Spectral Graph Theory and ts Applcatons September 16, 2004 Lecturer: Danel A. Spelman Lecture 5 5.1 Introducton In ths lecture, we wll prove the followng theorem: Theorem 5.1.1. Let G be a planar graph wth n vertces of maxmum degree d, and let λ 2 be the second-smallest egenvalue of ts Laplacan. Then, λ 2 8d n. The proof wll nvolve almost no calculaton, but wll use some specal propertes of planar graphs. However, ths proof has been generalzed to many planar-lke graphs, ncludng the graphs of wellshaped 3d meshes. Recently, Jon Kelner has generalzed t graphs of bounded genus. I begn by recallng two defntons of planar graphs. Defnton 5.1.2. A graph s planar f there exsts an embeddng of the vertces n IR 2, f : V IR 2 and a mappng of edges e E to smple curves n IR 2, f e : [0, 1] IR 2 such that the endponts of the curves are the vertces at the endponts of the edge, and no two curve ntersect n ther nterors. Defnton 5.1.3. A graph s planar f there exsts an embeddng of the vertces n IR 2, f : V IR 2 such that for all pars of edges (a, b) and (c, d) n E, wth a, b, c, and d dstnct, the lne segment from f(a) to f(b) does not cross the lne segment from f(c) to f(d). These defntons are equvalent. 5.2 Geometrc Embeddngs We typcally upper bound λ 2 by evdencng a test vector. Here, we wll upper bound λ 2 by evdencng a test embeddng. The bound we apply s: Lemma 5.2.1. For any d 1, v v j 2 λ 2 = mn v 1,...,v n IR d : P v =0 v 2. (5.1) 5-1
Lecture 5: September 16, 2004 5-2 Proof. Let v = (x, y,..., z ). We note that v v j 2 = (x x j ) 2 + (y y j ) 2 + + (z z j ) 2. Smlarly, v 2 = x 2 + y 2 + + z 2. It s now trval to show that λ 2 RHS: just let x = y = = z be gven by an egenvector of A+B+ +C λ 2. To show that λ 2 RHS, we apply my favorte nequalty: A +B + +C mn ( A A, B B,..., C ) C, and then recall that x = 0 mples (x x j ) 2 x2 λ 2. For an example, consder the natural embeddng of the square wth corners (±1, ±1). The key to applyng ths embeddng lemma s to obtan the rght embeddng of a planar graph. Usually, the rght embeddng of a planar graph s gven by Koebe s embeddng theorem, whch I wll now explan. I begn by consderng one way of generatng planar graphs. Consder a set of crcles {C 1,..., C n } n the plane such that no par of crcles ntersects n ther nterors. Assocate a vertex wth each crcle, and create an edge between each par of crcles that meet at a boundary. The resultng graph s clearly planar. Koebe s embeddng theorem says that every planar graph results from such an embeddng. Theorem 5.2.2 (Koebe). Let G = (V, E) be a planar graph. Then there exsts a set of crcles {C 1,..., C n } n IR 2 that are nteror-dsjont such that crcle C touches crcle C j f and only f (, j) E. Ths s an amazng theorem, whch I won t prove today. You can fnd a proof at http://math.mt.edu/~spelman/course/lect3.html Such an embeddng s often called a kssng dsk embeddng of the graph. From a kssng dsk embeddng, we obtan a natural choce of v : the center of dsk C. Let r denote the radus of ths dsk. We now have an easy upper bound on the numerator of (5.1): v v j 2 = (r + r j ) 2 2r 2 + 2r2 j. On the other hand, t s trcker to obtan a lower bound on v 2. In fact, there are graphs whose kssng dsk embeddngs result n (5.1) = Θ(1). These graphs come from trangles nsde trangles nsde trangles... Such a graph s depcted below:
Lecture 5: September 16, 2004 5-3 Graph Dscs We wll fx ths problem by lftng the planar embeddngs to the sphere by stereographc projecton. Gven a plane, IR 2, and a sphere S tangent to the plane, we can defne the stereographc projecton map, Π, from the plane to the sphere as follows: let s denote the pont where the sphere touches the plane, and let n denote the opposte pont on the sphere. For any pont x on the plane, consder the lne from x to n. It wll ntersect the sphere somewhere. We let ths pont of ntersecton be Π(x ). The fundamental fact that we wll explot about stereographc projecton s that t maps crcles to crcles! So, by applyng stereographc projecton to a kssng dsk embeddng of a graph n the plane, we obtan a kssng dsk embeddng of that graph on the sphere. Let D = Π(C ) denote the mage of crcle C on the sphere. We wll now let v denote the center of D, on the sphere. If we had v = 0, the rest of the computaton would be easy. For each, v = 1, so the denomnator of (5.1) s n. Let r denote the straght-lne dstance from v to the boundary of D. We then have v v j 2 (r + r j ) 2 2r 2 + 2r2 j. So, the denomnator of (5.1) s at most 2d r2. On the other hand, the area of the cap encrcled by D s at least πr 2. As the caps are dsjont, we have πr 2 4π, whch mples that the denomnator of (5.1) s at most 2d r 2 8d. Puttng these nequaltes together, we see that mn v 1,...,v n IR d : P v =0 v v j 2 v 2. 