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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?

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    $\begingroup$ 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 $$x^2 + y^2 = z^2 + w^2 .$$ $\endgroup$ Commented Nov 28, 2018 at 2:21

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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.

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  • $\begingroup$ 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 $\endgroup$ Commented Nov 28, 2018 at 17:18
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    $\begingroup$ 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. $\endgroup$ Commented Nov 28, 2018 at 18:50

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