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.
ItI show that it is the only two dimensional component. Proof:
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 want to understandnote that $\mathbb{R} Z \setminus b(S^2)$. We break this up into two sets:$b(\mathbb{C} S^2)$ is Zariski dense in $Z \setminus b(\mathbb{C} S^2)$$\mathbb{C} Z$, and $(b(\mathbb{C} S^2) \cap \mathbb{R}^3) \setminus b(S^2)$.
Note thatso the points of $\mathbb{C} Z \setminus b(\mathbb{C} S^2)$ isare a complex variety of complex dimension $\leq 1$. So its, and thus the real points, which make up $Z \setminus b(\mathbb{C} S^2)$, are of $\mathbb{R} Z \setminus b(\mathbb{C} S^2)$ have real dimension at most $\leq 1$$2$.
Now, considerIt remains to bound the dimension of $b(\mathbb{C} S^2) \cap \mathbb{R}^3$$(\mathbb{R} Z \cap b(\mathbb{C} S^2)) \setminus b(S^2)$. Suppose
Everywhere on $S^2$, we have the polynomial identity that the vectors $(x,y,z) \in \mathbb{C} S^2 \setminus S^2$$(x,y,z)$ and $b(x,y,z) \in \mathbb{R}^3$$(\nabla F)|_{b(x,y,z)}$ are parallel. Then $b(x,y,z) = b(\bar{x}, \bar{y}, \bar{z})$Therefore, where thethis identity remains true on $\bar{\phantom{Z}}$ means complex conjugate, since$\mathbb{C} S^2$. $b$ has real coefficients(The zero vector is parallel to every vector.)
Let $R$ be the set of points$(u,v,w) \in \mathbb{C} Z$ and suppose $(x,y,z) \in \mathbb{C} S^2$ for which$(\nabla F)^2|_{(u,v,w)}$ is nonzero, then there exists someat most two possible preimages for $(x', y', z') \in \mathbb{C} S^2$ with$(u,v,w)$ in $(x,y,z) \neq (x',y',z')$ and$\mathbb{C} S^2$: the points $b(x,y,z) = b(x',y',z')$$\pm \tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$. ThenMoreover, the identity $R$ is a complex variety of dimension$(x,y,z) \cdot (b(x,y,z) - b(-x,-y,-z)) = 2 (\mathrm{width})$ holds everywhere on $\leq 1$$\mathbb{C} S^2$, so we never have $b(R)$ is as well$b(x,y,z) = b(-x,-y,-z)$. Once again, the real pointsSo there is actually only one preimage of $b(R)$ have dimension$(u,v,w)$ in $\leq 1$$\mathbb{C} S^2$.
I can't rule outFurthermore, suppose that $(u,v,w)$ is real. Then the possibilityformula $\pm \tfrac{\nabla F}{\sqrt{(\nabla F)^2}}$ is manifestly real. We have shown that, at any point of $Z$ has some isolated points$\mathbb{R} Z$ where $\nabla F$ is nonzero, or even an isolated curveif that point is in $b(\mathbb{C} S^2)$, away fromthen it is in $b(S^2)$. Since $\nabla F$ vanishes along (at most) a curve, but this is enough to showconcludes 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 interioronly two dimension component of $b(\mathbb{C} S^2) \cap \mathbb{R}^3$ is $b(S^2)$. (And this is wheretrue for other complex varieties defined over $F$$\mathbb{R}$; it isn't special to the sphere.) But I did a crummy job explaining why $b$ is negativegenerically 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.