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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?

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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 $\lambda$.)

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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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