cardioid, deltoid, folium

The cardioid and the deltoid are two of my favorite curves. They arise in similar ways: one is an epicycloid, and the other is a hypocycloid. In a sense, each is the simplest non-trivial example of their respective type. They make excellent examples for calculus problems. But as I learned this week, they are actually the same curve.

This post is about the claim made in italics in the previous paragraph. Obviously I don’t mean that the classical constructions mentioned above (and described below) produce the same curves in the Euclidean plane. Rather, they are the same from the perspective of complex projective geometry. When I searched for this fact on Google after uncovering it for myself, I only found one mention of it, in a textbook from 1923 entitled An Introduction to Projective Geometry. I assume it was well-known at the time, and today is probably known to certain algebraic geometers, but it seems worth explicating for a larger audience.

First, the curves. Epicycloids and hypocycloids are both examples of roulettes, curves traced out by a point marked on one curve, which is free to move, as it rolls along another curve, which is fixed, without slipping. To generate an epicycloid or hypocycloid, both the fixed curve and the moving curve are circles; the difference is that for an epicycloid, the rolling circle is outside the fixed circle, and for a hypocycloid the rolling circle is on the inside. The shape of the epicycloid or hypocycloid is determined by the ratio of the circles’ radii. For an epicycloid, we can choose a 1:1 ratio, which means the marked point on the rolling circle makes contact with the fixed circle once as the outer circle completes a circuit. A hypocycloid cannot be constructed from circles whose radii have a 1:1 ratio, and a 2:1 ratio simply produces a line segment, so the simplest hypocycloid arises from a 3:1 ratio. The construction of these simplest examples is illustrated below. (These animations were created using a Desmos graph with the help of GIFsmos.) The first is called the cardioid (“heart-like”) and the second is the deltoid (“triangle-like”).

In both cases, the rolling circle is given a radius of 1, and in both cases the centers of the two circles remain at a distance of 2. By watching carefully, one can see that in both cases the marked point makes two revolutions around the center of the rolling circle. For the cardioid, these revolutions are counterclockwise, and so the cardioid can be parameterized by

$$(2\cos\theta + \cos2\theta, 2\sin\theta + \sin2\theta)\text.$$ In the case of the deltoid, the marked point’s revolutions are made clockwise, and so the deltoid can be parameterized by $$(2\cos\theta + \cos2\theta, 2\sin\theta - \sin2\theta)\text.$$ These formulas are very similar, but certainly not the same, and the pictures they produce are quite different. So how can I claim that the curves are the same?

Our first step toward understanding the claim involves switching to complex numbers. If we collect the $x$- and $y$-coordinates of the plane $\mathbb{R}^2$ into a single complex coordinate, then the parameterizations above become

$2e^{i\theta} + e^{2i\theta} \qquad$ and $\qquad 2e^{i\theta} + e^{-2i\theta}$. Now we want to extend to the complex plane $\mathbb{C}^2$ (note: I think of $\mathbb{C}$ as the complex line because it is one-dimensional as a complex vector space). A standard trick is to add a second coordinate that is conjugate to the first, which makes the parameterizations $\big(2e^{i\theta} + e^{2i\theta},2e^{-i\theta} + e^{-2i\theta}\big) \qquad$ and $\qquad \big(2e^{i\theta} + e^{-2i\theta},2e^{-i\theta} + e^{2i\theta}\big)$. Now let’s set $t = e^{i\theta}$ and allow $t$ to take on all complex values (except $0$, but we’ll take care of that later) instead of just values on the unit circle. At the same time, let’s label the parameterizations $\gamma_C$ and $\gamma_D$, with $C$ standing for cardioid and $D$ for deltoid. This gives us $\gamma_C(t) = \left(2t + t^2,\dfrac{2}{t} + \dfrac{1}{t^2}\right) \qquad$ and $\qquad \gamma_D(t) = \left(2t + \dfrac{1}{t^2},\dfrac{2}{t} + t^2\right)$. We still can see superficial similarities in these formulas, but not enough to conclude that they define equivalent curves. In order to see their equivalence, we need to see what’s happening at infinity, which means introducing some projective geometry.

