Géométrie Différentielle, TD 1 du 3 février 2017

[0,1]×]0,1[ /(0,y)∼(1,1−y)
[0,1] ×[0,1] /(0, y)∼(1, y)
(x, 0) ∼(x, 1)
[0,1] ×[0,1] /(0, y)∼(1,1−y)
(x, 0) ∼(1 −x, 1)
[0,1] ×[0,1] /(0, y)∼(1,1−y)
(x, 0) ∼(x, 1)
M1M2n(U1, ϕ1) (U2, ϕ2)M1
M2ϕiUiB(0,2) Rn
C{x∈Rn|1
2<kxk<2}
f:x7→ x
kxk2C
x
M1\ϕ−1
1(B(0,1/2)) qM2\ϕ−1 2(B(0,1/2))
ϕ−1
1(C)ϕ−1 2(C)ϕ2◦f◦ϕ−1 1
X
Mi\ϕ−1
i(B(0,1/2)) →X X
X M1M2
M1#M2
n>2M1#M2M1M2

M#SnM
RP2
k
k
RPn
x= [x0:x1:. . . :xn].
(x0, . . . , xn)Rn+1
xRPn
xix
i= 0, . . . , n UiRPnxi6= 0
fi:Ui→Rn
fi([x0:. . . :xn]) = (x0
xi
,...,xi−1
xi
,xi+1
xi
,...,xn
xi
).
fiRPn
Gk(V)k V
C∞
B n −k V UB
Gk(V)B A B
L(A, B)A B
ψA,B :L(A, B)→UB
f7→ (Id +f)(A)
ψA,B
ψ−1
A0,B0◦ψA,B
L(A, B)L(A0, B0)ψ−1
A0,B0◦ψA,B C∞
Gk(V)UB
ψA,B Gk(V)
ψA,B Gk(V)C∞
Gk(V)