Computing kneading sequences for renormalizations of Lorenz maps

Question

I am stuck trying to understand certain claims made in this paper, and for completeness I will reproduce some definitions from it.

A Lorenz map $f$ on $I = [0,1]$ is a monotone increasing function that is continuous except at a critical point $c\in (0,1)$, where it has a jump discontinuity, and $f(I \setminus c)\subset I.$ The branches $f_0:[0,c]\to I, f_1:[c,1]\to I$ are assumed to satisfy $$f_0(c) = 1, f_1(c) = 0$$ $$f_k(x) = \varphi_k(|c-x|^\alpha)$$ for some exponent $\alpha>0$, $C^2$-diffeomorphisms $\varphi_k, k=0,1.$

A Lorenz map $f$ is said to be renormalizable if there exist $p,q \geq 1$ such that $J = [f^p(0), f^q(1)]$ is contained in $I$ and has $c$ in its interior, and such that the first-return map restricted to $J$ is also a Lorenz map after rescaling. This rescaled map is known as the renormalization $Rf$ of $f$. The symbolic coding of the branches defines the combinatorics $w = (w_0, w_1)$ of the renormalization, i.e . if $J_k = J\cap[k,c)$ define $w_k$ as the finite word on $\{0,1\}$ such that $f^j(J_k)\subset[w_k(j),c)$. A Lorenz map is said to be of $(a,b)$-type if its combinatorics are of the form $(011\cdots1, 10\cdots0)$ with $a$ 1's and $b$ 0's in the first and second positions respectively.

Now suppose one has a once $(2,1)$-renormalizable map with combinatorics $(011,10).$ The paper suggests (pg. 13) that a twice $(2,1)$-renormalizable map will have combinatorics $(0111010, 10011)$ and this can be obtained by substituting 011 for 0 and 10 for 1 in the original combinatorics. My question: why does this rule work? And what would the combinatorial type of (say) a thrice $(2,1)$-renormalization be? I can't find any resources that handle these computations.

Answer 1

Perhaps this will not be a good answer, but, firstly you might have a look at this paper In particular, have a look at Chapter 4. Also, you might want to look at the work of Glendinning and Hall here

Now, the way I understand it is the following (maybe is not the best because I only understand the symbolic setting). Recall that the kneading invariant of a Lorenz map, say $(\alpha, \beta)$ is renormalisable if there exist two finite words $\omega = w_1 \ldots w_m \quad \text{and} \quad \nu = v_1 \ldots v_n$ and two sequences $\{M_i\}_{i=1}^\infty, \, \{N_j\}_{j=1}^\infty \subset \mathbb{N}\cup \{\infty\}$ such that:

1. $n+m > 3$;
2. $(\omega^\infty, \nu^\infty)$ is also a kneading invariant;
3. $\alpha = \nu \omega^{M_1}\nu^{M_2}\omega^{M_3}\nu^{M_4}\ldots;\\ \beta = \omega\nu^{N_1}\omega^{N_2}\nu^{N_3}\omega^{N_4}\ldots, $

If $(\alpha,\beta)$ is not renormalisable then we say that $(\alpha, \beta)$ is prime. Also, if $n+m = 3$ then $(\alpha,\beta)$ is called trivially renormalisable.

In the references I gave above (you might check also this), it is mentioned/proved that if a Lorenz map is renormalisable if the kneading invariant is renormalisable. So, you only need to worry about the combinatorics of the kneading invariant to understand the Lorenz map (of course, if you can define a kneading invariant). I think this might answer your first question.

Regarding the second question: having a (2,1) renormalisable map, then the only words that will help to renormalise are $(011, 10)$ (if it was 1,2 renormalisable, then you would have $(01, 101)$). In particular, there are certain conditions needed in order to have a pair $\omega, \nu$ to be candidates to renormalise a kneading invariant as I mentioned in the definition; namely, $(\omega^\infty, \nu^\infty)$ must be a kneading invariant of another Lorenz map, and in fact it has to be prime or trivially renormalisable (this is because when you renormalise the kneading invariant you need that the words $\omega$ and $\nu$ to have minimal length). You can check some extra conditions for the linear case in the works of Glendinning (here) and Barnsley et.al. (here).

Now, if you apply the substitution you are mentioning above, you will get:

$011 \to 0111010 \to 01110101001110011 \quad \text{and} \quad 10 \to 10011 \to 100110111010$. Observe that the final words can be written concatenating 011 and 10. Now, the words $011$ and $10$ have minimal length, in the sense that $((011)^\infty, (10)^\infty)$ is trivially renormalisable. Take the final word and apply the inverse substitution you mentioned. In the first case, you will get the pair $(011)^\infty,(10)^\infty$ and in the second case you need to perform the inverse substitution twice (it is straightforward from the way I wrote it). Now, you can get a different (2,1) renormalisation with the same combinatorics. For example, if you take the sequences $(01101)^\infty, (100100)^\infty$ (which are kneading invariants) you can get a new $(2,1)$ renormalisable pair. Namely, apply the substitutions again, i.e.

$01101 \to 011101001110 \to 01110101001110011011101010011 = \omega'$

and

$100100 \to 1001101110011011 \to 10011011101001110101001101110100111010 = \nu'.$

It is not difficult to check that, if you have two sequences $\alpha$ and $\beta$ that satisfy 3. using the words $\omega'$ and $\nu'$ you can apply twice the inverse substitution you mentioned above. This will tell you that your kneading invariant is $(2,1)$ twice renormalisable, although it could be renormalisable using different combinatorics. In conclusion, if you take any kneading invariant of a Lorenz map, if you apply $K$ times a substitution of combinatorial type $(m,n)$ the resulting pair must be a $K$ times renormalisable kneading invariant with $(m,n)$ combinatorics.

Hope I helped.