Assume that $\tilde{R}_{r}\neq R_{i}$ and
${\mathcal{H}}(r)=\\{(i,j_{1}),\cdots,(i,j_{K})\\}$ with $j_{k}<j_{k+1}$. Let
$(i,j_{k})\in{\mathcal{H}}(r)$ in this manner:

  * i)

$\tilde{R}_{r}\subset H^{i}_{j_{k}}\setminus\partial^{s}H^{i}_{j_{k}}$ if and
only if ${\mathcal{H}}(r)=\\{(i,j_{k})\\}$, i.e. $K=1$.

  * ii)

$H^{i}_{j_{k}}\subset\tilde{R}_{r}$ if and only if $j_{1}<j_{k}<j_{K}$.

  * iii)

The sub-rectangle $H^{i}_{j_{k}}$ contains only one horizontal boundary
component of $\tilde{R}_{r}$ if and only if $K>1$, i. e. the cardinality of
${\mathcal{H}}(r)$ is bigger than $1$ ( $\sharp({\mathcal{H}}(r))>1$ ). In
this case, $H^{i}_{j_{k}}$ contains the inferior boundary of $\tilde{R}_{r}$
if $k=1$ and the upper boundary of $\tilde{R}_{r}$ if $k=K$.

In case that $\tilde{R}_{r}=R_{i}$, for all $(i,j)\in{\mathcal{H}}(r)$,
$H^{i}_{j}\subset\tilde{R}_{r}$.

###### Proof.

Item $i)$. If $\tilde{R}_{r}\subset
H^{i}_{j_{k}}\setminus\partial^{s}H^{i}_{j_{k}}$, the only horizontal sub-
rectangle of $R_{i}$ that intersects the interior of $\tilde{R}_{r}$ is
$H^{i}_{j_{k}}$, so ${\mathcal{H}}(r)=\\{(i,j_{k})\\}$ and $K=1$.

Conversely, if ${\mathcal{H}}(r)=\\{(i,j_{1})\\}$, the only horizontal sub-
rectangle of $R_{i}$ that intersects the interior of $\tilde{R}_{r}$ is
$H^{i}_{j_{1}}$, then $\overset{o}{\tilde{R}_{r}}\subset H^{i}_{j_{1}}$. Since
the stable boundaries of $\tilde{R}_{r}$ are disjoint from the stable
boundaries of $H^{i}_{j_{k}}$, it is necessary that $\tilde{R}_{r}\subset
H^{i}_{j_{k}}\setminus\partial^{s}H^{i}_{j_{k}}$.

Item $ii)$. The horizontal sub-rectangles $H^{i}_{j_{k-1}}$ and
$H^{i}_{j_{k+1}}$ intersect the interior of $\tilde{R}_{r}$, and
$H^{i}_{j_{k-1}}$ is below $H^{i}_{j_{k}}$, and $H^{i}_{j_{k+1}}$ is above
$H^{i}_{j_{k}}$, then it is necessary that
$H^{i}_{j_{k}}\subset\tilde{R}_{r}$.

Conversely, if $H^{i}_{j_{k}}\subset\tilde{R}_{r}$ and $\tilde{R}_{r}\neq
R_{i}$, the contention is proper, which implies the existence of horizontal
sub-rectangles of $R_{i}$, $H^{i}_{j{k+1}}$ above $H^{i}_{j_{k}}$ and
$H^{i}_{j_{k-1}}$ below $H^{i}_{j_{k}}$. They intersect the interior of
$\tilde{R}_{r}$ and therefore, $(i,j_{k-1})$ and $(i,j_{k+1})$ are elements of
${\mathcal{H}}(r)$. This implies that $j_{1}\leq j_{k-1}<j_{k}<j_{k+1}\leq
j_{K}$.

Item $iii)$.If $H^{i}_{j_{k}}$ contains only the inferior boundary of
$\tilde{R}_{r}$, the index $j_{k}$ satisfies that for all
$j^{\prime}\in{\mathcal{H}}(r)$, $j^{\prime}\geq j_{k}$. Since the upper
boundary of $\tilde{R}_{r}$ is not contained in the interior of
$H^{i}_{j_{k}}$, there exists another $j^{\prime}>j$ such that the upper
boundary of $\tilde{R}_{r}$ is contained in the interior of
$H^{i}_{j^{\prime}}$. Consequently, the interior of $\tilde{R}_{r}$ intersects
the interior of $H^{i}_{j^{\prime}}$. Hence,
$(i,j^{\prime})\in{\mathcal{H}}(r)$, $K>1$ and $j_{k}=j_{1}$. The case where
$H^{i}_{j_{k}}$ contains only the upper boundary of $\tilde{R}_{r}$ is
entirely symmetric.

Finally if $\tilde{R}_{r}=R_{i}$, then
${\mathcal{H}}(r)=\\{(i,j)\in{\mathcal{H}}(T)\\}$ and then every horizontal
sub-rectangle $H^{i}_{j}$ is contained in $R_{i}=\tilde{R}_{r}$, therefore
$H^{i}_{j}\subset\tilde{R}_{r}$.

∎

####  5.1.7. The number of horizontal sub-rectangles in $\tilde{R}_{r}$

We will count the number of horizontal sub-rectangles within the Markov
partition ${\mathcal{R}}_{S(\underline{w})}$ in the intersection
$\tilde{R}_{r}\cap H^{i}_{j}$, and we will also establish how these sub-
rectangles are rearranged by $f$. For that reason we consider the three
different cases that we outlined in Proposition 5.2.

