σ a 2–form, σ ≠ 0 such that

σ ∧ σ = 0.

Then there exist γ, γ′ 1–forms such that σ = γ ∧ γ′.

Proof of Lemma 7.1 — Let us introduce the one–form

u(ξ)(η) = σ(ξ, η).

Because σ ≠ 0 there exists a ξ such that u(ξ) ≠ 0. Now it turns out that

u(ξ) ∧ σ = 0.

Indeed, we have:

(u(ξ) ∧ σ)(η_{1}, η_{2}, η_{3}) =
= 2[σ(ξ, η_{1})σ(η_{2}, η_{3}) + σ(ξ, η_{2})σ(η_{3}, η_{1}) + σ(ξ, η_{3})σ(η_{1}, η_{2})] =
= (σ ∧ σ)(ξ, η_{1}, η_{2}, η_{3}) = 0

Our claim now is that σ = u(ξ) ∧ γ′. Let α_{1} := u(ξ) and complete this form to a base α_{1}, α_{2}, …, α_{n}. Then

σ = σ^{ij}α_{i} ∧ α_{j}.

From α_{1} ∧ σ = 0, it follows that

∑_{i,j≠1} σ^{ij}α_{i} ∧ α_{j} ∧ α_{1} = 0,

from which it follows that σ^{ij} = 0 for all i, j > 1. Hence

σ = ∑_{i=1∨j=1} σ^{ij}α_{i} ∧ α_{j} = α_{1} ∧ γ′,

for some appropriate γ′.

Second step : By the first step dω = ω ∧ ω′ + γ ∧ γ′. Note that if γ or γ′ are null form, we have just the Frobenius condition. Nevertheless, as we have seen in the lemma 7.1, γ is not 0 if σ is not, and in our setting this is implied by (12). The purpose of this second step is to show that [ω, γ, γ′] = [dy_{1}, dy_{2}, dy_{3}], and to apply the Frobenius theorem. We first have to show that the linear span of (ω, γ, γ′) is a differential ideal: We claim that:

Span(ω, γ, γ′) = {α : α ∧ ω ∧ dω = 0} =: J.  (14)

Let us prove that the right hand side is a differential ideal. Let α ∈ J, then

0 = d [α ∧ ω ∧ dω] =
= dα ∧ ω ∧ dω + α ∧ d [ω ∧ dω] =
= dα ∧ ω ∧ dω + α ∧ dω ∧ dω.

On the other hand, we have:

dω ∧ dω = 2ω ∧ ω′ ∧ γ ∧ γ′ = −ω′ ∧ ω ∧ dω.

Hence

α ∧ dω ∧ dω = ω′ ∧ α ∧ ω ∧ dω = 0,

from which follows that dα ∧ ω ∧ dω = 0, namely J is a differential ideal.

Now we prove (14). It is easy to verify that Span(ω, γ, γ′) ⊂ J. For the converse let α ∈ J be a 1–form (the general case follows in the same way). Because ω, γ, γ′ are linearly independent (by construction) we can complete them into a base α_{1}, …, α_{n} in such a way that α_{1} = ω, α_{2} = γ and α_{3} = γ′. Let us write α as linear combination of these forms, then take the external product with ω ∧ dω. We obtain

\begin{aligned}0 & = α ∧ ω ∧ dω = (σ^{i}α_{i}) ∧ ω ∧ γ ∧ γ′ = \\ & = ∑_{i=4}^{n} σ^{i}α_{i} ∧ α_{1} ∧ α_{2} ∧ α_{3}. \end{aligned}

By this follows that σ^{i} = 0 for i = 4, …, n, so that

α = σ^{1}ω + σ^{2}γ + σ^{3}γ′.

We have proved that Span(ω, γ, γ′) is a differential ideal. By the Frobenius theorem it follows that, for an appropriate coordinate system (y_{1}, …, y_{n}), we have:

[ω, γ, γ′] = [dy_{1}, dy_{2}, dy_{3}].

Third step : By the preceding step, there is a decomposition of ω into a sum of three terms ω = λ_{1}dy_{1} + λ_{2}dy_{2} + λ_{3}dy_{3}. We want to reduce this decomposition to two terms. Let us suppose that λ_{1} ≠ 0, and consider the ideal generated by ω and dy_{1}. As before, we show that:

[ω, dy_{1}] = {α : α ∧ ω ∧ dy_{1} = 0} =: J′,

and J′ is a differential ideal. Using the Frobenius theorem again, we have J = [dz_{1}, dz_{2}], hence the conclusion.

