Poncelet’s closure theorem

3 Elliptic curve

3.1 Curve and points

To prove the Poncelet’s closure theorem for the two-circle configuration, we will construct a bijection between \(\mathrm{Dom}\) and a certain elliptic curve \(E\), and show that the function \(\mathrm{next}\) maps to an addition in the group of the elliptic curve. We define the curve \(E\) as follows

Definition 18
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\(E\) is an affine cubic curve defined by the following equation, plus a point at infinity as the group identity

\[ Y^2 = X\left(r^2X^2 + \left(1-u^2-r^2\right)X + u^2\right) \]

\(E_0\) is the set of non-singular points of \(E\) including the point at infinity, which forms an abelian group.

The curve \(E\) is not always an elliptic curve because it can be singular. This is characterized by

Lemma 19
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The curve \(E\) contains a singular point only in the following 4 cases

  • \(u + r = \pm 1\) (positive singular), and the singular point is \((-u / r, 0)\)

  • \(u - r = \pm 1\) (negative singular), and the singular point is \((u / r, 0)\)

Proof

This can be solved from the system of equations of \(E\) and the singular point.

Corollary 20
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The curve \(E\) is singular if and only if the two circles are tangent to each other

In general, the group law still holds for non-singular points on a singular cubic curve, so we don’t need to consider the singular case separately for now.

We define three special points \(o\), \(w\), and \(g\) on \(E_0\), which will be useful soon.

Definition 21
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\[ o = (0, 0) \]
Definition 22
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\[ w = \left(\frac{u^2}{r^2}, \frac{u^2}{r^3}\right) \]
Definition 23
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\[ g = \left(1, \frac{1}{r}\right) \]

It is easy to show that they have the following relation

Lemma 24
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\[ g = o - w \]

where the subtraction follows the group law of \(E\)

Proof

Use the formula for group addition on elliptic curves

3.2 Mapping

Next, we define the mapping \(f : E_0 \to \mathrm{Dom}\) as follows

Definition 25
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\[ f(X, Y) = (\mathrm{fPoint}(X, Y), \mathrm{fChord}(X, Y)) \]
\[ \mathrm{fPoint}(X, Y) = \begin{cases} \left[u + r : 0 : 1\right] & ((X, Y) = \infty )\\ \left[\begin{matrix} r^2 (u + r) X^2 + 2 r(1 - r^2 - ur) X + u^2(u + r): \\ -2r^2kY: \\ (rX + u)^2 \end{matrix}\right] & (\mbox{otherwise}) \end{cases} \]
\[ \mathrm{fChord}(X, Y) = \begin{cases} \left[1 : -k : u + r\right] & ((X, Y) = \infty )\\ \left[\begin{matrix} 2 uk ((u^2-r^2)^2 + 4u^2): \\ (r(u+r)^2 X - u((u+r)^2 - 2))((u^2 - r^2)^2 - 4 u^2): \\ 8u^2 k (u^2 - r^2) \end{matrix}\right] & (\mbox{if the next line yields $[0 : 0 : 0]$})\\ \left[\begin{matrix} -2r^2((u + r)^2 - 1)(rX - u)Y + (rX + u)(r^2(u + r)X^2 + 2r(1 - r(u + r))X + (u + r)u^2): \\ -k(2r^2(rX + u)Y + (rX - u)(r^2(u + r)X^2 + 2r(1 - r(u + r))X + (u + r)u^2)): \\ (rX + u)((rX - u)^2(u + r)^2 + 4urX) \end{matrix}\right] \end{cases} \]

As a reminder, \(k^2 = (u+r)^2 - 1\). It doesn’t matter whether we pick \(k\) or \(-k\), as changing the sign of \(k\) results in mirroring around the x-axis.

We should also note that when \(k = 0\), which corresponds to the positive singular case, this mapping becomes trivial.

Lemma 26
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If \(k = 0\), \(f\) becomes a constant function

\[ f(X, Y) = ([u + r : 0 : 1], [u + r : 0 : 1]) \]

In such case, \(f\) is not useful, but we can flip the sign of \(r\), which keeps both \(\mathrm{Dom}\) and \(E\) the same, and turn it into a negative singular case and get a useful \(f\). This has no impact on the next few lemma, but this will appear again when we need the surjectivity of \(f\).

On the other hand, when \(k \ne 0\), we can further restrict the range of \(f\)

Lemma 27
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If \(k \ne 0\), \(f(X, Y) \in \mathrm{dom_0}\)

With the map \(f\), we can make connection between the elliptic curve \(E\) and the two-circle configuration with the following theorems

Theorem 28
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For all \(p \in E_0\),

\[ f(o - p) = \mathrm{rChord}(f (p)) \]
Proof

Expand the formula for \(f\), \(\mathrm{rChord}\), and subtraction for elliptic curve.

Theorem 29
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For all \(p \in E_0\),

\[ f(w - p) = \mathrm{rPoint}(f (p)) \]
Proof

Expand the formula for \(f\), \(\mathrm{rPoint}\), and subtraction for elliptic curve.

Theorem 30
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For all \(p \in E_0\),

\[ f(p + g) = \mathrm{next}(f (p)) \]
Proof

Use \(g = o - w\) and \(\mathrm{next} = \mathrm{rChord} \circ \mathrm{rPoint}\).

