I'm reading the book Indra's Pearls and on page 138 it says:
"Let $W$ be a reduced word which we suppose for example doesn't end with $B$. If we let $L_w$ be that part of the limit set (dust) contained in the disk $D_w$, then we can express the splitting into finer levels by saying that $L_W = L_{Wa} \cup L_{WA} \cup L_{WB}$. Let's suppose also that $W$ doesn't begin with $A$. Then we can expand everything using $a$ to get $a(L_w) =L_{aw} = a(L_{wa}) \cup a(L_{wA}) \cup a(L_{wB})$. "
I think that there is a mistake here, namely that it should be that the reduced word $W$ DOES end with $B$. Below I'll give some more context for people who are unfamiliar with the book and explain my reasoning for why I think this is a mistake. I do have to note that the book is very consistent with this mistake if it is one, as I have seen this again on page 156.
We have four disjoint Schottky disks $D_a, D_b, D_A$ and $D_B$ arranged like this: http://prntscr.com/10drwmt and four möbius transformations $a, b, A=a^{-1}$ and $B=b^{-1}$ which each map the outside of the disk with label the inverse of the corresponding map to the inside of the disk with as label the corresponding map; and they map the inside of the disk with as label the inverse of the corresponding map to the outside of the disk with as label the corresponding map. So for example: the outside of the disk $D_A$ is mapped by $a$ to the inside of the disk $D_a$ and the inside of the disk $D_A$ is mapped to the outside of the disk $D_a$. The inverses $A$ and $B$ behave just like the maps $a$ and $b$ as for example the outside of the disk $D_a$ is mapped by $A$ to the inside of the disk $D_A$ and the inside of the disk $D_a$ is mapped to the outside of the disk $D_A$. Applying these maps after one another and focusing on what happens with certain starting points we see that points get mapped into smaller disks inside the original disks $D_a, D_b, D_A$ and $D_B$. For example the map $aB=a(B)$ first maps a point outside of disk $D_b$ into disk $D_B$, and then maps this point to the inside of disk $D_a$, so we see that this map sends points into a smaller disk which in the book is denoted as $D_{aB}\subset D_a$.
Combinations of the maps $a, b, A$ and $B$ are called words in the book, so for example $a(b(a(B(B(b)))))=abaBBb$ is a word. A reduced word is a word where all operations which 'do nothing' are filtered out. So $abaB$ is a reduced word of $abaBBb$ as the $Bb=Id$ has been filtered out.
So to come back to my confusion: if $W$ is a reduced word which does not end with $B$ then it could end with $a$ for example. So we could consider the word $W=a$. Filling this into what is stated in the book we get that for the part of the limit set contained in disk $D_a$ holds that $L_a=L_{aa}\cup L_{aA}\cup L_{aB}$, but $L_{aA}=a(L_A)$ is the image set of the part of the limit set which is contained in disk $D_A$ obtained by applying the map $a$ on it, which means that this image set is in the EXTERIOR of $D_a$. The image sets $a(L_B)$ and $a(L_a)$ on the other hand are in the interior of $D_a$. I think the authors were trying to say that for some word $W$ the parts of the limit set in disk $D_W$ equals the part of the limit set which is inside the three disks which are inside this disk $D_W$ (note that in the fragment above they state "splitting into finer levels"). This way it is possible to set up some way to find the Hausdorff dimension (this is not done explicitly in the book but they stated the Hausdorff dimension and claimed they found it via an algorithm by McMullen's which is based on the idea they did explain). Either way, I think this word $W$ should end explicitly in $B$ to make sure that this idea works and we don't get any cases where parts of the limit set not in $D_a$, like $a(L_A)$, are part of the given equation $a(L_w) =L_{aw} = a(L_{wa}) \cup a(L_{wA}) \cup a(L_{wB})$. I'm also thinking that this is a mistake as they didn't add an $L_{Wb}$ term and only for that term we would have the same problem if the word $W$ would end explicitly in $B$, namely we would then have that $B(L_b)$ is in the exterior of $D_B$ while the other parts of the equation are in the interior of $D_B$. The only logical conclusion why the authors excluded this case I can come up with is that the word $W$ does end in $B$.
Is my reasoning correct and is this indeed a mistake? Or is the book correct and have I made some mistake in my reasoning? Thanks in advance!
Yes, it should be "$W$ does end with $B$". The decomposition of $L_W$ consists of $L_{Wa}$, $L_{WA}$, $L_{WB}$, but not $L_{Wb}$. That is exactly right if $W$ ends with $B$, because $Wb$ would then end with $Bb$ so it would not be in reduced form, so we disallow it.