Why are rotation numbers not homomorphic?

Question

If $f,g$ are degree-1 monotone maps of the circle, why do we generally have $\rho(f\circ g)\neq\rho(f)+\rho(g)$?

I mean, you might say that we have no right to expect an equality. After all, it's not hard to come up with counterexamples. For example, if $f$ is the piecewise-affine map that interpolates $[0,\frac13]\mapsto[0,\frac23]$ and $[\frac13, 1]\mapsto[\frac23,1]$, then $\rho(f)=0$, yet $\rho(f+\frac13)=\frac12$, rather than $\frac13$. But I still feel like there ought to be something deeper; some satisfying, big-picture reason why a homomorphism identity fails.

Is the "real" reason that $f$ and $g$ tend to disrupt each others' invariant measures? If so, is there a lurking analogy with number theory, where adding $a+b$ tends to disrupt the prime factorizations of $a$ and $b$?

What's the smart way to think about this?

Google tells me that something like my question is addressed in "Structure Theorems for Groups of Homeomorphisms of the Circle" at https://arxiv.org/abs/0910.0218, but I'll be honest, that's too much to digest.

(For what it's worth, this question is motivated by reading chapter 1 of Zakeri's Rotation Sets and Complex Dynamics, and wondering if it's unfair that we have to work so hard to nail down lemmas that ought to be free, if only these things could behave a little better.)

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Edit: For completeness, here's the relevant part of the paper I mentioned.

Lemma 1.8 (Beklaryan and Margulis, [3, 21]). Let $G \leq \mathrm{Homeo}_+(S^1)$. Then the following alternative holds:

1. $G$ has a non-abelian free subgroup, or
2. the map $\mathrm{Rot} : G \to (\mathbb R/\mathbb Z, +)$ is a group homomorphism.

The heart of the proof of Lemma 1.8 is contained in the following lemma, which itself is proven using only on classical results (Poincare’s Lemma and the Ping-pong Lemma)...

I'm not sure that this lemma conveys the intuition that I'm looking for. It does tell us a lot about what the obstructions to a homomorphism look like. But I still feel like I'm missing a "why".

Answer 1

To pull out my own comment on the counterexample:

For example, if $f$ is the piecewise-affine map that interpolates $[0,\frac13]\mapsto[0,\frac23]$ and $[\frac13, 1]\mapsto[\frac23,1]$, then $\rho(f)=0$, yet $\rho(f+\frac13)=\frac12$, rather than $\frac13$.

What's going on there? Well:

Rotation by $\frac13$ has a bunch of period-$3$ cycles, of which the nicest to write down is the orbit of $0$, namely $0\mapsto\frac13\mapsto\frac23\mapsto0$. Then when we compose (in either order) that rotation with $f$, which maps $0\mapsto0$ and $\frac13\mapsto\frac23$, we create a "short-cut" across that period-$3$ cycle, leaving a period-$2$ cycle. So to try and translate this to a "why": It's possible for $f$ to act on another map's cycles by "skipping around" in a way that is potentially independent from $f$'s own time-average behavior.

(I had asked about invariant measures, but there's no particular need to bring in that complication. It's enough to look at finite cycles.)

After some more thought, another useful perspective is just that these functions (degree-1 monotone maps of the circle) have way too much freedom. Sure, a (rational) rotation number tells you about behavior on cycles, but the function might be stretching in any direction outside of the cycles. You can even have two maps with different fixed points, such that their composition has no fixed points.

Finally, it's worth keeping in mind that the degree of a general continuous function is the thing that's homomorphic. In some sense, rotation numbers are a kind of infinitesimal adjustment around degree $1$. And just like in algebra, where we generally have $(1+x)(1+y)\neq(1+x+y)$ unless there is some conspiracy between $x$ and $y$, we shouldn't expect the rotation numbers to add, either.