f(x,y,z,t)=(x^3+y-z, y^3+z-t, z^3+t-x, t^3+x-y)
La matrice jacobienne de f en (x,y,z,t) est :
J(f)(x,y,z,t)=
(3x² ... 1 ... -1 ... 0 )
( 0 ... 3y² ... 1 ... -1 )
( -1 ... 0 ... 3 z² ... 1 )
( 1 ... -1 ... 0 ... 3t² )
Cherchons le rang par Gauss
L1 <---> L4
( 1 ... -1 ... 0... 3t² )
( 0 ... 3y² ... 1 ... -1 )
( -1 ... 0 ... 3 z² ... 1 )
(3x² ... 1 ... -1 ... 0 )
L3<--- L3+L1 et L4<--- L4-3x².L1
( 1 ... -1 ... 0 ... 3t² )
( 0 ... 3y² ... 1 ... -1 )
( 0 ... -1 ... 3 z² ... 3t² +1 )
( 0 ... 1 +3x² ... -1 ... -9x²t² )
L2 <---> L3
( 1 ... -1 ... 0 ... 3t² )
( 0 ... -1 ...3 z² ... 3t² +1 )
( 0 ... 3y² ...1 ... -1 )
( 0 ...1 +3x² ... -1 ... -9x²t² )
L3 <--- L3+3y²L2 et L4<--- L4+(1+3x²)L2
( 1 ... -1 ... 0 ... 3t² )
( 0 ... -1 ... 3 z² ... 3t² +1 )
( 0 ... 0 ... 1 +3z² ... -1+3y²( 3t² +1) )
( 0 ... 0 ... -1 +3z²(1+3x²) ... -9x²t²+(3t^2 +1)(1+3x²) )
L4 <--- L4- (-1 +3z²(1+3x²)) / (1 +3z²) L3
( 1 ... -1 ... 0 ... 3t² )
( 0 ... -1 ... 3 z² ... 3t² +1 )
( 0 ... 0 ... 1 +3z² ... -1+3y²( 3t² +1) )
( 0 ... 0 ... 0 ... -9x²t²+(3t^2 +1)(1+3x²)- (-1 +3z²(1+3x²))(-1+3y²( 3t² +1)) / (1 +3z²) )
==> rg =4 ssi -9x²t²+(3t^2 +1)(1+3x²)- (-1 +3z²(1+3x²))(-1+3y²( 3t² +1)) / (1 +3z²) #0
dans ce cas f est un difféomorphisme de R^4 sur R^4
f(x,y,z,t)=(0,0,0,0) <==> x=y=z=t=0
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