I Know About Artin–Tits Groups

All I know about Artin–Tits groups

Or: Why type A is so much easier...

Daniel Tubbenhauer

The ﬁrst “published” braid diagram.

(Page 283 from Gauß’ handwritten notes, volume seven, 1830). ≤

Joint with David Rose April 2019 Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Let Γ be a Coxeter graph.

Artin 1925, Tits 1961++. The Artin–Tits group and its Coxeter group quotient∼ are given by∼ generators-relations:

AT(Γ) = bi bibjbi = bjbibj h || · · · {z } |··· {z }i mij factors mij factors

2 W(Γ) = σi σi = 1, σiσjσi = σjσiσj h | ···| {z } |··· {z }i mij factors mij factors

Artin–Tits groups generalize classical braid groups, Coxeter groups generalize polyhedron groups.

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6 Maybe some categorical considerations help? In particular, what can Artin–Tits groups tell you about ﬂavor two?

My failure. What I would like to understand, but I do not.

Artin–Tits groups come in four main ﬂavors. Question: Why are these special? What happens in general type?

Artin–Tits (braid) groups

Flavor one. Finite helps helps Flavor three. Map- and aﬃne types ping class groups

Example Example Many open problems, e.g. the word problem.

Flavor two. Con- ? Flavor four. Right ﬁguration spaces helps helps angled groups

Example Example Vanilla ﬂa- vor. ?????.

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 My failure. What I would like to understand, but I do not.

Artin–Tits groups come in four main ﬂavors. Question: Why are these special? What happens in general type?

Artin–Tits (braid) groups

Flavor one. Finite helps helps Flavor three. Map- and aﬃne types ping class groups

Example Example Many open problems, e.g. the word problem.

Flavor two. Con- ? Flavor four. Right ﬁguration spaces helps helps angled groups

Example Example Vanilla ﬂa- vor. ?????.

Maybe some categorical considerations help? In particular, what can Artin–Tits groups tell you about ﬂavor two?

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 Faithfulness? Theorem (handlebodyFaithfulness? faithfulness). Rouquier’s action is known to be faithful in quite a few cases: For all g, n, Rouquier’sfinite type action (Khovanov–Seidel,gives rise Brav–Thomas), to a family of faithful actions The Hecke action is known to be faithfulJ−K in very few cases, e.g. for Γ of rank 1, 2. affine type A (Gadbled–Thiel–Wagner), affine type C (handlebody). But there is “no way”b toq prove this in general. Thereℬ mightr(g, n) bey hope( to prove(Γ)), b this inb general.M. K S 7→ J K Example (seems to work). Hecke distinguishes the braids where Burau failed: Example (the whole point). Zigzag already distinguishes braids: Theorem (handlebody HOMFLYPT homology).

This action extends to a HOMFLYPT invariant for handlebody links. Mnemonic:

M M k M M k

b = & b M = & b = J K J Kℋ2 M M k M M k

Please stop!

Let Γ be a Coxeter graph. The following commuting diagram exists in any type:

AT(Γ) AT(Γ) [ ] J−K decat. − b( q(Γ)) q(Γ) K S H decat. b( q(Γ)) q(Γ) K Z B

Question. How does this help to study Artin–Tits groups?

Here (killing idempotents for the last row): q b q I Hecke algebra (Γ), homotopy category of Soergel bimodules ( (Γ)). H K S I Hecke action [ ], Rouquier complex . − q J−K I Burau representation (Γ), homotopy category of representations of zigzag algebras b( q(Γ)). B K Z

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 4 / 6 Faithfulness? Theorem (handlebody faithfulness). Rouquier’s action is known to be faithful in quite a few cases: For all g, n, Rouquier’sfinite type action (Khovanov–Seidel,gives rise Brav–Thomas), to a family of faithful actions J−K affine type A (Gadbled–Thiel–Wagner), affine type C (handlebody). b q Thereℬ mightr(g, n) bey hope( to prove(Γ)), b this inb general.M. K S 7→ J K Example (the whole point). Zigzag already distinguishes braids: Theorem (handlebody HOMFLYPT homology).

This action extends to a HOMFLYPT invariant for handlebody links. Mnemonic:

M M k M M k

b = & b M = & b = J K J Kℋ2 M M k M M k

Please stop!

Let Γ be a Coxeter graph. The following commuting diagram exists in any type:

AT(Γ) AT(Γ) Faithfulness? [ ] J−K decat. − The Hecke action is knownb to( beq(Γ)) faithful in very fewq cases,(Γ) e.g. for Γ of rank 1, 2. But thereK isS “no way” to prove thisH in general. decat. Example (seems to work).b( Heckeq(Γ)) distinguishes theq(Γ) braids where Burau failed: K Z B

Question. How does this help to study Artin–Tits groups?

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 4 / 6 Theorem (handlebodyFaithfulness? faithfulness). For all g, n, Rouquier’s action gives rise to a family of faithful actions The Hecke action is known to be faithfulJ−K in very few cases, e.g. for Γ of rank 1, 2. But there is “no way”b toq prove this in general. ℬr(g, n) y ( (Γ)), b b M. K S 7→ J K Example (seems to work). Hecke distinguishes the braids where Burau failed:

Theorem (handlebody HOMFLYPT homology).

This action extends to a HOMFLYPT invariant for handlebody links. Mnemonic:

M M k M M k

b = & b M = & b = J K J Kℋ2 M M k M M k

Please stop!

Let Γ be a Coxeter graph. The following commuting diagram exists in any type:

AT(Γ)Faithfulness? AT(Γ)

Rouquier’s action is known to be faithful in quite[ ] a few cases: J−K decat. − finite typeb( (Khovanov–Seidel,q(Γ)) Brav–Thomas),q(Γ) affine type A (Gadbled–Thiel–Wagner),K S affineH type C (handlebody). There might be hope to prove this in general. decat. b( q(Γ)) q(Γ) Example (the wholeK Z point). Zigzag alreadyB distinguishes braids:

Question. How does this help to study Artin–Tits groups?

