## Periodic Crystals

You can click and drag the slope of the line and watch the crystal
structures (bottom) and sequences (right) which form based on this.
Use the mouse wheel to zoom.

**Q1)** Can you make a periodic structure?
Try clicking on any of the grid points. Periodicity should be
easiest to see for points near the bottom left (the origin).

**Q2)** What property of a gradient determines whether it forms
periodic structures?
Rationality - i.e. the gradient being written as a fraction of
finite integers. Since the gradient is the number of steps up
divided by the number of steps along,
any line connecting two grid points has a rational gradient.
If we hit the point $(q,p)$ (starting from $(0,0)$), we get a
gradient of $\frac{p}{q}$, and the same holds in reverse:
a gradient of $\frac{p}{q}$ must hit $(q,p)$.

**Q3)** How does the gradient determine how many atoms
appear in a periodic crystal before it begins to repeat?
If we look at the
staircase representation, we can see the total number of atoms
is given by the number of steps up, plus the number along,
meaning there will always be $p+q$ atoms per repetition
(or 'in the unit cell').

If a line directly hits a lattice point, that fixes its gradient, so
at least *some* gradients can be represented by just a point.

**Q4)** Is this representation unique?
The fraction $p \over q$, corresponding to hitting the point
$(q,p)$, isn't unique - we could multiply top and bottom by
an integer, and find another point the line hits
($np \over nq$ and $(nq,np)$).

**Q5)** What additional requirement could you add to *make*
it unique?
If we want to find a *unique* point corresponding to any
gradient, we can always simplify $\frac{p}{q}$ by cancelling all
common factors between $p$ and $q$, corresponding to finding the
first point we hit as we move out from the origin.

**Q6)** Do all possible gradients correspond to a point on the
lattice in this way? What does this mean for forming non-periodic
crystals?
We know all rational gradients correspond to a point,
so the only remaining gradients to account for are
irrational ones (meaning they can't be written
$p \over q$, where $p$ and $q$ are finite integers - say like
$\pi$,
$\phi$,
$\sqrt{2}$, or
$\sqrt[7]{5}$)?

Our 'point-based' representation requires that we hit a point
aside from $(0,0)$. But hitting any other point $(q,p)$ forces
the gradient to be $\frac{p}{q}$. Since this
can't be the case for irrational numbers,
irrational numbers never hit a point, and can't
be represented in this way.

Since the graient needs to hit a point other than $(0,0)$
to be periodic, and these never do, irrational numbers
form non-periodic structures. This isn't enough to be a
quasicrystal though, so learning about the number of
consecutive repetitions of parts of a crystal will
therefore be a major focus for us.

We'll spend a lot of time investigating the link between
structures of different gradients. You can save an
anchor point by right-clicking, to which you can snap
the gradient later. Try right-clicking near a lattice
point, near the gradient, and far away from either. Anchor
points are also removed by right-clicking on them.

## Offsets and Words

Let's take a look at how the offset slider
affects the generated structure, and how structures
can be understood as a sequence of letters.

**Q1)** Can you spot the relationship between the pattern of
atoms/steps and the sequence of 'S's and 'L's on the right?
For gradients less than 1, we tend to label the atoms based
on whether they correspond to short (S) or long (L) gaps.
This corresponds to steps *up* on the staircase corresponding
to the letter S, and steps *along* corresponding to the
letter L.

**Q2)** Does this still hold if we increase the gradient past 1?
When the gradient is 1, using 'S' for short segments
and 'L' for long segments poses a
problem: steps up and along are both the same length.
For gradients greater than 1, steps up are now
the longer ones, and vice versa. In order to maintain
some sense of continuity, however, we remain labelling
steps up (such as the first point) with an S, even
for gradients greater than 1. Practically, the exact
spacing can be accounted for in other ways, and often
isn't used, so the letters are mostly arbitrary.
A more consistent definition is to, think of S as
steps up, and L as steps along.

