The Poncelet porism and Cayley's criterion
Table of Contents
Introduction
The Poncelet porism can be summarized as follows:
Consider two smooth conics \(C_1\) and \(C_2\) in the plane. There exists a closed polygon inscribed in \(C_1\) and circumscribed around \(C_2\) if and only if there exist infinitely many such polygons.
Here "inscribed" means that the vertices of the polygon lie on \(C_1\) and "circumscribed" means that its edges are tangent to \(C_2\). To construct such a polygon (if one exists) we start with a point \(A_1\) on the first conic \(C_1\) and draw a tangent to \(C_2\) through \(A_1\). The tangent generically intersects \(C_1\) in one more point \(A_2\). From \(A_2\) we then draw another tangent to \(C_2\) obtaining a third point \(A_3\) and so on. If after a finite number of steps we come back to the starting point \(A_1\), we have obtained a closed polygon.
The following figure shows an example in which the construction gives a pentagon.1 The starting point can be dragged around on \(C_1\) demonstrating that we indeed get an infinite number of pentagons independent of the starting point!
There are two main questions:
- Why does the construction not depend on the initial condition?
- How can we check (algebraically) if two given conics lead to a closed polygon and maybe even more, how can we construct pairs of such conics?
For the first question we will look at the proof of the porism. It turns out that there is an elliptic curve associated to the two conics and that one step in the geometric construction corresponds to a translation on the elliptic curve by a fixed amount. The second question is answered by a criterion found by Cayley.
What is elliptic about this?
The space of lines tangent to a given conic \(C \subset \mathbb{P}^2\) is a conic \(C^\star\) in the dual space \(\left(\mathbb{P}^2\right)^\vee\). If \(M_C\) is the \(3 \times 3\) matrix describing \(C\), then \(C^\star\) is described by the inverse \(M_C^{-1}\).2
We consider two non-degenerate conics \(C_1\) and \(C_2\) in the plane \(\mathbb{P}^2\) and study pairs \((p, \ell)\) of points \(p\) on \(C_1\) and lines \(\ell\) tangent to \(C_2\) that contain the point \(p\). As \(\ell\) is tangent \(C_2\), it correponds to a point in the dual, so \(\ell \in C_2^\star\). In summary, we are interested in the set
\begin{align} E = \left\{(p, \ell) \in C_1 \times C_2^\star : p \in \ell\right\} \subset C_1 \times C_2^\star. \end{align}We will now see how this set can be identified with the points on an elliptic curve.
Conics are rational curves so our conics \(C_1\) and \(C_2^\star\) can be parametrized by rational maps
\begin{align} p&: \mathbb{P}^1 \to \mathbb{P}^2, \quad s \mapsto M_p \, \nu(s) \\ \ell&: \mathbb{P}^1 \to \mathbb{P}^2, \quad s \mapsto M_{\ell} \, \nu(s) \end{align}where \(s = [s_0 : s_1]\) are the homogeneous coordinates on \(\mathbb{P}^1\), \(\nu(s) = [s_0^2 : s_0 s_1 : s_1^2]\) and \(M_p\) and \(M_\ell\) are invertible \(3 \times 3\) matrices. The images of \(\mathbb{P}^1\) under these maps are precisely the conics \(C_1\) and \(C_2^\star\). This gives us a bijection
\begin{align} f: \mathbb{P}^1 \times \mathbb{P}^1 \to C_1 \times C_2^\star, \quad (s, t) \mapsto (p(s), \ell(t)). \end{align}The preimage of \(E\) under this map is a curve in \(\mathbb{P}^1 \times \mathbb{P}^1\) given by the equation
\begin{align} F(s, t) = \nu(t) \left(M_{\ell}^T M_{p}\right) \nu(s) = 0. \end{align}Note that \(F(s, t)\) is of bidegree \((2, 2)\) so it describes a genus-one curve.3
Proof of the porism
A given line \(\ell\) tangent to \(C_2\) generically intersects the conic \(C_1\) in two points; similarly there are two tangents to \(C_2\) for a given point \(p\) on \(C_1\). This means that the set \(E\) has two involutions:
\begin{align} \tau_1: (p, \ell) \mapsto (p', \ell), \quad \tau_2: (p, \ell) \mapsto (p, \ell'), \end{align}where in the definition of \(\tau_1\) the point \(p' \in C_1\) is the other intersection point of \(\ell\) with \(C_1\). Likewise, in \(\tau_2\) the line \(\ell'\) is the other line tangent to \(C_2\) through the point \(p\).
