# What surfaces in $\mathbb R^3$ are such that every planar section (with more than 1 point) has nontrivial symmetry?

In $\mathbb R^3$ , the intersection of a plane and a sphere (e.g. $x^2 + y^2 + z^2 = 1$) is either empty, a single point, or a circle. All isometries of those circles are realized by isometries of the full sphere. In contrast, every plane intersects a cone (e.g. $x^2 + y^2 - z^2 = 0$) in a conic section which has reflection symmetries along one, two, or more axes. The one reflection is realized by an isometry of the cone, but the second, in general, is not.

What surfaces in $\mathbb R^3$ , or general subsets of $\mathbb R^3$ , are such that all non-empty planar intersections are either 1 point or have nontrivial symmetry? When are those symmetries not extensible to a symmetry of the full surface?

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I think in the sphere example you need reflections of the sphere as well as rotations. –  user31373 Jul 20 '12 at 18:21
Do all conic sections have two reflection symmetries? Seems like the parabolas do not. –  Thomas Andrews Jul 20 '12 at 18:28
A point has trivial symmetry? What do you mean by nontrivial? –  tomasz Jul 20 '12 at 18:29
@tomasz He wrote "such that all non-empty planar intersections are either 1 point or have nontrivial symmetry?" –  Thomas Andrews Jul 20 '12 at 18:30
@tomasz You are wrong, there are no double parabolas that can be gotten via conic sections. The parabolas are limit points of the ellipses. The hyperbolas are the intersections of a plane with both directions of the cone. –  Thomas Andrews Jul 20 '12 at 18:31

I suspect the answer is that the surface in $\mathbb R^3$ must be a homogeneous quadratic form in $(x,y,z)$ set equal to a constant, as in $$a x^2 + b y^2 + c z^2 + r y z + s z x + t x y = k.$$ For you, all of $a,b,c,r,s,t,k$ are real numbers. There is little benefit to allowing lower order terms as in $\alpha x + \beta y + \gamma z,$ a translation takes the result to center on the origin again. For that matter, if you know enough linear algebra, a rotation takes this surface into a diagonalized one, $$a_1 x_1^2 + b_1 y_1^2 + c_1 z_1^2 = k_1,$$ where some of $a_1,b_1,c_1,k_1$ may be positive, some $0,$ some negative. You might try graphing these with all $a_1,b_1,c_1,k_1 \in \{-1,0,1 \},$ there are 16 possibilities but there is repetition, so maybe take $a_1 \leq b_1 \leq c_1$ but any $k_1.$

Note that the intersection of any plane with this surface is, in coordinates appropriate for that plane, and orthogonal coordinates if we demand it, a quadratic in two variables $u,v$ call it $A u^2 + B u v + C v^2 + D u + Ev = F.$ This has symmetries, or may be a single point, or a line or pair of lines, and so on.

Well, these examples work. I do not expect there will be any others. For most planes through one of these surfaces, the symmetry of the figure within the plane will not extend to the whole surface. Finally, a proof that only these surfaces work would be pretty elaborate.

EDIT: user mjqxxxx gave some I had not considered, the one that is connected and new is: take any curve in the $xy$ plane that has a symmetry, a reflection or $180^\circ$ rotation or something. Then construct the cylinder over it, meaning take the same figure with arbitrary $z.$ Any plane slice preserves the symmetry. Now, my current opinion is that if we take something in the plane that has only a $120^\circ$ rotation symmetry, some kind of pinwheel, after making the cylinder most slanted planes through the cylinder will not preserve that symmetry. Need to thinnmmnnkk. That would be a nice result, though, connected examples are either full-dimensional quadratic things or cylinders over lower-dimensional examples. EDDITTT: more in comments below. This is getting out of hand. The second example that is essentially lower-dimensional is take any curve in the $xy$ plane with $y \geq 0$ and rotate it around the $x$-axis.

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For ellipse conic sections, the second non-extending symmetry is interesting. It does extend to a unique isometry of $\mathbb{R}^3$ which moves the cone. Perhaps this motion of the cone which preserves the planar section can be better understood when the cone is viewed as a complex surface in $\mathbb{C}^3$? –  Obidiah Jul 20 '12 at 20:23
An infinite family of parallel cylinders? Or indeed any surface with continuous translational symmetry along one axis and a lattice-translational symmetry along the other two. Every slice will inherit either the continuous or the lattice symmetry, depending on whether it is parallel to the continuous symmetry axis. –  mjqxxxx Jul 20 '12 at 20:26
Here's another generic class: any surface with continuous rotational symmetry around an axis. Every slice will have at least one reflection symmetry remaining. –  mjqxxxx Jul 20 '12 at 20:58
No, like an infinite vase or something. For instance, take any curve at all in the $xz$-plane and rotate it around the $z$-axis; the surface swept out will work. –  mjqxxxx Jul 20 '12 at 21:38
@mjqxxxx, that's a good one. So, as with a cylinder over a curve, a connected figure is generated by a mostly arbitrary but lower dimensional figure. –  Will Jagy Jul 20 '12 at 21:46