Surface maps into free groups

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1 Surfce mps into free groups lden Wlker Novemer 10, 2014

2 Free groups wedge X of two circles: Set F = π 1 (X ) =,. We write cpitl letters for inverse, so = 1. e.g. () 1 =

3 Commuttors Let x nd y e loops. The commuttor of x nd y is the loop so [x, y] = xyx 1 y 1 [, ] = ()()()() = The set of ll products of commuttors is clled the commuttor sugroup, denoted [F, F ]. Exmple: the loop is in the commuttor sugroup, ecuse it is product of commuttors: [, ][, ] = ()() =

4 Commuttors Some rndom fcts: 1. [x, y] 1 = [y, x]. We check: [x, y][y, x] = xyx 1 y 1 yxy 1 x 1 = e 2. z[x, y]z 1 = [zxz 1, zyz 1 ], since: [zxz 1, zyz 1 ] = zxz 1 zyz 1 zx 1 z 1 zy 1 z 1 = zxyx 1 y 1 z 1

5 Commuttors Let x z denote the signed numer of occurrences of the letter z in the word x, so = 1, nd = 2. Lemm For loop x F, we hve x [F, F ] iff x = x nd x = x. So [F, F ]. Esy proof. The eliniztion is H 1 (F ) = F /[F, F ]. word x is trivil in the elinztion, i.e. in [F, F ], iff x = x nd x = x.

6 The commuttor sugroup Lemm For loop x F, we hve x [F, F ] iff x = x nd x = x. Direct proof. y induction on the length of x. WLOG, ssume the lst letter is. Find n in x, nd write x = st, where s nd t re words. Then (st)[(t) 1, ] = (st)(t 1 )()(t)() = st So st = st([(t) 1, ]) 1 = st[, (t) 1 ] The word st is shorter thn x, nd still hs mtching numers of, nd,. y induction, st is product of commuttors, so x is.

7 The commuttor sugroup Question: if x = x nd x = x, then x [F, F ], so x is product of commuttors. Wht is the smllest numer? I.e. wht is tht smllest k so tht there exist y i nd z i so tht x = [y 1, z 1 ][y 2, z 2 ] [y k, z k ] We cll this k the commuttor length cl(x) of x.

8 Commuttor length Exmple (Culler) [x, y] 3 = [xyx 1, y 1 xyx 2 ][y 1 xy, yy] So, [x, y] 3 cn oviously e written s product of three commuttors, so cl([x, y] 3 ) 3. ut it cn secretly e written s product of two, so cl([x, y] 3 ) 2. Finding cl(x) is hrd prolem tht cn e solved using surfces.

9 Surfces Some surfces: Some surfces with oundry: The genus is the numer of holes.

10 Euler chrcteristic Euler chrcteristic χ(s) mesures the complexity of S. χ(s) = 2 2(genus) (# oundries) g = 0 # = 0 χ = 2 g = 1 # = 0 χ = 0 g = 0 # = 3 χ = 1 g = 1 # = 1 χ = 1 g = 7 # = 4 χ = 16

11 Euler chrcteristic Euler chrcteristic χ(s) mesures the complexity of S. χ(s) = 2 2(genus) (# oundries) g = 0 # = 0 χ = 2 g = 1 # = 0 χ = 0 g = 0 # = 3 χ = 1 g = 1 # = 1 χ = 1 In (ny) tringultion of S, χ(s) = V E + F : g = 7 # = 4 χ = 16 V E + F = = 1

12 More tringultions 2 = 2 2g p = V E + F = = 2 1 = 2 2g p = V E + F = = 1

13 Gluing polygons to get surfces

14 Gluing polygons to get surfces d c d c d c c c c d d d

15 Gluing polygons to get surfces with oundry

16 Gluing polygons to get surfces with oundry g 1 g g c 1 g c1 c p c p Gluing produces genus g surfce with p oundries.

17 Surfce mps into free group How cn surfce mp to wedge of two loops? Stretch the surfce to mke it skinny: The oundry of this surfce mps to the commuttor [, ].

18 Surfce mps into free group Suppose we hve once-punctured torus mpping into free group so tht the oundry mps to loop x F. z y Consider the two loops y nd z in the surfce. They mp into loops y nd z in F, nd the oundry of the surfce mps to x = [y, z].

19 Surfce mps into free group Lemm For x F, x is commuttor iff there is mp of once-punctured torus into F so tht the oundry mps to x.

20 Surfce mps into free group Lemm In generl, if S is surfce of genus g with one oundry component, then mp S F tking the oundry of S to x is equivlent to n expression of x s product of g commuttors.

21 Surfce mps into free group y 2 z 2 z 2 y 2 y 3 z 1 z 3 y 1 y 3 z 1 z 3 We cn lso see this y looking t gluing polygon. Here x = [y 1, z 1 ][y 2, z 2 ][y 3, z 3 ] x y 1

22 Finding surfce mps We re going to compute commuttor length y finding surfce mps. How cn we find mp from surfce to free group? y uilding it out of pieces Let us forget commuttors for now nd just try to find surfce mp with given oundry.