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 4 / 6 Faithfulness? Faithfulness? Rouquier’s action is known to be faithful in quite a few cases: The Hecke action isfinite known type to (Khovanov–Seidel, be faithful in very Brav–Thomas), few cases, e.g. for Γ of rank 1, 2. affine typeBut A (Gadbled–Thiel–Wagner), there is “no way” to prove affine this type in general. C (handlebody). There might be hope to prove this in general. Example (seems to work). Hecke distinguishes the braids where Burau failed: Example (the whole point). Zigzag already distinguishes braids:

Let Γ be a Coxeter graph. The following commuting diagram exists in any type:

AT(Γ) AT(Γ) Theorem (handlebody faithfulness). [ ] J−K − For all g, n, Rouquier’s actionq givesdecat. rise to aq family of faithful actions b( (Γ))− (Γ) K J K H S b q ℬr(g, n) y ( (Γ)), b b . K 7→ M Sdecat. J K b( q(Γ)) q(Γ) K Z B Theorem (handlebody HOMFLYPT homology). Question. How does this help to study Artin–Tits groups? This action extends to a HOMFLYPT invariant for handlebody links. Mnemonic: Here (killing idempotents for the last row): q M M k b q I Hecke algebra (Γ), homotopy category of Soergel bimodules ( (Γ)). H M M k K S I Hecke action [ ], Rouquier complex . b = − &q b = J−K & b = I Burau representation (Γ),M homotopy category ofℋ representations2 of zigzag b q B J K J K algebras ( (Γ)). M M k K Z M M k

Please stop!

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 4 / 6 Example (type A).

Braid cobordismsTheorem are movies (handlebody of braids. functoriality). E.g. some generators are group monoid birth For all g, n, Rouquier’s action gives rise to a family of functorial actions J−K b q ℬcob(g, n) y ( (Γ)), b b M , bcob isotopybcob M. , & , K & 7→, J K : 7→ J K S −−−→ −−−−−−→ (ℬcob(g, n) is the 2-category of handlebody braid cobordisms.)

invertible invertible non-invertible Invertible ones encode isotopies, non-invertible ones “more interesting” topology. Final observation. Theorem (well-known?). In all (non-trivial) cases I know “faithful functorial”. The Rouquier complex⇔ is functorial in types A,B = C and aﬃne C. Is there a general statement?

b q Rouquier 2004. The 2-braid group (Γ) is im( ) ( s (Γ)). ∼ AT J−K ⊂ K S If you have a configuration space picture for AT(Γ) one can define the category of braid cobordisms ℬcob(Γ) in four space. Fact (well-known?). For Γ of type A, B = C or affine type C we have

(Γ) = inv(ℬ (Γ)). AT cob Corollary (strictness). We have a categorical action

b q inv(ℬcob(g, n)) y ( (Γ)), b b , bcob bcob . K S 7→ J K 7→ J K

Question (functoriality). Can we lift to a categorical action J−K b q ℬcob(g, n) y ( (Γ))? K S

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 5 / 6 Theorem (handlebody functoriality).

For all g, n, Rouquier’s action gives rise to a family of functorial actions J−K b q ℬcob(g, n) y ( (Γ)), b b M , bcob bcob M. K S 7→ J K 7→ J K

(ℬcob(g, n) is the 2-category of handlebody braid cobordisms.)

Final observation. Theorem (well-known?). In all (non-trivial) cases I know “faithful functorial”. The Rouquier complex⇔ is functorial in types A,B = C and aﬃne C. Is there a general statement?

b q Rouquier 2004. The 2-braid group (Γ) is im( ) ( s (Γ)). ∼ ExampleAT (type A). J−K ⊂ K S

If you haveBraid a configuration cobordisms spaceare movies picture of braids. for AT(Γ) E.g.one some can generators define the are category of braid cobordisms ℬcob(Γ) in four space. group monoid birth Fact (well-known?). For Γ of type A, B = C or affine type C we have

(Γ) = inv(ℬcob(Γ)). isotopy , & , AT & , : −−−→ −−−−−−→ Corollary (strictness). We have a categorical action

b q inv(ℬcob(g, n)) y ( (Γ)), b b , bcob bcob . invertible invertible K S 7→ J Knon-invertible7→ J K Invertible ones encode isotopies, non-invertible ones “more interesting” topology. Question (functoriality). Can we lift to a categorical action J−K b q ℬcob(g, n) y ( (Γ))? K S

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 5 / 6 Theorem (handlebody functoriality).

For all g, n, Rouquier’s action gives rise to a family of functorial actions J−K b q ℬcob(g, n) y ( (Γ)), b b M , bcob bcob M. K S 7→ J K 7→ J K

(ℬcob(g, n) is the 2-category of handlebody braid cobordisms.)

Final observation.

In all (non-trivial) cases I know “faithful functorial”. ⇔ Is there a general statement?

b q Rouquier 2004. The 2-braid group (Γ) is im( ) ( s (Γ)). ∼ ExampleAT (type A). J−K ⊂ K S

(Γ) = inv(ℬcob(Γ)). isotopy , & , AT & , : −−−→ −−−−−−→ Corollary (strictness). We have a categorical action

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 5 / 6 Example (type A).

Braid cobordisms are movies of braids. E.g. some generators are

group monoid birth

isotopy , & , & , : −−−→ −−−−−−→

b q Rouquier 2004. The 2-braid group (Γ) is im( ) ( s (Γ)). ∼ AT J−K ⊂ K S If you have a configuration space picture for AT(Γ) one can define the category of Theorem (handlebody functoriality). braid cobordisms ℬcob(Γ) in four space. FactFor (well-known?). all g, n, Rouquier’sFor Γ actionof type A,gives B = riseC to or a affine family type of functorial C we have actions J−K b (Γ)q = inv(ℬcob(Γ)). ℬcob(g, n) y ( (Γ)), b b M , bcob bcob M. ATK S 7→ J K 7→ J K Corollary (strictness). We have a categorical action (ℬcob(g, n) is the 2-category of handlebody braid cobordisms.) b q inv(ℬcob(g, n)) y ( (Γ)), b b , bcob bcob . K S 7→ J K 7→ J K

Question (functoriality). Can we lift to a categorical action J−K b q ℬcob(g, n) y ( (Γ))? K S

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 5 / 6 Example (type A).

Braid cobordisms are movies of braids. E.g. some generators are

group monoid birth

isotopy , & , & , : −−−→ −−−−−−→

invertible invertible non-invertible Invertible ones encode isotopies, non-invertible ones “more interesting” topology.

Theorem (well-known?).

The Rouquier complex is functorial in types A,B = C and aﬃne C.

Question (functoriality). Can weFinal lift observation.to a categorical action J−K In all (non-trivial) casesb q I know ℬcob(g, n) y ( (Γ))? “faithful functorial”.K ⇔ S Is there a general statement?

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 5 / 6 Thanks for your attention!

Example. In the dihedral case these (un)foldings correspond to bicolorings: Proof?