**Q3)** Can you spot any rough pattern between the length of
runs of the same letter (known as that letter's power),
and the gradient?
You might have also seen that the letter S will have
a power of the gradient rounded either down or up.
This turns out to generalise to the powers of more
complicated structures, as we'll see later, and
is part of how we'll prove aperiodicity.

Since the structure of atoms is determined by this sequence of
letters, this sequence is quite important. Sometimes
referred to as the 'word' for a particular gradient.

**Q4)** Pick a simple rational number for the gradient - say
something like

$1 \over 3$.
By how much do you have to change
the offset before atoms begin to jump?

For a fraction simplified to $p \over q$, we need to increase
the offset to at least $1 \over q$ for the first hop to occur.
The graph of these distances is
known as
Thomae's function,
or the popcorn function. To a physicist's eye, this function
is truly an abomination, it has the property that,
despite being continuous almost everywhere, it's
never continuous over any finite range (it's discontinuous
at every rational number). It's a great example of
why calculus and number theory tend to be so disjoint -
good luck trying to write a differential equation
which describes *this*!
**Q5)** Are atoms conserved as we change the offset?

Atoms are, in fact, conserved (meaning this
can happen in real quasicrystals - it's called a 'phason').
From the perspective of the staircase, moving past a
point corresponds from changing an along-then-up (i.e. LS)
to an up-then-along (SL).
Now, try an irrational number - say something like
$\phi^{-1}$.

**Q6)** How far do you have to move the offset now, before atoms
begin to jump (and how do you think this would change if we
could zoom as far out as we liked)?
For irrational numbers, any finite movement,
however small, will correspond to a jump - it might just
be really far away! Compared to
rational numbers, where numbers of the form $p \over q$ change
after a shift by $1 \over q$, irrational numbers can
*roughly* be thought of as having infinite
$p$ and $q$, so ${1 \over q} = 0$.

**Q7)** Are the first atoms to move near to the origin,
or far away? Why?
Aside from the hop precisely at the origin,
for irrational numbers, the first hops are always
a long way away. A rough argument for this
is that a change in structure near the origin would
require, as we lift up our line, that it crosses
a point near the origin before any
further away, implying that the best approximation
to the gradient
is near the origin. However, this implies that
there even *exists* a best rational approximation to a
particular irrational number, and it should be fairly
easy to convince yourself that given any rational
approximation, we can always build another, better one.

**Q8)** If I suggest that you can view a change in offset
as moving our origin to a new latice point,
am I right or wrong? Does your answer depend on whether
or not the crystal is rational?
As it turns out, only *certain* offsets correspond
directly to a shift in our perspective. Specifically,
it's those offsets which have *just* caused an atom to
jump: our line now crosses through the atom which
*just* jumped, and it's therefore equivalent to our
origin. Hence, in the case of an irrational number,
where any finite movement produces jumps,
continuously changing the offset is equivalent to
pinging us wildly across our infinite,
non-periodic, crystal.

## Approximants

Remember that you can mark and un-mark lattice
points by right-clicking to place an anchor, and
right-clicking the anchor to remove it.
**Q1)** Given the structure of a particular irrational
number (say $\phi^{-1}$),
can you find a series of rational
gradients (and therefore periodic structures, known
as approximants) which tend towards
$\phi^{-1}$ (you can do
this roughly, visually, or mathematically)?
One answer to this is to base this off all the
truncated decimal expansions of
$\phi^{-1}$, so:
$${0 \over 1},{6 \over 10},{61 \over 100},{618 \over 1000},{61803 \over 100000},...$$
Which, cancelling common factors, gives:
$${0 \over 1},{3 \over 5},{61 \over 100},{309 \over 500},{61803 \over 100000},...$$

**Q2)** Is this sequence unique?
Series of rational numbers approaching
$\phi^{-1}$
are far from unique: sequences don't even
need to get closer at every step, so there
are lots of ways one can safely add or remove
points from a sequence without damaging
its convergence.
(That's not to say the points we could add are
*unconstrained* - if you added $\frac{1}{1}$
after every approximant to
$\phi^{-1}$, there
would still be parts of your sequence which
*wouldn't* get closer, breaking convergence.)

