We show that if a tuple of Euclidean reflections has a finite orbit under the
Hurwitz action of the Artin braid group, then the group generated by these
reflections is finite. Humphries has published a similar statement but his
proof is irremediably flawed. At the same time as correcting his proof, our
proof is much simpler that Dubrovin and Mazocco's proof for triples of
reflections.Comment: redige le 1-9-200

Michel, Jean

English

arXiv

rédigé le 1-9-2004We show that if a tuple of Euclidean reflections has a finite orbit under the Hurwitz action of the Artin braid group, then the group generated by these reflections is finite. Humphries has published a similar statement but his proof is irremediably flawed. At the same time as correcting his proof, our proof is much simpler that Dubrovin and Mazocco's proof for triples of reflections

Hal-Diderot

Hurwitz action on tuples of Euclidean reflectionsJean MichelTo cite this version:Jean Michel. Hurwitz action on tuples of Euclidean reflections. re´dige´ le 1-9-2004. 2004.<hal-00003068>HAL Id: hal-00003068https://hal.archives-ouvertes.fr/hal-00003068Submitted on 13 Oct 2004HAL is a multi-disciplinary open accessarchive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come fromteaching and research institutions in France orabroad, or from public or private research centers.L’archive ouverte pluridisciplinaire HAL, estdestine´e au de´poˆt et a` la diffusion de documentsscientifiques de niveau recherche, publie´s ou non,e´manant des e´tablissements d’enseignement et derecherche franc¸ais ou e´trangers, des laboratoirespublics ou prive´s.ccsd-00003068, version 1 - 13 Oct 2004HURWITZ ACTION ON TUPLES OF EUCLIDEANREFLECTIONSJ. MICHELThis note was prompted by the reading of [4], which purports to show that ifan n-tuple of Euclidean reflections has a finite orbit under the Hurwitz action ofthe braid group, then the generated group is finite. I noticed that the proof givenis fatally flawed 1; however, using the argument of Vinberg given in [3], I found ashort (hopefully correct) proof which at the same time considerably simplifies thecomputational argument given in [3]. This is what I expound below. I first recallall the necessary notation and assumptions, expounding some facts in slightly moregenerality than necessary.0.1. Hurwitz action.Definition. Given a group G, we call Hurwitz action the action of the n-strandbraid group Bn with standard generators σi on Gn given byσi(s1, . . . , sn) = (s1, . . . , si−1, si+1, ssi+1i , si+2, . . . , sn).The inverse is given by σ−1i (s1, . . . , sn) = (s1, . . . , si−1,sisi+1, si, si+2, . . . , sn).Here ab is b−1ab and ba is bab−1.This action preserves the product of the n-tuple. We need to repeat some re-marks in [4]. By decreasing induction on i one sees that σi . . . σn(s1, . . . , sn) =(s1, . . . , si−1, sn, ssni , . . . , ssnn−1). In particular if γ = σ1 . . . σn−1 we get γ(s1, . . . , sn) =(sn, s1, . . . , sn−1)sn whence, if c = s1 . . . sn, we get that γn(s1, . . . , sn) = (s1, . . . , sn)c.We also deduce that given any subsequence (i1, . . . , ik) of (1, . . . , n), there existsan element of the Hurwitz orbit of (s1, . . . , sn) which begins by (si1 , . . . , sik).Assume now that the Hurwitz orbit of (s1, . . . , sn) is finite. Then some power ofγ fixes (s1, . . . , sn), thus some power of c is central in the subgroup generated bythe si. Similarly, by looking at the action of σ1 . . . σk−1 on an element of the orbitbeginning by (si1 , . . . , sik) we get that for any subsequence (i1, . . . , ik) of (1, . . . , n)there exists a power of si1 . . . sik central in the subgroup generated by (si1 , . . . , sik).0.2. Reflections. Let V be a vector space on some subfield K of C. We call com-plex reflection a finite order element s ∈ GL(V ) whose fixed points are a hyperplane.If ζ (a root of unity) is the unique non-trivial eigenvalue of s, the action of s can bewritten s(x) = x− rˇ(x)r where r ∈ V and rˇ is an element of the dual of V satisfyingDate: 10th August, 2004.1The problem is in proposition 2.3, which is essential to the main theorem (1.1) of the paper.The argument given there is basically that if a Coxeter group has a reflection representationwhere the image of the Coxeter element is of finite order, then the image of that representation isfinite. However this is false: the Cartan matrix( 2 −1 −1−1 2 −l−1 −l 2)where l2 + l =√2 defines Euclideanreflections which give a representation of an infinite rank 3 Coxeter group, such that the imageof the Coxeter group is infinite but the image of the Coxeter element is of order 8 (personalcommunication of F.Zara).12 J. MICHELrˇ(r) = 1 − ζ. These elements are unique up to multiplying r by a scalar and rˇ bythe inverse scalar. We say that r (resp. rˇ) is a root (resp. coroot) associated to s.0.3. Cartan Matrix. If (s1, . . . , sn) is a tuple of complex reflections and if ri, rˇi arecorresponding roots and coroots, we call Cartan matrix the matrix C = {rˇi(rj)}i,j .This matrix is unique up to conjugating by a diagonal matrix. Conversely, a classmodulo the action of diagonal matrices of Cartan matrices is an invariant of theGL(V )-conjugacy class of the tuple. It determines this class if it is invertible andn = dimV . Indeed, this implies that the ri form a basis of V ; and in this basis thematrix si differs from the identity matrix only on the i-th line, where the opposedof the i-th line of C has been added; thus C determines the si.If C can