Pigeon-hole and double counting 147 From linear algebra we know that the trace equals the sum of the eigenval- ues. And here comes the trick: While A looks complicated, the matrix A2 is easy to analyze. We note two facts: Any row of A contains precisely p + 1's. This implies that p + 1 is an 1 + eigenvalue of A, since A1 = ( p 1, where 1 is the vector consisting 1) of 1's. For any two distinct rows vi, v, there is exactly one column with a 1 in both rows (the column corresponding to the unique subspace spanned by vi, v,). Using these facts we find where I is the identity matrix and J is the all-ones-matrix. Now, J has p + the eigenvalue p2 + (of multiplicity 1) and 0 (of multiplicity p2 + p). 1 Hence A2 has the eigenvalues p2 + 2p + 1 = (p+ 1)2 of multiplicity 1 and p of multiplicity p2 +p. Since A is real and symmetric, hence diagonalizable, we find that A has the eigenvalue p + 1 or -(p + and p2 + 1 ) p eigenvalues From Fact 1 above, the first eigenvalue must be p + 1. Suppose *fi. that fi and -fi has multiplicity r, multiplicity s, then But now we are home: Since the trace is an integer, we must have r = s, p + 1. so trace A = 0 6. Sperner's Lemma In 19 1 1, Luitzen Brouwer published his famous fixed point theorem: Bn - Every continuous function f: Bn of an n-dimensional ball to itse2fhas a$xed point (a point x E Bn with f (x) x ) . = For dimension 1, that is for an interval, this follows easily from the inter- mediate value theorem, but for higher dimensions Brouwer's proof needed some sophisticated machinery. It was therefore quite a surprise when in 1928 young Emanuel Sperner (he was 23 at the time) produced a simple combinatorial result from which both Brouwer's fixed point theorem and the invariance of the dimension under continuous bijective maps could be deduced. And what's more, Sperner's ingenious lemma is matched by an equally beautiful proof - it is just double counting.
148 Pigeon-hole and double counting We discuss Sperner's lemma, and Brouwer's theorem as a consequence, for the first interesting case, that of dimension n = 2. The reader should have no difficulty to extend the proofs to higher dimensions (by induction on the dimension). Sperner's Lemma. Suppose that some "big" triangle with vertices Vl, V2, V 3 is triangulated (that is, decomposed into a jinite number of "small" triangles that fit to- gether edge-by-edge). Assume that the vertices in the triangulation get "colors" from the set {1,2,3) such that V, receives the color i (for each i), and only the col- to Vj (for i # j), ors i and j are used for vertices along the edge from V, 1 2 2 1 while the interior vertices are colored arbitrarily with 1, 2 or 3. The triangles with three different colors Then in the triangulation there must be a small "tricolored" triangle, which are shaded has all three different vertex colors. Proof. We will prove a stronger statement: the number of tricolored triangles is not only nonzero, it is always odd. Consider the dual graph to the triangulation, but don't take all its edges 3 only those which cross an edge that has endvertices with the (different) - colors 1 and 2. Thus we get a "partial dual graph" which has degree 1 at all vertices that correspond to tricolored triangles, degree 2 for all triangles in which the two colors 1 and 2 appear, and degree 0 for triangles that do not have both colors 1 and 2. Thus only the tricolored triangles correspond to vertices of odd degree (of degree 1 ) . However, the vertex of the dual graph which corresponds to the outside of the triangulation has odd degree: in fact, along the big edge from Vl to V2, 1 : 2 2 1 ' 2 there is an odd number of changes between 1 and 2. Thus an odd number ,/ : of edges of the partial dual graph crosses this big edge, while the other big - - - - - - - - - - - - - ' edges cannot have both 1 and 2 occurring as colors. Now since