The strategy will be:

  1. (1)

For every $(i,j)\in{\mathcal{H}}(r)$, compute the number of horizontal sub-
rectangles contained within $\tilde{R}_{r}\cap H^{i}_{j}$ and call it
$\underline{h}(r,j)$.

  2. (2)

Label the rectangles within $\tilde{R}_{r}\cap H^{i}_{j}$ as
$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}$.

  3. (3)

Determine the function $\underline{\rho}_{(r,j)}$ that, when given $(r,J)$,
returns a value $(r^{\prime},l^{\prime})$ such that
$f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$.

  4. (4)

Calculate $H_{r}$ by summing $\underline{h}(r,j)$ over all $j$ such that
$(i,j)\in{\mathcal{H}}(r)$, which gives us the number of horizontal sub-
rectangles within $\tilde{R}_{r}$.

  5. (5)

Enumerate the horizontal sub-rectangles within $\tilde{R}_{r}$ by
$\\{\tilde{H}^{r}_{\underline{J}}\\}_{\underline{J}=1}^{H_{r}}$ and specify
how to determine the _relative position_ of a rectangle
$\tilde{H}^{r}_{\underline{J}}$. This means finding a unique index $j$ and a
unique $J$ such that $\tilde{H}^{r}_{\underline{J}}=\tilde{H}^{r}_{j,J}$.

  6. (6)

Finally, define $\rho_{S(\underline{w})}(r,\underline{J})$ as
$\rho_{(r,j)}(r,J)$, where $j$ and $J$ were determined in the previous step.

##### First case: $\tilde{R}_{r}\subset
H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$

In some sense, this is the simplest scenario because all the horizontal sub-
rectangles of $\tilde{R}_{r}$ are contained within $H^{i}_{j}$. This allows us
to directly compute the value for $H_{r}$ and the function
$\rho_{S(\underline{w})}$ when restricted to the horizontal sub-rectangles of
$\tilde{R}_{r}$.

######  Lemma 5.17.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$, with
$r=\tilde{r}(i,s)$. Suppose we have a pair of indices
$(i,j)\in{\mathcal{H}}(r)$ such that $\tilde{R}_{r}\subset
H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$. In this case $s\neq 1$ and $s\neq
O(i,\underline{w})+1$ and we take $I_{t_{1},\underline{w}}$ as the lower
boundary of $\tilde{R}_{r}$ and $I_{t_{2},\underline{w}}$ as the upper
boundary of $\tilde{R}_{r}$.

Assume that $\rho_{T}(i,j)=(k,l)$. Let
$s^{-}=\underline{s}(t_{1}+1,\underline{w})$ and
$s^{+}=\underline{s}(t_{2}+1,\underline{w})$. Then,
$w_{t_{1}+1}=w_{t_{2}+1}=k$ and the number of horizontal sub-rectangles of
${\mathcal{R}}_{S(\underline{w})}$ contained in the intersection
$\tilde{R}_{r}\cap H^{i}_{j}$ is given by the formula:

(5.10) $\underline{h}(r,j)=|s^{-}-s^{+}|.$

###### Proof.

Look at Figure 5.4. Since $\tilde{R}_{r}\subset
H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$, $\tilde{R}_{r}$ doesn’t share a
stable boundary component with $R_{i}$. This means $s\neq 1$ and $s\neq
O(i,\underline{w})+1$, and we can take the intervals $I_{t_{1},\underline{w}}$
and $I_{t_{2},\underline{w}}$ as described in the Lemma. Since
$f(\tilde{R}_{r})\subset f(H^{i}_{j})\subset R_{k}$, then
$w_{t_{1}+1}=w_{t_{2}+1}=k$, as we claimed. Then the number of horizontal sub-
rectangles of $\tilde{R}_{r}$ consists of all the rectangles in
${\mathcal{R}}_{S(\underline{w})}$ contained in $R_{k}$ and located between
the intervals $I_{t_{1}+1,\underline{w}}$ and $I_{t_{2}+2,\underline{w}}$. If
their relative positions are given by the numbers $s^{-}$ and $s^{+}$, then
the number of such rectangles is given by $|s^{-}-s^{+}|$ and we has proved
our Lemma.

∎

######  Definition 5.15.

The horizontal sub-rectangles of $\tilde{R}_{r}$ as described in Lemma 5.17
are labeled from the bottom to the top with respect to the orientation of
$\tilde{R}_{r}$ as:

$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}.$

######  Remark 5.5.

As established in Item $i)$ of Proposition 5.2, the set ${\mathcal{H}}(r)$
contains only one element. Therefore, $H_{r}=|s^{-}-s^{+}|$,

######  Lemma 5.18.

With the hypothesis of Lemma 5.17, assume that $\Phi_{T}(i,j)=(k,l,\epsilon)$
and $f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$. Then:

  * i)

If $\epsilon=\epsilon(i,j)=1$, then $s^{-}<s^{+}$,
$r^{\prime}=\tilde{r}(k,s^{-}+J)$ and $l=l^{\prime}$.

  * ii)

If $\epsilon=\epsilon(i,j)=-1$, then $s^{-}>s^{+}$,
$r^{\prime}=\tilde{r}(k,s^{-}-J+1)$ and $l=l^{\prime}$.

