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

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.

• Have you already tested some (nontrivial) examples? Jan 13 '17 at 17:21
• Mikhalkin arxiv.org/abs/math/0010087 shows that, if $X$ is a Harnack curve, then $(z_1, z_2) \mapsto (\mathrm{Re}(z_1), \mathrm{Re}(z_2))$ covers a closed region in $\mathbb{R}^2$, mapping $2 \to 1$ onto the interior and $1 \to 1$ onto the boundary. The preimage of the boundary is $X \cap \mathbb{R}^2$, and this disconnects $X$. So I just need to write down a function which has different signs on the two components of $X \setminus \mathbb{R}^2$ and take that as the third coordinate of the map. Abstract arguments show such a function exists, but I haven't found an explicit formula. Jan 13 '17 at 17:33
• Not quite what you are asking, but related questions were studied in the context of mininal surfaces. One result says that under certain assumptions on a region $D \subset \mathbb{R}^n, \ n \geq 3$ every bordered Riemann surface $M$ admits a continuous map to $\overline{D}$ which is is a conformal full complete proper minimal immersion on $M^\circ$, in MR3407187 Alarc\'on, A.Drinovec Drnov\v sek, B.; Forstneri\v c, F; L\'opez, F. J. Every bordered Riemann surface is a complete conformal minimal surface bounded by Jordan curves. Proc. Lond. Math. Soc. (3) 111 (2015), no. 4, 851–886 Jan 13 '17 at 18:22
• @Holonomia: For most algebraic curves $X\subset\mathbb{C}^2$, the union of the (complex) tangent lines to $X$ is all of $\mathbb{C}^2$, so radially projecting $X$ onto a $3$-sphere in $\mathbb{R}^4=\mathbb{C}^2$ will introduce some 'pinch' points in the image (generically speaking, the same number as the dual degree of the curve minus the number of points on the line at infinity). In general, to do what David wants to do, you'd at least have to choose a smooth nonvanishing section of the normal bundle of $X$ in $\mathbb{C}^2$, which is possible (since $X$ is not compact) but not easy. Jan 14 '17 at 11:05
• @Holonomia: I can't answer this for David. He'll have to do that. I just pointed out that this construction doesn't answer David's actual question. Jan 14 '17 at 17:09

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.

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...

• It might be worth pointing out, though, that, if $X\subset\mathbb{CP}^2$ is a smooth algebraic curve, then its normal bundle is not trivial, and this implies that there cannot be a smooth map $\phi:\mathbb{CP}^2\to\mathbb{R}^3$ that is both an injective immersion when restricted to $X$ and a submersion on an open neighborhood of $X$. Jan 14 '17 at 18:52

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/.

• Can you find the map $\phi$ ?
– j.c.
Jan 25 '17 at 19:02