Drawing elliptic curve Consider an elliptic complex curve in $\mathbb{C}^2$ given by equation $w^2 = (z-a)(z-b)(z-c)$ where $a,b,c$ are complex mutually distinct constants. It is a $2$-dimensional surface in $4$-dimensional space (If we talk about real dimensions). How to construct a topologically equivalent real surface in $\mathbb{R}^3 \cup \left\{ \infty \right\}$? I'm told that it is a torus glued to infinity in one point, but I don't know why.
My next question is if it is possible to construct numerically such topologically equivalent surface in $\mathbb{R}^3 \cup \left\{ \infty \right\}$ for given 2-dimensional surface in 4-dimensional $\mathbb{C}^2$ if for each fixed $\xi \in \mathbb{C}$ we know points $w(z)$ of intersection of complex line $z = \xi$ with our surface?
 A: Let $E^0 = \{ (z, w) \in \mathbb{C}^2 : w^2 = (z - a)(z - b)(z - c) \}$. First, we compactify $E^0$ by taking its closure in $\mathbb{C P}^2$; in analytic terms, what we are doing is gluing $E^0$ to $E^1 = \{ (u, v) \in \mathbb{C}^2 : u = (v - a u)(v - b u)(v - c u) \}$ along the coordinate transformation
$$\left\lbrace \begin{aligned}
u & = 1 / w \\
v & = z / w &
\end{aligned} \right.
\left\lbrace \begin{aligned}
z & = u / v \\
w & = 1 / u
\end{aligned}
\right.$$
and one may verify by hand that $E = E^0 \cup E^1$ is a compact Riemann surface, and that $E \setminus E^0$ is just the point $\{ (u, v) = (0, 0) \}$. (This is the so-called "point at infinity".) To avoid confusion, we will use homogenous coordinates on $E$:  $(x : y : z)$ means the point $(z, w) = (y/x, z/x)$ when $x \ne 0$, and the point $(u, v) = (x/z, y/z)$ when $z \ne 0$.
Now, we consider meromorphic functions on $E$. If we differentiate the defining equation, we get
$$2 w \, \mathrm{d} w = (3 z^2 - 2 (a + b + c) z + (a b + b c + c a)) \, \mathrm{d} z$$
thus $\mathrm{d} w \ne 0$ whenever $w = 0$. (This is because $a, b, c$ are distinct.) The inverse function theorem then implies $w$ is a coordinate function near the three points $P_1 = (1 : a : 0)$, $P_2 = (1 : b : 0)$, $P_3 = (1 : c : 0)$. Similarly,
$$(1 + (a + b + c) v^2 - 2 (a b + b c + c a) u v  + 3 a b c u^2) \mathrm{d} u = (3 v^2 - 2 (a + b + c) u v + (a b + b c + c a) u^2) \mathrm{d} v$$
so $\mathrm{d} v \ne 0$ at $(u, v) = (0, 0)$, hence $v = z / w$ is a coordinate function near the point $O = (0 : 0 : 1)$. On the other hand, $z$ is holomorphic away from $O$ and further computation gives the series expansions
\begin{align}
z & = a - \frac{1}{(a - b)(c - a)} w^2 + O(w^4) && \text{near } P_1 \\
z & = b - \frac{1}{(a - b)(b - c)} w^2 + O(w^4) && \text{near } P_2 \\
z & = c - \frac{1}{(b - c)(c - a)} w^2 + O(w^4) && \text{near } P_3 \\
\frac{1}{z} & = v^2 + O(v^4) && \text{near } O
\end{align}
Thus, $z$ defines a holomorphic map $E \to \mathbb{C P}^1$ ramified at $\{ P_1, P_2, P_3, O \}$ (with ramification index $2$ at each ramification point). Thus, $E \setminus \{ P_1, P_2, P_3, O \}$ is a double cover of $\mathbb{C P}^1 \setminus \{ a, b, c, \infty \}$ and by making branch cuts and gluing sheets one may conclude that $E$ must be homeomorphic to the torus $S^1 \times S^1$. 
But we can also see this explicitly. Recall that for any non-parallel $\omega_1, \omega_2$ in $\mathbb{C}$, we have a unique meromorphic function $\wp$ satisfying the following properties:


*

*For all integers $n_1, n_2$, we have $\wp (t + n_1 \omega_1 + n_2 \omega_2) = \wp (t)$.

*The only poles of $\wp$ are at the lattice points $n_1 \omega_1 + n_2 \omega_2$, and these are all double poles.

*We have $\lim_{t \to 0} \left[ \wp (t) - \frac{1}{t^2} \right] = 0$.
This is the Weierstrass $\wp$-function for the lattice $\Lambda = \mathbb{Z} \omega_1 + \mathbb{Z} \omega_2$. Because it is doubly-periodic, it can be regarded as a meromorphic function on the complex torus $\mathbb{C} / \Lambda \cong S^1 \times S^1$. It satisfies the differential equation
$$\wp' (t)^2 = 4 \wp(t)^3 - g_2 \wp(t) - g_3$$
for certain complex numbers $g_2$ and $g_3$ depending only on $\omega_1$ and $\omega_2$. By setting $z = \wp(t)$ and $2 w = \wp' (t)$, we get the equation
$$w^2 = z^3 - \frac{1}{4} g_2 z - \frac{1}{4} g_3$$
so by choosing $\omega_1$ and $\omega_2$ appropriately, we can obtain an unramified holomorphic (!) map $\mathbb{C} / \Lambda \to E$ of degree $1$ – in other words, an analytic isomorphism! By choosing appropriate real coordinates on $\mathbb{C} / \Lambda$ and inverting this isomorphism, we can then construct an embedding of $E$ into $\mathbb{R}^3$, if so desired – but this will in general only be smooth and not necessarily, say, conformal.
