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# INTRO TO THE STUDY OF SOCIETY SOC 302

UT

GPA 3.91

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This 5 page Class Notes was uploaded by Natalie Williamson on Sunday September 6, 2015. The Class Notes belongs to SOC 302 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/181414/soc-302-university-of-texas-at-austin in Sociology at University of Texas at Austin.

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Date Created: 09/06/15

ARTIN GROUPS 3MANIFOLDS AND COHERENCE C MCA GORDON Dedicated to Fico on the occasion of his 60th birthday 1 Introduction By a labeled graph we shall mean a nite non empty graph P without loops or multiple edges each of whose edges is labeled by an integer greater than or equal to 2 Let the vertices of P be 3132 sn and let the label on an edge with endpoints s and 57 be raw 2 2 De ne abm to be the word abab of length m Then the Artin group AP associated with the labeled graph P is the group with generators 31 32 3n and relations 3157 spam one for each edge of P In particular if mij 2 then the generators s and 57 commute Note also that if P is the disjoint union of graphs F1 and F2 then AP E AFL AM A 3 manifold group is a group that is isomorphic to 7r1M for some connected 3 manifold M Note that we do not assume that M is orientable or compact or without boundary Taking a connected sum shows that if G1 and G2 are 3 manifold groups then so is G1 G2 If P is a tree then AP is the fundamental group of the complement of a link L in SS where L is a connected sum of 2m torus links see Bru Thus AP is a 3 manifold group If P is a triangle with each edge labeled 2 then AP ZS 7r1T3 is also a 3 manifold group In this note we con rm the suspicion of Hermiller and Meier HM p143 that these are the only connected graphs whose Artin groups are 3 manifold groups Theorem 11 For an Artin group AT the following are equiualent 1 AP is a 3 manifold group 2 AP is uirtually a 3 manifold group 3 Each component of P is either a tree or a triangle with each edge labeled 2 The equivalence of 1 and 3 was proved by Droms D in the case of all right Artin groups or graph groups that is when all labels are 2 and by Hermiller and Meier in the case when all labels are even In Section 3 we make some additional remarks about coherence In particular we show that AutF2 and the braid group B4 are incoherent although neither has a subgroup of the form F2 x F2 The latter fact for B4 ws originally proved by Akimenkov A using different methods Partially supported by TARP grant 00365805192001 1 2 Artin groups and 3manifolds Recall that a group is coherent if every nitely generated subgroup is nitely presented The following is proved in HM Proposition 57ii Lemma 21 Hermiller and Meier Let P be a cycle of length at least 4 Then AP is incoherent If P is a labeled graph we shall say that P is of in nite or nite type according as the Coxeter group corresponding to the Artin group AP is in nite or nite We will use p q r to denote a triangle with edge labels p q and r The triangles of nite type are then 2 2 m 233 234 and 23 5 If P is a triangle the simplicial complex K0 de ned in CD2 is either a triangle or a 2 simplex according as P is of in nite or nite type The Main Conjecture of CD1 and CD2 therefore holds for AF by CD1 The following lemma is then a consequence of CD2 Corollary 142 and Corollary 225 Lemma 22 Charney and Davis Let P be a triangle i If P is of in nite type then AP has geometric dimension 2 and xAL 1 ii If P is of nite type then AP has geometric dimension 5 and xAL 0 For the three triangles 236 244 and 333 of Euclidean type also follows from the descriptions of AP given in Sq1 Lemma 23 Let L be a triangle of in nite type Then AL is incoherent Proof Let 4p AP a Z be the epimorphism that maps each generator 3139 to 1 By Me Proposition 51 and Corollary 53 K kerap is nitely generated Now AP has geometric dimension 2 Lemma 22 and hence K has geometric dimension 2 Suppose K were nitely presented Then K would be