2020
DOI: 10.1016/j.jsc.2019.02.012
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The BMM symmetrising trace conjecture for groups G4, G5, G6, G7, G8

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Cited by 12 publications
(10 citation statements)
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“…The assumption is known for k = 3 (see [12] or [14]). Moreover, after this work was completed, the author, together with Boura, Chlouveraki, and Karvounis, provided another proof for k = 3 and proved the assumption for k = 4 as well (see [2]).…”
Section: Preliminariesmentioning
confidence: 91%
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“…The assumption is known for k = 3 (see [12] or [14]). Moreover, after this work was completed, the author, together with Boura, Chlouveraki, and Karvounis, provided another proof for k = 3 and proved the assumption for k = 4 as well (see [2]).…”
Section: Preliminariesmentioning
confidence: 91%
“…Therefore, we prevent CHEVIE from extracting automatically these square roots, which may unavoidably be inconsistent with our expectations. We give the example a 2 − ab + b 2 = 0: gap> H:=Hecke(ComplexReflectionGroup (8),[[Mvp("x")^2,Mvp("y")^2,Mvp("z")^2,Mvp("t")^2]]);; gap> T:=CharTable(H).irreducibles;; gap> t:=List(T,i->List(i,j->Value(j,["y", -E(12)^11*Mvp("x")])));; gap> t [5]=t [1]+t [2]; true A different choice of the square root y permutes the characters φ 15 and φ 16 . This permutation depends on a specialization of the roots of the parameters, not of the parameters themselves.…”
Section: 2mentioning
confidence: 99%
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“…However, for the third generator s (as denoted in [BCCK], later in this paper we denote it by s 1 ), we only need to check that τ (s) = 0. Since s is written as a linear combination of elements of B(G 7 ) \ {1} (see [BCCK,§4.2.4]), we obtain that τ (s) = 0, as required. Schönnenbeck [Sch,Lemma 1.4.3] has shown that if Conjecture 4.1 is true, then it is enough to prove Conjecture 4.6 only "from the left".…”
Section: 1mentioning
confidence: 92%
“…), the basis of [BCCK,§4.1.1] that we used for proving the BMM symmetrising trace conjecture. Given the symmetric role played by the generators s 1 and s 2 in the presentation of H(G 4 ), replacing s 1 with s 2 and s 2 with s 1 inside B l s1 (G 4 ) and B r s1 (G 4 ) yields B l s2 (G 4 ) and B r s2 (G 4 ) respectively.…”
Section: 33mentioning
confidence: 99%