Reparametrizing swung surfaces over the reals

Abstract

Let $\mathbb {K}\subseteq \mathbb {R}$ be a computable subfield of the real numbers (for instance, $\mathbb {Q}$). We present an algorithm to decide whether a given parametrization of a rational swung surface, with coefficients in $\mathbb {K}(\mathtt {i})$, can be reparametrized over a real (i.e., embedded in $\mathbb {R}$) finite field extension of $\mathbb {K}$. Swung surfaces include, in particular, surfaces of revolution.

Appendix: The parametric variety of Weil

The parametric Weil construction and the theory of hypercircles and ultraquadrics, are tools developed in [3, 11]. Here we will consider the specific parametric variety of Weil $V$ associated to the parametrization ${\mathcal P}(s,t)$ defined in the proof of Theorem 4.2 and the map

Recall that, by construction, $\mathcal {P}^*$ carries real points of $V$ to real points of $\mathcal {S}_\mathbb {C}$.

The importance of this variety $V$ is that it encodes the fact that $\mathcal {S}_\mathbb {C}$ is real-defined or real parametrizable.

Theorem 6.1

Let $V$ be the parametric variety of Weil associated to $\mathcal {P}$. If $\mathcal {S}_\mathbb {C}$ is a real-defined surface then there is (at least) one surface $U$ that is an irreducible component of $V$ such that $\mathcal {P}^*:U\rightarrow \mathcal {S}_\mathbb {C}$ is a dominant map. Moreover, if $\tau (u,v)$ is a real parametrization of $U$, then $\mathcal {P}^*(\tau (u,v))$ is a real parametrization of $\mathcal {S}_\mathbb {C}$.

Proof

This is a direct consequence of Theorem 10 in [3]. $\square$

Note that, in Theorem 6.1, the surface $U$ needs not be real-defined. By Andradas et al. [3], Corollary 13, if $\mathcal {S}_\mathbb {C}$ is real-defined we know that there exists a real-defined surface $W$ such that $\mathcal {P}^*:W\rightarrow \mathcal {S}_\mathbb {C}$ is dominant, but $W$ needs not to be irreducible.

In our particular case we want to explore with more detail the surfaces $U_i$, those components of $V$ such that the map $\mathcal {P}^*:U_i\rightarrow \mathcal {S}_\mathbb {C}$ is dominant. Specially, we would like to understand the projections of such components into the $(t_0,t_1)$ and $(s_0,s_1)$ planes.

Theorem 6.2

If $\mathcal {S}_\mathbb {C}$ is a real surface, then there is a real irreducible surface $U$, a component of $V$, such that the map $\mathcal {P}^*:U\rightarrow \mathcal {S}_\mathbb {C}$ is dominant and $\mathcal {P}^*$ takes real points of $U$ to real points of $\mathcal {S}_\mathbb {C}$.

Proof

By Andradas et al. [3], $\mathcal {P}^*:V\rightarrow \mathcal {S}_\mathbb {C}$ is generically (over an nonempty open subset of $\mathcal {S}_\mathbb {C}$) finite to one. So, if $U_i$ is a component of $V$ of dimension different from 2, then $\mathcal {P}^*: U_i \rightarrow \mathcal {S}_\mathbb {C}$ is not dominant. Let $U'$ be the union of all the components $W$ of $V$ such that the map $\mathcal {P}^*:W\rightarrow \mathcal {S}_\mathbb {C}$ is not dominant. In particular, $U'$ contains all components of $V$ that are not surfaces. Then $\mathcal {P}^*(U')$ is contained in a 1-dimensional subset of $\mathcal {S}_\mathbb {C}$. Let $\{U_1,\ldots , U_k\}$ be the remaining components of $V$. Each $U_i$ is a surface and $\mathcal {P}^*:U_i\rightarrow \mathcal {S}_\mathbb {C}$ is dominant. By Theorem 6.1 there is at least one such surface $U_i$.

Consider now the set $\mathcal {S}'_\mathbb {C}= \mathcal {P}^*(V)-\mathcal {P}^*(U')\subseteq \mathcal {S}_\mathbb {C}$. This is a subset of $\mathcal {S}_\mathbb {C}$ that contains a non-empty open Zariski subset of $\mathcal {S}_\mathbb {C}$ (Shafarevich, Chapter 1, §5, Theorem 6). It follows that the set of real points of $\mathcal {S}'_\mathbb {C}$ is Zariski-dense in $\mathcal {S}_\mathbb {C}$.