Remark 7.2 Note that if n ≤ 4, ω ∧ dω ∧ dω as a 5–form is identically zero. So in this case the only condition on ω is (12).

More generally the following theorem holds

Theorem 7.2 Let ω a differential form such that there exists an open set U ⊂ ℝ^{n} such that

ω ∧ ∧_{i=1}^{k−1}dω ≠ 0 on U,

and

ω ∧ ∧_{i=1}^{k}dω = 0 on U.

Then there exists functions λ_{i}, V_{i}, i = 1, …, k such that

ω = ∑_{i=1}^{k} λ_{i}dV_{i}.

The proof of the theorem (7.2) is very similar to the one of the theorem 7.1. We have immediately the

Corollary 7.1 Every regular 1–form on ℝ^{n} is a linear combination of at most \frac{n}{2} gradients if n is even, \frac{n+1}{2} if n is odd.

It is now immediate to see that the Browning–Chiappori conditions are sufficient to reconstruct the utility functions of a given household policy ξ. Indeed, define the linear form ω by ω(π) = ∑ ξ^{i}(π)dπ_{i}. Write down the Browning - Chiappori condition:

∑ σ^{ij}dπ_{i} ∧ dπ_{j} = ∑ (\frac{∂ξ^{i}}{∂π_{j}} − ∑ π_{k} \frac{∂ξ^{i}}{∂π_{k}} ξ^{j}) dπ_{i} ∧ dπ_{j} = ∑ a^{i}b^{j}dπ_{i} ∧ dπ_{j}

Rewrite it as follows:

dω = a ∧ b + γ ∧ ω  (15)

with a = ∑ a^{i}dπ_{i}, b = ∑ b^{j}dπ_{j}, γ = ∑ π_{k} \frac{∂ξ^{i}}{∂π_{k}} dπ_{i}. Clearly, relation (15) implies that ω ∧ dω ∧ dω = 0, so the Pfaff condition holds, and lemma 6.1 applies.

Remark 7.3 There is an additional question which we have not touched, and it is the question of concavity. In microeconomic models, like the Browing-Chiappori one, it is not enough to find fuctions U_{1} and U_{2} satisfying the first-order necessary conditions: one would also like these functions to be concave, so that the necessary conditions are also sufficient, and the problem is completely solved. Mathematically, this translates into the following question: what additional assumptions are necessary in Pfaff’s theorem so that the functions f and g are concave, and the functions ϕ and ψ positive? After these lectures notes were complete, this question was answered in [10].

Up to now, we assumed that there are only public goods in the household. In other words, the utility functions are U_{1}(X) and U_{2}(X). In this section we deal with the opposite situation, when there are only private goods. In other words, the utility functions are U_{1}(x_{1}) and U_{2}(x_{2}). The household problem then is as follows :

(P_{p,μ,U_{1},U_{2}}) \{ \begin{matrix}max \{U_{1}(x_{1}) + μ(p)U_{2}(x_{2})\} , \\ p · (x_{1} + x_{2}) ≤ 1. \end{matrix}

Given a function p ↦ x(p), we want to find functions x_{1}, x_{2}, μ, U_{1}, U_{2} in such a way that x(p) = x_{1}(p)+x_{2}(p) and the pair (x_{1}(p), x_{2}(p)) is the solution of (P_{p,μ,U_{1},U_{2}}).

Our approach now will be different from the preceding one: we will seek directly U_{1}(x) and U_{2}(x), instead of V_{1}(p) and V_{2}(p). For this reason, we start from the inverse demand function x ↦ p(x), which is considered given.

