19
$\begingroup$

Given a surface $f(x,y,z)=0$, how could you determine that it's symmetric about some plane, and, if so, how would you find this plane.

The special case where $f$ is a polynomial is of some interest.

The question is somewhat related to this one: How to identify surfaces of revolution

Here's a simple 2D example to experiment with: $$ 27 x^3 + 108 x^2 y + 144 x y^2 + 64 y^3 - 80 x^2 + 120 x y - 45 y^2 - 200 x + 150 y - 125 = 0$$ This one is symmetric, as the following picture suggests:

picture

In fact, if we do the translation/rotation described by the substitution $u = \tfrac15(3x+4y)$, $v = \tfrac15(4x−3y) + 1$, then the curve is just $u^3 - v^2 = 0$, which is obviously symmetric about the line $v=0$.

But how would you discover this translation/rotation if I didn't tell you, and how would you do similar things in the 3D surface case?

Added after a few days of thought:

We can consider the surface as an object made of thin sheet metal. As such, it has a center of mass, provided it's bounded, and any plane(s) of symmetry must pass though this center of mass. The plane then has only two remained degrees of freedom, so may be easier to find.

The same sort of reasoning applies in the case of a bounded 2D curve. Again, any line of symmetry must pass through the curve's centroid, so it has only one remaining degree of freedom, namely its slope/angle.

For curves and surfaces given by implicit equations, I don't really know how to calculate centroids, but I expect this can be done.

Fabricated from Comments Below:

Several people suggested looking at highest-degree terms only. So, in my example, we just look at the equation $$ 27 x^3 + 108 x^2 y + 144 x y^2 + 64 y^3 = 0 $$ Putting $w = y/x$, this is roughly equivalent to $$ 64w^3 + 144w^2 + 108 w + 27 = 0 $$ But the polynomial on the left is just $(4w + 3)^3$, so we have a root $w = -3/4$, with multiplicity three. Is the repeated root an accident that happens only in this case, or will it always happen?? Anyway, the vector $(-4,3)$ gives us the normal to the line of symmetry, and that surely can not be an accident.

I don't really understand why this magic process works, but it looks very promising for the 2D curve case.

I don't know how to generalize to the 3D surface case, or to non-polynomial cases.

$\endgroup$
  • 1
    $\begingroup$ For a polynomial, taking the Laplacian reduces the order but can only increase the symmetry. Eventually the order reduces to 1 or 2, both of which cases are easy. This gives you some candidate symmetries. Maybe we can make an algorithm from this approach? $\endgroup$ – apt1002 Dec 5 '13 at 2:26
  • 1
    $\begingroup$ In this case we get $162x+216y+288x+384y-160-90=0$ which simplifies to $50(9x+12y-5)=0$, so a candidate plane is $9x+12y-5=0$. $\endgroup$ – apt1002 Dec 5 '13 at 2:35
  • 1
    $\begingroup$ @apt1002 -- Good progress. Thanks. The plane of symmetry is actually $4x - 3y +5=0$, according to my calculations. This is perpendicular to the plane you gave. I checked your computation of the Laplacian, and it seems to be correct. $\endgroup$ – bubba Dec 5 '13 at 3:11
  • 1
    $\begingroup$ It seems that the hard part is to find a rotation matrix that brings the symmetry plane parallel to axes. For this, only the terms of the polynomial of maximal degree are relevant. To see this consider very large $x$ and/or $y$. $\endgroup$ – apt1002 Dec 5 '13 at 3:51
  • 1
    $\begingroup$ @apt1002: That's a good idea. We can take only the highest-order terms and then, at least in 2D, replace $x$ and $y$ with $\cos\theta$ and $\sin\theta$ to get a function $g(\theta)$. The value of $\theta$ about which $g$ is symmetric is the only possible direction of the symmetry axis. (Note that $2\tan^{-1}(1/2) = \tan^{-1}(4/3)$.) $\endgroup$ – user856 Dec 11 '13 at 0:51
3
+50
$\begingroup$

Here is an answer to this question in the case of compact surfaces (without boundary); maybe these ideas can be used in the general case as well. Let $F$ be a compact surface in $R^3$, bounding a solid $S$. In the setting you are interested in, one will probably have $F=\{x: f(x)=c\}$ and $S=\{x: f(x)\le c\}$. I will also assume that $f$ is a polynomial (I do not think it is really necessary, but I use this assumption to detect surfaces of revolution).