23 Finding surfce mps Note we could sk for surfce with multiple oundry loops. We ll show how to uild surfces with ny desired oundries. The skinny surfce hs oundry + +.

24 Finding surfce mps leled ftgrph is grph with cyclic order on the incident edges t ech vertex. We will lwys drw it fttened up. leled ftgrph is ftgrph whose edges hve een leled: leled ftgrph induces mp of surfce with oundry into the free group. This mp tkes the oundry to. Theorem (Culler) Every mp of surfce with oundry into free group fctors through leled ftgrph.

25 Finding surfce mps Let us look for leled ftgrph with oundry +. The strips (rectngles) tht cn occur re leled with letter-inverse pir These re ll possile strips; the letters re

26 Finding surfce mps Pick rectngles tht contin every letter once, nd glue up: oundry is

27 How to mp leled ftgrphs We just uilt leled ftgrph. The lels instruct us how to get mp into the wedge of loops:

28 Compring skinny surfces There re multiple wys to pir up the letters to get ftgrphs with set oundry. oth of these pirings give surfces with oundry

29 Compring ftgrphs Recll tht for surfce S, χ(s) = 2 2g p, nd χ(s) = V E + F for ny tringultion. Lemm For ftgrph S uilt out of J junctions nd R rectngles, we hve χ(s) = J R. Proof. Euler chrcteristic is invrint under homotopy, so just homotope S to the grph with J vertices nd R edges.

30 Compring ftgrphs Therefore, we cn esily compute the genus of these two surfces χ(s) = 1 4 = 3 So g = ((1 χ)/2) = 2 χ(s) = 3 4 = 1 So g = ((1 χ)/2) = 1

31 Compring ftgrphs χ(s) = 1 4 = 3 So g = ((1 χ)/2) = χ(s) = 3 4 = 1 So g = ((1 χ)/2) = 1 The left surfce shows tht cn e written s product of two commuttors. The right shows it cn e written s single commuttor. (this is ovious, since [, ] = ). So cl() = 1.

32 Clrifiction Ech surfce cn mp into free group in mny wys. (For exmple, every commuttor corresponds to different mp of the sme once-punctured torus). Equivlently, there re mny leled ftgrphs which re ctully the sme underlying surfce. These leled ftgrphs give two distinct mps into free group of genus two surfce with two oundries. On the right, the oundries re +, nd on the left, +.

33 Commuttor length lgorithm to compute cl(x): 1. uild ll possile rectngles tht cn occur in ftgrph with oundry x. 2. Tke ll possile sucollections of the rectngles so tht every letter in x ppers exctly once. 3. For ech sucollection, glue up the rectngles, nd compute the genus of the surfce. 4. The smllest possile genus is cl(x). Exmple Recll, oviously cl([, ] 3 ) 3, nd eing clever, we showed cl([, ] 3 ) 2. Doing the lgorithm proves tht cl([, ] 3 ) = 2.

34 cl([, ] n ) Lemm (Culler) cl([, ] n ]) = n Proof: Every polygon in ftgrph with oundry [, ] n must hve vlence t lest 4. This is ecuse the order of the letters in mens vertices simply cn t close up until we see t lest four rectngles. The smllest mgnitude Euler chrcteristic is chieved when there re s mny vertices s possile, i.e. when every vertex hs vlence 4, so χ(s) V V /2 = 2n n = n for surfce with oundry [, ] n.

35 cl([, ] n ) Lemm (Culler) cl([, ] n ]) = n Proof continued: Therefore, for the genus g of surfce with oundry [, ] n, we hve g 1 + ( χ) = n If n is odd, we cn construct n explicit surfce with g = n+1 2

36 cl([, ] n ) Lemm (Culler) cl([, ] n ]) = n Proof continued: If n is even, note g must e n integer, so g n 2 one genus to relize g = n We just dd

37 Stle commuttor length For x [F, F ], we define the stle commuttor length cl(x n ) scl(x) = lim n n (Fct: this limit exists). We hve cl(x n ) ncl(x), so scl(x) cl(x). Exmple We sw tht cl([, ] n ) = n/2 + 1, so cl([, ] n ) n/2 + 1 scl([, ]) = lim = lim = 1 n n n n 2

38 Stle commuttor length Why should the limit scl(x) = lim n cl(x n )/n exist? Lemm The sequence cl(x n ) is sudditive, i.e. cl(x n+m ) cl(x n ) + cl(x m ) Lemm (Fekete) Let n e ny sudditive sequence ( n+m n + m ) with ll n positive. Then lim n n n exists, nd lim n n n = inf n n n.

