This family of relations is the intersection of what two manifolds

Question

when an infinite cone intersects a plane we get conics and we get the generic equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ But what happens if we start with a similar equation, and try to determine the surfaces, (I am assuming one would need to a plane) $$Ax^3+Bx^2y+Cxy^2+Dy^3+Ex^2+Fxy+Gy^2+Hx+Iy+J=0$$ Or if you show that such a manifold would be impossible to exist. (by showing that all 2D? manifolds either miss some curves in the family or add extra)

Answer 1

It's the image of the (affine) Veronese map $(x, y) \mapsto (x^3, x^2y, xy^2, y^3, x^2, xy, y^2, x, y)$. This is the graph of a map $\mathbf{R}^2 \to \mathbf{R}^7$, so is diffeomorphic to $\mathbf{R}^2$, in particular a manifold. Intersecting it with the hyperplane $Az_1 + \dots + I z_9 = -J$ (where $z_1, \dots, z_9$ are the coordinates on $\mathbf{R}^9$) and looking at the diffeomorphic preimage in $\mathbf{R}^2$ gives you exactly that equation.

If you go through the same construction for the degree $2$ curve you get a surface in $\mathbf{R}^5$ (intersected with a plane), not the cone in $\mathbf{R}^3$ you might expect. The issue is that in the $\text{cone} \cap \text{plane}$ picture, the plane does not come with a choice of coordinates so choosing different ones gives different equations. A different way of fixing this is to say that up to some change of coordinates, every degree $2$ curve in the plane is given by a simpler equation $$A(x^2+y^2) + 2Bxy + C(x^2 - y^2) + F = 0$$ and now this is given by intersecting the image of $(x, y) \mapsto (x^2 + y^2, 2xy, x^2 - y^2)$ with a plane in $\mathbf{R}^3$. You can check that the image is a cone, cut out by the equation $z_1^2 = z_2^2 + z_3^2$. You do lose something, this is no longer a manifold at $(0,0,0)$ and that it is a manifold away from this point is trickier to prove (implicit function theorem).

If you don't need the cone to be the standard cone then you can take the even "simpler" equation $$Ax^2 + Bxy + Cy^2 + F = 0.$$ Similarly in the degree $3$ case you can reduce to curves of the form $$Ax^3 + Bx^2y + Cxy^2 + Dy^3 + Ex^2 + Fxy + Gy^2 + J = 0$$ and you get the image of $(x,y) \mapsto (x^3, x^2y, xy^2, y^3, x^2, xy, y^2)$ in $\mathbf{R}^7$, but again this image is not a manifold at the origin. It is at all its other points, again using the implicit function theorem though the equations defining it might be tricky to figure out if you haven't seen them before.