The solid $S$ then has a Loewner-Jones ellipsoid, which is the unique ellipsoid $E$ of the least volume containing $F$ (in fact, every compact in $R^n$ with not contained in a hyperplane has such ellipsoid). This ellipsoid is also known as Jones ellipsoid and least volume ellipsoid. It was discovered by Loewner and then rediscovered by Jones, if I remember correctly. People also use the largest volume ellipsoid contained in $S$, that one will also work.

The key is that, in view of uniqueness of $E$, every symmetry of $F$ is also a symmetry of $E$. There is a vast literature in the computational math community describing various algorithms for finding $E$, here is just a random paper on this subject which I found by googling. Now, suppose you are lucky and your ellipsoid has distinct (beyond the margin of error in the algorithm you would be using) principal axes. Then $E$ has exactly 3 planes of symmetry which pass through its center $C$ and pairs of axes. Then you test if symmetries in these planes preserve $S$.

Now, suppose you are unlucky and $E$ has a rotational symmetry in its axis $A$ (but is not the sphere). By intersecting $S$ with the plane $P$ orthogonal to $A$, we now get a 2-dimensional problem: Given a (closed and bounded) curve $\Gamma$ in the plane and a center $C$, determine if $\Gamma$ has a reflection symmetry in a line $L$ (through $C$). This is relatively easy since each point of intersection of $\Gamma$ with $L$ is a critical point of the distance function from $C$, so you can find such points using Lagrange multipliers and then check if any of the corresponding symmetries preserve $F$. This method will fail if $\Gamma$ is a circle. Instead of intersecting $F$ with planes passing through $C$, we can use other planes $P_k$ orthogonal to $L$. Now, if too many (comparing to the degree of $f$) of these intersections with $F$ are circles, then $F$ is a surface of revolution with the axis $A$. (I think, it should follow from Bezout theorem, but this has to be checked.) In this case, of course, any plane through $A$ will work.

The remaining case when $E$ is a sphere is similar to the circular case in one dimension lower, you would be then looking for critical points of the distance function from the origin.

Remark: The argument with Lowener-Jones ellipsoid should also handle your second question, about surfaces of revolution, in the bounded case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your answer. I haven't fully absorbed it, yet, but a comment ... Your basic idea (as I understand it so far) is to replace the surface or its related solid with an ellipsoid, and focus on the planes/axes of symmetry of this ellipsoid. This seems reasonable. But I wonder if we can use the ellipsoid of inertia, instead of the Loewner-Jones ellipsoid. This might be easier to obtain, and would give a connection with eigenvalues and principal axes, which (I suspect) are the keys to questions about symmetry. $\endgroup$ – bubba Dec 13 '13 at 6:08
  • $\begingroup$ @bubba: Yes, any canonical ellipsoid would work, I just happened to know L-J better. $\endgroup$ – Moishe Kohan Dec 13 '13 at 14:55
  • $\begingroup$ There a connection with the special case of quadrics. For an ellipsoid (the only compact quadric?) the ellipsoid of inertia is the ellipsoid itself. I suppose the same is true for the L-J ellipsoid. $\endgroup$ – bubba Dec 14 '13 at 2:14
  • $\begingroup$ @bubba: Yes, the same is true for L-J ellipsoid (for obvious reason). Yes, ellipsoid is the only compact quadric. $\endgroup$ – Moishe Kohan Dec 14 '13 at 8:13
  • $\begingroup$ I gave you the bounty, but this still doesn't feel like the "right" answer, to me, so I didn't accept. I think a better answer could come from the approach mentioned in the comments by apt1002, Rahul Nahrain, and Blue. I tried to summarize their ideas in a few paragraphs I added at the end of the question, but it's still very foggy, for me. Thanks again for your efforts. $\endgroup$ – bubba Dec 15 '13 at 6:12
2
$\begingroup$

Here's a stab at it. You can translate your question into whether or not a system of certain polynomial equations has a real solution.