**Q3)** If I add the condition that we have to
get closer with *every* term added, and that we
never add a fraction with a higher denominator
if one with a lower denominator could do a
better job, how does that narrow down your choices?
Two sequences which satisfy this condition for
$\phi^{-1}$
can be expressed particularly nicely in terms of
the Fibonacci numbers.
$${1 \over 1},{1 \over 2},{2 \over 3},{3 \over 5},{5 \over 8},{8 \over 13},{13 \over 21},...$$
$${1 \over 1},{2 \over 3},{5 \over 8},{13 \over 21},...$$
In general, this condition will narrow down
your options a lot, but it will be far from
unique, so these aren't the only options you
could have picked. If you didn't find any sequences
like these, I'd suggest checking these out and
getting a sense of their properties. In particular,
the first one will be key later on, whilst the second
may provide an interesting lens with which to contrast
the decisions we take when introducing continued
fractions, later.

**Q4)** [Something of an extension, for those interested
in adding rigor to the idea of strings 'converging']:
In what sense are these good approximations,
and why might not all ways to define 'closeness to our
limit' agree on whether or not we converge?
When we say a list of things converges to something,
we mean that, as we move throughout a list, we can say
that all future values in the list have a 'distance to
our limit' less than some number, and that, as we move
throughout our list, that 'distance' drops to zero.
Basing approximants around convergent gradients
essentially asks the question 'Within the first
period of our approximant, what's the length of the
longest continuous subsequence which matches the limit?',
defining 'distance to the limit' as the inverse of this
answer. For an example, compare the respective limit and
approximant

**SLSLLSL**SLLSLLSLSLLSL...

**SLSLLSL**L|SLSLLSLL|...

The bar denotes the periods of the approximant, and
the bold section is the longest continuous subsequence.
It has 7 letters, and hence the 'distance' is $\frac{1}{7}$.
If, instead, we defined the 'distance' based around asking
'What proportion of the infinite periodic sequence and
the limit match?', the question of convergence might be
far less clear, and certainly would be beyond the scope
of this introduction.
It's these sorts of details us physicists like to ignore
which really wind up the mathematicians, so they're good
things to look out for.

**Q5)** How does the distribution of approximants
change for other irrational numbers (say
$\sqrt{2}$)?
For $\phi^{-1}$, it seems
like there's a nice (and maybe
even complete, in some sense, in the set that a
large class of other sets can be constructed from
it) set of approximants meeting the condition that
each gets progressively closer to
$\phi^{-1}$, given
earlier by:
$${1 \over 1},{1 \over 2},{2 \over 3},{3 \over 5},{5 \over 8},{8 \over 13},{13 \over 21},...$$
And, if you try to find these on the graph, you'll
find they alternate. For
$\sqrt{2}$,
however, if we try to
find something similar, we might find something like:
$$\frac{1}{1},\frac{3}{2},\frac{4}{3},\frac{7}{5},...$$
Which contains, close to the origin,
a point which breaks the pattern of alternation:
$(3,4)$, or $4 \over 3$. We'll see later that the sequence for
$\phi^{-1}$ generalises nicely
to other irrational numbers,
but should always remember that, *unlike* in the
case of $\phi^{-1}$, there in
general may also be other
numbers which *could* fit into our
sequence, with different properties. They won't cause
us any problems, but their existence is worth remembering.

## Moving Between Approximants

We can now construct a series of approximants to our
quasicrystal, but that doesn't leave us better off,
if the only way of calculating the next approximant
is by producing the staircase, or intersection construction,
for the approximant line. If we have to do that,
we might as well go right back to drawing the line
for $\phi^{-1}$ and avoid
calculations with any other
lines. However, what if we began by calculating an
incredibly simple approximant (say $\frac{1}{1}$
for $\pi$), and then just work
out the *changes*
as we move between successive approximants?

