Let C be a smooth algebraic curve in P2. That is, for some homogeneous f∈C[x0,x1,x2] we let C={x∈P2 : f(x)=0}. Describe C as a manifold via the usual open sets Ui={x∈P2 : xi≠0} and charts
φ0 : U0→C2, [x0:x1:x2]↦(x1x0,x2x0),φ1 : U1→C2, [x0:x1:x2]↦(x0x1,x2x1),φ2 : U2→C2, [x0:x1:x2]↦(x0x2,x1x2).
Let w=[w0:w1:w2]∈P2 for which f(w)=0. The Jacobian of C at w is then
Jw=[∂f∂x0|w : ∂f∂x1|w : ∂f∂x2|w]∈P2.
Assume that ∂f∂x0|w≠0 and pass to φ0(U0) to get the Jacobian to be
J0w=(∂f/∂x1|w∂f/∂x0|w , ∂f/∂x2|w∂f/∂x0|w)∈C2.
Assume that w0≠0, so the tangent line to φ0(C)⊂C2 at φ0(w)=(w1/w0,w2/w0) is
Tφ0(w)={φ0(w)+tJ0w : t∈C}⊂C2.
A vector orthogonal to the Jacobian J0w is
ˉJ0w=(−∂f/∂x2|w∂f/∂x0|w , ∂f/∂x1|w∂f/∂x0|w)∈C2,
so the space space normal to Tφ0(w) is given by
T⊥φ0(w)={φ0(w)+tˉJ0w : t∈C}⊂C2.
Example: Let C⊂P2 be the zero locus of f(x0,x1,x2)=x20+x1x2−x1x0. The Jacobian is J=[2x0−x1:x2−x0:x1], and as J=0 implies x0=x1=x2=0, but 0∉P2, the curve C is smooth. Consider two points w=[1:1:0],z=[2:1:−2]∈C, at which the Jacobian is
Jw=[1:−1:1],Jz=[3:−4:1].
Both w0 and z0 are non-zero, with φ0(w)=(1,0) and φ0(z)=(1/2,−1), giving the tangent and normal spaces to be
T(1,0)={(1,0)+t(−1,1) : t∈C},T(1/2,−1)={(1/2,−1)+s(−4/3,1/3) : s∈C},T⊥(1,0)={(1,0)+t(−1,−1) : t∈C},T⊥(1/2,−1)={(1/2,−1)+s(−1/3,−4/3) : s∈C}.
The two normal spaces intersect at (t,s)=(1/3,−1/2) at distances of 1/3⋅||(−1,−1)||=√2/3≈0.471 and 1/2⋅||(−1/3,−4/3)||=√17/3≈1.374 from the points φ0(w),φ0(z), respectively. Hence the conditioning number of C is at most √2/3.
Given a smooth projective curve and a finite set of points, this Sage code will calculate the conditioning number from that collection of points.
φ0 : U0→C2, [x0:x1:x2]↦(x1x0,x2x0),φ1 : U1→C2, [x0:x1:x2]↦(x0x1,x2x1),φ2 : U2→C2, [x0:x1:x2]↦(x0x2,x1x2).
Let w=[w0:w1:w2]∈P2 for which f(w)=0. The Jacobian of C at w is then
Jw=[∂f∂x0|w : ∂f∂x1|w : ∂f∂x2|w]∈P2.
Assume that ∂f∂x0|w≠0 and pass to φ0(U0) to get the Jacobian to be
J0w=(∂f/∂x1|w∂f/∂x0|w , ∂f/∂x2|w∂f/∂x0|w)∈C2.
Assume that w0≠0, so the tangent line to φ0(C)⊂C2 at φ0(w)=(w1/w0,w2/w0) is
Tφ0(w)={φ0(w)+tJ0w : t∈C}⊂C2.
A vector orthogonal to the Jacobian J0w is
ˉJ0w=(−∂f/∂x2|w∂f/∂x0|w , ∂f/∂x1|w∂f/∂x0|w)∈C2,
so the space space normal to Tφ0(w) is given by
T⊥φ0(w)={φ0(w)+tˉJ0w : t∈C}⊂C2.
Example: Let C⊂P2 be the zero locus of f(x0,x1,x2)=x20+x1x2−x1x0. The Jacobian is J=[2x0−x1:x2−x0:x1], and as J=0 implies x0=x1=x2=0, but 0∉P2, the curve C is smooth. Consider two points w=[1:1:0],z=[2:1:−2]∈C, at which the Jacobian is
Jw=[1:−1:1],Jz=[3:−4:1].
Both w0 and z0 are non-zero, with φ0(w)=(1,0) and φ0(z)=(1/2,−1), giving the tangent and normal spaces to be
T(1,0)={(1,0)+t(−1,1) : t∈C},T(1/2,−1)={(1/2,−1)+s(−4/3,1/3) : s∈C},T⊥(1,0)={(1,0)+t(−1,−1) : t∈C},T⊥(1/2,−1)={(1/2,−1)+s(−1/3,−4/3) : s∈C}.
The two normal spaces intersect at (t,s)=(1/3,−1/2) at distances of 1/3⋅||(−1,−1)||=√2/3≈0.471 and 1/2⋅||(−1/3,−4/3)||=√17/3≈1.374 from the points φ0(w),φ0(z), respectively. Hence the conditioning number of C is at most √2/3.
Given a smooth projective curve and a finite set of points, this Sage code will calculate the conditioning number from that collection of points.