8d n. Thus, we merely need to verfy that we can ensure that v = 0. (5.2)
Lecture 5: September 16, 2004 5-4 Note that there s enough freedom n our constructon to beleve that we could prove such a thng: we can put the sphere anywhere on the plane, and we could even scale the mage n the plane before placng the sphere. By carefully combnng these two operatons, t s clear that we can place the center of gravty of the v s close to any pont on the boundary of the sphere. It turns out that ths s suffcent to prove that we can place t at the orgn. 5.3 The center of gravty We need a nce famly of maps that transform our kssng dsk embeddng on the sphere. It s partcularly convenent to parameterze these by a pont ω nsde the sphere. For any pont α on the surface of the unt sphere, I wll let Π α denote the stereographc projecton from the plane tangent to the sphere at α. I wll also defne Π 1 α and Π α ( ) = α. We also defne the map that dlates the plane tangent to the sphere at α by a factor a: D a α. We then defne f ω (x ) def = Π ω/ ω (. To handle the pont α, I let Π 1 α ( α) =, D 1 ω ω/ ω ( )) Π 1 ω/ ω. For α S and ω = aα, ths map pushes everythng on the sphere to a pont close to α. As a approaches 1, the mass gets pushed closer and closer to α. Instead of provng that we can acheve (5.2), I wll prove a slghtly smpler theorem. The proof of the theorem we really want s smlar, but about just a few mnutes too long for class. We wll prove Theorem 5.3.1. Let v 1,..., v n be ponts on the unt-sphere. Then, there exsts a crcle-preservng map from the unt-sphere to tself. The reason that ths theorem s dfferent from the one that we want to prove s that f we apply a crcle-preservng map from the sphere to tself, the center of the crcle mght not map to the center of the mage crcle. To show that we can acheve v = 0, we wll use the followng topologcal lemma, whch follows mmedately from Brouwer s fxed pont theorem. In the followng, we let B denote the ball of ponts of norm less than 1, and the the sphere of ponts of norm 1. Lemma 5.3.2. If φ : B B be a contnuous map that s the dentty on S. Then, there exsts an ω B such that φ(ω) = 0. We wll prove ths lemma usng Brouwer s fxed pont theorem: Theorem 5.3.3 (Brouwer). If g : B B s contnuous, then there exsts an α B such that g(α) = α. Proof of Lemma 5.3.2. Let b be the map that sends z B to z / z. The map b s contnuous at every pont other than 0. Now, assume by way of contradcton that 0 s not n the mage of φ,
Lecture 5: September 16, 2004 5-5 and let g(z ) = b(φ(z )). By our assumpton, g s contnuous and maps B to B. However, t s clear that g has no fxed pont, contradcton Brouwer s fxed pont theorem. Lemma 5.3.2, was our motvaton for defnng the maps f ω n terms of ω B. Now consder settng φ(ω) = 1 f ω (v ). n The only thng that stops us from applyng Lemma 5.3.2 at ths pont s that φ s not defned on S, because f ω was not defned for ω S. To fx ths, we defne for α S { α f z α f α (z ) = α otherwse. We then encounter the problem that f α (z ) s not a contnuous functon of α. To fx ths, we set { 1 f dst(ω, z ) < 2 ɛ, and h ω (z ) = (2 dst(ω, z ))/ɛ otherwse. Now, the functon f α (z )h α (z ) s contnuous, because h α ( α) = 0. So, we may set φ(ω) def = f ω(v )h ω (v ) h ω/ ω (v ), whch s now contnuous and s the dentty map on S. So, for any ɛ > 0, we may now apply Lemma 5.3.2 to fnd an ω for whch φ(ω) = 0. It s a smple exercse to verfy that for ɛ suffcently small, the ω we fnd wll have norm bounded away from 1, and so h ω (v ) = 1 for all, n whch case f ω (v ) = 0, as desred.