7 and 8 and 9 etc. Uses root combinatorics of ADE diagrams I2(7) A6 and theI2(8) fact that eachA7 AT(Γ) ofI2 ﬁnite(9) A8 type can be embeddedFact. in types ADE. This gives AT(I2(n)) , AT(Γ) Fact. The symmetries are given by exchanging ﬂags. Example. Type B “unfolds”→ into type A: ⇔ Γ = ADE for n =Coxeter number. -1 -2 4 0 1 2 0 Example (SAGE; n = 9). LKB says it is true: 1 2

b0 and b1 and b2 7→ 7→ 7→ But there is also a diﬀerent way, discussed later.

Example.

Fact (type A embedding). From this perspective the type A embedding is just shrinkingNote. holes to points!

Br(g, n) is a subgroup of the usual braid group ℬr(g+n). For the proof it is crucial that D 2 and the boundary points of the braids g • are only deﬁned up to isotopy, e.g.

D 2 D 2 = = 3 • shrink 3 The “full7→ wrap”. • • • ∼

A visualization= exercise. one can always “conjugate cores to the left”. ⇒ This is useful to deﬁne ℬr(g, ). ∞

This case is strange – it only arises under conjugation: 1 1 n 2 1 2 1 n ... Theorem (Lambropoulou 1993). Example. ∼ ... b b 7→ For any link l in the genus g handlebody ℋg there is a ... type B˜n n 1 braid in ℬr(g, ) whose (correct!) closure is isotopic to l. ... − ∞ 4 1 1 n 2 1 2 1 n 0 1 2 ... n 2 By a miracle, one can avoid the special− relation This relationn Fact. involves three = ! 3 players and inverses. ℋg is given by a complement in the 3-sphere S by an open tubular 2 neighborhood of the embedded graph obtained D3 Bad! by gluing g + 1 unknotted “core” edges to two vertices. “Z/ Z” Z/2Z Currently, not much∞ seems to be• known,• • but I think the same story works.

S 3 S 3 S 3 However, this is where it seems to! end, e.g. genus g = 3 wants to be 01 1 n ∞ 2order order 2 ∞ 0 1∞ 2 ... n 1 n 1 − n ∞ 3 the 3-ball ℋ0 = D a torus ℋ1 ℋ2 03 Back In some sense this can not work; remember Tits conjecture.

My failure. What I would like to understand, but I do not. Lawrence 1989, Krammer 2000, Bigelow 2000 (Cohen–Wales 2000, Digne 2000).∼ Let Γ be of finite∼ type. There exists∼ a faithful action of AT(Γ)∼ ∼ Artin–Tits groups come in four main flavors. on a finite-dimensional vector space. Question: Why are these special? What happens in general type? Upshot: One can ask a computer program questions about braids! Artin–Tits (braid) groups Figure: The Coxeter graphs of finite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

Flavor one. Finite helps helps Flavor three. Map- and affine types ping class groups Examples. This gives a generator-relation presentation. Example Example Type A3 ! tetrahedron ! symmetric group S4. 3 Many open Type B3 ! Andcube/octahedron the braid relation! measuresWeyl group the angle(Z/2Z between) n S3 hyperplanes.. problems, e.g. the word problem. Type H3 ! dodecahedron/icosahedron ! exceptional Coxeter group. For I2(4) we have a 4-gon: Idea (Coxeter 1934++). Flavor two. Con- ? Flavor four. Right Fix a flag F . ∼ figuration spaces helps helps angled groups Fix a hyperplane H0 permuting Example Example Vanilla fla- the adjacent 0-cells of F . vor. ?????. 4 Fix a hyperplane H1 permuting • • • the adjacent 1-cells of F , etc. cos(π/4) Figure: SAGE in action: The Burau (TL) action is not faithful, the LKB is.

Maybe some categorical considerations help? Write a vertex i for each Hi. In particular, what can Artin–Tits groups tell you about ﬂavor two? Back Connect i, j by an n-edge for Hi,Hj having angle cos(π/n). Back Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6

Proof? Let Br(g, n) be the group defined as follows. Crisp–Paris 2000 (Tits conjecture). For all m > 1, the subgroup Theorem (Häring-Oldenburg–Lambropoulou 2002, Vershinin 1998). The group ℬr(g, n) of braid in a g-times punctures∼ disk D 2 [0,∼1]: mEssentially:∼ Relate the problem to the mapping class (Σ) group of a surface Σ, g bi AT(Γ) is free (up to “obvious commutation”).ℳ × h i ⊂ which acts on π1(Σ, boundary) via Dehn twist. The map m Generators. Braid and twist generators Then bi , AT(Γ) ℳ(Σ) y π1(Σ, boundary) acts faithfully. In finite type thish isi a→ consequence→ of LKB; in type A it is clear: 1 g 1 i i+1 n 1 i g 1 2 n ...... Two types of braidings, the usual ones and “winding around cores”, e.g. Example. The surface Σ is built from Γ by gluing annuli Ani: ...... & ... bi ! ti ! ...... 2 2 1 g 1 i i+1 n 1 i g 1 2 n D3 [0, 1] D3 [0, 1] ∗ × 7→ × An j• Relations. Reidemeister braid relations, type C relations and special relations, e.g. An An i j: ∗ i ∗ + ∗ j ∗ = = → = & Ani • • ∗ 6 ∗ Involves three players and inverses! & •

∗ the only “obviousDehn commutation” twist along the orchid curve:no relation = & = This should have told me something:∗ ∗ I will7→ ∗ come back∗ to this later. 7→ Back 1 1 b1t2b1t2 = t2b1t2b1 (b1t2b1− )t3 = t3(b1t2b1− )

is an isomorphism of groups Br(g, n) ℬr(g, n). →

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6

The Alexander closure on ℬr(g, ) is given by merging core strands at infinity. The Alexander closure on ℬr(g, ) is given by merging core strands at infinity. cos(π/4) twice on a line: ∞ Theorem (Lambropoulou∞ 1993). ∼ 1 4 4 2 typeCurrentlyCñ : 0 known1 (to2 the best... of myn knowledge).1 n 0 For any link l in the genus g handlebody ℋg there is a − braid in ℬr(g, ) whose (correct!) closure is isotopic to l. ∞ Genus type A type C