Note that it can be obtained by simply inverting the usual demand function p ↦ x(p)

Introducing Lagrange multipliers we get the following optimality conditions

\{ \begin{matrix}\frac{∂U_{1}}{∂x_{1}^{i}} (x_{1}) = λ(x_{1} + x_{2})p_{i}(x_{1} + x_{2}), \\ \frac{∂U_{2}}{∂x_{2}^{i}} (x_{2}) = \frac{λ(x_{1} + x_{2})}{μ(x_{1} + x_{2})}p_{i}(x_{1} + x_{2}). \end{matrix}  (16)

Note that all the functions here are unknown, with the exception of p(x). Introducing appropriate functions φ_{1}, φ_{2}, we can rewrite (16) as follows:

\{ \begin{matrix}φ_{1}(x) + φ_{2}(x) = x, \\ \frac{∂U_{1}}{∂x_{1}^{i}} (φ_{1}(x)) = λ(x)p_{i}(x), \\ \frac{∂U_{2}}{∂x_{2}^{i}} (φ_{2}(x)) = \frac{λ(x)}{μ(x)}p_{i}(x). \end{matrix}  (17)

Let us rewrite (17) in the language of differential form. Setting

α_{1} := ∑ p_{i}(x)dx_{1}^{i},

the first equation in (16) becomes

dU_{1} = \frac{∂U_{1}}{∂x_{1}^{i}} (x_{1})dx_{1}^{i} = λ(x)α_{1},

while the second one becomes

dU_{2} = \frac{∂U_{2}}{∂x_{2}^{i}} (x_{2})dx_{2}^{i} = \frac{λ(x)}{μ(x)}α_{2},

where

α_{2} := ∑ p_{i}(x)dx_{2}^{i}.

We now rewrite the full system (17) in a geometrical way. Introduce first the space ℝ^{2l+2} with coordinates (x_{1}^{i}, x_{2}^{i}, λ, μ), and in this space introduce the manifold M defined by:

M := {(x_{1}^{i}, x_{2}^{i}, λ, μ) : x_{1} = φ_{1}(x), x_{2} = φ_{2}(x)} ,  (18)

Denote by i_{M}^{∗} the pullback operator from ℝ^{2l+2} to M. System (17) can now be written:

\{ \begin{matrix}i_{M}^{∗}dU_{1} = λ(x)i_{M}^{∗}α_{1}, \\ i_{M}^{∗}dU_{2} = \frac{λ(x)}{μ(x)}i_{M}^{∗}α_{2}. \end{matrix}  (19)

So the new problem is to find a submanifold M of ℝ^{2l+2}, and functions λ, μ, U_{1}, U_{2} satisfying the following conditions:

(a) M has dimension l and is the graph of a function of x_{1} + x_{2}.

(b) i_{M}^{∗}α_{1} = λ(x)^{−1}i_{M}^{∗}dU_{1}.

(c) i_{M}^{∗}α_{2} = μ(x)λ(x)^{−1}i_{M}^{∗}dU_{2}.

By the Frobenius theorem conditions (b) and (c) are equivalent to the following ones

(b’) i_{M}^{∗}(α_{1} ∧ dα_{1}) = 0;

(c’) i_{M}^{∗}(α_{2} ∧ dα_{2}) = 0.

The Cartan - Kähler theorem, which we shall describe in the next section, enables us to prove the following result (see [6]):

Proposition 7.1 Assume that there are 1–forms β, β′, γ, γ′ such that:

\{ \begin{matrix}dα_{1} ∧ dα_{1} = β ∧ α_{1} ∧ dα_{1} + γ ∧ α_{2} ∧ dα_{2}, \\ dα_{2} ∧ dα_{2} = β′ ∧ α_{1} ∧ dα_{1} + γ′ ∧ α_{2} ∧ dα_{2}. \end{matrix}  (20)

Then we can find U_{1}, U_{2}, μ such that p(x) is the inverse demand function corresponding to P_{p,μ,U_{1},U_{2}}.

We have now fully solved the household problem (up to concavity). In the case of public goods, the Browning - Chiappori conditions are necessary and sufficient. In the case of private goods, conditions (20) are necessary and sufficient. They imply the Browning Chiappori ones, but they are stronger. Indeed, take l = 3 and set:

p_{1}(x) = \frac{γ_{1}}{x_{1}} , p_{2}(x) = \frac{γ_{2}}{x_{2}} , p_{3}(x) = \frac{γ_{3}}{x_{3}} , with γ_{1} + γ_{2} + γ_{3} = 1,

then conditions (20) are not satisfied, but the Browning - Chiappori ones are, as the reader will check.

Note, however, two things. The first one is that the problem is solved only locally, that is, the functions are defined in some unknown neighbourhhod of a prescribed point. The second one is that we have said nothing about the concavity of the functions U_{1} and U_{2}. For the solution to make economic sense, they have to be concave; the result in [10] then gives us additional conditions which have to be met.