39 Sudditive sequences (n side) Lemm (Fekete) Let n e ny sudditive sequence ( n+m n + m ) with ll n positive. Then lim n n n exists, nd lim n n n = inf n n n. Proof. The sequence n /n is ounded elow y 0, so L = inf n n n exists. Given ɛ > 0, pick m so m m < L + ɛ 2 (L is the inf). Let C = mx k<m k. Pick N > m so tht C N < ɛ 2. Now, given n > N, we write n = qm + r for r < m (quotient nd reminder), nd we hve So n n L n n = qm+r qm + r q m + r qm + r L < ɛ, s desired. m m + r N < L + ɛ 2 + ɛ 2

40 Stle commuttor length Stle commuttor length scl(x) mesures how mny commuttors, on verge, re required per copy of x in lrge power of x. This cn e smller thn cl(x): Exmple word x cl(x) scl(x) /4 2 29/ /619 [, ] 1 1/2 [, ] [, ] 3 3 3/2 [, ] [, ] 5 3 5/2 [, ] 6 4 3

41 Stle commuttor length mzing fct: it is possile to compute scl(x). In fct, it is esier to compute thn cl(x).

42 Efficient surfce mps Computing scl is relted to finding efficient surfce mps. Originl question: given x [F, F ], find the surfce mp into F whose oundry mps to x with the smllest genus. etter question: given x [F, F ], find the most efficient surfce mpping to F whose oundries ll mp to powers of x. Let us think out the ltter question, setting side scl for the moment.

43 Efficient surfce mps Wht does efficient men? We sy surfce S is dmissile for x if ll of the (potentilly mny) oundries of S re powers of x. Given n dmissile S, we let n(s) e the totl numer of copies of x ppering in S. Compute: inf S χ(s) 2n(S) over ll dmissile surfces S (there re infinitely mny). If surfce S hs smll χ(s) 2n(S), then it hs smll verge complexity per copy of x. (It s χ/2n not χ/n ecuse it doesn t relly mtter nd χ/2n is pproximtely the genus).

44 Efficient surfces x x x x 2 χ(s) = 2 2(3) 1 = 5 n(s) = 1 χ(s) 2n(S) = 5 2 = 2.5 χ(s) = 2 2(4) 3 = 9 n(s) = 4 χ(s) 2n(S) = 9 8 = So while the surfce on the right is more complicted, it is lso more efficient.

45 Efficient surfces We cn serch for efficient surfces not just for word x, ut for severl words x, y, z. In this cse, we require tht every oundry component is power of x, y, or z, nd we require tht the totl numer of copies of ech word is the sme (nd we denote it n(s)). x x y x x y 3 χ(s) = 6, n(s) = 1 χ(s) = 10, n(s) = 3 χ(s) 2n(S) = 3 χ(s) n(s) = 10 6 < 3

46 Efficient surfces Exmple: The est surfce which wrps round + once hs χ(s)/2 = 1. The most efficient surfce wrps n = 3 times round, nd hs χ(s)/(2n(s)) = 2/3.

47 Efficient surfces nd scl Theorem (Clegri) For x [F, F ], Over ll dmissile surfces. scl(x) = inf S χ(s) 2n(S) So if we cn find efficient surfces, we cn compute scl. ut there re infinitely mny dmissile surfces for x; how cn we compute the inf?

48 Finding efficient surfces ny leled ftgrph for + cn e roken into pieces: Every rectngle is one of the finitely mny possiilities. ut we need to understnd the junctions, which we ll cll polygons.

49 Polygons ny polygon cn e cut into tringles, nd there re only finitely mny kinds of tringles.

50 Polygons

51 Finding efficient surfces Hence: 1. There re finitely mny kinds of rectngles nd tringles which cn occur in ftgrph with oundry x. 2. Let V e the vector spce over Q spnned y the set of rectngles nd tringles for x. The conditions on vector which ensure tht we cn glue it up re liner. 3. If v V is vector recording how mny rectngles nd tringles we hve, nd it cn e glued up, then the Euler chrcteristic of the resulting ftgrph is independent of how we glue, nd it is liner in v. These fcts men tht Theorem (Clegri, W) For x [F, F ], the vlue of scl(x) is the solution to liner progrmming prolem whose size is polynomil in the length of x. Furthermore, most-efficient surfce exists (the inf is relized).

52 Liner progrmming exmple Here is ig ftgrph, giving mp of surfce with 4 oundery components into free group:

53 Visulizing efficient surfces Suppose we re given word x. For ny ftgrph S dmissile for x, we hve χ(s) = (# edges) (# vertices), nd we wnt to minimize χ(s) (# edges) (# vertices) = 2n 2n So for given oundry length, surfce with long edges is efficient (ecuse there will e fewer edges):

54 Exmple theorems Let x e long rndom word in [F, F ] with length n. Theorem (Clegri-W) Let F hve rnk k. For ny l < 1, there is C > 1, so tht with proility 1 O(n C ), there is trivlent ftgrph with oundry x whose verge edge length is log(n) l 2 log(2k 1) Further, for ny l > 1, there is C > 1 so tht with proility 1 O(n C ), there is no trivlent ftgrph with the given verge edge length. Note this implies scl(x) Theorem (Clegri-W) n log(2k 1) log(n) 6l. For ny L, there is c > 0 such tht with proility 1 O(e nc ) there exists trivlent ftgrph with oundry x nd with every edge of length t lest L.

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