Behind is our interpretation of the symbol $\sqrt{t}$:

• $\sqrt{1} = 1$
• $\sqrt{1} = -1$  (!)
• $\sqrt{1} = \pm1$
• $\sqrt{1} = \{1,-1\}$

Is the denominator zero in $\left(\displaystyle\frac{1}{\sqrt[6]{t}\sqrt[3]{t}-\sqrt{t}},\,t\right)$ ?

1. Yes
2. No, except for $t=0$
3. Depends on what the $\sqrt[n]{t}$ mean to you
4. I feel dizzy already

In the parabola example, $\mP(t) = (t,\sqrt{t})$:

• If $\sqrt{t}$ is either branch (a univalued function) then $\mP$ covers half of $x=y^2$.
• If $\sqrt{t}$ is both branches (a multivalued function) then $\mP$ covers all of $x=y^2$.

Both are reasonable paths.

Definition of radical parametrization

A radical parametrization is a tuple $(x_1,\ldots,x_r)$
of elements of a radical field extension of $\C(t_1,\ldots,t_n)$.

More explicitly...

\[\C(\ot) = \F_0 \subset \F_1 \subset \cdots \subset \F_m\]

where $\F_i = \F_{i-1}(\delta_i)$ with $\delta_i^{e_i}=\alpha_i\in\F_{i-1}$

and $\F_m\ni$ $x_1,\ldots,x_r$$x_1$$(\ot)$$,\ldots,x_r$$(\ot)$

We call the $x_i$ functions because
we think of the $\ot$ as variables.

We also ask that their Jacobian has rank $n$.

Parabola example

$\mP(t) = (t,\sqrt{t})$ Not well defined!

• Define the tower: $\C(t)\subset\C(t)(\delta_1)$ where $\delta_1^2=t$
• There are two possible $\delta_1$! Fix one by value, say \[\delta_1(+1)=+1\]
• Now $\mP(t) = (t,\delta_1)$ or $\mP(t) = (t,-\delta_1)$

\[(x(t),y(t)) = \left( t+\frac{1-\sqrt[4]{t^3+2t}}{\sqrt{t^2-\sqrt[3]{t-1}}} , \frac{\sqrt[4]{5\sqrt[3]{t-1}+1}}{t^3+5} \right)\]

\[(x(t),y(t)) = \left(t+\frac{1-\delta_4}{\delta_2},\frac{\delta_3}{t^3+5}\right)\]

• $\delta_1^3=t-1$, $\quad\delta_1(2)=1$
• $\delta_2^2=t^2-\delta_1$, $\quad\delta_2(2)=-\sqrt{3}$
• $\delta_3^4=5\delta_1+1$, $\quad\delta_3(2)=i\sqrt[4]{6}$
• $\delta_4^4=t^3+2t$, $\quad\delta_4(2)=\sqrt[4]{12}$

Incidence varieties

Idea: turn those $\delta$'s into variables that
we will relate to the $t$'s.

Incidence varieties

Given a radical parametrization $\mP$, express it as

\[x_i(\ot)=\frac{x_{iN}(\ot,\delta_1,\ldots,\delta_m)}{x_{iD}(\ot,\delta_1,\ldots,\delta_m)} \,,\quad \delta_i^{e_i}=\alpha_i\]

\[\alpha_1=\frac{\alpha_{1N}(\ot)}{\alpha_{1D}(\ot)}, \ \ldots, \ \alpha_m=\frac{\alpha_{mN}(\ot,\delta_1,\ldots,\delta_{m-1})}{\alpha_{mD}(\ot,\delta_1,\ldots,\delta_{m-1})}\]

We define the incidence variety $\BP$ associated to this representation as the zeroset of the polynomials

• $(\Delta_1)^{e_1}\cdot\alpha_{1D}(\OT) - \alpha_{1N}(\OT)$,
  $(\Delta_i)^{e_i}\cdot\alpha_{iD}(\OT,\Delta_1,\ldots,\Delta_{i-1})-\alpha_{iN}(\OT,\Delta_1,\ldots,\Delta_{i-1})$
• $X_j\cdot x_{jD}(\OT,\Delta_1,\ldots,\Delta_m) - x_{jN}(\OT,\Delta_1,\ldots,\Delta_m)$
• $Z\cdot \mathrm{lcm}(x_{1D},\ldots,x_{rD}) \cdot \mathrm{lcm}(\alpha_{1D}, \cdots, \alpha_{mD}) - 1$

in the ring $\C[\OT,\Delta_1,\ldots,\Delta_m,X_1,\ldots,X_r,Z]$

Example

\[ \mP = \left(t,\sqrt{1-t^2}\right) \]

\[ \delta^2 = 1-t^2, \ x = t, \ y=\delta \]

\[ \BP = V(\Delta^2-1+T^2, X-T, Y-\Delta, Z-1) \]

We can eliminate everything except $X$ and $Y$:

\[ Y^2+X^2=1 \]

Radical variety

The image of a radical parametrization is, intuitively, the evaluation of the $t$'s, the $\delta$'s, and the coordinates as functions of them.