The complex projective line $\mathbb{P}^1$, also known as the Riemann sphere, is obtained by adding a single point, labeled $\infty$, to the ordinary complex line $\mathbb{C}$. The points of $\mathbb{P}^1$ may be thought of as the “slopes” of lines through the origin in $\mathbb{C}^2$. Indeed, it is often useful to assign coordinates to $\mathbb{P}^1$ using non-zero vectors $(s,t)$ in $\mathbb{C}^2$, where two vectors correspond to the same point of $\mathbb{P}^1$ if they are scalar multiples of each other, $(s,t)\sim(\lambda s,\lambda t)$ if $\lambda\in\mathbb{C}\setminus\{0\}$. We write the equivalence class of $(s,t)$ as $[s:t]$; these are called homogeneous coordinates on $\mathbb{P}^1$. We can recover $\mathbb{P}^1$ as $\mathbb{C}\cup\{\infty\}$ by sending $[s:t]$ to the slope $t/s$ if $s \ne 0$; then $[0:1]$ is sent to $\infty$.

In a similar way, we can extend $\mathbb{C}^2$ to the complex projective plane $\mathbb{P}^2$ by adding points at infinity, and the most convenient way to do so is by homogenous coordinates. We start with non-zero vectors $(u,v,w)$ in $\mathbb{C}^3$ and consider $(\lambda u, \lambda v, \lambda w)$ to define the same point of $\mathbb{P}^2$ as $(u,v,w)$ if $\lambda\in\mathbb{C}\setminus\{0\}$. Then $[u:v:w]$ are homogeneous coordinates on $\mathbb{P}^2$. The points with $u\ne0$ correspond to points of the original complex plane $\mathbb{C}^2$, by sending $[u:v:w]$ to $(v/u,w/u)$. The points with $u=0$ constitute the new line at infinity, which is just a copy of $\mathbb{P}^1$ with coordinates $[0:v:w]$.

Now we can extend the cardioid and the deltoid to curves in $\mathbb{P}^2$, not just $\mathbb{C}^2$. We start with the parameterizations $\gamma_C$ and $\gamma_D$, append an initial coordinate of 1, then clear denominators (we can do this because of the equivalence that defines homogeneous coordinates). Then we get

$\gamma_C(t) = \big[t^2:2t^3 + t^4:2t + 1\big] \qquad$ and $\qquad \gamma_D(t) = \big[t^2:2t^3 + 1:2t + t^4\big]$. These allow for the possibility of $t = 0$, but apparently leave out the point at infinity $\infty$, so we make one more modification, replacing $t$ with $t/s$ and again clearing denominators to obtain $\gamma_C([s:t]) = \big[s^2 t^2:2st^3 + t^4:2s^3t + s^4\big] \qquad$ and
$\qquad \gamma_D([s:t]) = \big[s^2t^2:2st^3 + s^4:2s^3t + t^4\big]$. Here we see a feature characteristic of maps from one projective space to another, when homogeneous coordinates are used: each component of the map must be homogeneous of the same degree (in this case, four). By expressing the parameterizations of the cardioid and the deltoid in this way, we see that both curves touch the line at infinity at the two points $[0:1:0]$ and $[0:0:1]$, corresponding to $[0:1]$ and $[1:0]$, respectively, for the cardioid, and in the reverse order for the deltoid. Still this isn’t enough to show that the curves are the same! We need one more ingredient.

A projective transformation of $\mathbb{P}^1$ or $\mathbb{P}^2$ is induced by a linear transformation of the homogeneous coordinates. Readers who are already familiar with the Riemann sphere will recognize projective transformations of $\mathbb{P}^1$ as Möbius transformations (also known as fractional linear transformations): given $a,b,c,d\in\mathbb{C}$, we can convert $[s:t] \mapsto [as+bt:cs+dt]$ to a Möbius transformation in the coordinate $z = s/t$, where it becomes $z \mapsto \dfrac{az+b}{cz+d}$. The condition for this function to be invertible is $ad - bc \ne 0$, which is the same as the condition for the matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ to be invertible. In the same way, projective transformations of $\mathbb{P}^2$ arise from invertible linear transformations of $\mathbb{C}^2$. Two objects in $\mathbb{P}^1$ or $\mathbb{P}^2$ are called projectively equivalent if there is a projective transformation that carries one to the other. And now we can state precisely what was meant in the opening paragraph:

The cardioid and the deltoid are projectively equivalent in $\mathbb{P}^2$.

