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Monday 9:01 PM
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After the first announcement of a solution, an alternate approach has been given in the attached PDF. Can you compare this approach with the previous example and see if it retroactively helps explain the construction of the former?
ChatGPT said:
Worked for 26s
Yes. I checked the identities independently, and the two maps are mutually inverse polynomial isomorphisms.
Let
$$ X = \left{\right. \left(\right. a , b , c , d , e \left.\right) \in \mathbb{C}^{5} : a^{2} e - a b d + c b^{2} = 1 , a d + b c = 1 \left.\right} . $$
X={(a,b,c,d,e)∈C 5:a 2 e−ab d+c b 2=1,a d+b c=1}. Define $\Phi : \mathbb{A}^{3} \rightarrow X$Φ:A 3→X by
$$ b & = 1 + a y , \ c & = 1 - \frac{3}{2} a y + a^{2} z , \ d & = \frac{1}{2} y - a z + \frac{3}{2} a y^{2} - a^{2} y z , \ e & = - 2 z + 4 y^{2} - 4 a y z + 3 a y^{3} - 2 a^{2} y^{2} z . $$
b c d e=1+a y,=1−2 3a y+a 2 z,=2 1y−a z+2 3a y 2−a 2 yz,=−2 z+4 y 2−4 a yz+3 a y 3−2 a 2 y 2 z. Direct expansion gives
$$ a d + b c = 1 $$
a d+b c=1 and
$$ a^{2} e - a b d + c b^{2} = 1 $$
a 2 e−ab d+c b 2=1 identically in $a , y , z$a,y,z. Thus $\Phi$Φ is defined on all of $\mathbb{A}^{3}$A 3, including $a = 0$a=0.
Now define $\Psi : X \rightarrow \mathbb{A}^{3}$Ψ:X→A 3 by
$$ y & = 2 b d - a e , \ z & = 2 d^{2} + c e + 6 b d^{2} + 3 b c e - \frac{9}{2} e . $$
y z=2 b d−a e,=2 d 2+ce+6 b d 2+3 b ce−2 9e. Substituting the forward formulas into these expressions gives identically
$$ 2 b d - a e = y $$
2 b d−a e=y and
$$ 2 d^{2} + c e + 6 b d^{2} + 3 b c e - \frac{9}{2} e = z . $$
2 d 2+ce+6 b d 2+3 b ce−2 9e=z. Hence
$$ \Psi \circ \Phi = \left(id \right)_{\mathbb{A}^{3}} . $$
Ψ∘Φ=id A 3. Conversely, start with $\left(\right. a , b , c , d , e \left.\right) \in X$(a,b,c,d,e)∈X, define $y , z$y,z by the polynomial formulas above, and reconstruct
$$ b^{'} & = 1 + a y , \ c^{'} & = 1 - \frac{3}{2} a y + a^{2} z , \ d^{'} & = \frac{1}{2} y - a z + \frac{3}{2} a y^{2} - a^{2} y z , \ e^{'} & = - 2 z + 4 y^{2} - 4 a y z + 3 a y^{3} - 2 a^{2} y^{2} z . $$
b′c′d′e′=1+a y,=1−2 3a y+a 2 z,=2 1y−a z+2 3a y 2−a 2 yz,=−2 z+4 y 2−4 a yz+3 a y 3−2 a 2 y 2 z. Reducing $b^{'} - b , c^{'} - c , d^{'} - d , e^{'} - e$b′−b,c′−c,d′−d,e′−e modulo the two defining equations of $X$X gives zero in each case. Therefore
$$ \Phi \circ \Psi = \left(id \right)_{X} . $$
Φ∘Ψ=id X. So indeed
$$ \boxed{X \cong \mathbb{A}^{3}} $$
X≅A 3 by the explicit polynomial isomorphism
$$ \left(\right. a , y , z \left.\right) \leftrightarrow \left(\right. a , b , c , d , e \left.\right) $$
(a,y,z)⟷(a,b,c,d,e) displayed above.
