One last thing on this: the Right wants women to be afraid of men. They're basically saying "yes, all men!"
When they make jokes like "Watch out, women!" what they're really saying is that these dogs bite.
It's a flex of power and control. They weirdly want the predators and prey framing.
![Some commentators, however, continue to deny the possibility
of social science. Among them are academic lawyers such as Grant
Gilmore and Arthur Leff. Professor Gilmore wrote recently:
For two hundred years we have been in thrall to the eighteenthcentury hypothesis that there are, in social behavior and in
societal development, patterns which recur in the same way
that they appear to recur in the physical universe.
• . . [T]he hypothesis is itself in error. Man's fate will
forever elude the attempts of his intellect to understand it. The
accidental variables which hedge us about effectively screen
the future from our view. The quest for the laws which will](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:wlnye2inqqyh2onzpr2qo5mh/bafkreigik2pkg2zjbcslfdjlnoaygs2qih2pis7gogbcsc5v5l5khvhld4)
![The Jacobian determinant is indeed exactly
−
2
−2, a nonzero constant. Now the injectivity claim:
Scrutinizing whether this counterexample genuinely undermines the conjecture
Scrutinizing whether this counterexample genuinely undermines the conjecture
Check where the three points map
bash
python3 << 'EOF'
import sympy as sp
x, y, z = sp.symbols('x y z')
P = (1+x*y)**3 * z + y**2 * (1+x*y) * (4+3*x*y)
Q = y + 3*x*(1+x*y)**2 * z + 3*x*y**2 * (4+3*x*y)
R = 2*x - 3*x**2*y - x**3*z
pts = [(0,0,sp.Rational(-1,4)),
(1,sp.Rational(-3,2),sp.Rational(13,2)),
(-1,sp.Rational(3,2),sp.Rational(13,2))]
for p in pts:
sub = {x:p[0], y:p[1], z:p[2]}
img = (sp.simplify(P.subs(sub)), sp.simplify(Q.subs(sub)), sp.simplify(R.subs(sub)))
print(f"F{p} = {img}")
# Confirm the three source points are distinct
print("Distinct source points:", len(set(pts))==3)
EOF
Output
F(0, 0, -1/4) = (-1/4, 0, 0)
F(1, -3/2, 13/2) = (-1/4, 0, 0)
F(-1, 3/2, 13/2) = (-1/4, 0, 0)
Distinct source points: True
The Jacobian determinant is indeed
−
2
−2, a nonzero constant, and I've verified that three distinct points all map to the same output
(
−
1
/
4
,
0
,
0
)
(−1/4,0,0). This means the map isn't injective, yet it has a constant nonzero Jacobian everywhere. Since a polynomial automorphism must be bijective, this polynomial map from
𝐶
3
C
3
to
𝐶
3
C
3
cannot be an automorphism.
If this construction is correct, it would be a counterexample to the Jacobian conjecture — one of the most famous open problems in mathematics. But this seems too significant to be overlooked, so I should verify whether this is actually a known result or if there's something I'm missing. eed to be a counterexample to the Jacobian conjecture. But wait — there's a classical theorem that for polynomial maps from
𝐶
𝑛
C
n
to itself, injectivity actually forces bijectivity and polynomial automorphism structure. So the question becomes whether a non-injective polynomial map can even have a nonz…](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:wlnye2inqqyh2onzpr2qo5mh/bafkreia47s244swaamigujwj6riltbzppubcvktmajp5b4jp23w43b6dn4)
![This is a specific, checkable mathematical claim, so let me actually verify it rather than guess. Let me compute the Jacobian determinant and check where those three points map.
Verified mathematical calculations and mapped point convergence
Verified mathematical calculations and mapped point convergence
Compute Jacobian determinant of the claimed map
bash
pip install sympy --break-system-packages -q 2>/dev/null; python3 << 'EOF'
import sympy as sp
x, y, z = sp.symbols('x y z')
# The map from the tweet
P = (1+x*y)**3 * z + y**2 * (1+x*y) * (4+3*x*y)
Q = y + 3*x*(1+x*y)**2 * z + 3*x*y**2 * (4+3*x*y)
R = 2*x - 3*x**2*y - x**3*z
F = sp.Matrix([P, Q, R])
vars = sp.Matrix([x, y, z])
J = F.jacobian(vars)
detJ = sp.expand(J.det())
print("Jacobian determinant:", detJ)
print("Simplified:", sp.simplify(detJ))
EOF
Output
Jacobian determinant: -2
Simplified: -2
The Jacobian determinant came out to
−
2
−2, which is nonzero, so now I need to verify that these three points all map to the same location.
Done
The Jacobian determinant is indeed exactly
−
2
−2, a nonzero constant. Now the injectivity claim:
Scrutinizing whether this counterexample genuinely undermines the conjecture](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:wlnye2inqqyh2onzpr2qo5mh/bafkreidi2uzizbegwecaifkx7idpeatxv7sgmfsnosfga5z3wkrdchunjy)
