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Is this spirograph curve algebraic? I an only write it in polar coordinates:

$$ z = e^{i\theta}\left(\frac{7}{8} + \frac{1}{4} e^{6i\theta}\right) $$

and here is a picture. It is a six-sided rose-shaped curve, a hypotrochoid.

enter image description here

I read somewhere that all spirograph curves are algebraic, so this must be the solution of some polynomial equation $p(x,y) = 0$. Then I could ask questions about this curve as a Riemann surface.

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  • $\begingroup$ What kind of equation are you going to allow? Just polynomial equations? $\endgroup$ – Fly by Night Dec 15 '17 at 18:11
  • $\begingroup$ only polynomials $p(x,y)$ $\endgroup$ – cactus314 Dec 15 '17 at 18:24
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If we write $$x=(\cos{\theta})(-\frac78+16(\cos{\theta})^6-28(\cos{\theta})^4+14(\cos{\theta})^2)$$ $$y=(\sin{\theta})(\frac58+16(\cos{\theta})^6-20(\cos{\theta})^4+6(\cos{\theta})^2)$$

and use the relation $(\cos{\theta})^2+(\sin{\theta})^2-1=0$, in M2:

R=QQ[s,c,x,y,MonomialOrder=>Lex]
I=ideal(x-c*(-7/8+16*c^6-28*c^4+14*c^2),y-s*((5/8)+16*c^6-20*c^4+6*c^2),s^2+c^2-1)
gens gb I

we get:

$1099511627776x^{14}+7696581394432x^{12}y^2-1322849927168x^{12}+23089744183296x^{10}y^4-7937099563008x^{10}y^2+37580963840x^{10}+38482906972160x^8y^6-19842748907520x^8y^4+187904819200x^8y^2+ 16089350144x^8+38482906972160x^6y^8-26456998543360x^6y^6+375809638400x^6y^4+64357400576x^6y^2-211099320320x^6+23089744183296x^4y^{10}-19842748907520x^4y^8+375809638400x^4y^6+96536100864x^4 y^4+3252665450496x^4y^2-1567641600x^4+7696581394432x^2y^{12}-7937099563008x^2y^{10}+187904819200x^2y^8+64357400576x^2y^6-3223940235264x^2y^4-3135283200x^2y^2-5511240000x^2+1099511627776y^{14}- 1322849927168y^{12}+37580963840y^{10}+16089350144y^8+220674392064y^6-1567641600y^4-5511240000y^2-8303765625=0$

Visualized in geogebra using the ImplicitCurve command:

hypo

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  • $\begingroup$ holy number crunch, such a simple parametrizationg giving rise to this $\endgroup$ – mercio Dec 16 '17 at 13:06
  • $\begingroup$ @mercio that's what I don't understand $\endgroup$ – cactus314 Dec 16 '17 at 13:09
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You have given a parametrization of your curve by a circle, and a circle has genus $0$, so it is a genus $0$ curve. Looking at random lines intersecting it, you should expect its total degree to be at least $6$ in $x,y$. To account for the high degrees it must have a lot of nodes (you can already see $6$ of them but there should be at least $10$ if the degree is $6$)

Since it is invariant by reflexion around the axis, the equation can be written in terms of $x^2$ and $y^2$.
In fact it is invariant by the action of a diedral group of order $12$ so if you knew by heart the subring of $\Bbb R[x,y]$ invariant by that you could say even more.

writing $e^{i\theta} = c+is$ where $c^2+s^2=1$ you can write $x=f(c)$ and $y=sg(c)$ where $f$ and $g$ are some polynomials of degree $7$ and $6$.

Then $x^2$ and $y^2$ are two polynomials of degree $7$ in $c^2$, so they must have an algebraic relation.
The dimension of the space of polynomials of degree at most $14$ in two variables is greater than the dimension of the space of polynomials of degree at most $98$ in one variable, so the curve's equation has degree at most $28$ in $x,y$.

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  • $\begingroup$ It might not be so easy to pin down what the ring should be. We know there will be a $D_{12}=\mathbb{Z}_6 \ltimes \mathbb{Z}_2$ action. I also observe a number of bitangents while computing the intersections $\ell \cap C$. Perhaps this is something like a Lefschetz pencil? There is a fair bit of non-generic behavior in these curves. Here is some discussion of intersection multiplicity. $\endgroup$ – cactus314 Dec 15 '17 at 19:54
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Given $$ x=\frac{7}{8}\cos\theta+\frac{1}{4}\cos(7\theta),\qquad y=\frac{7}{8}\sin\theta+\frac{1}{4}\sin(7\theta) $$ we have $x^2+y^2=\frac{1}{64}\left(53+28\cos(6\theta)\right)$ and the given hypotrochoid is an algebraic curve since we may eliminate the $t$ variable in $$ \left\{\begin{array}{ccc} x& =& \frac{7}{8} T_1(t) + \frac{1}{4}T_7(t)\\x^2+y^2&=&\frac{53}{64}+\frac{7}{16}T_6(t).\end{array}\right.$$

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  • $\begingroup$ perhaps you can run it through a computer to eliminate $\theta$? $\endgroup$ – cactus314 Dec 15 '17 at 17:59
  • $\begingroup$ @cactus314: I am doing it, it appears to be quite time-consuming. $\endgroup$ – Jack D'Aurizio Dec 15 '17 at 18:00
  • $\begingroup$ does replacing $w = e^{i\theta}$ work... what is the meaning of that equation? I guess it is the "complexification" which is a different surface. $\endgroup$ – cactus314 Dec 15 '17 at 18:04
  • $\begingroup$ @cactus314: I thought you wanted a polynomial in the $x$ and $y$ variables vanishing over the points of your curve in the $xy$-plane. As written, $z$ is obviously a polynomial in $e^{i\theta}$. $\endgroup$ – Jack D'Aurizio Dec 15 '17 at 18:17
  • $\begingroup$ Eliminating $t$ from this system doesn't lead to the correct curve. The last suggestion in @mercio's answer does; see my answer. $\endgroup$ – Jan-Magnus Økland Dec 16 '17 at 12:21

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