Unlike their infinite non-periodic limit, rational approximants
also have the advantage of periodicity
- in order to understand an approximant's structure
*everywhere*, we only actually need to understand
a single period, even if it does get very large.

**Q1)** Consider slowly moving the graph from a
rational point near the origin to any other further
out: When, visually, will a given atom hop?
(Hint: This is similar to what we found when adjusting
the offset earlier).
The structure changes whenever the gradient line
crosses a lattice point.

**Q2)** Where does the structure change the least?
Is it close to the origin, or further away?
As we saw earlier when varying the offset, the number of
lattice points close to the line is the
lowest near the origin, meaning changes to the gradient
will cross the
fewest points, such that changes in the gradient
are understood most simply, there.

**Q3)** Focusing now on the first period of the
final structure, what determines how many changes
occur between the two lines (i.e., how many times
do atoms jump position as we move?).
The number of changes between the two lines
will depend slightly on if you're moving up
or down. If you're moving up (i.e. $1 \over 1$ to
$17 \over 16$, both approximations to
$\sqrt[12]{2}$, the
frequency multiplier for a semitone on a
typical piano), you will instantly push out
the staircase at all points corresponding to the
first approximant, $1 \over 1$
(i.e. $(n,n)$), whilst moving the gradient
downwards by a small amount makes no instant
change at these
points (this is all simply based on how the
staircase structure was defined). Otherwise,
the number of changes from a smaller
approximant to a larger approximant within
the first period is simply the number of
points between those lines, plus those points
possibly corresponding to the approximants themselves.

**Q4)** Consider our approximants to
$\phi^{-1}$, as
chosen above: how many points in the first
period (as well as *which* points) change as we
move between approximants (particularly
successive approximants) from the sequence below?
$${1 \over 1},{1 \over 2},{2 \over 3},{3 \over 5},{5 \over 8},{8 \over 13},{13 \over 21},...$$
This sequence has the property (as for *why*
this is the case, we'll find this out later) that
the only changes within the first period of the
larger approximation are those corresponding to the
approximants themselves - there are no points between
the lines until we get to much higher denominators
(again, we'll prove this later). Therefore, there should
be one change every time the line is moved, each time
at the site of one of the approximants.

**Q5)** Can you find another sequence where even
fewer points change between each step?
It turns out, at least one point always has to change
as we move from rational approximant to rational
approximant. However, we *can* find a sequence with
this minimum number of points changing, but which
converges faster. As an example, what if we choose
approximants which converge from above, rather than
alternating, such
as the other series we found earlier for
$\phi^{-1}$:
$${1 \over 1},{2 \over 3},{5 \over 8},{13 \over 21},...$$
It's worth noting the reason we're able to make
this gain to convergence, which we can see if we
move from $5 \over 8$ down to $8 \over 13$ and back up
to $13 \over 21$, whilst
paying attention to, rather than the first period
of $8 \over 13$, the first period of $13 \over 21$.
The act of moving
down to hit $8 \over 13$ changes one point, but moving back
up to $13 \over 21$ undoes that change straight after.

**Q6)** Two approximants for
$\sqrt{2}$
are given by $7 \over 5$,
and $17 \over 12$, do the changes as you move
between these two fit your intuition from
$\phi^{-1}$?
For the approximants of
$\sqrt{2}$
given, something
different does happen, although you may or
may not have already built this into your
model. The distance between the two approximants
is sufficiently large that two points corresponding to
the $7 \over 5$
approximant occur before $17 \over 12$ (specifically, the
points $(5,7)$ and $(10,14)$), meaning that rather than
one point hopping, as in the case of
$\phi^{-1}$, we see
two points hopping, instead.

## Proving we have Approximants

Having introduced continued fractions, and suggested
that the truncated continued fractions would form a
useful set of approximants, all that we have left to
do is to show that, in fact, they do. The property
we'd like to demonstrate is that gradients between
two of our approximants always have a more
complicated fraction.

