Does there exist an irreducible polynomial $f \in \mathbb{R}[x, y, z]$ such that:

$$ V := \{ (x, y, z) \in \mathbb{R}^3 : f(x, y, z) \leq 0 \} $$

is a solid of constant width with a finite symmetry group?

The analogous result is true in two dimensions, with an explicit degree-$8$ example given in this paper. If we revolve this curve about its axis of symmetry (or equivalently replace every instance of $y^2$ with $y^2 + z^2$), we would obtain a degree-$8$ surface of constant width depicted below:

axisymmetric solid of constant width

This has an infinite symmetry group isomorphic to $O(1)$, so does not answer the question. Similarly, the sphere:

$$ f(x, y, z) = x^2 + y^2 + z^2 - 1 \leq 0 $$

has symmetry group $O(3)$, which is again infinite. Does there exist an algebraic solid of constant width and finite automorphism group?


There exist many algebraic surfaces of constant breadth with no continuous symmetries, even ones with no symmetries at all. To see this, consider the properties of the support parametrization:

The support parametrization. Let $B\subset\mathbb{R}^3$ be a smooth, strongly convex surface, i.e., the outward Gauss map $u:B\to S^2$ is a diffeomorphism. Since it is a diffeomorphism, there is a smooth inverse $b:S^2 \to B$ that is also a diffeomorphism. Since, for $u\in S^2$, the (supporting) tangent plane to $B$ at $b(u)$ is orthogonal to $u$, it can be written in the form $x\cdot u = p(u)$, where $p: S^2\to \mathbb{R}$ is a smooth function on $S^2$. Then, of course, one has $b(u)\cdot u = p(u)$ and one also has $\mathrm{d}b\cdot u = 0$ (since the differential of $b$ at $u\in S^2$ has its image being the subspace parallel to the tangent plane to $B$ at $b(u)$, which, as we have seen, is perpendicular to $u$). The two equations $b(u)\cdot u = p(u)$ and $\mathrm{d}b\cdot u = 0$ then determine $b$ in terms of $p$: One has the vector equation $$ b = p\,u + \nabla p, $$ where $\nabla p$ is the gradient vector field of $p$ as a function on $S^2$. The function $p$ is called the support function of $B$, and the above parametrization is called the support parametrization. (Of course, there is nothing special about $3$ dimensions here, the support parametrization works for strongly convex hypersurfaces in all dimensions.)

Surfaces of Constant Breadth. Now, consider the two supporting tangent planes to $B$ that are perpendicular to $u$: They are the tangent plane at $b(u)$ and the tangent plane at $b(-u)$. They have equations $x\cdot u = p(u)$ and $x\cdot (-u) = p(-u)$ respectively. In particular, the distance between these two planes is $p(u)+p(-u)$. Thus, $B$ has constant breadth $2d$ if and only if its support function $p$ satisfies $p(u)+p(-u) = 2d$. In particular, the function $p_0=p-d$ must be an odd function on $S^2$, i.e., $p_0(-u) = -p_0(u)$ for all $u\in S^2$.

Conversely, if $p_0:S^2\to\mathbb{R}$ is an odd function, then $p = p_0+d$ for $d>0$ sufficiently large is then the support function of a strongly convex body of constant breadth $2d$. Alternatively, for all sufficiently small $\epsilon>0$, the function $p = 1 + \epsilon p_0$ is the support function of a strongly convex surface $B_\epsilon$ of constant breadth $2$.

An algebraic example. To construct an algebraic example, it suffices to take $p_0$ to be an odd algebraic function on $S^2$. Taking $p_0$ to be the restriction of a linear function in $\mathbb{R}^3$ to $S^2$ will only give a surface that is a translation of the round $S^2$, so we need to look at something else.

The next simplest choice would be $p_0 = xyz$, where $S^2$ is defined by $x^2+y^2+z^2=1$. For $\epsilon^2<1$, the surface $B_\epsilon=b_\epsilon(S^2)$ is a surface of constant breadth $2$ that is smooth, strongly convex, and algebraic. (When $\epsilon^2=1$, the surface $B_1$ is still strictly convex and algebraic and has breadth $2$, but it is not smooth. When $\epsilon^2>1$ the image $b_\epsilon(S^2)$ is not a convex surface.) When $\epsilon\not=0$, it is easy to show that $B_\epsilon$ has the symmetry of the function $xyz$ but no more symmetry than that, i.e., the symmetry group will have (finite) order $24$ (in fact, it is the symmetry group of the regular tetrahedron.) It is algebraic because it is parametrized by algebraic functions on the algebraic surface $S^2$. It is therefore (a component of) the zero locus of an irreducible polynomial function $f_\epsilon(x,y,z)$ on $\mathbb{R}^3$. A calculation using elimination theory (carried out by MAPLE) shows that $f_\epsilon$ has degree $20$ (when $\epsilon\not=0$) and, for most values of $\epsilon$, has more than 1000 terms.