In particular, there is no hidden restriction or exceptional locus at $a = 0$a=0; the apparent divisions in the preliminary formulas for $d$d and $e$e have removable singularities, and the final maps are polynomial everywhere.
You said:
If one composes Phi with the map F(a,b,c,d,e) = (ac, ad + bc, ae + bd, be) and then drops the ad+bc=1 term, can you confirm symbolically that one obtains a polynomial from (a,y,z) to C^3 with constant Jacobian?
ChatGPT said:
Yes. After composing $\Phi$Φ with
$$ F \left(\right. a , b , c , d , e \left.\right) = \left(\right. a c , \textrm{ } a d + b c , \textrm{ } a e + b d , \textrm{ } b e \left.\right) $$
F(a,b,c,d,e)=(a c,a d+b c,a e+b d,b e) and discarding the identically constant coordinate $a d + b c = 1$a d+b c=1, one gets the polynomial map
$$ G : \mathbb{C}_{a , y , z}^{3} \rightarrow \mathbb{C}^{3} $$
G:C a,y,z 3→C 3 given by
$$ G \left(\right. a , y , z \left.\right) = \left(\right. G_{1} , G_{2} , G_{3} \left.\right) , $$
G(a,y,z)=(G 1,G 2,G 3), where
$$ \boxed{G_{1} = a c = a - \frac{3}{2} a^{2} y + a^{3} z ,} $$
G 1=a c=a−2 3a 2 y+a 3 z,
$$ \boxed{G_{2} = a e + b d = \frac{1}{2} y - 3 a z + 6 a y^{2} - 6 a^{2} y z + \frac{9}{2} a^{2} y^{3} - 3 a^{3} y^{2} z ,} $$
G 2=a e+b d=2 1y−3 a z+6 a y 2−6 a 2 yz+2 9a 2 y 3−3 a 3 y 2 z, and
$$ \boxed{G_{3} = b e = - 2 z + 4 y^{2} - 6 a y z + 7 a y^{3} - 6 a^{2} y^{2} z + 3 a^{2} y^{4} - 2 a^{3} y^{3} z .} $$
G 3=b e=−2 z+4 y 2−6 a yz+7 a y 3−6 a 2 y 2 z+3 a 2 y 4−2 a 3 y 3 z. Equivalently,
$$ G_{1} & = a - \frac{3}{2} a^{2} y + a^{3} z , \ G_{2} & = \frac{1}{2} y - 3 a z + 6 a y^{2} - 6 a^{2} y z + \frac{9}{2} a^{2} y^{3} - 3 a^{3} y^{2} z , \ G_{3} & = - 2 z + 4 y^{2} - 6 a y z + 7 a y^{3} - 6 a^{2} y^{2} z + 3 a^{2} y^{4} - 2 a^{3} y^{3} z . $$
G 1G 2G 3=a−2 3a 2 y+a 3 z,=2 1y−3 a z+6 a y 2−6 a 2 yz+2 9a 2 y 3−3 a 3 y 2 z,=−2 z+4 y 2−6 a yz+7 a y 3−6 a 2 y 2 z+3 a 2 y 4−2 a 3 y 3 z. A direct symbolic differentiation gives
$$ \boxed{det \frac{\partial \left(\right. G_{1} , G_{2} , G_{3} \left.\right)}{\partial \left(\right. a , y , z \left.\right)} = - 1.} $$
det∂(a,y,z)∂(G 1,G 2,G 3)=−1. Thus $G$G is a polynomial map with constant nonzero Jacobian.
A quick consistency check at $a = 0$a=0 is
$$ G \left(\right. 0 , y , z \left.\right) = \left(\right. 0 , \frac{y}{2} , 4 y^{2} - 2 z \left.\right) , $$
G(0,y,z)=(0,2 y,4 y 2−2 z), whose Jacobian matrix at $a = 0$a=0 is triangular enough to give determinant $- 1$−1, consistent with the global calculation.
So this composition produces exactly the desired Keller map
$$ \boxed{G : \mathbb{C}^{3} \rightarrow \mathbb{C}^{3} , det D G \equiv - 1.} $$
G:C 3→C 3,det D G≡−1.