A generic rotation can be described by choosing an axis using two angles in spherical coordinates, and then choosing a third angle to rotate by. The matrix for such a rotation has entries that are quartic polynomials in six variables that represent the sines and cosines of the three angles.

If your surface has a plane of symmetry, then for the right choice of these three angles, followed by a translation in the $z$-axis direction, you should be left with a polynomial that has only even powers of $z$.

If you give names to the seven unknowns (six trig values and one translation value), and apply the generic rotation and translation to the defining equation of the surface, then we can isolate the coefficients of $z,xz,yz,x^2z,z^3,z^5,$ etc. and see if we can solve for them all simultaneously being $0$. We already have three quadratic relations on the six trig variables. If we get equations in the seven unknowns corresponding to $z,xz,yz,x^2z,z^3,z^5$, etc (there will only be finitely many) then we have a system of polynomial equations in seven unknowns.

Such a system is consistent/inconsistent in $\mathbb{C}$ according to whether certain algebraic conditions are met on the equations by Hilbert's nullstellensatz. I think a CAS would be able to determine the nullstellensatz criterion, but I would consult an expert. I'm not sure what can be done about $\mathbb{R}$.


In 2D with your example, it would go like this, using $c=\cos\theta$, $s=\sin\theta$, and $a$ for the translation variable in the $y$-direction:

$$27 x^3 + 108 x^2 y + 144 x y^2 + 64 y^3 - 80 x^2 + 120 x y - 45 y^2 - 200 x + 150 y - 125 = 0$$

Apply rotation, followed by translation:

$$27(xc+(y+a)s)^3 +108 (xc+(y+a)s)^2(xs-(y+a)c) +144(xc+(y+a)s)(xs-(y+a)c)^2 +64(xs-(y+a)c)^3 -80(xc+(y+a)s)^2 +120(xc+(y+a)s)(xs-(y+a)c) -45(xs-(y+a)c)^2 -200(xc+(y+a)s) +150(xs-(y+a)c)-125 = 0$$

I cheat and have a CAS expand:

$$27 a^3 s^3+81 a^2 c s^2 x+81 a^2 s^3 y+81 a c^2 s x^2+162 a c s^2 x y+81 a s^3 y^2+27 c^3 x^3+81 c^2 s x^2 y+81 c s^2 x y^2+27 s^3 y^3 -108 a^3 c s^2-216 a^2 c^2 s x-324 a^2 c s^2 y+108 a^2 s^3 x-108 a c^3 x^2-432 a c^2 s x y+216 a c s^2 x^2-324 a c s^2 y^2+216 a s^3 x y-108 c^3 x^2 y+108 c^2 s x^3-216 c^2 s x y^2+216 c s^2 x^2 y-108 c s^2 y^3+108 s^3 x y^2 +144 a^3 c^2 s+144 a^2 c^3 x+432 a^2 c^2 s y-288 a^2 c s^2 x+288 a c^3 x y-288 a c^2 s x^2+432 a c^2 s y^2-576 a c s^2 x y+144 a s^3 x^2+144 c^3 x y^2-288 c^2 s x^2 y+144 c^2 s y^3+144 c s^2 x^3-288 c s^2 x y^2+144 s^3 x^2 y -64 a^3 c^3-192 a^2 c^3 y+192 a^2 c^2 s x-192 a c^3 y^2+384 a c^2 s x y-192 a c s^2 x^2-64 c^3 y^3+192 c^2 s x y^2-192 c s^2 x^2 y+64 s^3 x^3 -80 a^2 s^2-160 a c s x-160 a s^2 y-80 c^2 x^2-160 c s x y-80 s^2 y^2 -120 a^2 c s-120 a c^2 x-240 a c s y+120 a s^2 x-120 c^2 x y+120 c s x^2-120 c s y^2+120 s^2 x y -45 a^2 c^2-90 a c^2 y+90 a c s x-45 c^2 y^2+90 c s x y-45 s^2 x^2 -200 a s-200 c x-200 s y +150xs-150yc-150ac-125=0 $$

