Ytukyg

I have been reading the Wikipedia page on the Quadrifolium there are two of them:

begin{eqnarray*}
r &=& sin 2theta \
(x^2 + y^2)^3 &=& 4 x^2 y^2
end{eqnarray*}

and it's $45^circ$-rotated version, which is also a rose curve:

begin{eqnarray*}
r &=& cos 2theta\
(x^2 + y^2)^3 &=& (x^2 - y^2)^2 \
end{eqnarray*}

These equations also have a $(x,y)$-parameterization, which can be found with some algebra.

begin{eqnarray*}
x &=& cos k theta cos theta = tfrac{1}{4}big[ e^{i(k+1)theta} + e^{i(k-1)theta} + e^{-i(k-1)theta} + e^{-i(k+1)theta} big] \
y &=& cos k theta sin theta
,= tfrac{1}{4}big[ e^{i(k+1)theta} - e^{i(k-1)theta} + e^{-i(k-1)theta} - e^{-i(k+1)theta} big] \
end{eqnarray*}

And $cos k (theta + phi)$ for a rotated version. Here $theta = 0, frac{pi}{4}$ and $k = 2$.

Here is the picture of the quadrifolium, a kind of rose curve, and when it's rotated.

The reason I am going to look at this question again, is because I wanted to study the rational paramterization. Let $X$ be our curve, with a singular point at the origin $(0,0)$. We're trying to describe $X(mathbb{Q})$.

The curve on the left right has no rational points.

Naively, if I divide one equation by the other, I get an expression for the tangent function. In fact, $frac{x}{y} = tan theta$ for all values of $theta in [0, 2pi]$ just as if it were a circle. Let's define a map:

$$ big[ (x,y) = (cos theta, sin theta) big] mapsto
Big[ ( (2x^2 - 1 )x, (1 - 2y^2) y) = ( (2 cos^2 theta - 1) cos theta , (1 - 2sin^2 theta)sin theta ) Big] $$

And we have that $LHS in mathbb{Q}$ if $RHS in mathbb{Q}$. Is this sufficient? Is this map birational, is this enough for a paramerization?

Here is even another strategy, since we have Fourier series (even just trigonometric series because there are only 4 terms).

$$ left[ begin{array}{c} x \ y end{array}right] =
left[ begin{array}{c}
e^{3itheta} + e^{itheta} + e^{-itheta} + e^{-3itheta} \
e^{3itheta} - e^{itheta} + e^{-itheta} - e^{-3itheta}end{array} right]
= left[ begin{array}{rrrr} 1 & 1 & 1 & 1 \
1 & -1 & 1 & -1 end{array} right]
left[ begin{array}{c} e^{3itheta} \ e^{itheta} \
e^{-itheta} \ e^{-3itheta} end{array} right] $$

Or we could even try a different approach using the triple-angle formulas of trigonometry:

begin{eqnarray*}
cos 3theta &=& 4 cos^3 theta - 3 cos theta \
sin 3theta &=& 3 sin theta - 4 sin^3 theta \
tan 3theta &=& frac{3 tan theta - tan^3 theta}{1 - 3 tan^2 theta}
end{eqnarray*}

Now these are reading as a degree-map from the circle to another algebraic variety (or scheme). In that case, why does the angle $3theta$ appear in a 4-fold symmetric curve?

These singularities could be studied by Netwton's method (e.g. Puiseux series or resolution of singularities). Unfortunatley I could never get the jargon straight about the exceptional divisor and so forth. E.g. the strategy here to rationalize the lemniscate seems to amount to computing a blow-up of the curve. At this point I am going to consult an algebraic geometry textbook.

Example What is the "derivative" or tangent space at the origin $(0,0)$ ?

You may want to find a parametrization like $(x, y) = (f_1(t), f_2(t))$ for some nonconstant rational functions $f_1(t), f_2(t)in mathbb{Q}(t)$. Until now, I can show you that there's no such parametrization with $f_1(t), f_2(t)in mathbb{Q}[t]$ (i.e. both are polynomial with rational coefficients).

If such parametrization exists, we should have $x^{2} - y^{2} = f(t)^{3}$ and $x^{2} + y^{2} = f(t)^{2}$ for some $f(t)in mathbb{Q}[t]$. Then
$$
y^{2} = frac{1}{2}f(t)^{2}(1-f(t)),
$$
so $frac{1}{2}(1-f(t)) = g(t)^{2}Leftrightarrow f(t) = 1-2g(t)^{2}$ for some $g(t)in mathbb{Q}[t]$. Then $$x^{2} = frac{1}{2}(f(t)^{2} + f(t)^{3}) =(1-2g(t)^{2})^{2}(1-g(t)^{2})$$
Now $mathrm{gcd}(1-2g(t)^{2}, 1-g(t)^{2}) = 1$, so we should have $1-g(t)^{2} = h_{1}(t)^{2}$ and $1-2g(t)^{2} =h_{2}(t)^{2}$ for some $h_{1}, h_{2}in mathbb{Q}[t]$. By the way, this give $2h_{1}(t)^{2} -h_{2}(t)^{2} = 1$, which is impossible if we see the first coefficients of $h_{1}, h_{2}$, unless they are constant.
I think we can use the same argument in case of rational functions, too. (I'm not sure)