###### Proof.

Item $i)$. If $\epsilon(i,j)=1$ then $f$ preserves the vertical orientation of
$\tilde{R}_{r}$, and then
$I_{t_{1}+1,\underline{w}}<I_{t_{2}+1,\underline{w}}$, so $s^{-}<s^{+}$.

The rectangle $\tilde{H}^{r}_{j,J}$ is in vertical position $J$ inside
$\tilde{R}_{r}\cap H^{i}_{j}$ and it is sent to the rectangle whose position
inside $R_{k}$ is at vertical position $J$ above the the rectangle in position
$s^{-}$ withing $R_{k}$. Therefore, $f(\tilde{H}^{r}_{j,J})$ is inside a
rectangle $\tilde{R}_{r^{\prime}}$ inside to $R_{k}$ whose position is equal
to $s^{-}+J$. In other words, $r^{\prime}=\tilde{r}(k,s^{-}+J)$.

Item $ii)$. If $\epsilon(i,j)=-1$ then $f$ inverts the vertical orientation of
$\tilde{R}_{r}$. Then, $I_{t_{1}+1,\underline{w}}>I_{t_{2}+1,\underline{w}}$
and $s^{-}>s^{+}$.

The rectangle $\tilde{H}^{r}_{j,1}$ that is positioned at place $1$ inside
$\tilde{R}_{r}\cap H^{i}_{j}$ is sent into to the rectangle
$\tilde{R}_{r^{\prime}}$, whose upper boundary is at position $s^{-}$,
therefore $r^{\prime}=s^{-}=s^{-}-(1-1)$. In general the upper boundary of the
rectangles $\tilde{R}\subset R_{k}$ determine its position inside $R_{k}$.
That means that $f(\tilde{H}^{r}_{j,J})$ is inside a rectangle
$\tilde{R}_{r^{\prime}}$ withing $R_{k}$ whose position is given by
$s^{-}-J+1$. In other words, $r^{\prime}=\tilde{r}(k,s^{-}-J+1)$.

Finally in bot cases, Proposition 5.1 implies that $l^{\prime}=l$. ∎

######  Definition 5.16.

With the hypothesis of Lemma 5.17, assume that $\Phi_{T}(i,j)=(k,l,\epsilon)$.
Then:

If $\epsilon(i,j)=1$ we define:

(5.11) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{-}+J),l).$

and if $\epsilon(i,j)=-1$:

(5.12) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{+}-J+1),l).$

######  Remark 5.6.

In this particular case the function $\rho_{S(\underline{w})}(r,J)$ is
determined by either Equation 5.12 or Equation 5.11. Towards the end of this
section, we will provide a unified approach of all the cases that we are going
to consider, for this reason we are keeping the notation $\tilde{H}^{r}_{j,J}$
even if we can change it by $\tilde{H}^{r}_{J}$ as the parameter $j$ is not
important anymore.

Figure 5.4. First case: $\tilde{R}_{r}\subset
H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$

##### Second case: $H^{i}_{j}\subset\tilde{R}_{r}$

######  Lemma 5.19.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$ be a rectangle, with
$r=\tilde{r}(i,s)$. Let $(i,j)\in{\mathcal{H}}(r)$ be any pair of indices such
that $H^{i}_{j}\subset\tilde{R}_{r}$ and assume that $\rho_{T}(i,j)=(k,l)$.
Then, the number of horizontal sub-rectangles of
${\mathcal{R}}_{S(\underline{w})}$ that are contained in the intersection
$\tilde{R}_{r}\cap H^{i}_{j}$ is given by the formula:

(5.13) $\underline{h}(r,j)=O(k,\underline{w})+1.$

###### Proof.

See Figure 5.5. The vertical sub-rectangle $V^{k}_{l}=f(H^{i}_{j})$ of
$R_{k}\in{\mathcal{R}}$ intersects to all the rectangles
$\tilde{R}_{r^{\prime}}\in{\mathcal{R}}_{S(\underline{w})}$ that are contained
in $R_{k}$, but by definition this quantity is equal to
$O(k,\underline{w})+1$. Even more, this number coincides with the number of
horizontal sub-rectangles of ${\mathcal{R}}_{S(\underline{w})}$ that are
contained inside $\tilde{R}_{r}\cap H^{i}_{j}$ and therefore:

(5.14) $\underline{h}(r,j)=O(k,\underline{w})+1,$

is the number of horizontal sub-rectangles of
${\mathcal{R}}_{S(\underline{w})}$ inside $\tilde{R}_{r}\cap H^{i}_{j}$.

∎

We proceed to label such horizontal sub-rectangles.

######  Definition 5.17.

The horizontal sub-rectangles of $\tilde{R}_{r}$ described in Lemma 5.19 are
labeled from the bottom to the top with respect to the vertical orientation of
$\tilde{R}_{r}$ as:

$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}.$

Figure 5.5. Second case: $H^{i}_{j}\subset\tilde{R}_{r}$

Let’s determine the image of any such horizontal sub-rectangle of
$\tilde{R}_{r}\cap H^{i}_{j}$.

######  Lemma 5.20.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$, with
$r=\tilde{r}(i,s)$. Let $(i,j)\in{\mathcal{H}}(r)$ be any pair of indices such
that $H^{i}_{j}\subset\tilde{R}_{r}$, and let $\tilde{H}^{r}_{j,J}$ be a
horizontal sub-rectangle of $\tilde{R}_{r}$. Assume that $\rho_{T}(i,j)=(k,l)$
and that $f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$.