Let $p=(p_1,p_2,p_3)$ be a real point of $\mathcal {S}'_\mathbb {C}$. Since $p\in \mathcal {P}^*(V)$, then $p=\mathcal {P}(a,b)$, for some $a=a_0+\mathtt {i}a_1,\,b=b_0+\mathtt {i}b_1,\,a_0,a_1,b_0,b_1\in \mathbb {R}$. Now

$$\begin{aligned} \frac{A_0(a_0,a_1)+\mathtt {i}A_1(a_0,a_1)}{A(a_0,a_1)}\cdot \frac{C_0(b_0,b_1)+\mathtt {i}C_1(b_0,b_1)}{C(b_0,b_1)}=\phi _1(a)\psi _1(a)=p_1\in \mathbb {R}, \end{aligned}$$

so

$$\begin{aligned} A(a_0,a_1)\ne 0, C(b_0,b_1)\ne 0 \end{aligned}$$

and

$$\begin{aligned} A_0(a_0,a_1)C_1(b_0,b_1)+A_1(a_0,a_1)C_0(b_0,b_1)=0. \end{aligned}$$

Analogously,

$$\begin{aligned} D(b_0,b_1)\ne 0, B(a_0,a_1)\ne 0 \end{aligned}$$

and

$$\begin{aligned} A_0(a_0,a_1)D_1(b_0,b_1)+A_1(a_0,a_1)D_0(b_0,b_1)=0. \end{aligned}$$

Thus, $(a_0,a_1,b_0,b_1)\in V\cap \mathbb {R}^4$. Moreover, $(a_0,a_1,b_0,b_1)\notin U'$, by our choice of $p$; and $(a_0,a_1,b_0,b_1)\in U_1\cup \cdots \cup U_k$. Therefore, we have proved that any real point of $\mathcal {S}'_\mathbb {C}$ comes from at least one real point in $(a_0,a_1,b_0,b_1)\in U_1\cup \cdots \cup U_k$.

If no $U_i$ were real, then the set of real points of each $U_i$ would be contained in a 1-dimensional subset $R_i$ of $U_i$. Then, the set of real points of $\mathcal {S}'_\mathbb {C}$ would be contained in $\mathcal {P}^*(R_1)\cup \cdots \cup \mathcal {P}^*(R_k)$, which is included in a dimension 1 subset of $\mathcal {S}'_\mathbb {C}$, contradicting the fact that this set is Zariski dense in $\mathcal {S}_\mathbb {C}$. So, there is at least one component $U_i$ that is real.

The fact that any real point of $U_i$ maps to a real point of $\mathcal {S}_\mathbb {C}$ follows from the definition of $V$ and $\mathcal {P}^*$. $\square$

With this result and bearing in mind the special shape of non planar swung surfaces, we can analyze the structure of the surfaces $U_i$ in this case: they turn out to be either planes, cylinders or tori. First, we need the following technical lemma:

Lemma 6.3

Consider the polynomial $f=C_0D_1-C_1D_0\in \mathbb {R}[s_0,s_1]$. If $f$ is identically zero, then $\mathcal {S}_\mathbb {C}$ is a real plane.

Proof

Since $\psi (t)=(\psi _1,\psi _2)$ is a proper parametrization of a curve, both components cannot be constants. Assume, without loss of generality, that $\psi _2$ is not constant, so $D_0$ and $D_1$ are not zero. Now, suppose that $C_0D_1-C_1D_0 = 0$. Then $C_0/D_0 = C_1/D_1 = k(s_0,s_1)$. But, then, $C_0+\mathtt {i}C_1 = k\cdot (D_0+\mathtt {i}D_1)$ and

$$\begin{aligned} \psi _1(s_0+\mathtt {i}s_1) = \frac{C_0+\mathtt {i}C_1}{C} = \frac{D_0+\mathtt {i}D_1}{D}\cdot \frac{k\cdot D}{C} = \psi _2(s_0+\mathtt {i}s_1)\cdot \frac{k\cdot D}{C} \end{aligned}$$

So, $\frac{k\cdot D}{C} = \psi _1(s_0+\mathtt {i}s_1)/\psi _2(s_0+\mathtt {i}s_1)$ is both an $\mathtt {i}$-analytic rational function (i.e., the expansion in terms of real and imaginary parts of the complex function $\psi _1(s)/\psi _2(s)$, after decomposing the variable $s$ in real and imaginary terms, cf. [14]) and a real rational function. By the well known Cauchy–Riemann conditions for analyticity (cf. [14]), $kD/C$ must be, then, a real constant $r$. Thus, $\psi _1 = r\psi _2$ and $\mathcal {S}_\mathbb {C}$ is the real plane $\{ry-x=0\}$ in $\mathbb {C}^3$. $\square$

Theorem 6.4

Let $\mathcal {S}_\mathbb {C}$ be a real swung surface, different from a plane, given by the parametrization $\mathcal {P}$. Let $U$ be any irreducible surface in $V$ such that $\mathcal {P}^*:U\rightarrow \mathcal {S}_\mathbb {C}$ is dominant. Then, there are irreducible curves $Z_1, Z_2\subseteq \mathbb {C}^2$ such that $U=Z_1\times Z_2$. Moreover, $U$ is real if and only if both $Z_1, Z_2$ are real.

