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Conic sections are algebraic curves that can be constructed from different cross-sections of a cone in $\mathbb{R}^3$. Do there exist other geometric shapes in $\mathbb{R}^3$ besides the cone whos cross-sections can yield algebraic curves analogous to conic sections? Are there higher dimensional equivalents for 3D curves or sheets in 3D that can be defined in by some geometric figure like a cone (or potential alternative geometric figures) in $\mathbb{R}^4$?

Maybe I can be a bit more concrete. Conic sections are an example of a two-variable second degree polynomial of the form $$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. $$ I can construct a similar two-variable higher-degree polynomials as such $$ \text{Curve}_n(x,y) = A_0x^n + A_2x^{n-1}y + \ldots + A_{n-1}xy^{n-1} + A_ny^n + \text{Constant} = 0 $$ Does there exist something analogous to a "cone" for such a polynomial? What if I add another variable $z$?

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    $\begingroup$ "Do there exist other geometric shapes in $\mathbb{R}^3$ besides the cone?" Are you friggin kidding me?!? $\endgroup$ Commented Dec 28, 2020 at 17:50
  • $\begingroup$ The way I phrased that sentence I agree is not clear. Thank you for bringing that up, I will change it @DavidG.Stork. I think a better way to ask is, are there other geometric shapes in $\mathbb{R}^3$ where cross-sections of that shape will give you curves like conic sections? $\endgroup$
    – firest
    Commented Dec 28, 2020 at 18:02
  • $\begingroup$ The remark by David remains: yes, there are geometric shapes in $\mathbb R^3$ with cross sections that are curves !! Your question is still too vague. $\endgroup$
    – user65203
    Commented Dec 28, 2020 at 19:54
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    $\begingroup$ You are probably right @YvesDaoust. Please note that I am not a mathematician nor did I ever get a degree in mathematics so asking questions that isn't my strong suite may inevitably lead to ambiguity. This ignorance is my fault. I guess, as far as I am aware, conic sections are the only kind of algebraic curve I have ever learned that are associated with some 3d object. This makes me wonder if there is anything special with the cone, or if there a whole class of similar curves that I am unaware of. $\endgroup$
    – firest
    Commented Dec 28, 2020 at 23:33

2 Answers 2

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Let us consider a torus (2nd figure) with a vertical axis of revolution with $c$ = main radius, $d$ = radius of revolving circle ($c>d$ for a non-self intersecting torus).

The sections by vertical planes at distance $f$ from the axis of revolution are called "spiric sections".

For certain combinations of $c,d,f$ (see Appendix below) we obtain Cassinian ovals (first figure). Recall: A Cassinian oval $(C_b)$ with foci $F_1, F_2$ is the set of points $M$ such that $MF_1.MF_2=b^2$.

Please note

    1. that this locus has 2 connected components if $b<\frac{F_1F_2}{2}$,
    1. That the lemniscate obtained in the limit case where $b=\frac{F_1F_2}{2}$ is a Booth's lemniscate (not in general a Bernoulli's lemniscate as I thought at first),
    1. the similarity with the definition of an ellipse (with product of lengths instead of their sum).

Reference: https://www.lucamoroni.it/toric-sections/

enter image description here

Appendix: When vertical sections of a torus are Cassinian ovals ?

The torus equation is

$$(\sqrt{x^2 + y^2} - d)^2 + z^2 = c^2$$

Expanding the square, isolating the square root and squaring again, we can give it the fourth degree implicit polynomial equation:

$$(x^2 + y^2 + z^2-c^2)^2=4d^2(x^2+y^2)\tag{1}$$

Besides, by definition, the Cassinian oval implicit equation in a vertical plane parallel to any plane with equation $y=k$ is

$$\sqrt{(x-a)^2+z^2}\sqrt{(x+a)^2+z^2}=b^2$$

Squaring this relationship and expanding it, we get:

$$(x^2+z^2+a^2-2ax)(x^2+z^2+a^2+2ax)=b^4$$

$$(x^2+z^2+a^2)^2-4a^2x^2=b^4\tag{2}$$

It remain to express that the section of (1) by vertical plane with equation $y=f$ has equation:

$$(x^2 + f^2 + z^2-c^2)^2=4d^2(x^2+f^2)\tag{3}$$

Therefore (2) and (3) can be identified under the following conditions:

$$f^2-c^2=a^2, \ \ a=d, \ \ b^2=2df.$$

meaning that if $c,d$ are given,

$$f=\sqrt{c^2+d^2} \ \ \text{and} \ \ b=\sqrt{2df}$$

Therefore, for a given torus, there is exactly one vertical section that can be a Cassinian oval ; moreover, due to the fact that $f>c$, this section has a single component.