  * i)

If $\epsilon_{T}(i,j)=1$, $r^{\prime}=\tilde{r}(k,J)$ and $l=l^{\prime}$.

  * ii)

If $\epsilon_{T}(i,j)=-1$, then
$r^{\prime}=\tilde{r}(k,O(k,\underline{w})+1-J+1)$ and $l=l^{\prime}$.

###### Proof.

Item $i)$. If $\epsilon(i,j)=1$, the vertical orientation is preserved by $f$
restricted to $\tilde{R}_{r}$, and the horizontal sub-rectangle
$\tilde{H}^{r}_{j,J}$ that is in position $J$ inside $\tilde{R}_{r}\cap
H^{i}_{j}$ is sent to the rectangle $\tilde{R}_{r^{\prime}}$ located in
position $J$ inside $R_{k}$. This means that $r^{\prime}=\tilde{r}(k,J)$ and
in view of Proposition 5.1, $l^{\prime}=l$.

Item $i)$. If $\epsilon(i,j)=-1$, the vertical orientation is invert by $f$
restricted to $\tilde{R}_{r}\cap H^{i}_{j}$, and the horizontal sub-rectangle
$\tilde{H}^{r}_{j,J}$ that is in position $J$ inside $\tilde{R}_{r}\cap
H^{i}_{j}$ is sent to the rectangle $\tilde{R}_{r^{\prime}}$ located in
position $J-1$ below the top rectangle of ${\mathcal{R}}_{S(\underline{w})}$
(for $J=1$ the position is $O(k,\underline{w}))+1)$, and this position is
equal to $O(k,\underline{w})+1-J+1$ inside $R_{k}$. This means that
$r^{\prime}=\tilde{r}(k,O(k,\underline{w})+1-J+1)$. Finally, by Proposition
5.1, $l^{\prime}=l$.

∎

######  Definition 5.18.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$, with
$r=\tilde{r}(i,s)$. Let $(i,j)\in{\mathcal{H}}(r)$ be a pair of indices such
that $H^{i}_{j}\subset\tilde{R}_{r}$, and let $\tilde{H}^{r}_{j,J}$ be an
horizontal sub-rectangle of $\tilde{R}_{r}\cap H^{i}_{j}$. Assume that
$\rho_{T}(i,j)=(k,l)$.

If $\epsilon(i,j)=1$ we set:

(5.15) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,J),l),$

If $\epsilon(i,j)=-1$ we set:

(5.16)
$\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,O(k,\underline{w})+1-J+1),l).$

##### Third case: $H^{i}_{j}$ contains only one horizontal boundary component
of $\tilde{R}_{r}$

We consider the case when $H^{i}_{j}$ contains the upper boundary of
$\tilde{R}_{r}$.

######  Lemma 5.21.

Consider $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$, where
$r=\tilde{r}(i,s)$, and assume that $O(i,\underline{w})>0$. Take a pair of
indices $(i,j)\in{\mathcal{H}}(r)$ such that $H^{i}_{j}$ exclusively contains
the _upper boundary_ of $\tilde{R}_{r}$. Let $I_{t,\underline{w}}$ be the
upper boundary of $\tilde{R}_{r}$. Let
$s^{+}=\underline{s}(t+1,\underline{w})$ and assume that
$\rho_{T}(i,j)=(k,l)$. Then, the number of horizontal sub-rectangles of
${\mathcal{R}}_{S(\underline{w})}$ that are contained inside
$\tilde{R}_{r}\cap H^{i}_{j}$ can be computed using one of the following
formulas:

  * •

If $\epsilon_{T}(i,j)=1$:

(5.17) $\underline{h}(r,j)=s^{+}.$

  * •

If $\epsilon_{T}(i,j)=-1$

(5.18) $\underline{h}(r,j)=O(k,\underline{w})+1-s^{+}.$

Figure 5.6. Third case: $H^{i}_{j}$ contains only one horizontal boundary
component of $\tilde{R}_{r}$

###### Proof.

Look the Figure 5.6 to get some pictorial intuition.

Case $\epsilon_{T}(i,j)=1$. The upper boundary of $\tilde{R}_{r}$ is mapped to
the stable segment $I_{t+1,\underline{w}}$, positioned at place $s^{+}$ inside
$R_{k}$ and the inferior boundary of $\tilde{R}_{r}\cap H^{i}_{j}$ coincides
with the inferior boundary of $H^{i}_{j}$ and since $\epsilon_{T}(i,j)=1$, $f$
preserves the vertical orientation of $\tilde{R}_{r}\cap H^{i}_{j}$.
Consequently, the image of the inferior boundary of $\tilde{R}_{r}\cap
H^{i}_{j}$ is contained in the inferior boundary of $R_{k}$. The number of
rectangles in ${\mathcal{R}}_{S(\underline{w})}$ between the segment
$I_{t+1,\underline{w}}$ and the inferior boundary of $R_{k}$ is given by
$s^{+}$ and this number is equal to the number of horizontal sub-rectangles of
$\tilde{R}_{r}\cap H^{i}_{j}$.

