Des formes bilinéaires en combinatoire Introduction 1 Un problème

2

k > 1k
n
a
G
a d (a)a
26d(a)< k G 26G < n
S=Pa∈G1
S=X
a
d(a) = X
G
G
a G a x
G G (x)a x G 6=G(x)
x y G G (x)6=G(y)G
x y d (a)>G
n > k (a, G)
a G
0< k −d(a)6k−G < n −G
G
n−G<d(a) k−d(a)
(a, G)
a G
S0=X
a6∈G
G
n−G<X
a6∈G
d(a)
k−d(a)=S00
G n −G
G G
S0=X
GX
a6∈G
G
n−G=X
G
(n−G)G
n−G=X
G
G

a k −d(a)
a a
S00 =X
aX
G|a6∈G
d(a)
k−d(a)=X
a
(k−d(a)) d(a)
k−d(a)=X
a
d(a)
n6k
a Ga
a a b
Ga∩Gb1
(Ga∩Gb)≥1a b
E m a
A1, . . . , AnE(Ai∩Aj) =
a i 6=j n 6m
E
{1,2, . . . , m}M m n
mij 1i Aj0
T=tMM n
tij Ai∩Aj
T n
M n 6m T
tij =a i 6=j
n > 1tii >a i Ai
Ai∩Aj
i j tii =tjj =a Ai
AjAi∩Aja
AiAji6=j

T
(x1, . . . , xn)7→ aÃn
X
i=1
xi!2 +
n
X
i=1
(tii −a)x2
i
R
FER
E=mF
ϕ(f, g) = X
x∈E
f(x)g(x)
Ai⊂EF
χi∈ F
Aiϕ(χi, χj)
Ai∩Aj
ϕ(χi, χj) = a i 6=j ai=ϕ(χi, χi)>a
χin6m
λ1, . . . , λnλ1χ1+. . . +λnχn= 0
χiλiai+
(λ−λi)a= 0 λ=λ1+. . . +λni
ai=a λ = 0 a > 0
i ai> a λi(ai−a) = −λa
λiλ
λi
ϕ
ϕ(χi, χj)
MtMM
χi
T
E
Ai⊂E

AiAi∩Aj
E N
(A1, . . . , Ak)
AiAi∩Aj
i6=j
AiAi∩Aj
i6=j
AiAi∩Aj
AiAi∩Aj
N
R Z/2Z
FZ/2Z
EZ/2ZA E
χ∈ F
2
V
K
V
ϕ:V×V→K
ϕ ϕ(x, y) = ϕ(y, x)
x y V
2
ϕ
x∈V ϕ y ∈V ϕ (x, y) = 0 ϕ
xy =(x+y)2
−(x−y)2
4