Automorphisms of Lattices. Application to Curves
Jacques Martinet
Universit´ e de Bordeaux, IMB
February 16, 2020 Luminy, March 2019
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 1 / 1
Automorphisms of Lattices. Application to Curves Jacques Martinet - - PowerPoint PPT Presentation
Automorphisms of Lattices. Application to Curves Jacques Martinet Universit e de Bordeaux, IMB February 16, 2020 Luminy, March 2019 Jacques Martinet (Universit e de Bordeaux, IMB) February 16, 2020 1 / 1 G -lattices Let E be an n
Universit´ e de Bordeaux, IMB
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 1 / 1
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Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 3 / 1
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 3 / 1
2, −e∗ 1).
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 4 / 1
2, −e∗ 1).
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 4 / 1
x , y x3 ) .
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Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 6 / 1
5
t −t−1 −t−1 t 2 t −t−1 −t−1 t 2 t −t−1 −t−1 t 2
1 0 0 −1 0 1 0 −1 0 0 1 −1
2; 0, −1 represent A4, − 1 2 its scaled
2, 0), Aut(Λ) = D10.
2, then check
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 7 / 1
5
t −t−1 −t−1 t 2 t −t−1 −t−1 t 2 t −t−1 −t−1 t 2
1 0 0 −1 0 1 0 −1 0 0 1 −1
2; 0, −1 represent A4, − 1 2 its scaled
2, 0), Aut(Λ) = D10.
2, then check
t−2 .
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 7 / 1
tP A(α(t)) P = 5 1−t−t2 2+t
1 0 −1 1 1
2, 0]: the fixed point of α.
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 8 / 1
tP A(α(t)) P = 5 1−t−t2 2+t
1 0 −1 1 1
2, 0]: the fixed point of α.
tP =−P shows the symplectic property:
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 8 / 1
tP A(α(t)) P = 5 1−t−t2 2+t
1 0 −1 1 1
2, 0]: the fixed point of α.
tP =−P shows the symplectic property:
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 8 / 1
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Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 9 / 1
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 9 / 1
9 · Gal(F9/F3), respectively; the latter group contains
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Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 11 / 1
t 0 −t t 2 t t 2 t −t 0 t 2
1 0 −1 0 1 −1 0
tP A(t)P = A(−t) , A(t) A(−t) = 2(2 − t2) I4 , and tP = −P .
4 with the order M of Hurwitz’s quaternions and shows the
e de Bordeaux, IMB) February 16, 2020 11 / 1
6 , B ←
6 .
6 .
n , n = p − 1, p an odd prime: 1 p TrQ(ζp)/Q(zz)
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 12 / 1
6 , B ←
6 .
6 .
n , n = p − 1, p an odd prime: 1 p TrQ(ζp)/Q(zz)
2
2
n
p
p ,
n
n ≥ 2i ;
4
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 12 / 1
2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2
6 (2 × S7)
6
3, − 1 3, − 1 3), representing A∗ 6 indeed the
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 13 / 1
2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2 t1 t2 t3 t3 t2 t1 2
6 (2 × S7)
6
3, − 1 3, − 1 3), representing A∗ 6 indeed the
1 − 4t1t2 − 3t2 2 − 3t1 − 8t2 − 3 = 0 ,
−6t2+40t+50, t2 = 17t2−20t−25 −6t2+40t+50 .
Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 13 / 1
n , for which
3, −e∗ 1 + e∗ 4, −e∗ 2 + e∗ 5, −e∗ 3 + e∗ 6, −e∗ 4, e∗ 1 − e∗ 5, e∗ 2 − e∗ 6}
3 · (−e∗ 1 + e∗ 4), u2 = e∗ 3 · (−e∗ 2 + e∗ 5) and
3 · (−e∗ 3 + e∗ 6) (of course, u3 = −u1 − u2 − 1), obtaining
1 −8t1t2+t2 2 +t1−2t2+1
t2
1 +3t1t2+4t2 2 +4t1+6t2−3
2t2
1 +6t2t1+t2 2 +t1+5t2+1
t2
1 +3t1t2+4t2 2 +4t1+6t2−3 .
7 ).
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Jacques Martinet (Universit´ e de Bordeaux, IMB) February 16, 2020 15 / 1
n . In this latter case, the Hurwitz bound shows that
n ),
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