We now turn to the general statement of the Cartan-Kähler theorem.

8 The Cartan–Kähler theorem

We now present the key result upon which our appraoch relies.This theorem, due to Cartan and Kähler, solves the following, general problem. Given a certain family of differential forms (not necessarily 1-forms, nor even of the same degree), a point x̄ and an integer m ≥ 1, can one find some m-dimensional submanifold M containing x̄ and on which all the given forms vanish ?

8.1 An introductory example

As an introduction, let us start from a simple version of our problem, namely the Cauchy-Lipschitz theorem for ordinary differential equations. It states that, given a point x̄ ∈ ℝ^{n} and a C^{1} function f, defined from some neighborhood U of x̄ into ℝ^{n}, there exists some ϵ > 0 and a C^{1} function φ: ]−ϵ, ϵ [⟶ U such that

\frac{dφ}{dt} = f (φ(t)) ∀t ∈ ] − ϵ, ϵ [  (21)

φ(0) = x̄

It follows that \frac{dφ}{dt}(0) = f(x̄). If f(x̄) = 0, the solution is trivial, φ(t) = x̄ for all t so we assume that f(x̄) does not vanish.

This theorem can be rephrased in a geometric way. Consider the graph M of φ :

M = { (t, φ(t) ) | − ϵ < t < ϵ }

This is a 1-dimensional submanifold of ] − ϵ, ϵ [×U . Now, let us introduce the 1-forms ω^{i} defined by :

ω^{i} = f^{i}(x)dt − dx^{i} , 1 ≤ i ≤ n − 1

Clearly φ solves the differential equation (21) if and only if, for any x = φ(t) on M, the ω^{i} all vanish on T_{x}M, the tangent space to M at x (remember that T_{x}M is spanned by the vector v(t) = (1, \frac{dφ_{1}}{dt} , …, \frac{dφ_{n}}{dt} ), and note that ω^{i} [v(t)] = f ^{i} (φ(t)) − \frac{dφ_{i}}{dt}). Equivalently, substituting x^{i} = φ^{i}(t) into formula (4) yields the pullbacks :

φ^{∗}ω^{i} = [f^{i} (φ(t)) − \frac{dφ^{i}}{dt} (t)] dt

which vanish if and only if φ solves equation (21).

So the Cauchy-Lipschitz theorem tells us how to find a 1-dimensional submanifold of ℝ × ℝ^{n} on which certain 1-forms vanish.

8.2 The general problem

The Cauchy-Lipschitz theorem deals with 1-forms of a specific nature. By extension, the fully general problem can formally be stated as follows.

Definition 8.1 Let ω^{k} , 1 ≤ k ≤ K, be differential forms of degree d_{k} on an open subset of ℝ^{n}, and M ⊂ ℝ^{n} a submanifold. We call M an integral manifold of the system :

ω^{1} = 0 , …, ω^{K} = 0  (22)

if, for any x on M, the ω^{i} all vanish on T_{x}M, the tangent space to M at x :

ω^{k} (ξ^{1}, …, ξ^{d_{k}}) = 0 1 ≤ k ≤ K  (23)

whenever x ∈ M and ξ^{i} ∈ T_{x}M for 1 ≤ i ≤ d_{k}.

An equivalent statement is that the pullbacks of the ω^{i} to M must all vanish.

Given a point x̄ ∈ ℝ^{n}, the Cartan-Kähler theorem will give necessary and sufficient conditions for the existence of an integral manifold containing x̄.

Necessary conditions are easy to find. Assume an integral manifold M ∋ x̄ exists, and let m be its dimension. Then its tangent space at x̄ , Tx̄M , is m-dimensional, and all the ωj ; x̄ must vanish on Tx̄M , because of formula (23). Any subspace E ⊂ Tx̄M with this property will be called an integral element of system (22) at x̄. The set of all m-dimensional integral elements at x̄ will be

Gm ; x̄ = \{E \begin{matrix}E ⊂ Tx̄M and dim E = m \\ ω1 ; x̄, …, ωK ; x̄ vanish on E \end{matrix} \}

Our first necessary condition is clear :

Gm ; x̄ ≠ ⊘  (24)

8.3 Differential ideals

To get the second one, let us ask a strange question : have we written all the equations ? In other words, does the system :

ω^{1} = 0, …, ω^{K} = 0  (25)

exhibit all the relevant information ?

