Juan González-Meneses Universidad de Sevilla. Institute for Mathematical Sciences. Singapore, June 27, 2007.

Transcription

1 7RZDUGVDSRO\QRPLDOVROXWLRQWR WKHFRQMXJDF\SUREOHP LQEUDLGJURXSV Juan González-Meneses Universidad de Sevilla Joint with -RDQ6%LUPDQ and 9*HEKDUGW. %UDLGV Institute for Mathematical Sciences. Singapore, June 27, 2007.

2 ,QWURGXFWLRQ Conjugacy problems Fix a group t &RQMXJDF\GHFLVLRQSUREOHP&'3 Given two elements y[, K t, determine whether they are conjugate. &RQMXJDF\VHDUFKSUREOHP&63 Given two conjugate elements y[, K t, find a conjugating element.

3 ,QWURGXFWLRQ Braid groups Braid group on Q strands (E. Artin, 1925)

4 ,QWURGXFWLRQ Positive elements 3RVLWLYHHOHPHQWV Braids in which every crossing is positive Positive elements determine a partial order in 4 Q. This order is: Invariant under left multiplication. A lattice order. (unique gcd s and lcm s)

5 ,QWURGXFWLRQ Simple elements 6LPSOHHOHPHQWV Positive elements in which every pair of strands cross at most once Simple elements of 4 Q Bij Permutations of Σ Q = Biggest simple element = +DOI WZLVW

6 ,QWURGXFWLRQ Word problem Garside (1969), Deligne (1972), Adyan (1984), Elrifai-Morton (1988), Thurston (1992). /HIW QRUPDOIRUP Simple elements Complement of a simple element The above is the left normal form of? if

7 ,QWURGXFWLRQ Left normal forms In general: Given simple elements y K Not in left normal form y X 0 simple simple In left normal form (left weighted) We call this procedure a ORFDOVOLGLQJ applied to y K. 5HPDUN: Possibly yx = ', or 0 = 1.

8 ,QWURGXFWLRQ Left normal forms Computation of a left normal form, given a product of simple elements: Apply all possible local slidings, until all consecutive factors are left weighted Left normal form: Maximal power of. Minimal number of factors. (canonical length)

9 &RQMXJDF\ SUREOHP *DUVLGH(OULIDL0RUWRQ%LUPDQ.R/HH)UDQFR*0*HEKDUGW (1969) (1988) (1998) (2003) (2005) &KDUQH\ (1992) Artin-Tits groups of spherical type are biautomatic.

10 &RQMXJDF\ SUREOHP The main idea Given an element?, compute the set of simplest conjugates of?. (in some sense) simplest can mean of minimal canonical length. But there are better choices? and and r are are conjugate their theircorresponding sets setscoincide.

11 &RQMXJDF\ SUREOHP Cyclings and decyclings Could? be simplified still more by a conjugation? Consecutive factors are left-weighted. What about? 5 and? 1? (OULIDL0RUWRQ (1988): &\FOLQJ RI?[ 'HF\FOLQJ RI?[ In this way,? 5 and? 1 can interact.

12 &RQMXJDF\ SUREOHP Cyclings and decyclings Using cyclings and decyclings, several kinds of sets have been defined: Summit sets Super summit sets Ultra summit sets Reduced super summit sets (Garside) (ElRifai-Morton) (Gebhardt) (S. J. Lee) But one can do better

13 1HZ LGHD Cyclic sliding If we just want to make interact? 5 and? 1, we can consider: (the prefix of the initial factor that should be added to the final factor to normalise the pair formed by both) Cyclic Cyclicsliding slidingof of?[?[ = conjugation by by simple simple

14 &\FOLF VOLGLQJ The set &6(?) &6(?) = Conjugates of? in a closed orbit under V =? sliding? sliding sliding?? Elements in CS(?): Have minimal canonical length. Are in a closed orbit under cycling. Are in a closed orbit under decycling.