Corollary 31
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For all \(p \in E_0\), \(n \in \mathbb {N}\),

\[ f(p + n g) = \mathrm{next}^{n}(f (p)) \]
Proof

By induction on the previous theorem

As another corollary, we show that \(f\) is injective except for positive-singular case

Corollary 32
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\(f\) is injective when \(k \ne 0\).

Proof

We first observe that the \(x/z\)-coordinate of \(\mathrm{fPoint}\) is quadratic and only depending on \(X\). Suppose \(\mathrm{fPoint}(X_1, Y_1) = \mathrm{fPoint}(X_2, Y_2)\) but \(X_1 \ne X_2\), then \(X_1\) and \(X_2\) must be two distinct roots of a certain quadratic equation. We then show from Vieta’s formulas that such pair of roots satisfies \((X_1, Y_1) + (X_2, Y_2) = o\), deducing the fact that \(\mathrm{rPoint}(\mathrm{fPoint}(X_1, Y_1)) = \mathrm{fPoint}(X_2, Y_2)\), making \(\mathrm{fPoint}(X_1, Y_1)\) a fixed point of \(\mathrm{rPoint}\). Such fixed point is only possible at the intersection of the two circles, but that gives us an overdetermined system of equations that leeds to contradiction.

Special cases needs to be discussed when \((X, Y) = o\) or \((X, Y) = \infty \), but it is easy to show the injectivity on these specific points with explicit calculation.

3.3 Inverse

We will also define the inverse mapping \(e : \mathrm{Dom_0} \to E_0\). We start with the following functions

\[ eN([x:y:z], [a:b:c]) = (kx + ((u + r) ^2 - 1) y) c ^2 + (2 u k a ^2 + u ((u + r) ^2 - 2) a b + (r (u + r) - 2) k a c + (2 - (u + r) (u + 2 r)) b c - u k c ^2) z \]
\[ eD([x:y:z], [a:b:c]) = (r ((u + r) a - c) ((u + r) b + k c)) z \]
\[ e_0([x:y:z]E) = \left(\frac{eN([x:y:z], E)}{eD([x:y:z], E)}, -\left(r \frac{eN([x:y:z], E)}{eD([x:y:z], E)} + u\right)^2\frac{y}{r^2kz} \right) \]

However, these functions hit \(0/0\) at several points. The full definition of \(e\) is

Definition 33
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For \(k \ne 0\), define

\[ e([x:y:z], [a:b:c]) = \begin{cases} \left(-\frac{u}{r}, -\frac{yuk}{xr^2}\right) & (z = 0)\\ \left(\frac{(u + r) ^2 y}{2 kr z}, -\frac{(u + r) ^2 y}{2 kr^2 z} \right) & ((u + r) a - c = 0, (u + r) b + k c \ne 0) \\ (0, 0) & ((u + r) a - c = 0, (u + r) b + k c = 0, u + r = 0, [a:b:c]=[1:k:0]) \\ \infty & ((u + r) a - c = 0, (u + r) b + k c = 0, u + r = 0, [a:b:c]\ne [1:k:0]) \\ \infty & ((u + r) a - c = 0, (u + r) b + k c = 0, u + r \ne 0, [a:b:c]=[u+r,0,1]) \\ w & ((u + r) a - c = 0, (u + r) b + k c = 0, u + r \ne 0, [a:b:c]\ne [u+r,0,1]) \\ \left(\frac{(u + r) ^2 (x+ (r - u) z)}{2 r z}, \frac{(u + r) ((u r + r ^2 - 1) (u + r)x + (r^4 + r^3 u - r^2 u^2 - r u^3 - u^2 - r^2) z) }{2 r ^2z} \right) & ((u + r) a - c \ne 0, (u + r) b + k c = 0) \\ e_0([x:y:z], [a:b:c]) & (\mbox{all other cases}) \end{cases} \]

We then verify that \(e\) is the right inverse of \(f\) on \(\mathrm{Dom_0}\):

Theorem 34
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When \(k \ne 0\), for all \((V, E) \in \mathrm{Dom_0}\),

\[ f(e(V, E)) = (V, E) \]
Proof

Just expand by definition and verify with a powerful computer.

And because \(f\) is injective, \(e\) is also the left inverse, making this a bijection

Corollary 35
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When \(k \ne 0\), for all \((X, Y) \in E_0\),

\[ e(f(X, Y)) = (X, Y) \]
Corollary 36

When \(k \ne 0\), \(f\) is a bijection between \(E_0\) and \(\mathrm{Dom_0}\)

3.4 Conclusion

Given the bijection, we give the complete correspondence of \(\mathrm{next}\) and \(g\)

Theorem 37

When \(k = 0\), \((\mathrm{next})^n(V, E) = (V, E)\) for \((V, E) \in \mathrm{Dom_0}\) if and only if \(n \cdot g = 0\).

Note that the LHS of iff depends on an arbitrarily chosen \((V, E) \in \mathrm{Dom_0}\), but the RHS doesn’t, so any two points in \(\mathrm{Dom_0}\) are also in iff transitively. From this, we conclude with the Poncelet’s closure theorem for the two-circle configuration

Theorem 38
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Given a two-circle configuration in field \(K\) with \(k = 0\), if for a natural number \(n\), and some \((V, E) \in \mathrm{Dom_0}\) it holds that \((\mathrm{next})^n(V, E) = (V, E)\), then the same holds for the same \(n\) and all \((V, E) \in \mathrm{Dom_0}\).

As mentioned before, the \(k = 0\) condition an be trivially eliminated by flipping the sign of \(r\).