I have a cubic surface in projective 3-space with equation [math]X_0^2 f_1(X_1,X_2,X_3) + X_0 f_2(X_1,X_2,X_3) + f_3(X_1,X_2,X_3)=0[/math] where f_i is a polynomial of degree i in the listed variables. The surface contains the point P=(1,0,0,0). Is it correct that the tangent cone at P is given by [math]2X_0 f_1 + f_2[/math]?
Tangent cone: X02f1(X1,X2,X3)+X0f2(X1,X2,X3)+f3(X1,X2,X3)=0
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blamocur
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blamocur
Elite Member
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- Oct 30, 2021
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First a disclaimer: I never heard of tangent cones before, but have recently found a definition in an SE post. My ignorance did not stop me, which means that you must take this post with a large grain of salt.
[math]f(X) = x_0^2 f_1(x_1,x_2,x_3) + x_0 f_2(x_1,x_2,x_3) + f_3(x_1,x_2,x_3)[/math][math]\frac{\partial f}{\partial x_0} = 2x_0 f_1 + f_2[/math]For [imath]k\geq 1[/imath]:
[math]\frac{\partial f}{\partial x_k} = x_0^2 \frac{\partial f_1}{\partial x_j} + x_0 \frac{\partial f_2}{\partial x_j} + \frac{\partial f_3}{\partial x_j}[/math]But because [imath]f_i[/imath] is homogenious of order [imath]i[/imath] we know that for [imath]i\geq 2[/imath]
[math]\frac{\partial f_i}{\partial x_k}(0,0,0) = 0[/math]which means that for [imath]k\geq 1[/imath]:
[math]\frac{\partial f}{\partial x_k}(1,0,0,0) = \frac{\partial f_1}{\partial x_k}(1,0,0,0) =[/math]Since [imath]f_1[/imath] is linear we must have [imath]\frac{\partial f_1}{\partial x_k} \neq 0[/imath] for some [imath]k[/imath] (unless [imath]f_1 \equiv 0[/imath], which I assume it is not ). This means that for our [imath]f[/imath] in the definition of the projective tangent cone we have [imath]k=1[/imath], and thus the cone is given by this linear equation:
[math]\sum_{j=0}^3 \frac{\partial f}{\partial x_j}(1,0,0,0) x_j = 0 = \frac{\partial f}{\partial x_0}(1,0,0,0) x_0 + \sum_{j=1}^3 \frac{\partial f_1}{\partial x_j}(1,0,0,0) x_j[/math]Now:
[math]\frac{\partial{f}}{\partial x_0}(1,0,0,0) = 2f_1(0) + f_2(0)[/math]And since [imath]f_1[/imath] is linear we know that
[math]\sum_{j=1}^3 \frac{\partial f_1}{\partial x_j}(1,0,0,0) x_j =f_1(x_1,x_2,x_3)[/math]Thus the cone equation is linear:
[math]\left(2f_1(0)+f_2(0) \right) x_0 + f_1(x_1,x_2,x_3) = 0[/math]
[math]f(X) = x_0^2 f_1(x_1,x_2,x_3) + x_0 f_2(x_1,x_2,x_3) + f_3(x_1,x_2,x_3)[/math][math]\frac{\partial f}{\partial x_0} = 2x_0 f_1 + f_2[/math]For [imath]k\geq 1[/imath]:
[math]\frac{\partial f}{\partial x_k} = x_0^2 \frac{\partial f_1}{\partial x_j} + x_0 \frac{\partial f_2}{\partial x_j} + \frac{\partial f_3}{\partial x_j}[/math]But because [imath]f_i[/imath] is homogenious of order [imath]i[/imath] we know that for [imath]i\geq 2[/imath]
[math]\frac{\partial f_i}{\partial x_k}(0,0,0) = 0[/math]which means that for [imath]k\geq 1[/imath]:
[math]\frac{\partial f}{\partial x_k}(1,0,0,0) = \frac{\partial f_1}{\partial x_k}(1,0,0,0) =[/math]Since [imath]f_1[/imath] is linear we must have [imath]\frac{\partial f_1}{\partial x_k} \neq 0[/imath] for some [imath]k[/imath] (unless [imath]f_1 \equiv 0[/imath], which I assume it is not ). This means that for our [imath]f[/imath] in the definition of the projective tangent cone we have [imath]k=1[/imath], and thus the cone is given by this linear equation:
[math]\sum_{j=0}^3 \frac{\partial f}{\partial x_j}(1,0,0,0) x_j = 0 = \frac{\partial f}{\partial x_0}(1,0,0,0) x_0 + \sum_{j=1}^3 \frac{\partial f_1}{\partial x_j}(1,0,0,0) x_j[/math]Now:
[math]\frac{\partial{f}}{\partial x_0}(1,0,0,0) = 2f_1(0) + f_2(0)[/math]And since [imath]f_1[/imath] is linear we know that
[math]\sum_{j=1}^3 \frac{\partial f_1}{\partial x_j}(1,0,0,0) x_j =f_1(x_1,x_2,x_3)[/math]Thus the cone equation is linear:
[math]\left(2f_1(0)+f_2(0) \right) x_0 + f_1(x_1,x_2,x_3) = 0[/math]