Affineg adds= 0 genus. Considerℬr(n the) = mapAT(An 1) ∼ − ˜ ˆ g = 1 ℬr(1, n) = Z n AT(An 1) = AT(An 1) ℬr(1, n) = AT(Cn) Fact. 1 1 n ∼2 − i∼ i+1 − 1 1∼ n 2 ...... ˜ g = 2 ℬr(2, n) ∼= AT(Cn) β 1 & β & β 2 is given by a complement in the 3-sphere 3 by an open tubular g 0 3 i 0 ℋg S ≥ 7→ ... 7→ 7→ ... neighborhood of the embedded graph obtained 1 1 n 2 i i+1 1 1 n 2 And some Z/2Z-orbifolds (Z/ Z =puncture): by gluing g + 1 unknotted “core” edges to two vertices. ∞ Genus type D type B S 3 S 3 S 3 = Allcockg = 01999. This gives an isomorphism of groups AT(C˜ ) ∼ ℬr(2, n). ∼ n −→ g = 1 ℬr(1, n) /2 = AT(Dn) ℬr(1, n) / = AT(Bn) Z Z ∼ Z ∞Z ∼ ˜ ˜ wrong closure wrong closure g = 2 ℬr(2, n) /2 /2 = AT(Dn) ℬr(2, n) / /2 = AT(Bn) correct closure correct closure Z Z×Z Z ∼ Z ∞Z×Z Z ∼ g 3 This is different from the classical Alexander closure. This is different from the classical Alexander closure. ≥ 3 (For orbifolds “genus” is just an analogy.) the 3-ball ℋ0 = D a torus ℋ1 ℋ2

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6

There is still much to do...

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 There is still much to do...

Example. In the dihedral case these (un)foldings correspond to bicolorings: Proof?

b0 and b1 and b2 7→ 7→ 7→ But there is also a diﬀerent way, discussed later.

Example.

Fact (type A embedding). From this perspective the type A embedding is just shrinkingNote. holes to points!

Br(g, n) is a subgroup of the usual braid group ℬr(g+n). For the proof it is crucial that D 2 and the boundary points of the braids g • are only deﬁned up to isotopy, e.g.

D 2 D 2 = = 3 • shrink 3 The “full7→ wrap”. • • • ∼

A visualization= exercise. one can always “conjugate cores to the left”. ⇒ This is useful to deﬁne ℬr(g, ). ∞

is an isomorphism of groups Br(g, n) ℬr(g, n). →

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6

Thanks for your attention!

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 This gives a generator-relation presentation. Fact. The symmetries are given by exchanging ﬂags. And the braid relation measures the angle between hyperplanes.

Fix a ﬂag F .

Fix a hyperplane H0 permuting the adjacent 0-cells of F .

Fix a hyperplane H1 permuting the adjacent 1-cells of F , etc.

Write a vertex i for each Hi. Connect i, j by an n-edge for Hi,Hj having angle cos(π/n).

Figure: The Coxeter graphs of ﬁnite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

Examples. Type A3 ! tetrahedron ! symmetric group S4. 3 Type B3 ! cube/octahedron ! Weyl group (Z/2Z) n S3. Type H3 ! dodecahedron/icosahedron ! exceptional Coxeter group. For I2(4) we have a 4-gon: Idea (Coxeter 1934++). ∼

Back This gives a generator-relation presentation.

And the braid relation measures the angle between hyperplanes.

Fix a hyperplane H0 permuting the adjacent 0-cells of F .

Fix a hyperplane H1 permuting the adjacent 1-cells of F , etc.

Write a vertex i for each Hi. Connect i, j by an n-edge for Hi,Hj having angle cos(π/n).

Figure: The Coxeter graphs of ﬁnite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

Examples. Type A3 ! tetrahedron ! symmetric group S4. Fact. The symmetries are given by exchanging3 ﬂags. Type B3 ! cube/octahedron ! Weyl group (Z/2Z) n S3. Type H3 ! dodecahedron/icosahedron ! exceptional Coxeter group. For I2(4) we have a 4-gon: Idea (Coxeter 1934++). Fix a ﬂag F . ∼ •

Back This gives a generator-relation presentation. Fact. The symmetries are given by exchanging ﬂags. And the braid relation measures the angle between hyperplanes.

Fix a hyperplane H1 permuting the adjacent 1-cells of F , etc.

Write a vertex i for each Hi. Connect i, j by an n-edge for Hi,Hj having angle cos(π/n).

Figure: The Coxeter graphs of ﬁnite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

•

Back This gives a generator-relation presentation. Fact. The symmetries are given by exchanging ﬂags. And the braid relation measures the angle between hyperplanes.

Write a vertex i for each Hi. Connect i, j by an n-edge for Hi,Hj having angle cos(π/n).

Figure: The Coxeter graphs of ﬁnite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

Fix a hyperplane H1 permuting the adjacent 1-cells of F , etc. •

Back This gives a generator-relation presentation. Fact. The symmetries are given by exchanging ﬂags. And the braid relation measures the angle between hyperplanes.

Connect i, j by an n-edge for Hi,Hj having angle cos(π/n).

Figure: The Coxeter graphs of ﬁnite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

Fix a hyperplane H1 permuting the adjacent 1-cells of F , etc. • • •

Write a vertex i for each Hi.

Back Fact. The symmetries are given by exchanging ﬂags.

Figure: The Coxeter graphs of ﬁnite type. (Picture from https://en.wikipedia.org/wiki/Coxeter_group.)

Examples. This gives a generator-relation presentation. Type A3 ! tetrahedron ! symmetric group S4. 3 Type B3 ! Andcube/octahedron the braid relation! measuresWeyl group the angle(Z/2Z between) n S3 hyperplanes.. Type H3 ! dodecahedron/icosahedron ! exceptional Coxeter group. For I2(4) we have a 4-gon: Idea (Coxeter 1934++). Fix a ﬂag F . ∼ Fix a hyperplane H0 permuting the adjacent 0-cells of F . 4 Fix a hyperplane H1 permuting • • • the adjacent 1-cells of F , etc. cos(π/4)

Write a vertex i for each Hi. Connect i, j by an n-edge for Hi,Hj having angle cos(π/n). Back Example. In the dihedral case these (un)foldings correspond to bicolorings: Proof?

7 and 8 and 9 etc. Uses root combinatorics of ADE diagrams I2(7) A6 and theI2(8) fact that eachA7 AT(Γ) ofI2 ﬁnite(9) A8 type can be embeddedFact. in types ADE. This gives AT(I2(n)) , AT(Γ) Example. Type B “unfolds”→ into type A: ⇔ Γ = ADE for n =Coxeter number. -1 -2 4 0 1 2 0 Example (SAGE; n = 9). LKB says it is true: 1 2

b0 and b1 and b2 7→ 7→ 7→ But there is also a diﬀerent way, discussed later.

Lawrence 1989, Krammer 2000, Bigelow 2000 (Cohen–Wales 2000, Digne 2000).∼ Let Γ be of ﬁnite∼ type. There exists∼ a faithful action of AT(Γ)∼ on a ﬁnite-dimensional∼ vector space.

