# Combinatorial interpretation of the identity $(f \circ f \circ f)(x) = x$ where $f(x) = 1/(1-x)$ for $x\in(-1,1)$

Let $$f(x) = 1/(1-x)$$. We can interpret this as the generating function of an infinite list of $$1$$s: $$(1, 1, 1, \cdots)$$.

Now, let's consider $$(f \circ f \circ f)(x)$$. We first compute $$(f \circ f)(x)$$, and use this to compute $$f \circ f \circ f$$.

\begin{align*} &(f \circ f)(x) = \frac{1}{1 - \frac{1}{1-x}} \\ &= \frac{1}{\frac{(1 - x) - 1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} = 1 - \frac{1}{x} \\ \end{align*}

\begin{align*} &(f \circ f \circ f)(x) = (f \circ f)(f(x)) = 1 - \frac{1}{f(x)} = 1 - (1 - x) = x \end{align*}

What is going on? It's cool that $$f^{\circ 3} = id$$, but I don't have an explanation for this. I lose the ability to interpret this as a generating function at step $$(2)$$: When we compute $$f \circ f= 1 - 1/x$$, it's unclear to me what this represents.

I have an explanation, but it's not very enlightening. We can consider $$f$$ as a mobius transform: $$f(z) = (0\cdot z + 1)/(-z+1)$$. This gives the matrix $$F$$ such that $$F^3 = I$$. So there should be some complex analytic explanation that I am unable to divine.

$$F = \begin{bmatrix} 0 & 1 \\ -1 &1 \end{bmatrix}; \quad F^3 = -I \underset{\small{\text{(projectively)}}}{\simeq} I$$

I was hoping to view either some Riemann sphere-based explanation (I don't know this very well) or a combinatorial explanation of the above phenomenon.

• $f(f(x)) = 1-1/x$ is not analytic near $x=0$, how would you interpret that as a generating function? – Martin R Jun 10 at 13:12
• Composition of generating functions does not deal well with constant term. For example, the representation by substitution that Combinatorial species offers for composition requires a zero constant term(Probably to avoid trying to put structure of what is not there). Check Bergeron's Combinatorial Species book. – Phicar Jun 10 at 15:07
• There might be something here if we categorify - something in the vein of seven trees in one (see Baez week 202 for a blog post about it). – runway44 Jul 18 at 3:43

Here are some interpretations in the context of Möbius transformations, i.e. of the holomorphic automorphism of the extended complex plane (or Riemann sphere). I'll use the notation $$f^{[n]}$$ for the $$n$$-th iteration of $$f$$.
A quick way to see that $$f^{} = id$$ is the following: The Möbius transformation $$f(z) = 1/(1-z)$$ maps the points $$0, 1, \infty$$ to $$1, \infty, 0$$, respectively, i.e. those three points are permuted cyclically. It follows that $$f^{}$$ fixes these points. Since a Möbius transformation is uniquely determined by the image of three distinct points, $$f^{} = id$$.
Another way is to use the classification of Möbius transformations. (See also John Olsen: The Geometry of Möbius Transformations for a nice overview). $$f(z) = \frac{0z + 1}{-1z + 1}$$ has $$\operatorname{tr}^2(f) = (0+1)^2 = 1$$, so that it is an elliptic transform. Elliptic transformations are conjugate to a rotation $$z \mapsto \lambda z$$ with the same (squared) trace: $$\lambda + \frac 1 \lambda + 2 = \operatorname{tr}^2(f) = 1 \\ \implies \lambda^2 + \lambda + 1 = 0 \\ \implies \lambda^3 = 1 \, , \, \lambda \ne 1 \, .$$ So $$f$$ is conjugate to a rotation by a third root of unity, and that implies $$f^{} = id$$.
That conjugation can also be computed explicitly. The fixed points of $$f$$ are roots of $$z^2-z+1=0$$ which are $$z_{1, 2} = \frac 12 (1 \pm i \sqrt 3) \, .$$ It follows that with $$T(z) = \frac{z-z_1}{z-z_2}$$ the “conjugate” Möbius transformation $$g = T \circ f \circ T^{-1}$$ has the fixed points $$0$$ and $$\infty$$ and therefore is a rotation: $$T(f(z)) = \lambda T(z)$$ for some constant $$\lambda \in \Bbb C$$. For $$z = \infty$$ we get $$-\frac{z_1}{z_2} = T(f(\infty)) = \lambda T(\infty) = \lambda \, ,$$ i.e. $$\lambda = \frac 12 (-1 + i\sqrt 3)$$. $$\lambda$$ is a third root of unity, so that $$T(f(f(f(z)))) = T(z) \implies f^{}= id \, .$$