But how do we find the projective equivalence? A clue may be found in one clear difference between the original curves drawn in the Euclidean plane, which niggled at me while I was trying to figure out their relationship. The deltoid clearly has three cusps, while the cardioid apparently only has one. If the curves are equivalent, where are the other cusps of the cardioid? The answer: on the line at infinity!

How can we tell? It’s time to apply some differential geometry and look at the tangent lines of these two curves. Returning to the parameterizations in terms of $t$, we find

$\gamma_C'(t) = \left(2 + 2t,-\dfrac{2}{t^2} - \dfrac{2}{t^3}\right) \qquad$ and $\qquad \gamma_D'(t) = \left(2 - \dfrac{2}{t^3},-\dfrac{2}{t^2} + 2t\right)$. Now a line in $\mathbb{C}^2$, with coordinates $(v,w)$, passing through $(a,b)$ in the direction $(s,t)$ has the equation $\begin{vmatrix} s & v - a \\ t & w - b \end{vmatrix} = 0$. Thus the tangent line to the cardioid at $\gamma_C(t)$ has the equation $$\begin{vmatrix} 2 + 2t & v - \big(2t + t^2\big) \\ -\frac{2}{t^2} - \frac{2}{t^3} & w - \big(\frac{2}{t} + \frac{1}{t^2}\big) \end{vmatrix} = 0$$ which, after some simplification, becomes $$wt^3 - 3t^2 - 3t + v = 0\text{.}$$ This is the line equation of the cardioid. In a similar fashion, we can find the line equation of the deltoid, which is $$t^3 - vt^2 + wt - 1 = 0\text{.}$$

Having the line equation of a curve, in terms of a parameter $t$, can be useful in several ways. As $t$ varies over $\mathbb{P}^1$, it produces all the tangent lines of the curve. (We’ll clarify what happens when $t = \infty$ in a moment.) But we can also let $(v,w)$ vary over $\mathbb{C}^2$ and find, for each point, which tangent lines of the curve pass through that point. Because the line equations of the cardioid and the deltoid are cubic polynomials in $t$, most points of $\mathbb{C}^2$ will lie on three tangent lines. Those points that lie on fewer than three tangent lines play a special role.

Let’s illustrate first with the deltoid. We’ll be looking at lots of cube roots, so let $\omega = e^{i\,2\pi/3}$; this means that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. When $(v,w)=(0,0)$, the line equation becomes $t^3 - 1 = 0$, so the tangent lines of the deltoid that pass through the origin correspond to the parameters $1$, $\omega$, and $\omega^2$. Indeed, the three points $\gamma_D(0) = (3,3)$, $\gamma_D(\omega) = (3\omega,3\omega^2)$, and $\gamma_D(\omega^2) = (3\omega^2,3\omega)$ are the three cusps of the deltoid. On the other hand, a point that belongs to the deltoid lies on tangent lines corresponding to at most two parameters (two of the points of tangency have “coalesced”). For example, when $(v,w)=(-1,-1)$, the line equation becomes $t^3 + t^2 - t - 1 = 0$, or $(t+1)^2(t-1) = 0$. At a cusp, all three tangent lines coincide: for example, when $(v,w)=(3,3)$, the line equation is $3t^3 - 3t^2 + 3t - 3 = 3(t-1)^3 = 0$. See the pictures below.

We can homogenize the line equation of the deltoid by replacing $t$ with $t/s$ and $(v,w)$ with $(v/u,w/u)$ and clearing denominators to obtain:

$$ut^3 - vst^2 + ws^2t - us^3 = 0\text.$$ When $[s:t] = [1:0]$ or $[0:1]$ (remember, this second point in homogeneous coordinates corresponds to $t=\infty$), we get the same equation of the tangent line, $u = 0$. This is the equation of the line at infinity, so the line at infinity is tangent to the deltoid at both $[0:0:1]$ and $[0:1:0]$! A line that is tangent to a curve at two points is called a bitangent.

The cardioid also has a bitangent, which is easier to see: when $t = \omega$ or $t = \omega^2$, respectively, the line equation of the cardioid becomes $w - 3\omega^2 - 3\omega + v = 0$ or $w - 3\omega - 3\omega^2 + v = 0$, both of which are equivalent to $v + w = 3$. The visible cusp occurs at $(-1,-1)$, where the line equation becomes $(t + 1)^3 = 0$. For an example of more generic behavior, look at $(-3,-3)$, where the line equation becomes $3t^3 + 3t^2 + 3t + 3 = 0$, or $3(t + 1)(t + i)(t - i) = 0$. See pictures below.

