4D quadratic sections

Higher-dimensional geometry (previously "Polyshapes").

4D quadratic sections

Postby quickfur » Thu Nov 17, 2011 10:16 pm

To revisit an old topic, which i believe may have been discussed before:

In 2D, all the quadratic manifolds (curves) can be obtained from the cone (hence, conic sections). Can the 3D quadratic manifolds be obtained from a 4D cone? If so, which cone would it be, the spherical cone?

And does this generalize to higher quadratics as well?
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Re: 4D quadratic sections

Postby Keiji » Thu Nov 17, 2011 10:22 pm

Can you give a list of the 3D quadratic manifolds?
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Re: 4D quadratic sections

Postby quickfur » Thu Nov 17, 2011 10:32 pm

They would be:

- the sphere
- the ellipsoid (treated differently from the sphere because in 2D, you get a circle from cutting a cone horizontally and an ellipse from cutting the cone at an angle shallower than its slope)
- the circular/elliptical paraboloid (basically a lathe of a parabola)
- the hyperbolic paraboloid (saddle surface)
- hyperboloid of 1 sheet (hyperbola lathed around the line that divides its two pieces)
- hyperboloid of 2 sheets (hyperbola lathed around the line intersecting its two pieces)
- (double) cone
- extrusions of circle/ellipse/parabola/hyperbola
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Re: 4D quadratic sections

Postby Marek14 » Thu Nov 17, 2011 10:42 pm

Well, if we ignore the degenerate manifolds (like groups of planes which are not curved, or cartesian products like cylinders), there are three basic kinds of quadratics:

1. elliptic: x^2 + y^2 + z^2 = a. For a positive, you get an ellipsoid. For a = 0, you get a single point, and for a < 0, you get an empty set.

2. parabolic: in 3D, these are x^2 + y^2 - z = 0 (elliptical paraboloid) and x^2 - y^2 - z = 0 (hyperbolic paraboloid)

3. hyperbolic: in 3D, they can be all classed as x^2 + y^2 - z^2 = a. For a positive, it's one-part hyperboloid, for a negative it's two-part, and for a = 0, it's cone.

Now, in 4D, there are two distinct types of cones, x^2 + y^2 + z^2 - w^2 = 0 and x^2 + y^2 - z^2 - w^2 = 0. My guess is that to get all 3D manifolds, you have to slice both. x-, y- and z-slices of first (spherical) cone give you either 3D cone (for coordinate = 0) or 2-part hyperboloid; w-slices give you either single point or ellipsoid. Slicing the second cone in any coordinate direction gives either cone or one-part hyperboloid. I presume that paraboloids are obtained in the same way as in 3D -> 2D, by slicing parallel to the side of the cone, but I don't know the exact transformation. My hunch is that you can eliminate two coordinates whose squares have opposite signs and replace them with a single coordinate in linear term, and in this case spherical cone would lead to elliptical paraboloid and the other one to hyperbolical paraboloid - however this is a pure conjecture on my part.
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Re: 4D quadratic sections

Postby Mrrl » Thu Nov 17, 2011 10:49 pm

Consider the projective 3D space. You have much less classes there:

Spheres (surfaces with positive curvature): include sphere, ellipsoid, elliptic parabaloid, hyperboloid of 2 sheets;
Cylinders (surfaces with zero curvature): include double cone and all kinds of extrusions;
Saddles (surfaces with negative curvature): include hyperboloid of 1 sheet and hyperbolic paraboloid.

When you embed an object from projective space to affine space (by selecting of the ideal plane), you really build the intersection of 4D cone with some hyperplane. So you need at least three kinds of cones:
- spherical cone: w^2=x^2+y^2+z^2;
- saddle cone: w^2+z^2=x^2+y^2;
- cylindrical cone: w^2=x^2+y^2

I'm not sure that it's enough to get all quadrics up to metrical equivalence, but at least you can get all affine types of quadrics. Two intersecting planes may be got by section of cylindrical cone, but tho parallel planes are missing (like two parallel lines in 2D is a quadric that is not any conic section).
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Re: 4D quadratic sections

Postby quickfur » Fri Nov 18, 2011 12:27 am

Hmm, you're right, mrrl, even in 2D you do have the parallel lines quadric, which is not generated by cone sections.

Anyway, I started digging in my old notes, and found a little analysis for classifying all quadrics in n dimensions that I did many years ago. Basically i reduce any n-dimensional quadric to a standard form via an affine transformation. Under such a transformation, we can separate all variables of a quadratic polynomial in n variables to get a sum of polynomials in each variable. Furthermore, we can eliminate all except maybe 1 linear term (e.g., x^2+y+z=K can be reduced to x^2+y=K'). If there is a linear term, then it's a "parabolic" manifold; otherwise, it's some combination of spherical and/or hyperbolic.

The rest of the classification is then based on the "signature" of the polynomial, basically the configuration of positive or negative terms for each squared variable. For example, x^2+y^2-z^2 has signature (++-) and x^2-y^2-z^2 has signature (+--). The order of signs in the signature doesn't matter, so we can always write the +'s first. There's an interesting relationship between quadrics if we consider the operation of negating the signature. For example, the negation of (++-) is (--+), which is the same as (+--). So the hyperboloid of 1 sheet is, in some sense, "conjugate" or "dual" of the hyperboloid of 2 sheets.

A cylinder is basically a lower-dimensional quadric embedded in a higher dimension (since the unspecified variable gives it the freedom of extrusion).

The combination of signature and type (parabolic/non-parabolic) defines a class of quadrics, and may be homogenous or not (constant term is 1 or 0). Self-dual signatures give rise to two members, homogenous and non-homogenous; non-self-dual signatures, if considered as a pair with their dual, gives 3 members: homogenous, non-homogenous, and dual non-homogenous.

There are also 3 types of degenerates: product of two coincident hyperplanes (which gives a perfect square (ax+y)^2=0); product of two parallel hyperplanes, and product of two non-parallel hyperplanes.

By counting the number of possibilities in each dimension, I obtained the number of quadrics in each dimension as P(n) = n^2+n-1. If we exclude the degenerates, then the number of quadrics is Q(n) = 0 for n=1, and Q(n) = n^2+n-3 for n>1. Q(1)=0 because 1D only has the degenerate quadric (x/a)^2=1. (I'm not 100% confident about Q(n) for n>1, though, I think i may have made a mistake in solving the recurrence.)

Anyway, the 4D quadrics are classified thus:

Category 1a: (++++): the 3-ellipsoid: only 1 member.

Category 1b: three types: (+++-), (++--), and (+---).
(+++-) is dual to (+---), and gives 3 types of quadrics: w^2+x^2+y^2-z^2=1, w^2-x^2-y^2-z^2=1 (dual to each other), and the homogenous case: w^2+x^2+y^2-z^2=0 (shape of the light cone).
(++--) is self-dual; giving us two members: w^2+x^2-y^2-z^2=0 and w^2+x^2-y^2-z^2=1.
So there are 5 members in category 1b.

Category 2: parabolics, with signature (+++) and (++-), giving us w^2+x^2+y^2-z = 0, and w^2+x^2-y^2-z=0.

Category 3: cylinders of 3D quadrics: there are 11 of these, 2 of which are degenerate.

So in total, there are 17 kinds of quadrics in 4D, or 19 if you include the two degenerates.
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