# Can all quadric surfaces be obtained by cutting a 4-dimensional cone?

In my high school multivariable calculus class, we recently learned of quadric surfaces. Since they appeared to be a generalization of conic sections to 3 dimensions, I wondered if they could be generated by finding the intersection of a 3-dimensional plane with a 4-dimensional cone similarly to how a conic section is the intersection of a 2-dimensional plane with a 3-dimensional cone. What follows is my attempt at experimenting with this idea. Although I was able to obtain some interesting results, I was not entirely successful in my endeavour which is why I am posting here.

I was able to deduce that a fourth dimensional equivalent of a circular cone is:

$$x^2+y^2+z^2=w^2$$

With this, I was able to get a few quadric surfaces with minor effort by simply plugging in a constant for one of the variables. For example, by setting $$w=3$$, I got the equation of a sphere:

$$x^2+y^2+z^2=9$$

And by setting $$z=3$$, I was able to derive the equation of a hyperboloid of two sheets:

$$x^2+y^2+9=w^2$$ $$w^2-x^2-y^2=9$$

I wanted to see if I could get the rest of the quadric surfaces by picking less mundane planes than just those aligned with the axes. So then I attempted to define a 3-dimensional plane in a 4-dimensional space with a point and a normal. I remembered that one definition for a plane is:

$$\vec{n} \cdot (\vec{P_0} - \vec{P})=0$$

Where $$\vec{n}$$ is the normal of the plane, $$\vec{P_0}$$ is the initial point, and $$\vec{P}$$ is any arbitrary point that lies on the plane. Armed with this knowledge, I plugged in $$\vec{n}=\langle A,B,C,D \rangle$$ and $$\vec{P}=\langle x,y,z,w \rangle$$ to arrive at:

$$Ax+By+Cz+Dw=\vec{n} \cdot \vec{P_0}$$

Since the cone should have rotational symmetry (admittedly I am not sure what this means in four dimensions so this could be a faulty assumption), I figured that any point would yield the same results as long as it does not lie on the plane $$w=0$$. With this knowledge, I picked the point $$(1,0,0,1)$$ because it seemed to be easy to work with.

Then, I tried various normals that I thought would yield interesting results. With this method, I was able to generate most quadric surfaces (including elliptic paraboloids). However, no matter what I tried, I was not able to generate a hyperbolic paraboloid or a hyperboloid of one sheet. Am I not picking the right normals for this? Or is my reasoning flawed? Could it be the case that it is actually not possible to generate these quadric surfaces by cutting a 4-dimensional cone with a plane?

• Unlike in three dimensions, in which there is effectively only a single type of cone, in four dimensions there are essentially two inequivalent types of cone: The "Lorentzian" one, which in suitable coordinates can be written as $$x^2 + y^2 + z^2 = w^2 ,$$ and the "neutral" one, which can be written as $$x^2 + y^2 = z^2 + w^2 .$$ Commented Nov 28, 2018 at 2:21

## 1 Answer

In dimension $$3$$, then, there's essentially only one cone, namely, the zero locus of a nondegenerate, indefinite quadratic form $$Q$$. Sylvester's Law of Inertia says that by making choosing suitable linear coordinates (and possibly replacing $$Q$$ by its negative, which doesn't change the zero locus), we can always put the cone in standard form: $$x^2 + y^2 - z^2 = 0 .$$

In dimension $$4$$, however, there are two inequivalent cones, with standard forms $$w^2 + x^2 + y^2 - z^2 = 0 \qquad \textrm{and} \qquad w^2 + x^2 - y^2 - z^2 = 0 .$$ (The corresponding quadratic forms respectively have signature $$(3, 1)$$ and $$(2, 2)$$, sometimes called Lorentzian and neutral, respectively.)

We can now recover the two missing quadric surfaces as hyperplane sections of the neutral cone, $$C$$:

• Intersecting a neutral cone $$C$$ with any hyperplane with spacelike normal vector $${\bf n}$$ (i.e., $$Q({\bf n}) > 0$$) and not passing through the origin gives the a hyperboloid of one sheet. For example, taking the standard form and the hyperplane $$\{z = 1\}$$, and using $$(w, x, y)$$ as coordinates on the hyperplane gives the standard form $$w^2 + x^2 - y^2 = 1 .$$

• Instead intersecting $$C$$ with a suitable hyperplane parallel to any $$2$$-plane contained in the cone (and again not passing through the origin) gives a hyperbolic paraboloid. For example, taking the standard form and the hyperplane $$\{w = z + a\}$$, $$a \neq 0$$, gives the surface $$w = z + a, x^2 + 2 a z + a^2 = y^2$$. If we use $$(x, y, z)$$ as coordinates on the hyperplane, applying the affine change $$x, y, z' := -(2 a z + a^2)$$ of coordinates there puts the equation in the standard form $$z' = x^2 - y^2$$ of a hyperbolic paraboloid.

• Thank you very much for this helpful answer! It seems that my primary mistake was assuming that there is only one type of four dimensional cone Commented Nov 28, 2018 at 17:18
• You're welcome, I'm glad you found it useful. And yes, this is a good example of geometric intuition that fails in higher dimensions. In general, on $\Bbb R^n$ there are $\left\lceil\frac{n - 1}{2}\right\rceil$ cones up to the equivalence relevant here. Commented Nov 28, 2018 at 18:50