Edit: Another example, this time with a cubic curve called the folium of Descartes, with equation:

$$x^3+y^3-3xy=0$$

that can be considered as the intersection of surface defined by $$z=-\operatorname{atan}(x^3+y^3-3xy)$$ with plane $z=0$ (see Fig.) (with the interest that the surface is bounded ($-\dfrac{\pi}{2}<z<\dfrac{\pi}{2}$).

We could as well have taken $z=x^3+y^3-3xy$ or $z=|x^3+y^3-3xy|$, etc.

enter image description here

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  • $\begingroup$ Thank you. I had not considered something like the torus and will dedicate some time studying it (your explanation is a good starting point). I have a(two) follow up question(s) if you don't mind. Can all planar curves be classified as cross-sections of different 3d objects like the torus or cone? And how many 3d objects like this exist? Please forgive me if my question is somewhat ambiguous or ill-posed. I am not a mathematician by training. I just think this stuff is cool. $\endgroup$
    – firest
    Commented Dec 28, 2020 at 23:51
  • $\begingroup$ To your first question, one can say "yes". If we consider planar curves defined by an implicit equation $ f(x,y)=0$, one can associate to it the surface with equation $z=f(x,y)$: the curve will be the intersection of this surface with the horizontal plane with equation $z=0$. To your second question : yes, it general it can be done in multiple ways. I will attempt to give examples as an edit to my answer. $\endgroup$
    – Jean Marie
    Commented Dec 29, 2020 at 0:12
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    $\begingroup$ Nice diagrams too. If you do not mind asking, how do you get these? $\endgroup$
    – Prags
    Commented Dec 30, 2020 at 7:19
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    $\begingroup$ @Prags No problem: I use Matlab. $\endgroup$
    – Jean Marie
    Commented Dec 30, 2020 at 8:54
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A 3D generalization of the conic curves are the quadric surfaces, of general equation

$$Ax^2+By^2+Cz^2+2Dxy+2Eyz+2Fzx+2Gx+2Hy+2Iz+J=0.$$

There are nine different types, not counting the degenerate ones (I would not count the sphere as different of the ellipsoid).

enter image description here

They can be seen as sections of an hypercone by an hyperplane,

$$\begin{cases}Ax^2+By^2+Cz^2+Jw^2+2Dxy+2Eyz+2Fzx+2Gxw+2Hyw+2Izw=0,\\w=1.\end{cases}$$

A cross-section of a quadric by a plane is a conic. Nothing new.

The intersection of two quadrics is a 3D quartic curve, forming a hair-raisingly complex family.

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    $\begingroup$ ... but I am afraid that those intersections of quadrics that are planar (as asked by the OP) are still conic curves. $\endgroup$
    – Jean Marie
    Commented Dec 28, 2020 at 23:12
  • $\begingroup$ @JeanMarie: the OP did not specifically request planar curves in the $\mathbb R^4$ setting (he was more focused on conic-like) and I did mention that sections of quadrics by planes are conics. So I question the relevance of your comment. $\endgroup$
    – user65203
    Commented Dec 29, 2020 at 8:23
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    $\begingroup$ The adjective "planar" is mentionned in the title. $\endgroup$
    – Jean Marie
    Commented Dec 29, 2020 at 10:19
  • $\begingroup$ @JeanMarie: do you deny that I wrote "A cross-section of a quadric by a plane is a conic. Nothing new." ? $\endgroup$
    – user65203
    Commented Dec 29, 2020 at 10:39
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    $\begingroup$ I deny nothing! I just wanted to say that the intersection of two quadrics $Q_1$ and $Q_2$ is either a non planar curve, or, if it is planar, and if we call $P$ the corresponding plane, $Q_1 \cap Q_2=Q_1 \cap P$ (the "nothing new" case) $\endgroup$
    – Jean Marie
    Commented Dec 29, 2020 at 11:30

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