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by: Trudie Stamm V
Trudie Stamm V
GPA 3.82


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This 4 page Class Notes was uploaded by Trudie Stamm V on Monday September 7, 2015. The Class Notes belongs to L A 1 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/181818/l-a-1-university-of-texas-at-austin in Liberal Arts at University of Texas at Austin.


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Date Created: 09/07/15
DISKS IN THE HEISENBERG GROUP ERIC KATERMAN 1 OUTLINE 1 Motivation a History i Dehn ii Cannon iii Thurston 7 formalize the notion of seeing the Cayley gmph b Automatic Groups i x notation G fg G lt SlR gtA S U S l Aquot A G ii De nition 1 machines iii De nition H pictures iv easy examples v other examples vi closure properties vii nonexamples viii Q nilpotent groups 2 Lemmas a short representatives proof by machines de nition I b small disks proof by pictures de nition H 3 Heisenberg group a H lt a7 77lla7 l7 17lawl7B77 gt b C lt WWW 117972y7517972I7y1727 717972 zy2 l gt implies that H E H RSG c use G acting on Z3 6 R3 to construct Cayley graph of H in R3 d show slides of empty F With w1 w2 e in general yzarea of disk D spanning wn has area close to n3 f elementary spanning surfaces correspond to conjugates of relators i area of S and its conjugates S 1 so need at least n3 of them to ll D so minimal is at least cubic therefore not automatic by 2 iii note combinatorial argument implies that minimal is cubic A V C V 2 MOTIVATION Much of this section is adapted from Far92 21 History In the 1910s Dehn solved the word problem for surface groups using the geometry of H2 In 1984 Cannon extended this to cocompact discrete groups of hyperbolic isometries The main thrust of Cannonls paper is that one can see the Cayley graphs for such groups ie there is a way to take a picture of the graph in 2 ERIC KATERMAN a nite ball around 0 and a nite machine that dictates how to stick copies of that nite piece together Example TU To formalize this notion use nitestate automata FSA De nition 1 Let G be a nitely generated group with generating set S and let A SUS l a1an We have a map Aquot A G denoted by w gt gt w G is an automatic group if 1 there is an FSA M over A such that 7r L A G is onto 2 the following languages are regular 0 L uv uv E LM andu5 o La1 uvuvELM andum o 0 La uvuvE LM andum This de nition is equivalent to the following G is an automatic group i 1 there is an FSA M such that 7r LM A G is onto 2 there is a constant k such that uv E LMdw 55 1 implies that u and v satisfy the hfellow traveller property Easy examples nite groups Z F2 drawn on side board Other examples hyperbolic groups abelian groups braid group mapping class group Closure properties direct product free product free product with amalgama tion over a nite subgroup nite index subgroups and supergroups quasiconvex subgroups Nonexamples SLnZn 2 3 nonsilly Baumslag Solitar groups Question nilpotent groups 3 LEMMAS Lemma 31 short representatives Let G be automatic with automatic structure A L There there is a constant N such that any w E A is equivalent to some w E L where lw l S lel n0 where no is the length of an accepted representative of proof by machines First we claim that there is a constant M such that if w E Lg E G with dw a g S 1 then there is a representative u E L with a g such that S M To see this choose M bigger than the number of states in any FSA in de nition I Let u E L such that H g so w u or uw is accepted by one of the ME for some I E A U e If gt M then Mr undergoes more than M transitions after reading w so we could shorten u by removing a loop from the path of transitions in Mac Now we take N M and induct on lf 0 w is the empty word and so it represents the identity and thus w N w E L with w E L and w no lel n0 lf gt 0 then w uz for some I E A and lt Apply lH to u so there is a u N u with u E L and lu l S 71 no But the claim implies that there isawordwELwithUwandlw lSlu lNSleln0 D DISKS IN THE HEISENBERG GROUP 3 De nition 2 Suppose w E L represents the trivial element Then w ll viriilvil with ri E Rvi E FA Say that a representation of w has combinatorial area areaw n De ne maxareaw S m 1 the isoperimetric function for lt AlR gt Note if 45 is an isoperimetric function for a di erent pre sentation of the same group then S C1 C2i for some constants C1 C2 so is bounded by say a polynomial then 45 will be bounded by some polynomial as well For example if 45 is bounded by a quadratic polynomial then we say that G satis es a quadratic isoperimetric inequality Lemma 32 Automatic groups have quadratic isoperimetric inequality proof by picture Goal given any representative w of the trivial word we must show that the combinatorial area of some disk D with boundary w has area bounded above by a quadratic polynomial in Take N no as in Lemma 3 1 and nd u such that u wt and lull S lel no for each t Also for all t ut and TH differ by a generator say I By de nition ll dw uts u13 S k for all ts so the loop utztutll decomposes into no more than lel no loops of length at most 2h 2 Therefore D may be decomposed into no more than lwllel 710 NW2 nolwl disks of length at most 2h 2 Consider the presentation G lt AR gt where w E l S 2h 2 Then the isoperimetric function with respect to this presentation is bounded by a quadratic polynomial in lwl namely lel2n0l D Note compare with hyperbolic groups which satisfy a linear isoperimetric in equality 4 THE HEISENBERG GROUP We return to our question of whether nilpotent groups are automatic It turns out that the answer is negative De ne H lt a mlla h l amt 57 gt Theorem 41 The Heisenberg group H is not automatic Proof We begin by observing that H E H RSG where G lt azyz I 1yzy zyz zy1zyzyz z y 2 l gt is a group of isometries of Rgl picture To see this note that H is just the HNN extension given by the maps Lgo Z2 A Z2 where L is inclusion and 801 17909 157 SO H lta tla 1tat 1at t 1a gt lta7 ytlla7 l17la7tl17lt75lagt We now use the action of G on Z3 6 R3 to build the Cayley graph in Rgl 4 ERIC KATERMAN We now consider the word wn an na 2na n nani We claim that any disk spanning it in 1quot must consist of at least n3 elementary spanning surfaces i e disks corresponding to conjugates of relators Which are just images of 5152 53 under G where 51 spans a h 1 5392 spans my and 53 spans 57 In order to bound the combinatorial area from below just estimate the area of spanning disks from belowi Projecting to the yz plane we nd that 51 has projectional area 12 52 has projectional area 0 and 53 has projectional area 1 But the projectional area of D is n3 so D must consist of at least n3 elementary spanning disks and therefore the minimal isoperimetric inequality of H is at least cubici In particular H does not satisfy a quadratic isoperimetric inequality so it cannot be automatici D


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