Upshot: One can ask a computer program questions about braids!

Figure: SAGE in action: The Burau (TL) action is not faithful, the LKB is.

Back Example. In the dihedral case these (un)foldings correspond to bicolorings:

7 and 8 and 9 etc.

I2(7) A6 I2(8) A7 I2(9) A8 Fact. This gives AT(I2(n)) , AT(Γ) → ⇔ Γ = ADE for n =Coxeter number.

Example (SAGE; n = 9). LKB says it is true:

Lawrence 1989, Krammer 2000, Bigelow 2000 (Cohen–Wales 2000, Digne 2000).∼ Let Γ be of ﬁnite∼ type. There exists∼ a faithful action of AT(Γ)∼ on a ﬁnite-dimensional∼ vector space.

Proof? Upshot: One can ask a computer program questions about braids! Uses root combinatorics of ADE diagrams and the fact that each AT(Γ) of ﬁnite type can be embedded in types ADE.

Example. Type B “unfolds” into type A:

-1 -2 4 0 1 2 0

1 2 b0 and b1 and b2 7→ 7→ 7→ But there is also a diﬀerent way, discussed later. Figure: SAGE in action: The Burau (TL) action is not faithful, the LKB is.

Back Proof?

Uses root combinatorics of ADE diagrams and the fact that each AT(Γ) of ﬁnite type can be embedded in types ADE.

Example. Type B “unfolds” into type A:

-1 -2 4 0 1 2 0 Example (SAGE; n = 9). LKB says it is true: 1 2

b0 and b1 and b2 7→ 7→ 7→ But there is also a diﬀerent way, discussed later.

Lawrence 1989, Krammer 2000, Bigelow 2000 (Cohen–Wales 2000, Digne 2000).∼ Let Γ be of ﬁnite∼ type. There exists∼ a faithful action of AT(Γ)∼ ∼ on a ﬁnite-dimensionalExample. In the dihedral vector casespace. these (un)foldings correspond to bicolorings:

Upshot: One can ask a computer program questions about braids! 7 and 8 and 9 etc.

I2(7) A6 I2(8) A7 I2(9) A8 Fact. This gives AT(I2(n)) , AT(Γ) → ⇔ Γ = ADE for n =Coxeter number.

Figure: SAGE in action: The Burau (TL) action is not faithful, the LKB is.

Back Proof?

Uses root combinatorics of ADE diagrams and the fact that each AT(Γ) of ﬁnite type can be embedded in types ADE.

Example. Type B “unfolds” into type A:

-1 -2 4 0 1 2 0

1 2

b0 and b1 and b2 7→ 7→ 7→ But there is also a diﬀerent way, discussed later.

Upshot: One can ask a computer program questions about braids! 7 and 8 and 9 etc.

I2(7) A6 I2(8) A7 I2(9) A8 Fact. This gives AT(I2(n)) , AT(Γ) → ⇔ Γ = ADE for n =Coxeter number.

Example (SAGE; n = 9). LKB says it is true:

Figure: SAGE in action: The Burau (TL) action is not faithful, the LKB is.

Back Proof?

Essentially: Relate the problem to the mapping class ℳ(Σ) group of a surface Σ, which acts on π1(Σ, boundary) via Dehn twist. m Then bi , AT(Γ) ℳ(Σ) y π1(Σ, boundary) acts faithfully. h i → →

Example. The surface Σ is built from Γ by gluing annuli Ani:

∗ An j• An An i j: ∗ i ∗ + ∗ j ∗ = An → • • ∗ i ∗ •

∗ Dehn twist along the orchid curve:

∗ ∗ 7→ ∗ ∗

Crisp–Paris 2000 (Tits conjecture). For all m > 1, the subgroup ∼ bm AT(Γ) is free (up to “obvious commutation”). h i i ⊂ In ﬁnite type this is a consequence of LKB; in type A it is clear:

& = = 6

the only “obvious commutation” no relation

This should have told me something: I will come back to this later.

Back Proof? Crisp–Paris 2000 (Tits conjecture). For all m > 1, the subgroup mEssentially:∼ Relate the problem to the mapping class (Σ) group of a surface Σ, bi AT(Γ) is free (up to “obvious commutation”).ℳ h i ⊂ which acts on π1(Σ, boundary) via Dehn twist. m Then bi , AT(Γ) ℳ(Σ) y π1(Σ, boundary) acts faithfully. In ﬁnite type thish isi a→ consequence→ of LKB; in type A it is clear:

Example. The surface Σ is built from Γ by gluing annuli Ani:

∗ An j• An An i j: ∗ i ∗ + ∗ j ∗ = = → = & Ani • • ∗ 6 ∗ •

∗ the only “obviousDehn commutation” twist along the orchid curve:no relation

This should have told me something:∗ ∗ I will7→ ∗ come back∗ to this later.

Back Proof?

This works in general: For each right-angled Γ there exists a Γ0 such that i W(Γ0) ∼= AT(Γ) o (Z/2Z) . Corollary.

Tits’ reﬂection representation gives a faithful action on a ﬁnite-dimensional R-vector space.

This is the only case where I know that the Artin–Tits group embeds into a Coxeter group.

Recall. Right-angled means m 2, . ij ∈ { ∞} Fact (well-known?). Let Γ be of right-angled type. There exists a faithful action of AT(Γ) on a ﬁnite-dimensional R-vector space.

Example. Γ = I ( ), the inﬁnite dihedral group. 2 ∞

∞

∞ ∞ ∞ I ( ) Γ 2 ∞ 0 Deﬁne a map AT(Γ) W(Γ0),s ss,t tt. → 7→ 7→ Crazy fact: This is an embedding, and actually

2 W(Γ0) ∼= AT(Γ) o (Z/2Z) . Thus, via Tits’ reﬂection representation, it follows that AT(Γ) is linear.

Back Recall. Right-angled means m 2, . ij ∈ { ∞} Fact (well-known?). Let Γ be of right-angled type. There exists a faithful action of AT(Γ) on a ﬁnite-dimensional R-vectorProof? space.

This works in general: Example. Γ = I2( ), the inﬁnite dihedral group. For∞ each right-angled Γ there exists a Γ0 such that i W(Γ0) ∼= AT(Γ) o (Z/2Z) . ∞

∞ Corollary. ∞ ∞ I ( ) Γ 2 ∞ 0 Tits’ reflection representation gives a faithful action Define a map on a finite-dimensional R-vector space. AT(Γ) W(Γ0),s ss,t tt. This is the→ only case where7→ I know7→ that Crazy fact: This isthe an Artin–Tits embedding, group and embeds actually into a Coxeter group. 2 W(Γ0) ∼= AT(Γ) o (Z/2Z) . Thus, via Tits’ reflection representation, it follows that AT(Γ) is linear.

