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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.

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    $\begingroup$ $f(f(x)) = 1-1/x$ is not analytic near $x=0$, how would you interpret that as a generating function? $\endgroup$ – Martin R Jun 10 at 13:12
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    $\begingroup$ 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. $\endgroup$ – Phicar Jun 10 at 15:07
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    $\begingroup$ 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). $\endgroup$ – runway44 Jul 18 at 3:43
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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 $f^{[3]} = id$ 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 $f^{[3]}$ fixes these points. Since a Möbius transformation is uniquely determined by the image of three distinct points, $f^{[3]} = id$.

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 $f^{[3]} = id$.

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 $$ T(f(f(f(z)))) = T(z) \implies f^{[3]}= id \, . $$

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  • $\begingroup$ Thank you for pointing me to the reference related to the mobius transform! This is a good explanation from the complex analysis perspective :D. I am still holding out hope for a purely combinatorial explanation, however. $\endgroup$ – Siddharth Bhat Jun 10 at 7:10

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