Case $\epsilon_{T}(i,j)=-1$. The upper boundary of $\tilde{R}_{r}\cap
H^{i}_{j}$ is equal to the upper boundary of $\tilde{R}_{r}$ and is mapped by
$f$ to the stable segment $I_{t+1,\underline{w}}$, positioned at place $s^{+}$
inside $R_{k}$. The inferior boundary of $\tilde{R}_{r}\cap H^{i}_{j}$
coincides with the inferior boundary of $H^{i}_{j}$ , but since
$\epsilon_{T}(i,j)=-1$, $f$ inverts the vertical orientation of
$\tilde{R}_{r}\cap H^{i}_{j}$. Consequently, the image of the inferior
boundary of $\tilde{R}_{r}\cap H^{i}_{j}$ is contained in the upper boundary
of $R_{k}$. The number of rectangles in ${\mathcal{R}}_{S(\underline{w})}$
between the segment $I_{t+1,\underline{w}}$ and the upper boundary of $R_{k}$
is given by $O(k,\underline{w})+1-s^{+}$ and this number is equal to the
number of horizontal sub-rectangles of $\tilde{R}_{r}\cap H^{i}_{j}$.

∎

The next lemma addresses the situation in which the rectangle $H^{i}_{j}$ just
contains the inferior boundary of $\tilde{R}_{r}$. Its proof follows a similar
approach to that of Lemma 5.21, so we won’t delve into further details.

######  Lemma 5.22.

Consider $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$, where
$r=\tilde{r}(i,s)$, and assume that $O(i,\underline{w})>0$. Take a pair of
indices $(i,j)\in{\mathcal{H}}(r)$ such that $H^{i}_{j}$ just contains the
lower boundary of $\tilde{R}_{r}$. Let $I_{t,\underline{w}}$ be the lower
boundary of $\tilde{R}_{r}$, and denote
$s^{+}=\underline{s}(t+1,\underline{w})$. In this scenario, the number of
horizontal sub-rectangles of ${\mathcal{R}}_{S(\underline{w})}$ that are
contained in $\tilde{R}_{r}\cap H^{i}_{j}$ can be determined using one of the
following formulas:

  * •

If $\epsilon_{T}(i,j)=1$:

(5.19) $\underline{h}(r,j)=O(k,\underline{w})+1-s^{+}.$

  * •

If $\epsilon_{T}(i,j)=-1$

(5.20) $\underline{h}(r,j)=s^{+}.$

######  Definition 5.19.

The horizontal sub-rectangles of $\tilde{R}_{r}$ described in Lemma 5.21 and
Lemma 5.22 are labeled from the bottom to the top with the vertical
orientation of $\tilde{R}_{r}$ as:

$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}.$

######  Lemma 5.23.

With the hypothesis of Lemma 5.21, assume that
$\Phi_{T}(i,j)=(k,l,\epsilon_{T}(i,j))$ and
$f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$. Suppose that
$H^{i}_{j}$ just contains the upper boundary of $\tilde{R}_{r}$, then we have
the following situations:

  * i)

If $\epsilon_{T}(i,j)=1$, then $r^{\prime}=\tilde{r}(k,J)$ and $l=l^{\prime}$

  * ii)

If $\epsilon_{T}(i,j)=-1$, then
$r^{\prime}=\tilde{r}(k,O(k,\underline{w})+1-J+1)$ and $l=l^{\prime}$.

###### Proof.

Item $i)$. When $\epsilon_{T}(i,j)=1$ $f$ preserves the vertical orientation
restricted to $\tilde{R}_{r}\cap H^{i}_{j}$. Consequently, the horizontal sub-
rectangle $\tilde{H}^{r}_{j,J}$ that is located at position $J$ within
$\tilde{R}_{r}\cap H^{i}_{j}$, is send to the rectangle
$\tilde{R}_{r^{\prime}}$ that is located at position $J$ over the rectangle
that is in the position $s^{+}$ within $R_{k}$, therefore
$r^{\prime}=\tilde{r}(k,s^{+}+J)$.

Item $ii)$. When $\epsilon_{T}(i,j)=-1$, $f$ inverts the vertical orientation
of $\tilde{R}_{r}\cap H^{i}_{j}$. Consequently, the horizontal sub-rectangle
$\tilde{H}^{r}_{j,J}$ that is located at position $J$ within
$\tilde{R}_{r}\cap H^{i}_{j}$, is send to the rectangle located at position
$J-1$ below the upper rectangle in $R_{k}$, this is given by
$O(k,\underline{w})+1-(J-1)$ withing $R_{k}$. Therefore,
$r^{\prime}=\tilde{r}(k,O(k,\underline{w})+1-J+1)$.

In bot cases Proposition 5.1 implies that $l^{\prime}=l$.

∎

The proof of the next lemma is the same, but in this case we consider the case
when $H^{i}_{j}$ contains the inferior boundary of $\tilde{R}_{r}$.

######  Lemma 5.24.

With the hypothesis of Lemma 5.22, assume that
$\Phi_{T}(i,j)=(k,l,\epsilon_{T}(i,j))$ and
$f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$. Suppose that
$H^{i}_{j}$ just contains the inferior boundary of $\tilde{R}_{r}$, then we
have the following situations:

  * i)

If $\epsilon_{T}(i,j)=1$, then $r^{\prime}=\tilde{r}(k,s^{+}+J)$ and
$l=l^{\prime}$

  * ii)

If $\epsilon_{T}(i,j)=-1$, then $r^{\prime}=\tilde{r}(k,s^{+}-J+1)$ and
$l=l^{\prime}$.