be chosen Hermitian (resp. symmetric), such a choice is then uniqueup to conjugating by a diagonal matrix of norm 1 elements of K (resp. of signs).If C is Hermitian (which implies that the si are of order 2), then the sesquilinearform given by tC is invariant by the si (if the si are not of order 2, but the matrixobtained by replacing all elements on the diagonal of tC by 2’s is Hermitian, thenthe latter matrix defines a sesquilinear form invariant by the si).0.4. Coxeter element. We keep the notation as above and we assume that the riform a basis of V . We recall a result of [2] on the “Coxeter” element c = s1 . . . sk.If we write C = U +V where U is upper triangular unipotent and where V is lowertriangular (with diagonal terms−ζi, thus V is also unipotent when si are of order 2),then the matrix of c in the ri basis is −U−1V (to see this write it as Us1 . . . sn = −Vand look at partial products in the left-hand side starting from the left). As U isof determinant 1, we deduce that χ(c) = det(xI + U−1V ) = det(xU + V ) whereχ(c) denotes the characteristic polynomial ; in particular det(C) = χ(c) |x=1; onealso gets that the fix-point set of c is the kernel of C, equal to the intersection ofthe reflecting hyperplanes.0.5. The main theorem. The next theorem implies the statement given in [4]([4, 1.1] considers Euclidean reflections with the ri linearly independent; if the riare chosen of the same length this implies that C is symmetric, and as C is thenthe Gram matrix of the ri it is invertible):Theorem. Let (s1, . . . , sn) be a tuple of reflections in GL(Rn) which have an as-sociated Cartan matrix symmetric and invertible. Assume in addition that theHurwitz orbit of the tuple is finite. Then the group generated by the si is finite.Proof. In the next paragraph, we just need that (s1, . . . , sn) is a tuple of complexreflections with a finite Hurwitz orbit and with the ri a basis of V .A straightforward computation shows than an element of GL(V ) commutes tothe si if and only if it acts as a scalar on the subspaces generated by {ri}i∈I whereI is a block of C (i.e., a connected component of the graph with vertices {1, . . . , n}and edges (i, j) for each pair such that either Ci,j or Cj,i is not zero). The finitenessof the Hurwitz orbit implies that for any subsequence (i1, . . . , ik) of (1, . . . , n), thereexists a power of si1 . . . sik which commutes to si1 , . . . , sik . This power acts thusas a scalar on each subspace generated by the rij in a block of the submatrix ofC determined by (i1, . . . , ik). As the determinant of each sij on this subspace isa root of unity, the scalar must be a root of unity. Thus, the restriction of eachsi1 . . . sik to the subspace < ri1 , . . . , rik > generated by the rij is of finite order.HURWITZ ACTION 3We use from now on all the assumptions of the theorem. Thus the si are order2 elements of O(C), the orthogonal group of the quadratic form defined by C.Also, χ(c) is a polynomial with real coefficients. As c is of finite order, any realroot of χ(c) is 1 or −1. This implies that χ(c) |x=1 is a nonnegative real number,and thus detC also. The same holds for any principal minor of C, since such aminor is χ(c′) |x=1 where c′ is the restriction of some si1 . . . sik to < ri1 , . . . , rik >.The quadratic form defined by C is thus positive, and as detC 6= 0 it is positivedefinite (cf. [1, §7, exercice 2]).We now digress about the Cartan matrix of two reflections s1 et s2. Such amatrix is of the form(2 ab 2). If a = 0 and b 6= 0 or a 6= 0 and b = 0 then s1s2 is ofinfinite order. Otherwise, the number ab is a complete invariant of the conjugacyclass of (s1, s2) restricted to < r1, r2 >, and s1s2 restricted to this subspace is offinite order m if and only if there exists k prime to m such that ab = 4 cos2 kpi/m.Since C is symmetric and since the restriction of sisj to < ri, rj > is of finiteorder, there exists prime integer pairs (ki,j ,mi,j) such that Ci,j = ±2 coski,jpi/mi,j .If K is the cyclotomic subfield containing the lcm(2mi,j)-th roots of unity, and ifO is the ring of integers of K, we get that all coefficients of C lie in O. It follows,if G is the group generated by the si, that in the ri basis we have G ⊂ GL(On).We now apply Vinberg’s argument as in [3, 1.4.2]. Let σ ∈ Gal(K/Q). Thenσ(C) is again positive definite: all arguments used to prove that C is positivedefinite still apply for σ(C): it is real, symmetric, invertible and the Hurwitz orbitof (σ(s1), . . . , σ(sn)) is still finite. Since G ⊂ O(C), which is compact, the entriesof the elements of G in the ri basis are of bounded norm. Since O(σ(C)) is alsocompact for any σ ∈ Gal(K/Q), we get that entries of elements of G are elements ofO all of whose complex conjugates have a bounded norm. There is a finite numberof such elements, so G is finite. References[1] N.Bourbaki, “Alge`bre”, Chap. 9, Hermann , 1959.[2] A.J.Coleman, “Killing and the Coxeter transformation of Kac-Moody algebras”, Invent.Math. 95(1989) 447–478.[3] B.Dubrovin and M.Mazzocco, “Monodromy of certain Painleve´-VI transcendents and reflec-tion groups”, Invent. Math. 141(2000), 55–147.[4] S. P. Humphries, “Finite Hurwitz braid group actions on sequences of Euclidean reflections”,J. Algebra 269 (2003), 556–588.

Hurwitz action on tuples of Euclidean reflections

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Hurwitz action on tuples of Euclidean reflections

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