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• Lecture 1: Sperner, Brouwer, Nash.

Lecture 1: Sperner, Brouwer, Nash.

Philippe Bich, PSE and University Paris 1 Pantheon-Sorbonne,France.

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 1. Simplex

A n-simplex (or simplex of dimension n) is (x0, ...xn) ={n

i=0 ixi : (0, ..., n) Rn+1+ :

ni=0 i = 1}, where

x0, ..., xn are independent vectors (of some vector space...)If x0, ..., xn is the canonical basis of Rn+1, (x0, ...xn) issimply denoted n.A face of (x0, ...xn) is any simplex (xi1 , ...xik ) where1 i1 < ... < ik nThe points x0, .., xn are called vertices of (x0, ...xn).The faces of (x0, ...xn) with two vertices are called edges.

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 1. Simplex

For any v =n

i=0 ixi (x0, ...xn), (v) = {i : i 6= 0}.A simplicial subdivision of (x0, ...xn) is any collection ofsimplices 1, ...,m such that(1) mi=1i = .(2) for every 1 i , j n, i j is either empty or acommon face of i and j .EXERCISE 1: Prove that if some k above has adimension < n, then i 6=k i = Exemples....

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 2. proper coloring

Consider a simplex , with a simplicial subdivision 1, ...,m,and let V the union of the vertices of 1, ...,m. A propercoloring of is, by definition, any function

c : V {0,1, ...,n}

such thatv V , c(v) (v).

Example...

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Consider a simplex , with a simplicial subdivision1, ...,m, let V be the union of the vertices of 1, ...,m,and let be a proper coloring of .We will say that a simplex i in the simplicial subdivision isfully colored if

{(v) : v i} = {0,1, ...,n}.

Sperners LemmaThe number of simplices i in the simplicial subdivision thatare fully colored is odd (thus nonempty).

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Example for n = 2 (there are three points x0, x1, x2, thus 3colors). A simplicial subdivision is defined by the "small"triangles as follows:

Here, 23 simplices in the subdivision. Only 3 simplices are fullycolored

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Emanuel Sperner (German Mathematician, 1905-1980).Lemma written when he was 22 years old (in his PHD).

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Proof by induction.

For n = 1: two possible colorations, and the simplices inthe subdivision are the small edges.

PROOF=EXERCISE 2!

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Proof by induction.

Assume Sperners Lemma true for n 1, and let with aproper coloring (and a corresponding simplicialsubdivision).Call C the set of fully colored nsimplices in thesubdivision.Call A the set of Almost-fully colored simplices in thesubdivision, i.e. colored {0,1, ...,n 1}.Call n1 the face of whose vertices are colored{0,1, ...,n 1}. The subdivision defined on definesnaturally (by intersection) a subdivision of n1(EXERCISE 3!), and the induced coloring on n1 isproper. Call B the set of (n 1)simplices of the inducedsubdivision on n1 which are colored {0,1, ...,n 1}.

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Proof by induction.

We now define a Graph G = (V ,E) (recall, in general, it isdefined by a set of nodes V , and a set of edges E , an edgebeing a subset of V with two elements (which is a way tolink two nodes: we say that such nodes are adjacent):The set of nodes is chosen equal to V = A B C.To choose the set of edges of the graph, we will decidethat two elements in V are adjacent if and only if theyshare a (n 1)-face which is colored {0,1, ...,n 1}.Recall that in a graph, the degree deg(v) of a node v Vis the number of edges containing v .If v B, deg(v) = 1.If v C, deg(v) = 1.If v A, deg(v) = 2.

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 3. Sperners Lemma

Thus, vV

deg(v) = Card(B) + Card(C) + 2Card(A).

But in any graph, the sum of the degrees of the nodes isalways even (EXERCISE 4!)And by induction, the cardinal of B is odd.Thus the cardinal of C is odd. This finishes the proof.

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 4. Brouwer theorem

Brouwer TheoremEvery continuous mapping f : n n, where n is an-simplex, admits a fixed point, i.e. there exists x n suchthat f (x) = x .