Proof

Consider the two projections $\pi _1:\mathbb {C}^4\rightarrow \mathbb {C}^2,\,\pi _2:\mathbb {C}^4\rightarrow \mathbb {C}^2$, so that $\pi _1(t_0,t_1,s_0,s_1)=(t_0,t_1)$ and $\pi _2(t_0,t_1,s_0,s_1)=(s_0,s_1)$. Let $Z_i$ be the Zariski closure of $\pi _i(U),\,i=1,2$. Clearly, $Z_1,\,Z_2$ are irreducible varieties of $\mathbb {C}^2$. If $\dim (Z_i)$ were 0, then $\mathcal {P}^*(U)$ would not be dense in $\mathcal {S}_\mathbb {C}$, contradicting the hypothesis. If $U=Z_1\times Z_2$, then it is clear that $U$ is real if and only if $Z_1$ and $Z_2$ are real. Since always $U\subseteq Z_1\times Z_2$ and both varieties are irreducible, to prove the theorem, it suffices to show that they have the same dimension, i.e., that $dim(Z_i)\le 1,\,i=1,2$.

Since $\mathcal {S}_\mathbb {C}$ is not a plane, $\phi _2(t)$ is not a constant, so, by Recio et al. [14], $B_1(t_0,t_1)$ is not a constant and $Z_1\subseteq \{B_1(t_0,t_1)=0\}$ has dimension at most 1.

Now, since $\psi =(\psi _1,\psi _2)$ is a curve, one of the components is not a constant. Assume, without loss of generality, that $\psi _1$ is not constant. Then, neither $C_0$ nor $C_1$ are constants.

Now, we distinguish three cases. First, if $A_0\equiv 0$ in $U$, then $A_1\not \equiv 0$ in $U$, because $\mathcal {P}^*(U)$ is dense in $\mathcal {S}_\mathbb {C}$. Since $A_0C_1+A_1C_0\equiv 0$ in $U$, it must happen that $C_0\equiv 0$ in $U$, yielding $Z_2\subseteq \{C_0=0\}$ and, thus, $\dim (Z_2)\le 1$.

Analogously, if $A_1\equiv 0$ in $U$, then $A_0\not \equiv 0$ in $U$ and $C_1\equiv 0$ in $U$. Hence $Z_2\subseteq \{C_1=0\}$ and $\dim (Z_2)\le 1$.

Finally, assume that neither $A_0$ nor $A_1$ are zero in $U$, then

$$\begin{aligned} A_0A_1(C_0D_1-C_1D_0) = A_0D_1(A_1C_0+A_0C_1)-A_0C_1(A_1D_0+A_0D_1) \end{aligned}$$

is zero in $U$. It follows that $C_0D_1-C_1D_0\equiv 0$ in $U$ and $Z_2\subseteq \{C_0D_1-C_1D_0=0\}$. Since $\mathcal {S}_\mathbb {C}$ is not a plane, $C_0D_1-C_1D_0$ is not identically zero (in $\mathbb {C}^2$) by Lemma 6.3 and, thus, $\dim (Z_2)\le 1$. $\square$

Finally, we show another technical result:

Lemma 6.5

Let $U\subseteq \mathbb {C}^{n+m}$ be a real irreducible variety such that $U=U_1\times U_2$ is the Cartesian product of two irreducible varieties $U_1\subseteq \mathbb {C}^n,\,U_2\subseteq \mathbb {C}^m$. Let $F(\overline{x}, \overline{y})\in \mathbb {R}(U)$ be a real rational function (i.e., $F(p)\in \mathbb {R}$, for any real point where $F$ is defined) such that it has two different representations $F(\overline{x}, \overline{y})=G(\overline{x}) = H(\overline{y})$. Then $F$ is a real constant function equal to some $c\in \mathbb {R}$.

Proof

Let $p_{x_0}\in U_1$ be a point such that $G(p_{x_0})=c$ is defined. The fiber $\{p_{x_0}\}\times U_2\subseteq U$ is isomorphic to $U_2$ and, for any $p=(p_{x_0},p_y)\in \{p_{x_0}\}\times U_2$, we have that $F(p)=H(p_y)=G(p_{x_0})=c$. Hence $H$ is constant in $U_2$ and $c=H(\overline{y})=F(\overline{x},\overline{y})$ is constant in $U$. Since both $F$ and $U$ are real, $c\in \mathbb {R}$. $\square$