The answer is negative. To see why, recall that M is a submanifold of ℝ^{n}, and denote by φ_{M} : M → ℝ^{n} the standard embedding : φ_{M}(x) = x for all x ∈ M. M is an integral manifold of system (25) if :

φ_{M}^{∗} ω^{1} = 0, …, φ_{M}^{∗} ω^{K} = 0  (26)

But we know that exterior differentiation is natural with respect to pullbacks, that is, d commutes with φ_{M}^{∗}. So (26) implies :

φ_{M}^{∗}(dω^{1}) = 0, …, φ_{M}^{∗}(dω^{K}) = 0

In other words, M is also an integral manifold of the larger system :

\{ \begin{matrix}ω^{1} = 0, …, ω^{K} = 0 \\ dω^{1} = 0, …, dω^{K} = 0 \end{matrix}  (27)

which is different from (25). So integral elements of (27) will be different from integral elements of (25), and it is not clear which ones we should be working with.

To resolve this quandary, we shall assume that systems (25) and (27) have the same integral elements. In other words, the second equations in (27) must be algebraic consequences of the first ones. The precise statement for this is that the ω^{k}, 1 ≤ k ≤ K, must generate a differential ideal:

∀k, dω^{k} = Σ α_{j}^{k} ∧ ω^{j}  (28)

Our second necessary condition is that the ω^{k}, 1 ≤ k ≤ K, must generate a differential ideal.

Note that if the given family {ω^{k} | 1 ≤ k ≤ K} does not satisfy this condition, the enlarged family {ω^{k} , dω^{k} | 1 ≤ k ≤ K} certainly will (because ddω^{k} = 0). So the condition that the ω^{k} must generate a differential ideal can be understood as saying that the enlargement procedure has already taken place.

8.4 A counter example

It turns out that conditions (24) and (28) are almost sufficient. All we have to do is to replace (24) by a slightly stronger condition.

To see that (24) is not sufficient, let us give a simple example. Consider two functions f and g from ℝ^{n−1} into itself, with f(0) = g(0) = v ≠ 0. Define α^{i}(x, t) and β^{i}(x, t) , 1 ≤ i ≤ n − 1 , by

α^{i}(x, t) = f^{i}(x)dt − dx^{i}
β^{i}(x, t) = g^{i}(x)dt − dx^{i}

and consider the exterior differential system in ℝ^{n} :

α^{i}(x, t) = 0 , β^{i}(x, t) = 0 , 1 ≤ i ≤ n − 1

The α^{i} and the β^{i} generate a differential ideal, and there is an integral element at 0, namely the line carried by (1, N), so G_{0}^{1} ≠ ⊘. However, finding an integral manifold of the initial system containing 0 amounts to finding a common solution of the two Cauchy problems :

\frac{dx}{dt} = f(x) , x(0) = 0  (29)
\frac{dx}{dt} = g(x) , x(0) = 0  (30)

which does not exist in general. So, though (24) is fulfilled, it is still impossible to integrate.

The counter example can also be restated in a more traditional way. Consider (29) and (30) as standard PDE, that have to be solved locally around 0. The natural first step is to linearize this system at 0 - which, incidentally, is exactly what the α^{i} and β^{i} do. Now, it turns out that the linearized version, at 0, does have a solution - namely

\frac{dx}{dt}(0) = v

So we are in a case where the linearized problem has a solution, but the latter do not locally extend. Why is that ? Clearly, the problem is that the equality f(x) = g(x) - which is necessary for the existence of a solution - holds at x = 0, but not in the neighbourhood; the technical translation being that 0 is not an ordinary point. The EDC interpretation, in this basic example, is quite simple. Consider the space E(x, t) spanned by the forms α^{i}(x, t) and β^{i}(x, t). At any point but (0, 0), the dimension of E(x, t) is 2. At (0, 0), however, we have that dim E(0, 0) = 1. In other words, at (0, 0) we have a discontinuous drop in the dimension of the space. This is why (0, 0) is not an ordinary point.