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 4. Brouwer theorem

L.E.J. Brouwer (Dutch mathematician, 1881-1966.)Apart his mathematical results (topology!) he is the father ofintuitionism, a philosophy on how we should foundMathematics. In particular, Brouwer was an "opponent" to theformalism point of view of Hilbert.In particular, Intuitionism, close to constructivism, refuses"non-constructive" proofs (e.g., no contradiction proof!).

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 4. Brouwer theorem

ProofEach element v of n can be represented through his components (v0, ..., vn), recalling v =

ni=0 vi xi

where the xi are the vertices of the simplex n . The mapping f itself can be represented through itscomponents f = (f0, f1, ..., fn), where f (v) =

ni=0 fi (v)xi .

Consider some arbitrary simplicial subdivision. Define a coloring by

v =n

i=0

vi xi , (v) {i (v) : fi (v) vi}.

It is possible (EXERCISE 5!), and is defines a proper coloring (EXERCISE 6!).

From Sperner, there exists some simplex ((0), (1), ..., (n)) of the subdivision that is fully colored, thatis fi ((i)) (i)i for every i (up to a permutation, we can assume (0) has a color 0, (1) has a color 1,...

Now, we can consider for every k a simplicial subdivision (called k -subdivision) such that the diameter ofeach simplex of the subdivision tends to zero when k tends to +.

k given, there is (k (0), k (1), ..., k (n)) of the k -subdivision that is fully colored, that isfi (

k (i)) k (i)i for every i = 0, ..., n and every k 0

Extracting subsequences, we can assume that the sequence of simplices (k (0), k (1), ..., k (n)) tends

to some v (when k +), and at the limit, by continuity, fi (v) vi for every i = 0, 1, ..., n. Easily, v is

a fixed-point of f .

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 5. Nash theorem

DefinitionA Game G = (X1, ...,XN ,u1, ...,uN) is a collection of playersi = 1, ...,N, and for each player i , a strategy space Xi , and apayoff function ui : Ni=1Xi R.

TheoremConsider a game G = (X1, ...,XN ,u1, ...,uN) where all strategyspaces are simplices, all payoff functions are continuous, andfor every i , ui is concave with respect to the strategy xi of playeri (the strategies of players other than i being fixed). Then thereexists a Nash equilibrium x = (x1, ..., xN) X1 ... XN , whichmeans:

i = 1, ...,N, xi Xi ,ui(xi , xi) ui(xi , xi)

where ui(xi , xi) denotes the payoff of player i if he plays xi ,and the other players j 6= i plays xj .

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 5. Nash theorem

John Nash (American mathematician 1928-2015).One of the greatest genius of mathematics according to somemathematicians. Won Nobel prize in Economics and Abelmedal (like Nobel prize in mathematics).Was ill (schizophrenia ?) during 25 years, then he recovered!recently killed with his wife in a car crash.See the film "A Beautiful Mind" by Ron Howard (Russel Croweis Nash).

Philippe Bich Lecture 1: Sperner, Brouwer, Nash

• 5. Nash theorem (proof from Geanakoplos)

ProofLemma: if f : X R is concave (X being a convex set) then for everyy X , x f (x) x y2 is strictly concave.Let X = X1 ... XN .Consider for every player i the mapping bi : X Xi defined by

bi (x1, ..., xN) = argmaxyiXi ui (yi , xi ) xi yi2.

From the previous lemma, it is well defined, and easily, it is continuous(for continuity, let ui (yi , xi ) = ui (yi , xi ) xi yi2. Assume thatbi = argmaxyiXi ui (yi , xi ) and that argmaxyiXi ui (yi , x

ni ) converge to

some bi 6= bi for some sequence xni converging to xi . Then find acontradiction (EXERCISE 7!)) .

From Brouwer, the mapping x (b1(x), ..., bN(x)) has some fixed-pointx , which is a Nash equilibrium (EXERCISE 8!: Hint: assumeui (xi , xi ) > ui (xi , xi ) for some xi , and find a contradiction consideringxi + (1 )xi ).

Philippe Bich Lecture 1: Sperner, Brouwer, Nash