######  Definition 5.20.

With the hypothesis of Lemma 5.21, assume that $\Phi_{T}(i,j)=(k,l,\epsilon)$.
Suppose that $H^{i}_{j}$ just contains the upper boundary of $\tilde{R}_{r}$:

  * i)

If $\epsilon_{T}(i,j)=1$, we define:

(5.21) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,J),l).$

  * ii)

if $\epsilon_{T}(i,j)=-1$ we take:

(5.22)
$\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,O(k,\underline{w})+1-J+1),l).$

######  Definition 5.21.

With the hypothesis of Lemma 5.22, assume that
$\Phi_{T}(i,j)=(k,l,\epsilon_{T}(i,j))$. Suppose that $H^{i}_{j}$ just
contains the lower boundary of $\tilde{R}_{r}$, then:

  * i)

If $\epsilon_{T}(i,j)=1$, we define:

(5.23) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{+}+J),l).$

  * ii)

if $\epsilon_{T}(i,j)=-1$ we take:

(5.24) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{+}-J+1),l).$

####  5.1.8. The number $H_{r}$

######  Corollary 5.4.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S(\underline{w})}$ with
$r=\tilde{r}(i,s)$, then the number $H_{r}$ of horizontal sub-rectangles of
the Markov partition $(f,{\mathcal{R}}_{S(\underline{w})})$ that are contained
in $\tilde{R}_{r}$ is given by:

(5.25) $H_{r}=\sum_{\\{j:(i,j)\in{\mathcal{H}}(r)\\}}\underline{h}(r,j).$

####  5.1.9. The function $\rho_{S(\underline{w})}$ in the geometric type
$T_{S(\underline{w})}$

Figure 5.7. The relative position of $\tilde{H}^{r}_{\underline{J}}$ for
$\underline{J}=\underline{6}$

######  Definition 5.22.

The sub-rectangles of the geometric Markov partition
${\mathcal{R}}_{S(\underline{w})}$ contained in $\tilde{R}_{r}$ are labeled
as:

$\\{H^{r}_{\underline{J}}\\}_{\underline{J}=1}^{H_{r}}$

from the bottom to the top with respect to the vertical orientation of
$\tilde{R}_{r}$.

We’ve explained how to compute $\underline{\rho}_{(r,j)}(r,J)$ for the three
distinct cases. Now, we need to integrate them into a unified function,
$\rho_{S(\underline{w})}$. The initial step in this process is to find a pair
$(j,J)$ than, once given $r=\tilde{r}(i,s)$ and
$\underline{J}\in\\{1,\cdots,H_{r}\\}$,
$\tilde{H}^{r}_{j,J}=\tilde{H}^{r}_{\underline{J}}$. We suggest to look the
Figure 5.7 to get some intuition about our procedure and definitions below.

######  Lemma 5.25.

For each $\underline{J}\in\\{1,\cdots,H_{r}\\}$, there exists a unique index
$j_{(\underline{J})}$ such that $(i,j)\in{\mathcal{H}}(r)$ and
$\tilde{H}^{r}_{\underline{J}}\subset H^{i}_{j}\cap\tilde{R}_{r}$. The index
$j_{(\underline{J})}$ is determined by satisfy the next inequalities:

(5.26) $\sum_{\\{j^{\prime}:(i,j^{\prime})\in{\mathcal{H}}(r)\text{ and
}j^{\prime}<j_{(\underline{J})}\\}}\underline{h}(j^{\prime},r)<\underline{J}\leq\sum_{\\{j^{\prime}:(i,j^{\prime})\in{\mathcal{H}}(r)\text{
and }j^{\prime}\leq j_{(\underline{J})}\\}}\underline{h}(j^{\prime},r).$

###### Proof.

This is immediate. ∎

######  Definition 5.23.

Given $\underline{J}\in\\{1,\cdots,H_{r}\\}$ we call the index
$j_{(\underline{J})}$ that was determined in Lemma 5.25 the _horizontal
parameter_ of $\underline{J}$.

######  Definition 5.24.

Let $r=\tilde{r}(i,s)$, $\underline{J}\in\\{1,\cdots,H_{r}\\}$, and let
$j_{(\underline{J})}$ be the horizontal parameter of $\underline{J}$. The
_relative index_ of $H^{r}_{J}$ inside of $\tilde{R}_{r}\cap
H^{i}_{j_{\underline{J}}}$ is given by:

(5.27)
$J_{(\underline{J})}:=\underline{J}-\sum_{j^{\prime}<j_{({\underline{J})}}}\underline{h}(r,j^{\prime}).$

######  Definition 5.25.

Let $\underline{J}\in\\{1,\cdots,H_{r}\\}$ the _relative position_ of
$\underline{J}$ inside $\tilde{R}_{r}$ is the pair
$(j_{(\underline{J})},J_{(\underline{J})})$ of the horizontal parameter of
$\underline{J}$ and the relative index of $H^{r}_{J}$ inside of
$\tilde{R}_{r}\cap H^{i}_{j_{(\underline{J}})}$

######  Lemma 5.26.

The rectangle $\tilde{H}^{r}_{j_{(\underline{J})},J_{(\underline{J})}}$ and
the rectangle $\tilde{H}^{r}_{\underline{J}}$ are the same.