Another, less intuitive, example goes as follows. Given a smooth function f(x, y) of two variables, find two functions of one variable, ϕ(x) and ψ(y), such that f(x, y) = ϕ(x) + psi(y). This can be written as an exterior differential system in the submanifold M of ℝ^{4} defined by the equation

M = {x, y, ϕ, ψ | ϕ + ψ = f(x, y)}  (31)

The corresponding exterior differential system is simply:

dϕ ∧ dx = 0  (32)
dψ ∧ dy = 0  (33)
dx ∧ dy ≠ 0  (34)

The reader will check easily that the system is closed, and that there is an integral element at every point. In addition, there is no clear singularity, as in the preceding example. And yet, if \frac{∂^{2}f}{∂x∂y} ≠ 0 there is no integral manifold. integral manifold

In general, what we need is a regularity condition which will exclude such pathological situations. Technically, this condition will state that the relevant equality holds true at ordinary points. Of course, this requires a general definition of the concept of an ordinary point. This is quite easy in the case of 1-forms, but may be more complex in general. This is the topic of the next subsection.

8.5 Regularity conditions and ordinary points

If all the ω_{k}(x) are 1-forms, the regularity condition is clear enough: the dimension of the space spanned by the ω_{k}(x), 1 ≤ k ≤ K, in T_{x}^{∗}ℝ^{n}, should be constant on a neighborhood of x̄ (which is obviously not the case in the counterexample above). Note that, locally, this dimension can only increase, that is, the codimension can only decrease.

If some of the ω_{k} have higher degree, the regularity condition is more complicated, because several codimensions are involved. However, the idea is the same : all these codimensions should be constant in a neighborhood of x̄, which amounts to saying that they should have the lowest possible value at x̄. This is expressed in the following.

Pick a point x̄ ∈ ℝ^{n} ; from now on, we work in the tangent space V = T_{x}ℝ^{n}. Let E ⊂ V be an m-dimensional integral element at x̄. Pick a basis ᾱ_{1}, …, ᾱ_{n} of V^{∗} such that :

E = {ξ ∈ V | ⟨ξ, ᾱ_{i}⟩ = 0 ∀i ≥ m + 1 }

For n′ ≤ n, denote by I(n′, d) the set of all ordered subsets of {1, …, n′} with d elements. Denote by d_{k} the degree of ω_{k}. For every k, writing ω_{k}(x̄) in the ᾱ_{i} basis, we get

ω_{k}(x̄) = ∑_{I∈I(n,d_{k})} c_{I}^{k} ᾱ_{i_{1}} ∧ … ∧ ᾱ_{i_{d_{k}}} .

In this summation, it is understood that I = {i_{1}, …, i_{d_{k}}}. Since ω^{k}(x̄) vanishes on E, each monomial must contain some ᾱ_{i} with i ≥ m + 1. Let us single out the monomials containing one such term only. Regrouping and rewriting, we get the expression :

ω^{k}(x̄) = ∑_{J∈I(m,d_{k}−1)} β̄_{J}^{k} ∧ ᾱ_{j_{1}} ∧ … ∧ ᾱ_{jd_{k}−1} + remainder

where β̄_{J}^{k} is a linear combination of the ᾱ_{i} for i ≥ m+1 ; and all the monomials in the remainder contain ᾱ_{i} ∧ ᾱ_{i′} for some i > i′ ≥ m + 1.

Define an increasing sequence of linear subspace H_{0}^{∗} ⊂ H_{1}^{∗} ⊂ … ⊂ H_{M}^{∗} ⊂ V^{∗} as follows :

H_{m}^{∗} = Span[ β̄_{J}^{k} | 1 ≤ k ≤ K, J ∈ I(m, d_{k} − 1)}
H_{m−1}^{∗} = Span[ β̄_{J}^{k} | 1 ≤ k ≤ K, J ∈ I(m − 1, d_{k} − 1)}
H_{0}^{∗} = Span[ β̄_{J}^{k} | 1 ≤ k ≤ K, J ∈ I(0, d_{k} − 1)}

The latter is just the linear subspaces generated by those of the ω^{k}(x̄) which happen to be 1-forms. We define an increasing sequence of integers 0 ≤ c_{0}(x̄, E) ≤ ... ≤ c_{m}(x̄, E) ≤ n by :

c_{i}(x̄, E) = dim H_{i}^{∗}

We are finally able to express Cartan’s regularity condition. Denote by ℙ^{m}(ℝ^{n}) the set of all m-dimensional subspaces of ℝ^{n} with the standard (Grassmannian) topology : it is known to be a manifold of dimension m(n − m). Denote by G^{m} the set of all (x, E) such that E is an m-dimensional integral element at x. Note that G^{m} is a subset of ℝ^{n} × ℙ^{m}(ℝ^{n}).