**Q1)** By editing the coefficients of the
continued fraction, convince yourself that, as you
move down the fraction, increasing the coefficients
alternately increases and decreases the value of
the continued fraction. Can you work out why this is?
Increasing the first (or, I guess, zeroth)
coefficient of the continued fraction has a
fairly obvious effect - we're just increasing
a number out the front of the fraction:
$$a_0+m+\frac{1}{a_1+...}$$
will always increase as $m$ increases -
which holds even if $m$ isn't an integer.
If we add $m$ to $a_1$, which gives us:
$$a_0+\frac{1}{a_1+m+...}$$
increasing it increases something we're dividing by,
and so decreases the result. What if we go deeper?
Increasing $a_2$ gives:
$$a_0+\frac{1}{a_1+\frac{1}{a_2+m+...}}$$
Increasing $m$ increases the amount we're
dividing by in the fraction $\frac{1}{a_2+m+...}$,
decreasing it and $a_1+\frac{1}{a_2+m+...}$,
which in turn increases $\frac{1}{a_1+\frac{1}{a_2+m+...}}$.
And if we add it deeper, the pattern repeats -
every layer deeper we move, this same pattern
happens, and adding something to the coefficient
has the same alternating effect.

## Proving we have Approximants (cont.)

Remainder mode replaces a term in the continued
fraction with a remainder $r$ which you can
edit to move between $0$ and $\infty$.

**Q2)** What's special about the value of the
continued fraction at $r=0$ and $r=\infty$?
Consider:
$$F_2(r)=a_0+\frac{1}{a_1+\frac{1}{a_2+r}}$$
When $r=0$, this just gives us
the second convergent fraction,
$F_2$. However, when $r \rightarrow \infty$,
$a_2+r \rightarrow \infty$, meaning that
$\frac{1}{a_2+r} \rightarrow 0$, leaving us with
$F_2(\infty) \rightarrow a_0+\frac{1}{a_1+0}=F_1$.

In other words, increasing $r$ sweeps out all
gradients between the two approximants.

When developing this, we had to choose a scaling
for the slider for $r$, over these next two questions,
consider why we picked the distribution of points we did.

**Q3)** Does the resulting gradient for any other
$r$ ever have a simpler fractional form (i.e. first
hitting a point closer to the origin than those
for $r=0$ and $r=\infty$)?
Values of $r$ aside from $0$ and $\infty$ can
*never* result in a simpler form.
If $r$ has a fractional part, it simply adds
extra depth to the continued fraction,
forcing us to expand more terms, increasing
complexity. If $r$ is, instead, a positive
integer, increasing it increases the very
first term we multiply up, increasing the first
denominator and all which follow. No matter
what we pick, the resulting fraction is
more complicated.

**Q4)** As we move the remainder between $0$ and $\infty$,
what changes occur in the structure (as usual,
deduced from the first period) of the
approximant we added the remainder to?
As the slider is moved from $0$ to $\infty$,
it never passes through a point between the
two approximants aside from points corresponding
to the approximants themselves, meaning the *only*
change is the instant we decrease from $\infty$
(in general, depending on the order of approximant
at which we dropped in the remainder, we might
instead get a change when we hit the point $r=0$).

## Periodicity and Continued Fraction Coefficients

**Q1)** As we hop from approximant to approximant,
how can we work out the structure of the next
approximant from the previous, as well as from
the continued fraction?
We've already discussed this a bit, but now we
have convergents available on which to base our
approximants, we can be a little more specific.
Most points don't shift as we move from approximant
$(q_1,p_1)$ to approximant $(q_2,p_2)$, aside from points
corresponding to one or other of the approximants
(either $(q_2,p_2)$, itself, or $(nq_1,np_1)$, for integer $n$
such that $nq_1 \lt q_2$, as we found previously). Aside
from these points (which we'll cover later), the
structure of the next approximant is given by
repeating the previous structure a certain number
of times:
$$W_{n+1} \approx \textrm{repeat} \; W_n \; \textrm{for} \; L_{n+1} \; \textrm{letters}$$
These approximant points are very significant
in the early iterations, if you check this for
yourself, but once the length of the word starts
getting larger, they become less and less significant.
In terms of scale, since the length of $W_n$ only
ever increases with $n$, which means that deeper
coefficients control the maximum power (number
of consecutive repetitions) of subsequences at
ever larger scales.

