Putting algebraic curves in $\mathbb{R}^3$

Question

Let $X \subset \mathbb{C}^2$ be a smooth algebraic curve. Thinking of $\mathbb{C}^2$ as $\mathbb{R}^4$, is there a smooth map $\phi: \mathbb{C}^2 \to \mathbb{R}^3$ so that $\phi: X \to \mathbb{R}^3$ is a closed injective immersion? I mean this question in two senses:

• Does the map $\phi$ exist at all?

• Can we actually write down some simple formula for it, in terms of coordinates $x_1+i y_1$, $x_2+i y_2$ on $\mathbb{C}^2$ and the defining equation $F(x_1+iy_1,x_2+iy_2)=0$ of $X$?

The motivation for this question is that I'd like to make some pretty pictures of algebraic curves.

Answer 1

Taking this as two questions, the second being more interesting than the first, I can at least answer the first (less interesting) question. The answer is 'Yes, such maps $\phi:\mathbb{C}^2\to\mathbb{R}^3$ do exist, though they (necessarily) depend on the smooth curve $X$'.

Given a smooth algebraic curve $X\subset\mathbb{C}^2$, suppose that $X$ is defined as the zero locus of an reduced polynomial $F(z,w)$ (i.e., $F$ has no multiple factors). Then, by the assumption that $X$ is smooth (by which I assume that David means 'smooth, embedded'), the polynomials, $F$, $F_z$ and $F_w$ have no common zeros. In fact, for each $x\in X$, the vector $N(x)=\bigl(\,\overline{F_z(x)},\,\overline{F_w(x)}\,\bigr)$ is nonzero and (unitarily) orthogonal to the (complex) tangent line to $X$ at $x$, since the $1$-form $\mathrm{d}F = F_z\,\mathrm{d}z+F_w\,\mathrm{d}w$ vanishes when pulled back to the curve $X$. Let $U(x)= N(x)/|N(x)|$ be the corresponding unit vector.

Now, because $X$ is smooth and algebraic, outside a compact set, it is asymptotic to a finite set of lines, and it is not difficult to see that there is a positive function $e:X\to (0,1)$ such that the mapping $S:X\times\Delta(1)\to\mathbb{C}^2$ (where $\Delta(r)\subset\mathbb{C}$ is the disk of radius $r>0$ about $0$) defined by $$ S(x,t) = x + t\,e(x)U(x) $$ is an injective diffeomorphism. (If all of the asmptotic lines of $X$ are distinct, one can even take $e$ to be a (suitably small) constant.)

Now, we also know that $X$ is a compact (oriented) Riemann surface with a (nonzero) finite number of points removed. (In fact, $X$ has no compact components.) As such, there exists a smooth, closed embedding $\psi: X\to \mathbb{R}^3$ with the property that the normal 'tube' of radius $1$ around $\psi(X)$ is also smoothly embedded. Let $u:X\to S^2$ be a unit normal vector field for the immersion $\psi$ and extend $\psi$ to $\psi:X\times [-\tfrac12,\tfrac12]\to\mathbb{R}^3$ by setting $$ \psi(x,t) = \psi(x) + t\,u(x) $$ for $|t|\le \tfrac12$. Now, I claim that there is a smooth map $\phi:S\bigl(X\times \Delta(\tfrac12)\bigr)\to \mathbb{R}^3$ that satisfies $$ \phi\bigl(S(x,t)\bigr) = \psi\bigl(x,\mathrm{Re}(t)\bigr) $$ whenever $|t|\le\tfrac12$ and that it is a smooth submersion on an open neighborhood of $X\subset S\bigl(X\times \Delta(\tfrac12)\bigr)$ that is injective and immersive on $X$ itself.

Now, extend $\phi$ smoothly any way one likes beyond the set $S\bigl(X\times \Delta(\tfrac12)\bigr)\subset\mathbb{C}^2$. (One can even require that $\phi$ take the complement of $S\bigl(X\times \Delta(1)\bigr)$ to a single point of $\mathbb{R}^3$.)

Unfortunately, there may be no simple recipe for choosing $\psi$, which is what one would really need to get an affirmative answer to the second question.

Answer 2

Henry Segerman has asked a similar question about curves in $\mathbb{CP}^2$ and then managed to produce the sculpture (of the Klein quartic). So perhaps he will have something interesting to say...

Answer 3

Consider the surface in $\mathbb{R}^4$ with equations \begin{align*} x_1^2+x_2^2+x_3^2+x_4^2 &=1 \\ (3x_3^2-2)x_4-\sqrt{2}(x_1^2-x_2^2)x_3 &=0. \end{align*}

This is conformally isomorphic to the normalization of the hyperelliptic curve $$w^2=z^5-(a^2+a^{-2})z^3+z, $$ for a unique value of $a\in(0,1)$ which is approximately $0.0983562$. It is also conformally isomorphic to the quotient of the unit disc by a certain Fuchsian group which again depends on a single parameter $b\in(0,1)$ which is approximately $0.8005319$. There is a very long story behind all this, which is spelled out in my memoir "Uniformization of embedded surfaces" at https://arxiv.org/abs/1607.06433. There is also a large body of associated Maple code and pictures which can be downloaded from the arxiv, or more conveniently from https://neilstrickland.github.io/genus2/.