Back Example.

Fact (type A embedding).

Br(g, n) is a subgroup of the usual braid group ℬr(g+n).

= = The “full7→ wrap”.

A visualization= exercise.

Let Br(g, n) be the group deﬁned as follows.

Generators. Braid and twist generators

1 g 1 i i+1 n 1 i g 1 2 n ...... & ... bi ! ti ! ...... 1 g 1 i i+1 n 1 i g 1 2 n Relations. Reidemeister braid relations, type C relations and special relations, e.g. Involves three players and inverses!

= & =

1 1 b1t2b1t2 = t2b1t2b1 (b1t2b1− )t3 = t3(b1t2b1− )

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Fact (type A embedding).

Br(g, n) is a subgroup of the usual braid group ℬr(g+n).

= = The “full7→ wrap”.

A visualization= exercise.

Let Br(g, n) be the group deﬁned as follows.

Example. Generators. Braid and twist generators

= & =

1 1 b1t2b1t2 = t2b1t2b1 (b1t2b1− )t3 = t3(b1t2b1− )

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Fact (type A embedding).

Br(g, n) is a subgroup of the usual braid group ℬr(g+n).

= = 7→

A visualization exercise.

Let Br(g, n) be the group deﬁned as follows.

Example. Generators. Braid and twist generators

= & = =

1 1 b1t2b1t2 = t2b1t2b1 (b1t2b1− )t3 = t3(b1t2b1− )

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Example.

The “full wrap”.

Let Br(g, n) be the group deﬁned as follows.

Generators. Braid and twist generators

1 g 1 i i+1 n 1 i g 1 2 n Fact (type A embedding)...... & ... bi ! ti ! ...... Br(g, n) is1 a subgroupg 1 i i+1 of then usual braid group1 i ℬgr(g1+n2). n Relations. Reidemeister braid relations, type C relations and special relations, e.g. Involves three players and inverses! = = 7→

= & = A visualization exercise.

1 1 b1t2b1t2 = t2b1t2b1 (b1t2b1− )t3 = t3(b1t2b1− )

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 1 / 6 Theorem (H¨aring-Oldenburg–Lambropoulou 2002, Vershinin 1998). ∼ ∼ The map

From this perspective the type A embedding is just shrinkingNote. holes to points!

For the proof it is crucial that 7→D 2 and the boundary points of the braids g • are only deﬁned up to isotopy, e.g.

D 2 D 2 3 • shrink 3 • • • ∼

one can always “conjugate cores to the left”. ⇒ 7→ This is useful to deﬁne ℬr(g, ). ∞

is an isomorphism of groups Br(g, n) ℬr(g, n). →

The group ℬr(g, n) of braid in a g-times punctures disk D 2 [0, 1]: g ×

Two types of braidings, the usual ones and “winding around cores”, e.g.

D 2 [0, 1] D 2 [0, 1] 3 × 3 ×

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6 From this perspective the type A embedding is just shrinkingNote. holes to points!

For the proof it is crucial that D 2 and the boundary points of the braids g • are only deﬁned up to isotopy, e.g.

D 2 D 2 3 • shrink 3 • • • ∼

one can always “conjugate cores to the left”. ⇒ This is useful to deﬁne ℬr(g, ). ∞

Theorem (H¨aring-Oldenburg–Lambropoulou 2002, Vershinin 1998). The group ℬr(g, n) of braid in a g-times punctures∼ disk D 2 [0,∼1]: g × The map

Two types of braidings, the usual ones and “winding around cores”, e.g.

D 2 [0, 1] D 2 [0, 1] 3 × 7→ 3 ×

7→

is an isomorphism of groups Br(g, n) ℬr(g, n). →

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6 Theorem (H¨aring-Oldenburg–Lambropoulou 2002, Vershinin 1998). ∼ ∼ The map

Note.

For the proof it is crucial that 7→D 2 and the boundary points of the braids g • are only deﬁned up to isotopy, e.g.

D 2 D 2 3 • 3 • • • ∼

one can always “conjugate cores to the left”. ⇒ 7→ This is useful to deﬁne ℬr(g, ). ∞

is an isomorphism of groups Br(g, n) ℬr(g, n). →

The group ℬr(g, n) of braid in a g-times punctures disk D 2 [0, 1]: g ×

Two types of braidings,From thisthe perspectiveusual ones theand type “winding A embedding around cores”, e.g. is just shrinking holes to points! D 2 [0, 1] D 2 [0, 1] 3 × 3 ×

& shrink

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6 Theorem (H¨aring-Oldenburg–Lambropoulou 2002, Vershinin 1998). ∼ ∼ The map

From this perspective the type A embedding is just shrinking holes to points!

7→

shrink

7→

is an isomorphism of groups Br(g, n) ℬr(g, n). →

The group ℬr(g, n) of braid in a g-times punctures disk D 2 [0, 1]: g ×

Two types of braidings, the usual ones and “winding around cores”, e.g. Note.

D 2 [0, 1] D 2 [0, 1] For the proof it is crucial3 × that D 2 and the boundary points3 × of the braids g • are only deﬁned up to isotopy, e.g.

D 2 & D 2 3 • 3 • • • ∼

one can always “conjugate cores to the left”. ⇒ This is useful to deﬁne ℬr(g, ). ∞

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 2 / 6 Theorem (Lambropoulou 1993). ∼

For any link l in the genus g handlebody ℋg there is a braid in ℬr(g, ) whose (correct!) closure is isotopic to l. ∞

Fact.

3 ℋg is given by a complement in the 3-sphere S by an open tubular neighborhood of the embedded graph obtained by gluing g + 1 unknotted “core” edges to two vertices.

S 3 S 3 S 3

3 the 3-ball ℋ0 = D a torus ℋ1 ℋ2

The Alexander closure on ℬr(g, ) is given by merging core strands at inﬁnity. ∞

wrong closure correct closure This is diﬀerent from the classical Alexander closure.

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 Fact.

3 ℋg is given by a complement in the 3-sphere S by an open tubular neighborhood of the embedded graph obtained by gluing g + 1 unknotted “core” edges to two vertices.