the number of odd vertices in any finite graph is even (by equa- tion (4)), we find that the number of small triangles with three different colors (corresponding to odd inside vertices of our dual graph) is odd. With this lemma, it is easy to derive Brouwer's theorem. Proof of Brouwer's fixed point theorem (for 2). Let A be the tri- A - n = o), angle in R3 with vertices el = (1,0, e2 = (0,1, and e3 = (0,0,1). o), A has a fixed point, It suffices to prove that every continuous map f : since A is homeomorphic to the two-dimensional ball B2. We use 6(7) to denote the maximal length of an edge in a triangulation 7 . . . . One can easily construct an infinite sequence of triangulations TI, 7 2 , of A such that the sequence of maximal diameters 6(7k) converges to 0. Such a sequence can be obtained by explicit construction, or inductively, for example by taking 7k+1 to be the barycentric subdivision of Tk. For each of these triangulations, we define a 3-coloring of their vertices v < v := min{i : f (v), by setting X(v) that is, X(v) is the smallest index i i ) , is negative. Assuming that f has such that the i-th coordinate off (v) v - no fixed point, this is well-defined. To see this, note that every v E A lies
Pigeon-hole and double counting 149 2 3 = 1, hence xi in the plane xl + + # v, x 2 vi = 1. So iff (v) then at least one of the coordinates of f (v) v must be negative (and at least one - must be positive). Let us check that this coloring satisfies the assumptions of Sperner's lemma. First, the vertex ei must receive color i, since the only possible negative component of f (ei) ei is the i-th component. Moreover, if v lies on the - 0, so the i-th component off (v) v cannot edge opposite to ei, then u, = - be negative, and hence v does not get the color i. Spemer's lemma now tells us that in each triangulation Tk there is a tri- colored triangle {vk:' , vkZ2, with X(vk:') v " ~ ) = i. The sequence of points ( v " ' ) ~ > ~ need not converge, but since the simplex A is compact some subsequence has a limit point. After replacing the sequence of tri- angulations Tk by the corresponding subsequence (which for simplicity we also denote by '&) we can assume that (vkZ1)k converges to a point E A. Now the distance of vkT2 and vkz3 from vkZ1 v is ' at most the mesh length 6 ( 3 ) , which converges to 0. Thus the sequences (vk2) and ( v " ~ ) converge to the same point v. But where is f (v)? We know that the first coordinate f (vkZ1) is smaller than that of vk" for all k. Now since f is continuous, we derive that the first coordinate off (v) is smaller or equal to that of v. The same reasoning works for the second and third coordinates. Thus none of the coordinates of f (v) v is positive - and we have already seen that this contradicts - # v. the assumption f (v) 0 References [ I ] L. E. J. BROUWER: ~ b e r Abbildungen von Mannigfaltigkeiten, Math. An- nalen 71 (1912), 97-1 15. [2] W. G. BROWN: On graphs that do not contain a Thomsen graph, Canadian Math. Bull. 9 (1966), 281-285. [3] P. ERDBS, A. RENYI & V. SOS: On a problem ofgraph theory, Studia Sci. Math. Hungar. 1 (1966), 215-235. [4] P. ERDBS & G. SZEKERES:A combinatorialproblem in geometry, Cornpositio Math. (1935). 463-470. [5] S. HOSTEN & W. D. MORRIS: The order dimension of the complete graph, Discrete Math. 201 (1999), 133- 139. [6] I. REIMAN: Uber ein Problem von K. Zurankiewicz, Acta Math. Acad. Sci. Hungar. 9 (1958), 269-273. [7] J. SPENCER: Minimal scrambling sets of simple orders, Acta Math. Acad. Sci. Hungar. 22 (197 l), 349-353. [8] E. SPERNER: Neuer Beweis fur die Invarianz der Dimensionszahl und des Gebietes, Abh. Math. Sern. Hamburg 6 (1928), 265-272. [9] W. T. TROTTER: Cornbinatorics and par ti all)^ Ordered Sets: Dimension Theory John Hopkins University Press, Baltimore and London 1992.
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