Here is an illustration of the action of these involutions:
Each involution has four fixed points. For \(\tau_2\) these are precisely the four intersection points of \(C_1\) and \(C_2\) and the tangents at these points. For \(\tau_1\) the fixed points are the four bitangents corresponding to the four intersection points of \(C_1^\star\) and \(C_2^\star\).
In the following graphic the four intersection points and the tangents to \(C_2\) are marked in blue. The four points on \(C_1\) on which the tangents to \(C_2\) are also tangent to \(C_1\) are shown in a brownish orange. (This figure is unfortunately not interactive.)
The composition
\begin{align} j = \tau_2 \circ \tau_1: (p, \ell) \mapsto (p', \ell') \end{align}corresponds to making one step in the geometric construction of the polygon. The condition that the polygon closes for arbitrary initial condition is precisely that after \(n\) applications of \(j\) we get back to the starting point, so
\begin{align} j^n = \operatorname{id}. \label{eq:jn-eq-id} \end{align}To study the map \(j\) one can go the torus \(\mathbb{C} / \Lambda\) associated to the elliptic curve discussed above. On the torus, any involution \(\tau : \mathbb{C} / \Lambda \to \mathbb{C} / \Lambda\) with at least one fixed point takes the form
\begin{align} \tau : z \mapsto (-z + u) \mod \Lambda \end{align}where \(u\) is some arbitrary complex number. It follows that the map \(j\) acts on the torus as
\begin{align} j: z \mapsto (\tau_2 \circ \tau_1)(z) = z + (u_2 - u_1) \mod \Lambda. \end{align}This is simply a translation by the fixed complex number \(w = u_2 - u_1\) that is determined by the conics \(C_1\) and \(C_2\).
At the level of the torus the condition \eqref{eq:jn-eq-id} for a closed polygon becomes
\begin{align} \forall z \in \mathbb{C} / \Lambda: \quad j^n(z) = z + n w \equiv 0 \quad \mod \Lambda. \end{align}This means that the point \(w\) has to be a point of order \(n\), so \(n w \equiv 0 \mod \Lambda\). Note that this condition is completely independent of the initial condition! It is only a property of \(w\) which in turn is determined by the conics.
Cayley criterion
From the proof of Poncelet's porism it follows that there are infinitely many \(n\)-gons inscribed in \(C_1\) and circumscribed around \(C_2\) if and only if a particular point on the associated elliptic curve is of order \(n\). For an elliptic curve given as a cubic in the projective plane, a criterion due to Cayley gives the following necessary and sufficient criterion:
Let \(E\) be a cubic in the projective plane \(\mathbb{P}^2\) given by the equation
\begin{align} y^2 z - (x - \lambda_1 z) (x - \lambda_2 z) (x - \lambda_3 z) = 0, \label{eq:cayley-criterion-curve} \end{align}where the \(\lambda_i \in \mathbb{C}^\star\) are distinct. Choose the point \([0 : 1 : 0]\) as the neutral element for the group law on \(E\) making it an elliptic curve. Consider the following expansion around \(x = 0\):
\begin{align} \sqrt{(x - \lambda_1) (x - \lambda_2) (x - \lambda_3)} = \sum_{k=0}^{\infty} a_k x^k. \end{align}Then the point \([0 : a_0 : 1]\) is a point of order \(n\) on \(E\) if and only if (depending on whether \(n\) is even or odd)
\begin{align} \begin{vmatrix} a_2 & \cdots & a_{m+1} \\ \vdots & \ddots & \vdots \\ a_{m+1} & \cdots & a_{2m} \end{vmatrix} = 0 \quad (n = 2m + 1) \quad \text{or} \quad \begin{vmatrix} a_3 & \cdots & a_{m+1} \\ \vdots & \ddots & \vdots \\ a_{m+1} & \cdots & a_{2m-1} \end{vmatrix} = 0 \quad (n = 2m). \end{align}
The original paper by Cayley appears to be this one, but a proof in more modern language can be found in this paper by Griffiths and Harris. An interesting point made in Cayley's paper is that the condition for an \(n\)-gon has to include all the conditions for smaller \(m\)-gons if \(m\) divides \(n\). For example, the condition for a hexagon has to factor into the condition for a triangle and a new condition for a "proper" hexagon. The reason for this is that given a triangle we can form a degenerate hexagon where some of the vertices are the same.