S 3 S 3 S 3

3 the 3-ball ℋ0 = D a torus ℋ1 ℋ2

The Alexander closure on ℬr(g, ) is given by merging core strands at inﬁnity. Theorem (Lambropoulou∞ 1993). ∼

For any link l in the genus g handlebody ℋg there is a braid in ℬr(g, ) whose (correct!) closure is isotopic to l. ∞

wrong closure correct closure This is diﬀerent from the classical Alexander closure.

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 The Alexander closure on ℬr(g, ) is given by merging core strands at inﬁnity. Theorem (Lambropoulou∞ 1993). ∼

For any link l in the genus g handlebody ℋg there is a braid in ℬr(g, ) whose (correct!) closure is isotopic to l. ∞

Fact.

3 ℋg is given by a complement in the 3-sphere S by an open tubular neighborhood of the embedded graph obtained by gluing g + 1 unknotted “core” edges to two vertices.

S 3 S 3 S 3

wrong closure correct closure This is diﬀerent from the classical Alexander closure.

3 the 3-ball ℋ0 = D a torus ℋ1 ℋ2

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 3 / 6 cos(π/3) on a line:

type An 1 : 1 2 ... n 2 n 1 − − −

The classical case. Consider the map

1 i i+1 n

β ...... braid rel.: = i 7→ 1 i i+1 n

∼= Artin 1925. This gives an isomorphism of groups AT(An 1) ℬr(0, n). ∼ − −→

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 4 / 6 cos(π/4) on a line:

type C : 0 4 1 2 ... n 1 n n −

The semi-classical case. Consider the map

1 2 n 1 i i+1 n

β ... & β ...... braid rel.: = 0 7→ i 7→ 1 2 n 1 i i+1 n

= Brieskorn 1973. This gives an isomorphism of groups AT(C ) ∼ ℬr(1, n). ∼ n −→

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 5 / 6 This case is strange – it only arises under conjugation: 1 1 n 2 1 2 1 n ...... Example.

Currently knownb (to the best ofb my knowledge). 7→ ... type B˜n n 1 − Genus type A ... type C 4 1 1 n 2 1 2 1 n 0 1 2 ... n 2 By a miracle, one can avoid the special− relation g = 0 ℬr(n) ∼= AT(An 1) − This relation g = 1 ℬr(1, n) = AT(A˜n 1) = AT(Aˆn 1) ℬn r(1, n) = AT(Cn) ∼ Z n − ∼ − ∼ involves three ˜ g = 2 = ! ℬr(2, n) ∼= AT(Cn) players and inverses. g 3 ≥ 2 D3 Bad! And some Z/2Z-orbifolds (Z/ Z =puncture): “Z/ Z” ∞ Z/2Z Currently,Genus not much∞ seemstype Dto be• known,• • but I think thetype same B story works. g = 0 However, this is where it seems to! end, e.g. genus g = 3 wants to be g = 1 ℬr(1, n)1 /2 = AT(Dn) ℬr(1, n) / = AT(Bn) 0Z Z ∼ Z ∞Z ∼ 1 ˜ n ˜ g = 2 ℬr(2, n) /2 /2 = AT(Dn) ℬr(2, n) / /2 = AT(Bn) Z∞ Z×Z Z ∼ Z ∞Z×Z Z ∼ 2order order 2 g 3 ∞ 0 1∞ 2 ... n 1 n ≥ 1 − n (For∞ orbifolds “genus” is just an analogy.) 03 Back In some sense this can not work; remember Tits conjecture.

cos(π/4) twice on a line:

type C˜ : 01 4 1 2 ... n 1 n 4 02 n −

Aﬃne adds genus. Consider the map

1 1 n 2 i i+1 1 1 n 2 ...... β01 & βi & β02 7→ ... 7→ 7→ ... 1 1 n 2 i i+1 1 1 n 2

= Allcock 1999. This gives an isomorphism of groups AT(C˜ ) ∼ ℬr(2, n). ∼ n −→

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 Example. Currently known (to the best of my knowledge). type B˜n n 1 − Genus 4 type A type C 0 1 2 ... n 2 g = 0 ℬr(n) = AT(An 1) − ∼ − g = 1 ℬr(1, n) = AT(A˜n 1) = AT(Aˆn 1) ℬn r(1, n) = AT(Cn) ∼ Z n − ∼ − ∼ ˜ g = 2 ! ℬr(2, n) ∼= AT(Cn) g 3 ≥ 2 D3 And some Z/2Z-orbifolds (Z/ Z =puncture): “Z/ Z” ∞ Z/2Z Currently,Genus not much∞ seemstype Dto be• known,• • but I think thetype same B story works. g = 0 However, this is where it seems to! end, e.g. genus g = 3 wants to be g = 1 ℬr(1, n)1 /2 = AT(Dn) ℬr(1, n) / = AT(Bn) 0Z Z ∼ Z ∞Z ∼ 1 ˜ n ˜ g = 2 ℬr(2, n) /2 /2 = AT(Dn) ℬr(2, n) / /2 = AT(Bn) Z∞ Z×Z Z ∼ Z ∞Z×Z Z ∼ 2order order 2 g 3 ∞ 0 1∞ 2 ... n 1 n ≥ 1 − n (For∞ orbifolds “genus” is just an analogy.) 03 Back In some sense this can not work; remember Tits conjecture.

This case is strange – it only arises under conjugation: 1 1 n 2 1 2 1 n cos(π/4) twice on a line: ...... 1 4 4 2 type C˜n : 0 1b 2 ... b n 1 n 0 ... 7→ − ... 1 1 n 2 1 2 1 n Aﬃne adds genus.ByConsider a miracle, the one map can avoid the special relation This relation 1 1 n 2 i i+1 involves three 1 1 n 2 ... = ... β01 & βi players& andβ02 inverses. 7→ ... 7→ 7→ ... 1 1 n 2 i i+1 Bad! 1 1 n 2

= Allcock 1999. This gives an isomorphism of groups AT(C˜ ) ∼ ℬr(2, n). ∼ n −→

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 Example. Currently known (to the best of my knowledge). type B˜n n 1 − Genus 4 type A type C 0 1 2 ... n 2 g = 0 ℬr(n) = AT(An 1) − ∼ − g = 1 ℬr(1, n) = AT(A˜n 1) = AT(Aˆn 1) ℬn r(1, n) = AT(Cn) ∼ Z n − ∼ − ∼ ˜ g = 2 ! ℬr(2, n) ∼= AT(Cn) g 3 ≥ 2 D3 And some Z/2Z-orbifolds (Z/ Z =puncture): “Z/ Z” ∞ Z/2Z Genus ∞ type D • • • type B g = 0 However, this is where it seems to! end, e.g. genus g = 3 wants to be g = 1 ℬr(1, n)1 /2 = AT(Dn) ℬr(1, n) / = AT(Bn) 0Z Z ∼ Z ∞Z ∼ 1 ˜ n ˜ g = 2 ℬr(2, n) /2 /2 = AT(Dn) ℬr(2, n) / /2 = AT(Bn) Z∞ Z×Z Z ∼ Z ∞Z×Z Z ∼ 2order order 2 g 3 ∞ 0 1∞ 2 ... n 1 n ≥ 1 − n (For∞ orbifolds “genus” is just an analogy.) 03 Back In some sense this can not work; remember Tits conjecture.