The homogeneous version of the cardioid’s line equation is

$$wt^3 - 3ust^2 - 3us^2t + vs^3 = 0\text.$$ When $[s:t] = [1:0]$, this becomes the $w$-axis $v = 0$, and when $[s:t] = [0:1]$, we get $w = 0$. In each of these cases, we see that only one tangent line passes through the point, just as we saw for the cusps of the deltoid. So we have identified the three cusps of the cardioid—$[1:-1:-1]$, $[0:0:1]$, and $[0:1:0]$. The tangent lines through all three of these cusps pass through the origin in $\mathbb{C}^2$, with homogeneous coordinates $[1:0:0]$.

We now have enough information to show the equivalence of the cardioid and the deltoid. To define a projective transformation from $\mathbb{P}^1$ to itself, we need to specify where three points go; to define a projective transformation from $\mathbb{P}^2$ to itself, we need to specify the images of four points, no three of which are collinear. We’ll show how to transform the line equation of the deltoid into the line equation of the cardioid via pullback.

We’re looking for projective transformations $f : \mathbb{P}^1 \to \mathbb{P}^1$ and $g : \mathbb{P}^2 \to \mathbb{P}^2$ such that $\gamma_D \circ f = g \circ \gamma_C$. Starting with $f$, we require

$f\big([1:0]\big) = [1:\omega]$,   $f\big([0:1]\big) = [\omega:1]$,   and   $f\big([1:-1]\big) = [1:1]$, so that the parameters of the cardioid’s cusps are sent to those of the deltoid’s cusps. This can be accomplished by defining $$f\big([s:t]\big) = [s - \omega t : \omega s - t]\text.$$ Meanwhile, $g$ needs to satisfy $g\big([0:0:1]\big) = [1:3\omega:3\omega^2]$,  $g\big([0:1:0]\big) = [1:3\omega^2:3\omega]$,
$g\big([1:-1:-1]\big) = [1:3:3]$,   and   $g\big([1:0:0]\big) = [1:0:0]$, which is accomplished by $$g\big([u:v:w]\big) = [ 3u+v+w : 3\omega^2 v + 3\omega w : 3\omega v + 3\omega^2 w ]\text.$$ Now substitute the components of $f$ and $g$ into the variables of the deltoid’s line equation, expand, and simplify. The result is the line equation of the cardioid. You can calculate this by hand, or just let SageMath do it for you:

xxxxxxxxxx

 

1

var('U','V','W','S','T');

2

w = e^(2*pi*i/3);

3

line_eqn = (3*U+V+W)*(w*S-T)^3 - 3*w*(w*V+W)*(S-w*T)*(w*S-T)^2 + 3*w*(V+w*W)*(S-w*T)^2*(w*S-T) - (3*U+V+W)*(S-w*T)^3;

4

line_eqn.expand().simplify()

Language: Sage Gap GP HTML Macaulay2 Maxima Octave Python R Singular

---

Messages

One of the curves mentioned in the title of this post has been conspicuously absent so far: the folium of Descartes. This is another favorite curve of mine, invariably given in my calculus classes as an exercise in implicit differentiation. Its equation is $x^3 + y^3 = xy$.

So what’s the connection between this curve and the others? Well, if we extract the coefficients from the deltoid’s line equation and use them to define a new curve $\gamma_F$, we get $$\gamma_F\big([s:t]\big) = [ s^3 - t^3 : st^2 : -s^2 t ]\text,$$ which parameterizes $$v^3 + w^3 = uvw\text,$$ the homogeneous version of the folium’s equation. This means that the folium is dual to the deltoid (and thus also to the cardioid)! The tangent lines of the cardioid/deltoid have been converted into points of the folium, and likewise points of the cardioid/deltoid become tangent lines of the folium. Just as each point of $\mathbb{C}^2$ lies on three tangent lines of the cardioid/deltoid, counted with multiplicity, each line of $\mathbb{C}^2$ intersects the folium at three points, counted with multiplicity. The bitangent of the deltoid and cardioid has been converted into a point of self-intersection. If we look at points of the form $[1:v:\bar{v}]$, then the threefold symmetry of the folium is revealed (the three asymptotic directions correspond to the three tangent lines that pass through the origin, which as we saw are the tangent lines at the cusps).