of type FP Bro p199 and so xK would be de ned We would then have MAP xKXZ 0 Bro p250 St2 compare G contradicting Lemma 22 D In W Wall asked whether a group of the form F 0 F where F and F are free and C has nite index in F and F is coherent This was answered negatively by Gersten G who showed that the double of a free group of rank 2 2 along a subgroup of nite index 2 3 is always incoherent We remark that Lemma 23 also provides examples which are not doubles since Squier has shown Sq1 that A23 6 and A3 33 can each be expressed as a free product with amalgamation F 0 F where rankF 4 rank F 3 and C has index 2 in F and index 3 in F Lemma 24 A23 3 and A23 4 are incoherent Proof Since A2 3 4 embeds in A2 3 3 as a subgroup of nite index La it is enough to show that A2 3 4 is incoherent One way to do this is to use the fact that A3 3 3 embeds 2 in A23 4 KP together with Lemma 23 Another argument is that the commutator subgroup A of A2 3 4 is nitely generated but since H2A E Z not nitely presented Sq2 D Note that A2 2 m Am x Z where Am is the Artin group of a single edge with label m Since Am is a 3 manifold group it is coherent by Sc it is also easy to show this directly and hence A2 2 m is coherent Lemma 25 A2 2771 m gt 2 and A2 3 5 are not virtually E manifold groups Proof Let G be a nitely generated group with an epimorphism 4p G a Z such that kerap is nitely generated and let H be a subgroup of G of nite index Then 4p induces an epimorphism 11 H a Z where kerzJ has nite index in kerap Now suppose that H is a 3 manifold group Since H is nitely generated it is the fundamental group of a compact 3 manifold Sc Therefore since kenJ is nitely generated by Stl kenJ is a Q mam39fold group ie it is either free or the fundamental group of a closed surface Hence kerap is Virtually a 2 manifold group Suppose A2 2 m Am x Z is Virtually a 3 manifold group Then by the above dis cussion Am has a subgroup B of nite index such that B is either free or the fundamental group of a closed orientable surface Since Am is the fundamental group of a compact orientable irreducible 3 manifold M whose boundary consists of tori xAm xM wa 0 Hence xB O implying that B is isomorphic to either Z or Z x Z But if m gt 2 this contradicts the fact that Am contains a non abelian free group Now let A A2 3 5 and let 4p A a Z be abelianization Then A kerap is nitely generated by Me Suppose A is Virtually a 3 manifold group Then there is a 2 manifold subgroup B of A of nite index By a standard transfer argument H2B Q a H2A Q is surjective But dim H2A Q 7 Sq2 whereas dim H2BQ 1 D Proof of Theorem 1 Clearly 1 implies 2 and 3 implies 1 we must show that 2 implies A subgraph F0 of P is full if every edge of P whose vertices are in F0 is an edge of P0 We recall the basic fact Le that if F0 is a full subgraph of a labeled graph P then the homomorphism APO a AP induced by the inclusion map To C P is injective Let P be a connected labeled graph and suppose that AP is Virtually a 3 manifold group By Sc 3 manifold groups and hence Virtual 3 manifold groups are coherent Also a subgroup of a Virtual 3 manifold group is clearly a Virtual 3 manifold group It follows from Lemma 21 that P is ohordal ie has no full subgraph that is a cycle of length 2 4 By Lemmas 24 and 25 any triangle in P has all labels equal to 2 lfP is not a tree or a 2 2 2 triangle then P has a full subgraph F0 of one of the forms shown in Figure 1 where all unlabeled edges are understood to have label 2 see 3 6 ii iii FIGURE 1 FIGURE 2 In cases and ii APO A x Z where in case A Z3 and in case ii A is the Artin group of a tree with two edges each labeled 2 Since APO is virtually a 3 manifold group by assumption A has a subgroup B of nite index that is either free or the fundamental group of a closed orientable surface Since xA O we have xB O and hence B E Z or Z x Z which is clearly impossible In case iii m odd let 4p APO a Z be the epimorphism de ned by 4pa ltpb 1 Lpc 4pd 0 By Me kerap is nitely generated Hence kerap has a subgroup B of nite index that is