Currently, not much seems to be known, but I think the same story works. = Allcock 1999. This gives an isomorphism of groups AT(C˜ ) ∼ ℬr(2, n). ∼ n −→

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 Example. Currently known (to the best of my knowledge). type B˜n n 1 − Genus 4 type A type C 0 1 2 ... n 2 g = 0 ℬr(n) = AT(An 1) − ∼ − g = 1 ℬr(1, n) = AT(A˜n 1) = AT(Aˆn 1) ℬn r(1, n) = AT(Cn) ∼ Z n − ∼ − ∼ ˜ g = 2 ! ℬr(2, n) ∼= AT(Cn) g 3 ≥ 2 D3 And some Z/2Z-orbifolds (Z/ Z =puncture): “Z/ Z” ∞ Z/2Z Genus ∞ type D • • • type B g = 0 ! g = 1 ℬr(1, n) /2 = AT(Dn) ℬr(1, n) / = AT(Bn) Z Z ∼ Z ∞Z ∼ g = 2 r(2, n) 1 AT(D˜ ) r(2, n)n AT(B˜ ) ℬ Z/2Z Z/2Z ∼= n ℬ Z/ Z Z/2Z ∼= n order× order 2 ∞ × g 3 ∞ ≥ 1 n (For orbifolds “genus” is just an analogy.) Back

Currently, not much seems to be known, but I think the same story works. = Allcock 1999. This gives an isomorphism of groups AT(C˜ ) ∼ ℬr(2, n). However,∼ this is where it seems to end, e.g. genus g = 3nwants−→ to be 01

∞ 02 1 2 ... n 1 n ∞ − ∞ 03 In some sense this can not work; remember Tits conjecture.

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 This case is strange – it only arises under conjugation: 1 1 n 2 1 2 1 n ...... Example.

b b 7→ ... type B˜n n 1 ... − 4 1 1 n 2 1 2 1 n 0 1 2 ... n 2 By a miracle, one can avoid the special− relation This relationn involves three = ! players and inverses. 2 D3 Bad! “Z/ Z” Z/2Z Currently, not much∞ seems to be• known,• • but I think the same story works.

However, this is where it seems to! end, e.g. genus g = 3 wants to be 01 1 n ∞ 2order order 2 ∞ 0 1∞ 2 ... n 1 n 1 − n ∞ 03 Back In some sense this can not work; remember Tits conjecture.

cos(π/4) twice on a line:

typeCurrentlyC˜ : 0 known1 4 1 (to2 the best... of myn knowledge).1 n 4 02 n − Genus type A type C

Aﬃneg adds= 0 genus. Considerℬr(n the) = mapAT(An 1) ∼ − ˜ ˆ g = 1 ℬr(1, n) = Z n AT(An 1) = AT(An 1) ℬr(1, n) = AT(Cn) 1 1 n ∼2 − i∼ i+1 − 1 1∼ n 2 ...... ˜ g = 2 ℬr(2, n) ∼= AT(Cn) gβ013 & βi & β02 ≥ 7→ ... 7→ 7→ ... 1 1 n 2 i i+1 1 1 n 2 And some Z/2Z-orbifolds (Z/ Z =puncture): ∞ Genus type D type B = Allcockg = 01999. This gives an isomorphism of groups AT(C˜ ) ∼ ℬr(2, n). ∼ n −→ g = 1 ℬr(1, n) /2 = AT(Dn) ℬr(1, n) / = AT(Bn) Z Z ∼ Z ∞Z ∼ ˜ ˜ g = 2 ℬr(2, n) /2 /2 = AT(Dn) ℬr(2, n) / /2 = AT(Bn) Z Z×Z Z ∼ Z ∞Z×Z Z ∼ g 3 ≥ (For orbifolds “genus” is just an analogy.)

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6 This case is strange – it only arises under conjugation: 1 1 n 2 1 2 1 n ......

Currently knownb (to the best ofb my knowledge). ... 7→ ... Genus 1 1typen A 2 1 2 1 n type C By a miracle, one can avoid the special relation g = 0 ℬr(n) ∼= AT(An 1) − This relation g = 1 ℬr(1, n) = AT(A˜n 1) = AT(Aˆn 1) ℬr(1, n) = AT(Cn) ∼ Z n − ∼ − ∼ involves three ˜ g = 2 = ℬr(2, n) ∼= AT(Cn) players and inverses. g 3 ≥ Bad! And some Z/2Z-orbifolds (Z/ Z =puncture): ∞ Currently,Genus not much seemstype Dto be known, but I think thetype same B story works. g = 0 However, this is where it seems to end, e.g. genus g = 3 wants to be g = 1 ℬr(1, n)1 /2 = AT(Dn) ℬr(1, n) / = AT(Bn) 0Z Z ∼ Z ∞Z ∼ ˜ ˜ g = 2 ℬr(2, n) /2 /2 = AT(Dn) ℬr(2, n) / /2 = AT(Bn) Z∞ Z×Z Z ∼ Z ∞Z×Z Z ∼ g 3 02 1 2 ... n 1 n ≥ ∞ − (For∞ orbifolds “genus” is just an analogy.) 03 In some sense this can not work; remember Tits conjecture.

cos(π/4) twice on a line: Example. 1 4 4 2 type Cñ : 0 1 2 ... n 1 n 0 type Bñ − n 1 − 4 0 1 2 ... n 2 Affine adds genus. Consider the map − n 1 1 n 2 i i+1 1 1 n 2 ... ! ... β01 & βi & β02 7→ ... 7→ 2 7→ ... 1 1 n 2 Di3 i+1 1 1 n 2 “ / ” /2 Z ∞Z • • • Z Z

∼= Allcock 1999. This gives an isomorphism of groups AT(C˜n) ℬr(2, n). ∼ ! −→

1 n order order 2 ∞ 1 n

Back

Daniel Tubbenhauer All I know about Artin–Tits groups April 2019 6 / 6