Problème n Partie A : Z-bases du réseau - Académie de Nancy-Metz

n◦
Z R
R2
R2d
R2
O=0
0e1=1 0e2=0 1
Z2R2R
R
−4
−3
−2
−1
1
2
3
4
5
6
7
−4−3−2−1 1234567
0
Z
B= (e0
1, e0 2)R2BZ
R
e0
1, e0 2∈ R
XRX=ae0
1+be0 2a, b ∈Z
C= (e1, e2)R2CZR
e0
1=a1
b1e0
2=a2
b2R2
A=a1a2
b1b2.
X∈R2x, y ∈RX=xe0
1+ye0 2
X=Ax
y.
(e0
1, e0 2)ZR
(a1, a2, b1, b2)∈Z4

(x1, x2, y1, y2)∈Z4
x1e0
1+y1e0 2=e1x2e0 1+y2e0 2=e2.
B=x1x2
y1y2AB =I2
det(A)∈ {−1,1}
(a1, a2, b1, b2)∈Z4det(A)∈ {−1,1}
A A−1
(e0
1, e0 2)ZR
e0
1=a1
b1R
e0
1ZRa1b1
a1b1
e0
2R(e0 1, e0 2)ZR
ZR7
10
f:R2−→ R2C= (e1, e2)R2
A
f(R)⊆ R A
f(R) = R
Imf
f
f−1(R)⊆ R
A A−1
det(A)∈ {−1,1}
A
det(A)∈ {−1,1}
(f(e1), f(e2)) Z
R
f(R) = R

G f R2f(R) = RG0
f G f(O) = O
G G0
G
f∈G0f
A f R2
XR1O
f(e1)f(e2)
1
0,−1 0,0 1,0 −1.
A
H=
1 0
0 1,−1 0 0 1,1 0 0−1,−1 0 0−1,
0 1
1 0,0−1 1 0 ,0 1 −1 0,0−1 −1 0 
.
s1s2C
A1=0 1
1 0A2=−1 0 0 1.
s1s2
s1◦s2s2◦s1
s1s2G0
f
R2R2H f G0
G0
tx
yt∈Gx y∈ R
f∈G t0−f(O)t0
G g =t0◦f G0
f G f =t◦g t
Rg G0

TR2
0
0
1
2
0
1
2
1
2
C
1
2
1
2
−1
2
1
2
−1
2
−1
2
1
2
−1
2
.
−0.5 0.5
−0.5
0.5
0
C=[
g∈G0
g(T).
g1g2G0
g1(T)g2(T)
X∈R2tXX
R2=[
X∈R
tX(C).
X Y R
tX(C)tY(C)
R2=[
f∈G
f(T).
f1f2G
f1(T)f2(T)

k
tk:
R2−→ R2
x
y7→ x+k ysk:
R2−→ R2
x
y7→ −x+k −y
tksk
k l tk◦slsk◦tlsk◦sltk◦tl
H={tk, sk|k∈Z}H G
F=[
f∈H
f(T),
T F
−0.5 0.5 1.1.5 2.2.5
−0.5
0.5
0
G
n◦
R
f]0,+∞[
]0,+∞[
x y
f(xf(y)) = yf(x).
f]1,+∞[A
x>1f(x)6A
IRϕ I I
ϕ I x I
ϕ(ϕ(x)) = x.
R R