To apply the criterion, we have to change the description of the elliptic curve which was above presented as a curve of bidgree \((2, 2)\) in \(\mathbb{P}^1 \times \mathbb{P}^1\). We consider the pencil of conics \(C_x = x C_1 + C_2\). There are four points that are common to all members of the pencil, namely the four intersection points \(p_0, p_1, p_2, p_3\) of \(C_1\) and \(C_2\). For a given member \(C_x\) of the pencil we draw a tangent line to \(C_x\) through one of these points, say \(p_0\). This line intersects the conic \(C_2\) in one more point \(p(x)\). From \(p(x)\) there are now as before two tangents to \(C_1\). There are four exceptional points: If \(p(x)\) coincides with one of the intersection points \(p_i\), there is only one tangent. Since the conic \(C_1\) on which \(p(x)\) moves is birational to \(\mathbb{P}^1\), we have a double cover of \(\mathbb{P}^1\) branched over four points. This is an elliptic curve and it is isomorphic to the one constructed above.
Here is an illustration of the pencil of conics, the point \(p(x)\) that moves on \(C_1\) and the two tangents from it to \(C_2\). The slider changes the value of \(x\); for \(x = 0\) we get the conic \(C_2\), for \(x \to \infty\) we get \(C_1\). Note that the tangent lines are not real if \(p(x)\) lies inside \(C_2\), so they are not always visible in the picture.
A equation for the curve just described in words is given by
\begin{align} y^2 = \det\left(x M_1 + M_2\right), \label{eq:cubic-p2-poncelet} \end{align}where \(M_1\) and \(M_2\) are the \(3 \times 3\) matrices that describe \(C_1\) and \(C_2\) respectively. The determinant is a polynomial of degree \(3\) in \(x\), so it has three roots \(x_i\) that correspond to three of the intersection points \(p_i\) of the conics. The fourth intersection point corresponds to \(x = \infty\).
The upshot is that the curve given by equation \eqref{eq:cubic-p2-poncelet} is of the form \eqref{eq:cayley-criterion-curve} in Cayley's criterion. This gives us a way to check if two given conics \(C_1\) and \(C_2\) lead to a closed polygon.
We can also try to go in the other direction: If we fix the first conic \(C_1\) and take a simple equation such as
\begin{align} \lambda_1 x^2 + \lambda_2 y^2 + \lambda_3 = 0 \label{eq:simple-conic-equation} \end{align}for the second one4, we can compute the condition for a polygon with \(n\) as a function of the \(\lambda_i\). This condition will be a polynomial in the \(\lambda_i\) so it describes a curve in the two-dimensional projective space of conics of the form \eqref{eq:simple-conic-equation}. Each point on this curve corresponds to a closed \(n\)-gon. For the purpose of obtaining nice pictures one should also impose some extra conditions that make sure the conics and the tangent lines are real.