Ignoring terms with even powers of $y$:

$$81 a^2 s^3 y+162 a c s^2 x y+81 c^2 s x^2 y+27 s^3 y^3 -324 a^2 c s^2 y-432 a c^2 s x y+216 a s^3 x y-108 c^3 x^2 y+216 c s^2 x^2 y-108 c s^2 y^3 +432 a^2 c^2 s y+288 a c^3 x y-576 a c s^2 x y-288 c^2 s x^2 y+144 c^2 s y^3+144 s^3 x^2 y -192 a^2 c^3 y+384 a c^2 s x y-64 c^3 y^3-192 c s^2 x^2 y -160 a s^2 y-160 c s x y -240 a c s y-120 c^2 x y+120 s^2 x y -90 a c^2 y+90 c s x y -200 s y -150yc=0 $$

Starting with the relation between $c$ and $s$, and then grouping coefficients of $y$, $xy$, etc setting them equal to $0$:

$$\begin{align} c^2+s^2&=1\\ 81 a^2 s^3 -324 a^2 c s^2 +432 a^2 c^2 s-192 a^2 c^3-160 a s^2-240 a c s-90 a c^2-200 s -150c&=0&(y)\\ 162 a c s^2 -432 a c^2 s +216 a s^3 +288 a c^3-576 a c s^2+384 a c^2 s-160 c s-120 c^2+120 s^2+90 c s&=0&(xy)\\ 81 c^2 s -108 c^3 +216 c s^2-288 c^2 s+144 s^3-192 c s^2&=0&(x^2 y)\\ 27 s^3-108 c s^2+144 c^2 s-64 c^3&=0&( y^3) \end{align}$$

And now the question is equivalent to asking if this system of polynomials in three variables has a solution. Again, I think this is something that a CAS can do, but I'd consult an expert first.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Presumably, you mean $c^2+s^2=1$ in your last set of equations. $\endgroup$ – Blue Dec 12 '13 at 16:17
  • 1
    $\begingroup$ One can eliminate $s$ from the very last equation: Use $s^2=1-c^2$ to reduce the $s$-degree to $1$, then solve for $s$ and substitute into $s^2+c^2=1$ (effectively computing a resultant). The polynomial then factors as $(5c-3)^3(5c+3)^3$. It turns out that $(c,s) = (3/5 k,4/5k)$, with $k\in\{1,-1\}$, solves all the $a$-less polynomials. The remaining one reduces to $1+ak=0$, so that $a = -1/k=-k$. $\endgroup$ – Blue Dec 12 '13 at 16:44
  • $\begingroup$ @Blue Re $1\ne0$, thanks! I was thinking that to be general, I didn't want to go into any specifics about how to solve this system. Actually, I think it is beyond a CAS to solve systems like this of higher degree without using numerical methods. But I'm curious to have an expert chime in on whether or not a CAS can use the nulstellensatz criterion to establish if there is a solution over $\mathbb{C}$, and then also if there is a way to make that work over $\mathbb{R}$ too. $\endgroup$ – alex.jordan Dec 12 '13 at 18:30
  • $\begingroup$ @Blue -- As you can see, your solution is consistent with the info given by Leonard and Rahul in their comments following the question. As it should be, of course. $\endgroup$ – bubba Dec 13 '13 at 6:17
  • $\begingroup$ @alex -- so, the approach is to apply a (parameterized) rotation, and then look for parameter values that produce the desired result when the rotation is applied. In this case the desired result is disappearance of odd-degree terms like $x$, $xy$, $x^2y$. I wonder if a different way of parameterizing rotation matrices (something different from angles) would make the algebra simpler. Quaternions?? $\endgroup$ – bubba Dec 13 '13 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.