You said:
The original counterexample to the Jacobian conjecture is $$ F(z_1,z_2,z_3) = \Big((1+z_1 z_2)^3 z_3 + z_2^2 (1+z_1z_2) (4+3z_1z_2), z_2 + 3 z_1 (1+z_1z_2)^2 z_3 + 3 z_1 z_2^2 (4+3z_1z_2), 2 z_1 - 3 z_1^2 z_2 - z_1^3 z_3\Big)$$. Can you see how this relates to the map G?
ChatGPT said:
Worked for 22s
Yes. They are the same map, up to a linear change of source coordinates, a reversal of the output coordinates, and two output rescalings.
Set
$$ \boxed{a = z_{1} , y = z_{2} , z = - \frac{1}{2} z_{3} .} $$
a=z 1,y=z 2,z=−2 1z 3. Recall
$$ G = \left(\right. G_{1} , G_{2} , G_{3} \left.\right) , $$
G=(G 1,G 2,G 3), where
$$ G_{1} = a - \frac{3}{2} a^{2} y + a^{3} z . $$
G 1=a−2 3a 2 y+a 3 z. Under this substitution,
$$ $$
2 G 1=2 z 1−3 z 1 2z 2−z 1 3z 3, which is exactly the third coordinate of the original counterexample.
Likewise, direct expansion gives
$$ \boxed{2 G_{2} = z_{2} + 3 z_{1} \left(\right. 1 + z_{1} z_{2} \left.\right)^{2} z_{3} + 3 z_{1} z_{2}^{2} \left(\right. 4 + 3 z_{1} z_{2} \left.\right) ,} $$
2 G 2=z 2+3 z 1(1+z 1z 2)2 z 3+3 z 1z 2 2(4+3 z 1z 2), the second coordinate, and
$$ \boxed{G_{3} = \left(\right. 1 + z_{1} z_{2} \left.\right)^{3} z_{3} + z_{2}^{2} \left(\right. 1 + z_{1} z_{2} \left.\right) \left(\right. 4 + 3 z_{1} z_{2} \left.\right) ,} $$
G 3=(1+z 1z 2)3 z 3+z 2 2(1+z 1z 2)(4+3 z 1z 2), the first coordinate.
Consequently, writing $F_{o r i g}$F orig for the displayed counterexample,
$$ \boxed{F_{o r i g} \left(\right. z_{1} , z_{2} , z_{3} \left.\right) = \left(\right. G_{3} , \textrm{ }\textrm{ } 2 G_{2} , \textrm{ }\textrm{ } 2 G_{1} \left.\right) \left(\right. z_{1} , z_{2} , - \frac{z_{3}}{2} \left.\right) .} $$
F orig(z 1,z 2,z 3)=(G 3,2 G 2,2 G 1)(z 1,z 2,−2 z 3). Equivalently, define the linear maps
$$ A \left(\right. z_{1} , z_{2} , z_{3} \left.\right) = \left(\right. z_{1} , z_{2} , - \frac{1}{2} z_{3} \left.\right) $$
A(z 1,z 2,z 3)=(z 1,z 2,−2 1z 3) and
$$ B \left(\right. u_{1} , u_{2} , u_{3} \left.\right) = \left(\right. u_{3} , 2 u_{2} , 2 u_{1} \left.\right) . $$
B(u 1,u 2,u 3)=(u 3,2 u 2,2 u 1). Then
$$ \boxed{F_{o r i g} = B \circ G \circ A .} $$
F orig=B∘G∘A. This also checks the Jacobian constants. Since
$$ det A = - \frac{1}{2} , det B = - 4 , det D G = - 1 , $$
det A=−2 1,det B=−4,det D G=−1, we obtain
$$ det D F_{o r i g} = \left(\right. - 4 \left.\right) \left(\right. - 1 \left.\right) \left(\right. - \frac{1}{2} \left.\right) = \boxed{- 2} , $$
det D F orig=(−4)(−1)(−2 1)=−2, which agrees with the Jacobian of the original formula.
So the factorization construction has recovered the original map exactly, modulo these elementary linear normalizations.
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