either free or the fundamental group of a closed orientable surface Since kerap contains 001 E Z x Z we must have B E Z x Z But this contradicts the fact that kerap also contains the commutator subgroup of Am which is a non abelian free group In case iii m even de ne 4p APO a Z by ltpb 1 4pa Lpc 4pd O as in Then kerap is nitely generated and contains both Z x Z and a non abelian free group giving a contradiction as before D 3 Coherence It is natural to ask which Artin groups AP are coherent For graph groups ie when all edge labels are 2 this has been answered by Droms D AP is coherent if and only if P is chordal For the general case it is necessary to be able to answer the following question Question 31 Is A23 5 coherent If at most one of pqr is even the homomorphism 4p Ap q r a Z in the proof of Lemma 23 is abelianization so that proof shows that if p q r is of in nite type then the commutator subgroup of Ap q r is nitely generated but not nitely presented However as pointed out in Sq2 the commutator subgroup of A23 5 is nitely presented the same argument applies to A23 lf A2 3 5 is incoherent one can show that an Artin group AP is coherent if and only if P is chordal every complete subgraph of P with 3 or 4 vertices has at most one edge label gt 2 and P has no full subgraph of the form shown in Figure 2 where m gt 2 and unlabeled edges are understood to have label 2 If A2 3 5 is coherent the characterization would be more complicated Let Fn denote the free group of rank n A popular way of showing that a group is incoherent is to show that it has a subgroup isomorphic to F2 x F2 which is well known 4 to be incoherent see eg For example AutF3 has such a subgroup F13 and hence AutFn is incoherent for n 2 3 For n 2 we have Theorem 32 1 AutF2 239s incoherent 2 F2 X F2 does not embed in AutF2 Let B denote the n strand braid group Note that B is coherent for n g 3 Since A23 3 E B4 we see by Lemma 24 that B is incoherent for n 2 4 Also since the center of F2 x F2 is trivial by Part 2 of Theorem 32 and the proof of Part 1 below we recover the result of Akimenkov A that F2 x F2 does not embed in B4 It follows see the proof of Lemma 24 that the incoherent groups A2 3 4 and A3 3 3 also do not contain an F2 x F2 It is shown in Ma that F2 x F2 does embed in B for n 2 5 Proof of Theorem 32 1 Let ZB4 denote the center of 134 then B4ZB4 is isomorphic to an index 2 subgroup of AutF2 DFG Now A3 33 embeds in B4 KP and since it is a free product with amalgamation of the form F4 9410 F3 Sq1 it has trivial center and hence embeds in AutF2 Since A3 3 3 is incoherent by Lemma 23 AutF2 is also incoherent 2 There is a short exact sequence 14 F2 L AutF2 L GL2Z1 where is conjugation by 9 Mapping GL2Z onto PSL2Z E Z2 Z3 gives the related sequence 1gtkerpgtAutF2 42294234 1 where ker7r has index 4 in ker p In particular ker p is virtually free Suppose H lt AutF2 where H 64 B x 39y 6 H1 x H2 E F2 x F2 We claim that either pH1 1 or pH2 1 For if not then writing 64 p64 etc we may assume that 64 1 34 7 By the Kurosh Subgroup Theorem any abelian subgroup of Z2 Z3 is cyclic Therefore 64 z 645 y say Then we have 1 64 mp yq for some p q E Z It follows that my has a non trivial center and hence again by the Kurosh Subgroup Theorem my 2 say Therefore 7y 5 lt is cyclic implying that ker p O H2 34 1 Similarly ker p O H1 34 1 This gives Z x Z lt ker p contradicting the fact that kerp is virtually free We may assume then without loss of generality that pH2 1 Then le1 is injective otherwise we would have Z x H2 lt ker p again contradicting the fact that ker p is virtually free It follows that 7rlH1 is injective Also H2 lt ker p and therefore ker7r H2 has nite index in H2 Hence ker7r H2 4G where G lt F2 has rank 2 2 Note that since F2 has trivial center 4p 6 AutF2 commutes with 49 9 6 F2 if and only if 4pg 9 Hence if 4p 6 H1 then G lt Fixltp Since rank G 2 2 it follows from the Scott Conjecture BH that Fix4p has rank 2 Furthermore by CT there is a basis a b of 5

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