###### Proof.

The rectangle $\tilde{H}^{r}_{\underline{J}}$ within $\tilde{R}_{r}$ holds
position $\underline{J}$ among all its horizontal sub-rectangles. If we know
which horizontal sub-rectangle $H^{i}_{j}$ (or ${\mathcal{R}}$ ) contains
$\tilde{H}^{r}_{\underline{J}}$, we can determine the index $\underline{J}$ as
the sum of all the total of horizontal sub-rectangle of $\tilde{R}_{r}$ that
are contained in $H^{i}_{j^{\prime}}\cap\tilde{R}_{r}$ with $j^{\prime}<j$ and
then determine the position $J$ of $\tilde{H}^{r}_{\underline{J}}$ inside
$H^{i}_{j}\cap\tilde{R}_{r}$.

Equation 5.26 specifies that $\tilde{H}^{r}_{\underline{J}}$ is not included
in the horizontal sub-rectangles $H^{i}_{j^{\prime}}$ for
$j^{\prime}<j_{(\underline{J})}$ but is contained in
$H^{i}_{j_{(\underline{J})}}$. Even more, equation 5.27 said that
$\tilde{H}^{r}_{\underline{J}}$ holds the position $J_{(\underline{J})}$
inside $H^{i}_{j_{(\underline{J})}}$. Thus, we can conclude that:

$\tilde{H}^{r}_{j_{(\underline{J})},J_{(\underline{J})}}=\tilde{H}^{r}_{\underline{J}}$

as was claimed. ∎

The relative position of $\tilde{H}^{r}_{\underline{J}}$ inside the rectangle
$\tilde{R}_{r}\cap H^{i}_{j}$ allows us to apply the corresponding formula to
this $J_{(\underline{w})}$ in order to compute the label of the vertical sub-
rectangle of ${\mathcal{R}}_{S(\underline{w})}$ of its image.

######  Corollary 5.5.

Let $r=\tilde{r}(i,s)$ and $\underline{J}\in\\{1,\cdots,H_{r}\\}$. Let
$(j_{(\underline{J})},J_{(\underline{J})})$ the relative position of
$\underline{J}$ as was given in Definition 5.25. Then the function
$\rho_{S(\underline{w})}$ is given by the formula:

(5.28)
$\rho_{S(\underline{w})}(r,\underline{J}):=\underline{\rho}_{(r,j_{(\underline{J})})}(r,J_{(\underline{J})}).$

###### Proof.

This is just a direct consequence of our previous constructions and the Lemma
5.26. ∎

####  5.1.10. The orientation function $\epsilon_{S(\underline{w})}$

######  Lemma 5.27.

Let $r=\tilde{r}(i,s)$ and $\underline{J}\in\\{1,\cdots,H_{r}\\}$. Let
$j_{(\underline{J})}$ the horizontal parameter of $\underline{J}$ (Definition
5.23). Then:

(5.29)
$\epsilon_{S(\underline{w})}(r,\underline{J}):=\epsilon_{T}(i,j_{(\underline{J})}).$

###### Proof.

Since $\tilde{H}^{r}_{\underline{J}}\subset H^{i}_{j_{(\underline{J})}}$, the
reversal or preservation of the horizontal orientation only depends on
$\epsilon_{T}(i,j)$. This gives the equality. ∎

####  5.1.11. Conclusion

######  Corollary 5.6.

Let $T$ be a geometric type in the pseudo-Anosov class with a binary incidence
matrix denoted as $A:=A(T)$. Given a periodic code
$\underline{w}\in\Sigma_{A}$, there is algorithm to compute the geometric
type:

$T_{S(\underline{w})}:=\\{N,\\{H_{r},V_{r}\\}_{r=1}^{N},\Phi_{S(\underline{w})}:=(\rho_{S(\underline{w})},\epsilon_{S(\underline{w})})\\}.$

in therms of the given geometric type $T$ and the code $\underline{w}$.

###### Proof.

Each parameter in $T_{S(\underline{w})}$ has been rigorously determined
through the following steps:

  1. (1)

The number $N$ was calculated in Corollary 5.1.

  2. (2)

The values of $V_{r}$ were established in Corollary 5.2. 5.2

  3. (3)

The formula for $H_{r}$ was provided in Corollary 5.4

  4. (4)

The permutation $\rho_{S(\underline{w})}$ was obtained in Corollary 5.5

  5. (5)

The function $\epsilon_{S(\underline{w})}$ was established in Lemma 5.27

It’s important to note that each of these results is derived from an explicit
algorithm that solely relies on the combinatorial information of $T$ and
$\underline{w}$. ∎

###  5.2. Cut along a family of periodic codes

Let $T$ be a geometric type in the pseudo-Anosov class with a binary incidence
matrix $A:=A(T)$. Consider a generalized pseudo-Anosov homeomorphism
$f:S\rightarrow S$ with a geometric Markov partition ${\mathcal{R}}$ of
geometric type $T$. Suppose we have a finite family of non-$s$-boundary
periodic codes:

${\mathcal{W}}=\\{\underline{w}^{1},\cdots,\underline{w}^{Q}\\}\in\Sigma_{A}.$

In this section, we aim to generalize the procedure for cutting along the
stable intervals determined by a single periodic code to cutting along the
family of stable intervals determined by the family ${\mathcal{W}}$.