Definition. Let (x̄, Ē) ∈ G^{m} - that is, Ē is an m-dimensional integral element at x̄. We say that (x̄, Ē) is ordinary if there is some neighborhood U of (x̄, Ē) in ℝ^{n} × ℙ^{m}(ℝ^{n}) such that G^{m} ∩ U is a submanifold of codimension

c_{0}(x̄, Ē) + … + c_{m−1}(x̄, Ē)

If all the ω^{k} are 1-forms, denote by d(x) the dimension of the space spanned by the ω^{k}(x). Then c_{i}(x, E) = d(x) for every i, and (x̄, Ē) is ordinary if G^{m}∩U is a submanifold of codimension md(x̄) in ℝ^{n} ×ℙ^{m}(ℝ^{n}). This implies that, for every x in a neighborhood of x̄, the set of E ∈ G_{x}^{m} (integral elements at x) has codimension md(x̄) in ℙ^{m}(ℝ^{n}). It can be seen directly to have codimension md(x). So d(x) = d(x̄) in a neighborhood of x̄ : this is exactly the regularity condition we wanted for 1-forms.

In the general case, if (x̄, Ē) is ordinary, the numbers c_{i} will also be locally constant :

c_{i}(x, E) = c_{i}(x̄, Ē) = c_{i} ∀(x, E) ∈ U.

The (non-negative) numbers

s_{0} = c_{0}
s_{i} = c_{i} − c_{i−1} for 1 ≤ i < m
s_{m} = n − m − c_{n−1}

are called the Cartan characters. We shall use them later on.

8.6 The main result

We are now in a position to state the Cartan-Kähler theorem :

Theorem 8.1 (Cartan-Kähler) Consider the exterior differential system :

ω^{k} = 0 , 1 ≤ k ≤ K

Assume that the ω^{k} are real analytic and that they generate a differential ideal. Let x̄ be a point and let Ē be a regular integral element at x̄. Then there is a real analytic integral manifold M, containing x̄ and such that :

Tx̄M = \overline{E} .

This result is due to Elie Cartan in the case when the ω^{k} have degree 1 or 2, and was extended by Kähler to the general case; the reader may refer to [5] or [4] for a proof. Nothing should come as a surprise in this statement, except the real analyticity. It comes from the very generality of the Cartan-Kähler theorem. Indeed, every system of partial differential equations can be written as an exterior differential system.

To conclude, let us mention the question of uniqueness. There is no uniqueness in the Cartan-Kähler theorem : there may be infinitely many analytic integral manifolds going through x̄ and having Ē as a tangent space at x̄. However, the theorem describes in a precise way (not given here) the set

M_{U} = \{M \begin{matrix}M is an integral manifold \\ and there exists (x, E) ∈ U \\ such that x ∈ M and T_{x}M = E \end{matrix}  \} ,

where U is a suitably chosen neighborhood of (x̄, Ē). Loosely speaking, each M in M_{U} is completely determined by the (arbitrary) choice of s_{m} analytic functions of m variables, the s_{m} being the Cartan character.

9 Markets

In this concluding section we will give some indications about markets. It is assumed that there are n consumers and l goods, and all the goods are private. Each consumer has a certain amount of non-monetary resources ω_{i} ∈ ℝ_{+}^{l} and he has to solve the following problem:

(P_{i,p}) \{ \begin{matrix}max U_{i}(x_{i}), \\ p · x_{i} ≤ p · ω_{i}. \end{matrix}

The quantity x_{i}(p) − ω_{i} is called excess demand of i-th agent whereas Z(p) := ∑_{i} x_{i}(p) − ∑_{i} ω_{i} is called excess demand. Finally the quantity Ω := ∑_{i} ω_{i} is the total quantity of goods in the market. If Z ≡ 0 then the market is said at equilibrium.