15 &\FOLF VOLGLQJ The set &6(?) Every Every two two elements in in &6(?) &6(?) are are FRQQHFWHG by by VLPSOH VLPSOHHOHPHQWV. r V V V V V V F &6(?) An arrow is a minimal simple element if it cannot be decomposed as a product of smaller arrows.

16 &\FOLF VOLGLQJ The set &6(?) Every Every two two elements in in &6(?) &6(?) are are FRQQHFWHG by by PLQLPDOVLPSOHHOHPHQWV. One can compute a GLUHFWHGJUDSK: 9HUWLFHVElements in &6(?). $UURZV minimal simple elements.

17 &RQMXJDF\ SUREOHP A polynomial solution? One can solve the conjugacy problems (CDP & CSP) by computing the above sets. But if one needs a polynomial solution, one must solve the following: # times times one onemust mustslide slide? to toobtain obtainan anelement in in &6(?)? &6(?)? How Howbig bigcan &6(?) &6(?) be? be? Sometimes exponential size!

18 &RQMXJDF\ SUREOHP Bounding the size of &6(?) In random examples of big canonical length, DOORIWKHP satisfy: *HEKDUGW This happens for z[ = 3,...,8 and Remark: $OOthese examples are are SVHXGR$QRVRY and and ULJLG ULJLG

19 *HRPHWULF DSSURDFK 1LHOVHQ7KXUVWRQ FODVVLILFDWLRQ Braids in 4 Q can be seen as automorphisms of the z-times puncturted disc

20 *HRPHWULF DSSURDFK 1LHOVHQ7KXUVWRQ FODVVLILFDWLRQ 3HULRGLF%UDLGV = (Roots of, for some E) = 5HGXFLEOH %UDLGV = Preserve a family of disjoint, closed curves

21 *HRPHWULF DSSURDFK 1LHOVHQ7KXUVWRQ FODVVLILFDWLRQ 3HULRGLF%UDLGV = (Roots of, for some E) = 5HGXFLEOH %UDLGV = Preserve a family of disjoint, closed curves 3VHXGR$QRVRY %UDLGV = Preserve two transverse measured foliations scaling the measure of and the measure of by by

22 *HRPHWULF DSSURDFK,GHDIRU VROYLQJ WKH &63LQSRO\QRPLDO WLPH 3HULRGLFFDVH Change of Garside structure. Use Artin-Tits groups of type B. 5HGXFLEOHFDVH Use the reduction curves to split the problem into several simpler ones. 3VHXGR$QRVRYFDVH Take powers to simplify the problem.

23 3HULRGLF EUDLGV 3URSHUWLHV %HVWYLQD (1999) After &R[HWHU(1934) In In general, #(USS(?)) is is exponential in in z. z. %HVVLV'LJQH0LFKHO (2002) The The centralizer of of? is is either either % Q or Q or the the braid braid group group of of an an annulus.

24 3HULRGLF EUDLGV 3URSHUWLHV.pUpNMDUWy (1919), (LOHQEHUJ (1934): Every Every periodic braid braid is is conjugate to to a power power of of either either δ or or ε. ε.

25 3HULRGLF EUDLGV $SRO\QRPLDO DOJRULWKP %LUPDQ*HEKDUGW*0 (2006) Idea: change of Garside structure! &DVH %LUPDQ.R/HH (1998): There is a Garside structure of 4 Q whose Garside element is precisely δ. With this structure:

26 3HULRGLF EUDLGV &RQMXJDWHV RI SRZHUV RI G $OJRULWKP Input: Two braids [ and V.

27 3HULRGLF EUDLGV &RQMXJDWHV RI SRZHUV RI H Braid group of the annulus, with Q strands. = Artin-Tits group of type % Q = {Braids on Q strands, invariant under rot(180º)} (Bessis-Digne-Michel, 2002)

28 3HULRGLF EUDLGV &RQMXJDWHV RI SRZHUV RI H Input:

29 5HGXFLEOHEUDLGV 3URSHUWLHV A reducible braid α preserves a family of curves, called a UHGXFWLRQV\VWHP. %LUPDQ/XERW]N\0F&DUWK\ (1983) There is a FDQRQLFDOUHGXFWLRQV\VWHP It can always be simplified by an automorphism η (i.e., by a conjugation).

