2017
DOI: 10.1017/s0305004117000287
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Prenilpotent pairs in the E10 root lattice

Abstract: Tits has defined Kac-Moody groups for all root systems, over all commutative rings with unit. A central concept is the idea of a prenilpotent pair of (real) roots. In particular, writing down his group presentation explicitly would require knowing all the Weyl-group orbits of such pairs. We show that for the hyperbolic root system E 10 there are so many orbits that any attempt at direct enumeration is impractical. Namely, the number of orbits of prenilpotent pairs having inner product k grows at least as fast … Show more

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Cited by 4 publications
(9 citation statements)
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“…So q = p − (α − 2α ′ ) p · (α − 2α ′ ) /6. Calculation reveals d(p, q) = cosh −1 4 3 · 3 + km − k 2 − m 2 4 − k 2 By applying ∂/∂m to the radicand and using m ≤ 0 and k ≥ 3, one checks that the right side is decreasing as a function of m. Therefore d(p, q) is at least what one would get by plugging in 0 for m: d(p, q) ≥ cosh −1 4 3 · 3 − k 2 4 − k 2 By differentiating one shows that the right side is decreasing in k. So d(p, q) is larger than the limit as k → ∞, which is cosh −1 4/3. This is a contradiction, proving the claim.…”
Section: Lemma 6 Suppose a Is Simply Laced And Hyperbolic Then Distmentioning
confidence: 98%
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“…So q = p − (α − 2α ′ ) p · (α − 2α ′ ) /6. Calculation reveals d(p, q) = cosh −1 4 3 · 3 + km − k 2 − m 2 4 − k 2 By applying ∂/∂m to the radicand and using m ≤ 0 and k ≥ 3, one checks that the right side is decreasing as a function of m. Therefore d(p, q) is at least what one would get by plugging in 0 for m: d(p, q) ≥ cosh −1 4 3 · 3 − k 2 4 − k 2 By differentiating one shows that the right side is decreasing in k. So d(p, q) is larger than the limit as k → ∞, which is cosh −1 4/3. This is a contradiction, proving the claim.…”
Section: Lemma 6 Suppose a Is Simply Laced And Hyperbolic Then Distmentioning
confidence: 98%
“…His definition, by generators and relations, is very complicated. Even enumerating his relations in non-affine examples is difficult and in some cases impracticable ( [10], [4]).…”
Section: Introductionmentioning
confidence: 99%
“…A similar common question is whether M occurs as a primitive sublattice of L. Under the same conditions as in the previous paragraph, this reduces to the problem of building a suitable candidate for the orthogonal complement of M ⊗ Z p in L ⊗ Z p , for each prime p. The case of odd p is no longer trivial, but still the p = 2 case usually dominates the analysis. See [2] for an extended calculation of this sort.…”
Section: The Larger Picturementioning
confidence: 99%
“…(3) their scales differ by a factor of 4 and they both have type I. Consider as well the fine symbol got by negating the signs of these terms, and also changing both from givers to receivers or vice-versa in case (2). Then the two fine symbols represent isometric lattices.…”
Section: Fine Symbolsmentioning
confidence: 99%
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