By Garrett Birkhoff

This vintage, written by means of younger teachers who turned giants of their box, has formed the knowledge of recent algebra for generations of mathematicians and is still a necessary reference and textual content for self learn and faculty classes.

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Proof. Let G be a non-planar graph and suppose dim(G) :::; 3. Let {L1> L 2, L 3} be a realizer of G. For a vertex v of G let Vi be the position of v in L i . Define an embedding ¢ of Gin IR3 by v ----t ¢(v) = (81V1,82V2,83V3), where the 8i are scalars which will be fixed later. For an edge e = {u, v} let ei = max( Ui, Vi) and embed e by e ----t ¢( e) (81 (e1 + ~), S2(e2 + ~), 83(e3 + ~ Note that by the definition of a realizer we have ». (*) ¢(V)i < ¢(e)i for i = 1,2,3 if and only if vEe. Adjust the 8i such that under the orthogonal projection 7r to the plane Xl + X2 + X3 = 0 all points in ¢(V U E) project to distinct points and these points are in general position.

However, with more effort an improvement by a factor of almost two is possible. The improvement is obtained by applying the probabilistic proof technique to an inequality of the form cr(G) ~ tm - (G) + 3t)n. 1) This inequality is known to be valid for t ~ 5. 5n/m. 1) only for t = 3 and obtain a slightly weaker constant. In the computation we use the probability p = 6n/m. 1) is based on another nice extremal problem. A drawing is said to be k-restricted if every edge is crossed by at most k other edges.

W) it follows that Wi 2: vi 2: Vi. Hence, for every W there is a coordinate i such that Wi 2: [ap]i. This proves that ap is on the surface Sv. For v E F we have (Vb V2, V3) S ap by definition and Vi 2: [ap]i if i is such that F c Ri (v). Therefore, v E F is on the border of 'V CXF. · Considering the counterclockwise neighbor it follows that [ap]i-I > Vi-I. Since there is a vertex v E F with Vi = [ap]i for i = 1,2,3 it can be concluded that ap is a maximum of Sv· 0 28 2 Schnyder Woods or How to Draw a Planar Graph?

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