**Q2)** Thinking about your answer to the
previous question, how do the coefficients of a
continued fraction relate to the number of times
we find repetitions of different lengths?
When we create $W_{n+1}$, the previous word ($W_n$)
will repeat $a_{n+1}$ times, before being cut-off
part way through the next repetition, meaning
that $W_n$ will repeat $a_{n+1}$ times consecutively.
In fact, it turns out it can actually repeat
more than this, but only twice more at most,
corresponding to getting lucky with the
structures which end up surrounding it on
both sides, which, although they do break
off the periodicity, may allow a single
additional repetition on each side. For
intuition, I'd suggest messing around with:
$$W_{n+1}=W_nW_{n-1}$$
$$W_0=\textrm{LS}$$
$$W_1=\textrm{LSL}$$
where you can find that LSL will be repeated at
most 3 times, for the same reason as above.
We'll find out later that this particular example
is actually equivalent to a quasicrystal
with a gradient of $\phi^{-1}$.

**Q3)** Aperiodicity requires that no part
of a structure, on any scale, repeats an arbitrarily
large number of times. Why does this make the
structure generated by
$\phi^{-1}$ aperiodic? The continued
fraction for $\frac{e-2}{3-e}$ is given by
$$\frac{e-2}{3-e}=2+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{1+\frac{1}{6+...}}}}}}$$
Is the structure generated by $e$ aperiodic?
$\phi^{-1}$ is nice and aperiodic - its
continued fraction contains only coefficients
of 1, meaning that at most we can ever find
3 consecutive repetitions of any part of it.
The fact that it has such an upper bound
means that we've found a quasicrystal (it
happens to be a hugely relevant quasicrystal,
too, central to patterns in the Penrose tiling).

$\frac{e-2}{3-e}$
is not aperiodic, since it
has arbitrarily large coefficients. Hence,
whatever upper bound someone proposes for
consecutive repetitions, by looking down the
continued fraction, I can find a coefficient
larger than that, corresponding to more consecutive
repetitions. As a result, there's no upper bound
on the number of repetitions in the structure of
$\frac{e-2}{3-e}$,
meaning its resulting structure
isn't aperiodic.

**Q4)** Why might it sometimes be useful to view
rational numbers as having a continued fraction
truncated by a coefficient of $\infty$ (see below)?
$$3+\frac{1}{7+\frac{1}{7+\frac{1}{\infty+...}}}=3.14$$
It doesn't matter what follows the $\infty$
(as long as it's positive or finite, anyway,
which it will be for all continued fractions),
since $\frac{1}{\infty+x}$ is definitely $0$. The reason this
might be intuitively *useful* is that periodicity
at particular scales is determined by continued
fraction coefficient. A coefficient of $\infty$ for
rational numbers makes it clear they repeat an
infinite number of consecutive times by the same
rules we use to judge maximum consecutive
repetitions for irrational gradients.

## Point Flipping

We've seen a few times now that,
as we slowly change gradient from one approximant to
the next, the points
corresponding to approximants will experience
a flip (say from up-then-right (SL) to
right-then-up (LS)).