Algebraically, the radical variety defined by $\mP$ is

\[ \VP = \overline{\pi_\mP(\CP)} \]

where $\pi_\mP(\OT,\overline\Delta,\overline{X},Z)=(\overline{X})$

defined as

$\VP$ coincides with $\overline{\Ima(\mP)}$, so it does not depend on the representation of $\mP$.

Tricky example

$\mP=(\sqrt[4]{t^2},t)$ where $\sqrt[4]1=1$ (!)

Take the tower $\C(t)\subset \C(\sqrt[4]{t^2})$ (degree 2 extension)

$\BP=V(\Delta^4-T^2,X-\Delta,Y-T,Z-1)$

$V(X^4-Y^2) = V(X^2-Y) \cup V(X^2+Y)$

$\Ima(\mP)$ is half of $V(X^2-Y)=\VP$

The "conjugate parametrizations" $(i^k\sqrt[4]{t^2},t)$, $k=1,2,3$, cover the other three half-parabolas.

Maybe you think that we should not allow
roots that are not simplified...

$\sqrt[4]{t^2} \quad \rightsquigarrow \quad \Delta^4-T^2$

$\sqrt{t} \quad \rightsquigarrow \quad \Delta^2-T$

 So do we, sometimes...

Can you simplify? 

Tracing index

Looking at the fibres of $\pi_\mP$ is a reasonable alternative to looking at the fibres of $\mP$.

The tracing index of $\mP$ is the degree of the map $\pi_P$, that is, the generic cardinality of $\pi_\mP^{-1}(p)$ for $p\in\VP$.

There is a dense subset of $\Ima(\mP)$ whose fibres by $\pi_P$ have cardinality equal to the tracing index of $\mP$.

Tracing index: example

$\mP=(t^2,\sqrt{t^2+1})$. Then $\VP=V(Y^2-X-1)$ and $\CP=\BP=V(\Delta^2-T^2-1,X-T^2,Y-\Delta,Z-1)$

Every point $(a,b)\in\VP$ with $a\neq0$ has two preimages in $\CP$, namely $T=\pm\sqrt{a}$, so the tracing index of $\mP$ is two.

On the other hand, $(1,\sqrt2)\in\VP$ has two preimages by $\mP$, but $(1,-\sqrt2)\in\VP$ has no preimage. Indeed, only half of the points of $\VP$ are covered by $\mP$.

Tower variety

$\VT=\overline{\Ima(\psi)}$ also encodes relevant information
about the parametrization.

The radical map

\[\mP\colon\quad \C^n\rightarrow\VP\colon\quad\quad \OT\mapsto\overline{X}\]

is then "lifted" to the rational map

\[\mR\colon\quad \VT\rightarrow\VP\colon\quad (\OT,\overline{\Delta})\mapsto\overline{X}\]

Properties in dimension one

\[ \mathrm{genus}(\VT)\geq\mathrm{genus}(\VP) \]

\[\ \Downarrow \]

\[ \VT \mbox{ rational curve} \ \Rightarrow\ \VP \mbox{ rational curve} \]

Additionally, if the tracing index of $\mP$ is one,

\[ \VT \mbox{ rational curve} \ \Leftrightarrow\ \VP \mbox{ rational curve} \]

Just by knowing which roots appear in a parametrization ($\VT$) you may be able to say things about the parametrization ($\VP$).

For example, any curve parametrized by rational functions in $t$ and $\sqrt[n]{t}$ is automatically rational, because $V(\Delta^n-T)$ is rational.

Properties in general dimension

Probably there are similar results
related to the plurigenera.

If the tracing index of $\mP$ is one,

\[ \VT \mbox{ rational} \ \Leftrightarrow\ \VP \mbox{ rational} \]

If $\VT$ is unirational, then $\VP$ is unirational. Furthermore, if $\mT(\ot)$ is a rational parametrization of $\VT$ then $\mP(\pi_\mathbb{T}\circ\mT(\ot))$ is a rational parametrization of $\VP$ where $\pi_\mathbb{T}(\OT,\overline{\Delta}) = (\OT)$.

Example

Let $\mP=\left(\sqrt[3]{t},\sqrt{1-\sqrt[3]{t^2}}\right)$.

$\VT=V(\Delta_1^3-T,\Delta_2^2-1+\Delta_1^2)$ is rational.parametrized as

\[ \mT(t)=\left( \left(\frac{2t}{1+t^2}\right)^3,\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2} \right) \]

We obtain the rational reparametrization

\[ \mP\left(\left(\frac{2t}{1+t^2}\right)^3\right) = \left( \frac{2t}{1+t^2},\frac{1-t^2}{1+t^2} \right). \]

Reparametrization algorithm

Assume that an algorithm that decides the existence
of rational parametrizations is known.

• INPUT: a radical parametrization $\mP$
• OUTPUT: one of
• A rational reparametrization of $\mP$
• "$\VP$ cannot be parametrized rationally"
• "No answer"

1. Compute $\VT$ and parametrize it with the known algorithm.
2. If $\VT$ is parametrized by $\mT(\ot)$ then
   RETURN $\pi_\mathbb{T}\circ\mT(\ot)$.
3. If $\VT$ cannot be parametrized rationally, compute the tracing index $m$ of $\mP$. If $m=1$, RETURN "$\VP$ cannot be parametrized rationally", ELSE RETURN "No answer".