30 5HGXFLEOHEUDLGV 3URSHUWLHV One can then decompose the disc ; along

31 5HGXFLEOHEUDLGV 3URSHUWLHV 7XEXODUEUDLG,QWHULRUEUDLGV

32 5HGXFLEOHEUDLGV 3URSHUWLHV 7XEXODUEUDLG,QWHULRUEUDLGV

33 5HGXFLEOHEUDLGV 3URSHUWLHV We can simplify the interior braids of an orbit of tubes.

34 5HGXFLEOHEUDLGV 3URSHUWLHV *0 (2003): This allows to split the CSP into simpler CSP s. ( if one knows the reducing curves! )

35 3VHXGR$QRVRY EUDLGV 3URSHUWLHV A JHQHULFbraid is is always always pseudo-anosov Unproved? *0:LHVW (2004): Its Its centralizer is is isomorphic to to %LUPDQ *HEKDUGW*0 (2006): A small small power power of of it it is is conjugate to to a ULJLGEUDLG. """

36 8VLQJSRZHUVWRGHWHFWFRQMXJDF\ *0 (2003): The Eth root of a braid is unique up to conjugacy. And if the braid is pseudo-anosov, the root is unique. &RUROODU\? and and r are are conjugate if if and and only only if if so so are are? E[ E[ and and r E We can solve the CDP by using powers. &RUROODU\ If If? and and r are are pseudo-anosov, then then the the FRQMXJDWLQJHOHPHQWV of of (?r[) (?r[) and and of of (? (? E,, r E )) coincide We can solve the CSP, LQWKHSVHXGR$QRVRYFDVH, by using powers.

37 5LJLG HOHPHQWV 'HILQLWLRQ? is ULJLG if (? is in normal form as a cyclic word.) 7KHRUHP (Birman, Gebhardt, GM, GM, 2006) 2006) If If? is isconjugate to toa rigid rigidelement, and and,, then thenuss(?) is isthe theset setof ofrigid rigidconjugates of of?.?. 7KHRUHP (Gebhardt, GM, GM, 2007) 2007) If If? is isconjugate to toa rigid rigidelement, then then &6(?) &6(?) is isthe theset setof ofrigid rigidconjugates of of?.?. (DVLHUFRPELQDWRULFV

38 5LJLG HOHPHQWV One Onecan caneasily easilydetermine determineififan anelement elementisisinin&6( &6(??).). Orbits Orbitsunder undersliding slidingare aretrivial. trivial. How many conjugate rigid elements there can be? 2SHQ SUREOHP &RQMXJDF\ SUREOHP

39 %DFNWR UHGXFLEOHHOHPHQWV If one can determine the UHGXFLQJ FXUYHV, one can split the conjugacy problem. If the reducing curves are VWDQGDUG (round circles) they are very easy to find. 7KHRUHP (Gebhardt, GM, GM, 2007) 2007) Up Uptoto taking takinga small smallpower powerifif necessary, if ifr is isa UHGXFLEOHEUDLG belonging to to &6(?), &6(?), then: then: Either Eitherits itsuhgxflqj FXUYHV FXUYHVare are VWDQGDUG, Or Orr is isuljlg.

40 &RQFOXVLRQ &RQMXJDF\ VHDUFK SUREOHP LQEUDLG JURXSV For SHULRGLF EUDLGV, we know a polynomial solution (Birman, Gebhardt, GM) For UHGXFLEOHEUDLGV, either one easily finds the reducing curves, or one can restrict to the case of ULJLG EUDLGV. For SVHXGR$QRVRY braids, one can also restrict to the ULJLG case. The Themain mainopen openproblem is istoto find finda polynomial solution to tothe the conjugacy search searchproblem for foruljlg ULJLGEUDLGV,