**Q1)** What happens to these points when
we move along by two approximants? (If you
have been keeping track of what rounding up,
rather than down, when generating a continued
fraction, does, then think how this might
relate to that).
As we move from, say,
$\frac{2}{3}$ to
$\frac{3}{5}$ in the
convergents of $\phi^{-1}$, we notice all the
structure of $\frac{2}{3}$
is retained perfectly,
aside from a single letter change corresponding
to us hitting $\frac{3}{5}$.
However, the second we
move back up from $\frac{3}{5}$
to start moving towards
$\frac{5}{8}$, that point
gets crossed again, this
time in the other direction. Intuitively,
we could imagine that, rather than hitting
$\frac{3}{5}$, we merely skim
close to it, before
returning back, never quite touching it. As
a result, flipping points corresponding to
the approximants of $\frac{3}{5}$
never actually make a
difference. And the same will be true of
any point: since we always return closer to
$\phi^{-1}$ afterwards,
meaning that the largest
number of points we could ever be dealing
with is simply $a_{n+1}$, the number of points
equivalent to our current convergent before
the next. As we find larger and larger
structures, these at most $a_{n+1}$ points
make up less and less of our overall
structure, meaning we can, hope to be able
to ignore it. (To be more precise, the
first point where occurs tends towards
being infinitely far away for irrational
numbers).

**Q2)** Argue why, when we repeat this process
many times, the number of points remaining affected
should be bounded for quasicrystals, and that
thus we should, roughly, find every other
structure to be *increasingly* similar to:
$$W_{n+1} = \textrm{repeat} \; W_n \; \textrm{for} \; L_{n+1} \; \textrm{letters}$$
Following on from the previous question, we simply
repeat the previous structure $W_n$ until we reach
the end of the first period of the next
structure, which is at $L_{n+1}$.
$$W_{n+1} = \textrm{repeat} \; W_n \; \textrm{for} \; L_{n+1} \; \textrm{letters}$$
The only difficulty arises from, as we've seen a
few times now, points corresponding to either the
current or previous convergent. Fortunately, we
just showed that the problems due to these points
don't accumulate, and so we only need to worry
about points corresponding to either this or the
previous convergent - in quasicrystals, the
distance between these two (and hence the number
of points corresponding to the simpler approximant)
is bounded, and so the proportion of such points
is $0$ as we tend to $\infty$.

**Q3)** Given $L_{n+1}=a_{n+1}L_n+L_{n-1}$,
this corresponds to:
$$W_{n+1} = W_n^{a_{n+1}} \; (\textrm{truncate} \; W_n \; \textrm{after} \; L_{n-1} \; \textrm{letters})$$
Why does this then give:
$$W_{n+1} = W_n^{a_{n+1}}W_{n-1}$$
For this final step, note that, since:
$$W_{n+1} = W_n^{a_{n+1}} \; (\textrm{truncate} \; W_n \; \textrm{after} \; L_{n-1} \; \textrm{letters})$$
then (just using $n-1$):
$$W_n = W_n^{a_n} \; (\textrm{truncate} \; W_{n-1} \; \textrm{after} \; L_{n-2} \; \textrm{letters})$$
Substituting one into the other:
$$W_{n+1} = W_n^{a_{n+1}} \; (\textrm{truncate} \; (W_n^{a_n} \; (\textrm{truncate} \; W_{n-1} \; \textrm{after} \; L_{n-2} \; \textrm{letters})) \; \textrm{after} \; L_{n-1} \; \textrm{letters})$$
This might not look much better, until we
note that we're truncating something which
begins with $W_{n-1}$ after $L_{n-1}$ letters. That's
exactly the length of $W_{n-1}$, and hence that
whole truncation just gives $W_{n-1}$:
$$W_{n+1} = W_n^{a_{n+1}}W_{n-1}$$

**Q4)** In the case of
$\phi^{-1}$, why might the
resulting lattice be called the 'Fibonacci lattice'?
For $\phi^{-1}$,
after the initial $a_0=0$, all
$a_n$ are equal to 1, meaning that we have:
$$W_{n+1} = W_nW_{n-1}$$
or, in other words, the next word is the
concatenation of the previous two, just as,
for the Fibonacci numbers, the next number
is the *sum* of the previous two (this also
means the length is always a Fibonacci number).