Example 1: Quadrangle
Here is an example of a closed quadrangle where \(C_1\) is a hyperbola:
The equations for the conics in this example are:
\begin{align} C_1&: \quad 4 x^2 + y^2 - 8 x y - 4 = 0 \\ C_2&: \quad x^2 + \frac{1}{16} y^2 - \frac{1}{5} = 0. \end{align}The expansion of the square root in the Cayley criterion is
\begin{align} \sqrt{\det\left(x M_1 + M_2\right)} = \sqrt{\frac{1}{80} (20 x + 1) (192 x^2 - 20 x - 1)} = \gamma \left(1 + 20 x - 96 x^2 - 4608 x^4 + \mathcal{O}(x^5)\right), \end{align}where \(\gamma = \sqrt{\det M_2} = \frac{1}{\sqrt{-80}}\). Note how the coefficient of \(s^3\) is zero as required for a quadrangle.
Example 2: Hexagon
Finally here is an example for a hexagon:
The conics are given by the equations:
\begin{align} C_1&: \quad x^2 + x y + y^2 - 5 = 0 \\ C_2&: \quad x^2 + \left(-\frac{8}{19} \sqrt{7} + \frac{12}{19}\right) y^2 - 4 = 0 \end{align}The expansion of the square root in the Cayley criterion is
\begin{align} \begin{split} \sqrt{\det(x M_1 + M_2)} = \gamma \Bigg( &1 + \frac{3 - \sqrt{7}}{4} x - \frac{10 + 5 \sqrt{7}}{16} x^2 - \frac{55 + 20 \sqrt{7}}{128} x^3 \\ &\; -\frac{250 + 95 \sqrt{7}}{512} x^4 - \frac{1165 + 440 \sqrt{7}}{2048} x^5 + \mathcal{O}(x^6) \Bigg) \end{split} \end{align}where \(\gamma = \sqrt{\det(M_2)}\). One can check that
\begin{align} \left(- \frac{55 + 20 \sqrt{7}}{128}\right) \left(- \frac{1165 + 440 \sqrt{7}}{2048}\right) - \left(-\frac{250 + 95 \sqrt{7}}{512}\right)^2 = 0 \end{align}which is the condition for a hexagon.
References
- P. Griffiths, J. Harris: On Cayley's explicit solution to Poncelet's porism
- P. Griffiths, J. Harris: A Poncelet theorem in space (alternative non-broken download link)
- A Cayley: On the porism of the in-and-circumscribed polygon
Footnotes
The equations for the conics are
\begin{align} C_1&: \quad x^2 + 2 x y + 3 y^2 - 5 = 0, \\ C_2&: \quad x^2 + \rho y^2 - \frac{1}{5} = 0, \end{align}where \(\rho \approx 0.5289815161400692\) is one of the roots of the following polynomial:
\begin{align} 729 - 36936 t + 720576 t^2 - 7884416 t^3 + 51800064 t^4 - 175251456 t^5 + 191102976 t^6. \end{align}The \(3 \times 3\) matrix \(M_C\) contains the coefficients of the equation for \(C\), so
\begin{align} C: \quad \sum_{i,j=0}^{3} x_i M_{ij} x_j = 0, \end{align}if \([x_0 : x_1 : x_2]\) are homogeneous coordinates for the projective plane.
The genus \(g\) of a smooth curve of bidegree \((d_1, d_2)\) in \(\mathbb{P}^1 \times \mathbb{P}^1\) is computed by
\begin{align} g = (d_1 - 1) (d_2 - 1). \end{align}This follows from the adjunction formula and a proof can for example be found in the book by Hartshorne on algebraic geometry in chapter V, proposition 1.5.
In fact using the coordinate transformations on \(\mathbb{P}^2\), we can always fix the matrix for \(C_1\) to be the identity matrix and the matrix for \(C_2\) to be \(\operatorname{diag}(\lambda_1, \lambda_2, \lambda_3)\). However, this does not lead to nice pictures, because \(C_1\) does not have any real points.