The problem we would like to solve is, as usual, the following: given a function Z : ℝ^{l} → ℝ^{l} is this an excess demand? The answer to this problem was given by Sommerschein ([23], [24], [25]), Mantel ([17], [18], [19]) and Debreu ([11]) in a series of papers around 1974 and it is as follows: if Z is a continuous function on ℝ^{l}, n ≥ l, such that p · Z(p) ≡ 0, then Z is an excess demand function. Moreover the utility functions U_{i} can be chosen to be concave. Note that up to now we have not treated the question of concavity of utility functions, as it was required by the second question given in the first section. Another fact which it is important to say about the previous result is that the answer is not local, but global (in all ℝ^{l}).

Here we want to study a different question, which is called to find the However, another question remained open since that time, and it is the

question of markets (not excess) demand functions. This means that each consumer has a fixed amount of money which he spends, and one observes the corresponding demand.

(P̃_{i,p}) \{ \begin{matrix}max U_{i}(x_{i}), \\ p · x_{i} ≤ 1. \end{matrix}

The optimal consumptions x_{i}(p) are called individual market demands and summing up we get the collective market demand function X(p) = ∑_{i} x_{i}(p). We will prove now the following

Theorem 9.1 (Chiappori–Ekeland) Let p̄ ∈ ℝ^{l}, n ≥ l, U a neighbourhood of p̄, X : U → ℝ^{l} real analytic such that

p · X(p) = 1, ∀p ∈ U.

Then there exists utility functions U_{i} : ℝ^{l} → ℝ, x̂_{i} : U → ℝ^{l}, real analytic for i = 1, …, n, such that:

(i) X(p) = ∑_{i} x̂_{i}(p);

(ii) x̂_{i}(p) solves (P̃_{i,p});

(iii) the U_{i} are concave in a neighbourhood of x̂_{i}(p).

Proof — We start by translating the conditions (i),(ii),(iii) in the language of the differential forms. We will discover a problem of the type to find an integral manifold.

Let us introduce the indirect utilities functions:

V_{i}(p) = U_{i}(x̂_{i}(p)) = max {U_{i}(x_{i}) + λ_{i}(1 − p · x_{i})} .

Then

∇V_{i}(p) = λ_{i}(p)x̂_{i}(p),

and in terms of the known function X we have the equation

X(p) = − ∑_{i=1}^{n} \frac{1}{λ_{i}(p)}∇V_{i}(p), p ∈ ℝ^{l},  (35)

or, in other words, once again the problem to decompose a given function as a linear combination of n gradients. Moreover, by the constraint p · x̂_{i}(p) = 1 there will be the equations

p · ∇V_{i}(p) = −λ_{i}(p), i = 1, …, n.  (36)

Finally the third conditions of the theorem 9.1 is equivalent to saying that the matrix of the second derivatives ^{(}^{1}^{)} ∇∇V_{i}(p) is positive definite. Now we introduce the manifold V in ℝ^{l} × ℝ^{ln} × ℝ^{n}:

V := \{(x^{i}, p_{i}^{k}, λ_{k}) : x_{i} = − ∑_{k=1}^{n} \frac{1}{λ_{k}} p_{i}^{k}, ∑_{i=1}^{l} x_{i}p_{i}^{k} = −λ_{k}, ∀i, k\} .

The meaning of the manifold V is clear. Note that V is defined by n + l equation. Now we would like to have:

(1) p_{i}^{k} as functions of (x_{1}, …, x_{n});

(2) there exists functions V_{k} such that p_{i}^{k} = \frac{∂V_{k}}{∂x_{i}}.

In the language of differential forms (2) is equivalent to

dV_{k} = ∑_{i=1}^{l} p_{i}^{k}dx_{i},

and by the Poincare lemma^{(}^{2}^{)}

∑_{i=1}^{l} dp_{i}^{k} ∧ dx_{i} = 0.

1Remember that we are using the indirect formulation.

2Remember that we are in a local framework.

Hence all the previous conditions are equivalent to the following problem: find an n–dimensional integral manifold for the system

\{ \begin{matrix}∑_{i=1}^{l} dp_{i}^{k} ∧ dx_{i} = 0, k = 1, …, n, \\ dx^{1} ∧ ⋯ ∧ dx^{n} ≠ 0. \end{matrix}  (37)

which is solved by applying the Cartan–Kähler theorem. We refer to [7] for the full proof.

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