The Second Eigenvalue of Planar Graphs

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1 Spectral Graph Theory Lecture 20 The Second Egenvalue of Planar Graphs Danel A. Spelman November 11, 2015 Dsclamer These notes are not necessarly an accurate representaton of what happened n class. The notes wrtten before class say what I thnk I should say. I sometmes edt the notes after class to make them way what I wsh I had sad. There may be small mstakes, so I recommend that you check any mathematcally precse statement before usng t n your own work. These notes were last revsed on November 11, Overvew Spectral Graph theory frst came to the attenton of many because of the success of usng the second Laplacan egenvector to partton planar graphs and scentfc meshes [DH72, DH73, Bar82,?, Sm91]. In ths lecture, we wll attempt to explan ths success by provng, at least for planar graphs, that the second smallest Laplacan egenvalue s small. One can then use Cheeger s nequalty to prove that the correspondng egenvector provdes a good cut. Theorem 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. I begn by recallng two defntons of planar graphs. Defnton 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 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). 20-1

2 Lecture 20: November 11, These defntons are equvalent Geometrc Embeddngs We typcally upper bound λ 2 by evdencng a test vector. evdencng a test embeddng. The bound we apply s: Here, we wll upper bound λ 2 by Lemma For any d 1, λ 2 = mn v 1,...,v n IR d : v =0 (,j) E v v j 2 v 2. (20.1) Proof. Let v = (x, y,..., z ). We note that v v j 2 = (x x j ) 2 + (y y j ) (z z j ) 2. Smlarly, (,j) E (,j) E v 2 = x 2 + (,j) E y z 2. (,j) E 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 (,j) E (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 (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 beautful proof n the book Combnatoral Geometry by Agarwal and Pach.

3 Lecture 20: November 11, 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 (20.1): 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 (20.1) = Θ(1). These graphs come from trangles nsde trangles nsde trangles... Such a graph s depcted below: 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 (20.1) 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 + 2r 2 j. So, the denomnator of (20.1) 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 (20.1) s at most 2d r 2 8d.

4 Lecture 20: November 11, Puttng these nequaltes together, we see that mn v 1,...,v n IR d : v =0 (,j) E v v j 2 v 2. 8d n. Thus, we merely need to verfy that we can ensure that v = 0. (20.2) 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 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. To handle the pont α, I let Π 1 α ( α) =, and Π α ( ) = α. We also defne the map that dlates the plane tangent to the sphere at α by a factor a: Dα. a We then defne f ω (x ) def = Π ω/ ω ( 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 (20.2), 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 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.

5 Lecture 20: November 11, Lemma 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 (Brouwer). If g : B B s contnuous, then there exsts an α B such that g(α) = α. Proof of Lemma 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, 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 , 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 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 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.

6 Lecture 20: November 11, Further progress Ths result has been mproved n the followng... References [Bar82] Earl R. Barnes. An algorthm for parttonng the nodes of a graph. SIAM Journal on Algebrac and Dscrete Methods, 3(4): , [DH72] W. E. Donath and A. J. Hoffman. Algorthms for parttonng graphs and computer logc based on egenvectors of connecton matrces. IBM Techncal Dsclosure Bulletn, 15(3): , [DH73] W. E. Donath and A. J. Hoffman. Lower bounds for the parttonng of graphs. IBM Journal of Research and Development, 17(5): , September [Sm91] Horst D. Smon. Parttonng of unstructured problems for parallel processng. Computng Systems n Engneerng, 2: , 1991.

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