TITLE: Algebraic surface of constant width?
QUESTION [15 upvotes]: 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:

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?

REPLY [14 votes]: 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.