It is not obvious that the zero locus of $f_\epsilon$ has no other (real) components other than $B_\epsilon$. Thus, I don't know that the interior of $B_\epsilon$ (which makes sense as long as $\epsilon^2\le 1$) can be characterized as the set where $f_\epsilon$ is negative. However, I don't know why you want this; it doesn't seem natural to me. Surely, you only really need it to be algebraic. In any case, one has an explicit parametrization of $B_\epsilon$ using any explicit parametrization of $S^2$. It is easy to draw pictures using graphical software. Here are what the surfaces $B_{1/2}$ and $B_1$ look like:

epsilon = 1/2 epsilon = 1

Added comment: David Speyer has shown that $f_\epsilon$ cannot change sign anywhere other than across the image $b_\epsilon(S^2)$, since, as he shows, this is the only two-dimensional component of the real locus of $f_\epsilon=0$.

More generally, taking $p = 1+p_0$, where $p_0 = \lambda_1\lambda_2\lambda_3$ and where each $\lambda_i$ is a linear function of $x$, $y$, and $z$, produces a $7$-parameter family of distinct algebraic surfaces of constant breadth $2$, and the generic one of these has no nontrivial symmetries. (In fact, one only gets continuous symmetries when $p_0 = \lambda^3$ for some linear function $\lambda$.)

  • 2
    $\begingroup$ I left an answer addressing the point in your last paragraph. In short, the interior of $B$ is the region where $F$ is negative, although I can't rule out that $F$ also drops down to $0$ without becoming negative at some other points. $\endgroup$ – David E Speyer Oct 31 '15 at 4:24
  • $\begingroup$ @DavidSpeyer: Thanks! That's a nice observation. $\endgroup$ – Robert Bryant Oct 31 '15 at 9:08

This is a comment on Robert Bryant's answer, addressing the point in his last paragraph. It is substantially rewritten from a previous answer, because I realized that the best way to address the Bryant's question below is to rewrite the answer.

Bryant constructs constant width surfaces as the image of the sphere $S^2 = \{ (x,y,z) : x^2+y^2+z^2=1 \}$ under a polynomial map $b: S^2 \to \mathbb{R}^3$. He points out that the image will be a connected component of $F=0$ for some polynomial $F$, but says that he does not know whether it is the only component.

I show that it is the only two dimensional component.

Let $\mathbb{C} S^2$ be the complex solutions to $x^2+y^2+z^2=1$. Let $\mathbb{R} Z$ be the real points of $F=0$, and $\mathbb{C} Z$ the complex points. We note that $b(\mathbb{C} S^2)$ is Zariski dense in $\mathbb{C} Z$, so the points of $\mathbb{C} Z \setminus b(\mathbb{C} S^2)$ are a complex variety of complex dimension $\leq 1$, and thus the real points of $\mathbb{R} Z \setminus b(\mathbb{C} S^2)$ have real dimension at most $2$.

It remains to bound the dimension of $(\mathbb{R} Z \cap b(\mathbb{C} S^2)) \setminus b(S^2)$.

Everywhere on $S^2$, we have the polynomial identity that the vectors $(x,y,z)$ and $(\nabla F)|_{b(x,y,z)}$ are parallel. Therefore, this identity remains true on $\mathbb{C} S^2$. (The zero vector is parallel to every vector.)

Let $(u,v,w) \in \mathbb{C} Z$ and suppose $(\nabla F)^2|_{(u,v,w)}$ is nonzero, then there at most two possible preimages for $(u,v,w)$ in $\mathbb{C} S^2$: the points $\pm \tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$. Moreover, the identity $(x,y,z) \cdot (b(x,y,z) - b(-x,-y,-z)) = 2 (\mathrm{width})$ holds everywhere on $\mathbb{C} S^2$, so we never have $b(x,y,z) = b(-x,-y,-z)$. So there is actually only one preimage of $(u,v,w)$ in $\mathbb{C} S^2$.

Furthermore, suppose that $(u,v,w)$ is real. Then the formula $\pm \tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$ is manifestly real. We have shown that, at any point of $\mathbb{R} Z$ where $\nabla F$ is nonzero, if that point is in $b(\mathbb{C} S^2)$, then it is in $b(S^2)$. Since $\nabla F$ vanishes along (at most) a curve, this concludes the proof.

A previous version of this answer explained more generally that, if a map $b: \mathbb{C} S^2 \to \mathbb{C}^3$ is generically injective, then the only two dimension component of $b(\mathbb{C} S^2) \cap \mathbb{R}^3$ is $b(S^2)$. (And this is true for other complex varieties defined over $\mathbb{R}$; it isn't special to the sphere.) But I did a crummy job explaining why $b$ is generically injective and, once I fixed that, it was easier to directly point out that $\tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$ is real when defined.

  • $\begingroup$ One question: It's not clear to me why the complex variety $R$ has dimension at most $1$. I see it this way: we know that $b:\mathbb{C}S^2\to\mathbb{C}Z$ is an (algebraic) map between complex surfaces in $\mathbb{C}^3$, and I agree that the image of $b$ is Zariski dense, so that the complement $\mathbb{C}Z\setminus b(\mathbb{C}S^2)$ is a variety of complex dimension at most one. However, I don't see why $b$ can't have degree higher than $1$, so that it multiply covers some Zariski dense subset of $\mathbb{C}Z$. If this happens, then $R$ will be Zariski dense in $\mathbb{C}S^2$, won't it? $\endgroup$ – Robert Bryant Oct 31 '15 at 16:08
  • $\begingroup$ Thanks for the explanation of why $b$ is generically injective (which I couldn't find in your original answer). That clears up my concern. $\endgroup$ – Robert Bryant Nov 1 '15 at 13:05

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