######  Convention 5.1.

For the remainder of this section, we will maintain fixed:

  * •

The family of non $s$-boundary periodic codes
${\mathcal{W}}=\\{\underline{w}^{1},\cdots,\underline{w}^{Q}\\}$.

  * •

The geometric type
$T=\\{n,\\{h_{i},v_{i}\\}_{i=1}^{n},\Phi_{t}:=(\rho_{T},\epsilon_{T})\\}$ and
its associated incidence matrix $A:=A(T)$, with the assumption that $A$ is
binary.

  * •

A generalized pseudo-Anosov homeomorphism $f:S\rightarrow S$. that have a
geometric Markov partition ${\mathcal{R}}=\\{R_{i}\\}_{i=1}^{n}$ with
geometric type $T$.

####  5.2.1. The Markov partition ${\mathcal{R}}_{S({\mathcal{W}})}$.

Lemma 5.1 holds for every code $\underline{w}^{q}\in{\mathcal{W}}$, and we can
define the set of intervals that are contained into the same rectangle $R_{i}$
of ${\mathcal{R}}$.

######  Definition 5.26.

For every $i\in\\{1,\cdots,n\\}$ the set of stable intervals determined by the
family ${\mathcal{W}}$ that are contained within $R_{i}$ is given by:

(5.30)
${\mathcal{I}}(i,{\mathcal{W}}):=\cup_{q=1}^{Q}{\mathcal{I}}(i,\underline{w}^{q}).$

######  Lemma 5.28.

Let $\tilde{R}$ be the closure of a connected component of
$\overset{o}{R_{i}}\setminus\cup{\mathcal{I}}(i,{\mathcal{W}})$. Then
$\tilde{R}$ is a horizontal sub-rectangle of $R_{i}$.

###### Proof.

The argument is the same as in Lemma LABEL:Lemm:_Stable_intervals_of_codes.
Cutting $\overset{o}{R_{i}}$ along the finite family of stable intervals in
${\mathcal{I}}(i,{\mathcal{W}})$ produces connected components whose closures
are horizontal sub-rectangles of $R_{i}$.

∎

######  Definition 5.27.

The family, which includes all the rectangles as described in Lemma 5.28,
corresponding to all the rectangles $R_{i}$ in ${\mathcal{R}}$, is denoted as:
${\mathcal{R}}_{S({\mathcal{W}})}$.

######  Lemma 5.29.

The family of rectangles ${\mathcal{R}}_{S({\mathcal{W}})}$ is a Markov
partition for $f$.

###### Proof.

The interiors of two distinct horizontal sub-rectangles within $R_{i}$ do not
intersect, and a horizontal sub-rectangle from $R_{i}$ does not overlap with
the interior of a horizontal sub-rectangle from $R_{j}$ if $j\neq i$.
Consequently, the interiors of two different rectangles in
${\mathcal{R}}_{{\mathcal{W}}}$ are disjoint.

The union $\cup{\mathcal{R}}_{S({\mathcal{W}})}$ coincides with the union of
the rectangles in ${\mathcal{R}}$, but ${\mathcal{R}}$ is Markov partition for
$f$, therefore $\cup{\mathcal{R}}_{S({\mathcal{W}})}=\cup{\mathcal{R}}=S$.

The stable boundary of ${\mathcal{R}}_{S({\mathcal{W}})}$ consists of two
types of stable intervals: those that are part of the stable boundary of
${\mathcal{R}}$ and whose image is contained in
$\partial^{s}{\mathcal{R}}\subset\partial^{s}{\mathcal{R}}_{S({\mathcal{W}})}$,
and those of the form $I_{t,\underline{w}^{q}}$ for some
$\underline{w}^{q}\in{\mathcal{W}}$ that, as stated in Lemma 5.2:
$f(I_{t,\underline{w}^{q}})\subset
I_{t+1,\underline{w}^{q}}\subset\partial^{s}{\mathcal{R}}_{S({\mathcal{W}})}$.
Therefore, $\partial^{s}{\mathcal{R}}_{S({\mathcal{W}})}$ is $f$-invariant.

Since the unstable boundary of ${\mathcal{R}}_{S({\mathcal{W}})}$ coincides
with the unstable boundary of ${\mathcal{R}}$,
$\partial^{u}{\mathcal{R}}_{S({\mathcal{W}})}$ is $f^{-1}$-invariant.

This proves that ${\mathcal{R}}_{S({\mathcal{W}})}$ is a Markov partition of
$f$. ∎

####  5.2.2. The geometrization of ${\mathcal{R}}_{S({\mathcal{W}})}$

Let’s choose an orientation for every rectangle in
${\mathcal{R}}_{S({\mathcal{W}})}$ and label its elements.

######  Definition 5.28.

If the rectangle $\tilde{R}\in{\mathcal{R}}_{S({\mathcal{W}})}$ is contained
within the rectangle $R_{i}\in{\mathcal{R}}$, we assign to $\tilde{R}$ the
vertical and horizontal orientations that are consistent with the respective
vertical and horizontal orientations of $R_{i}$.

######  Definition 5.29.

We label the rectangles in $\mathcal{R}_{S({\mathcal{W}})}$ using the
_lexicographic order_. To do this, we begin by numbering the horizontal sub-
rectangles of $R_{i}$ that belong to $\mathcal{R}_{S({\mathcal{W}})}$ from the
bottom to the top, denoting them as $\\{\tilde{R}_{i,s}\\}_{s=1}^{N_{i}}$,
where $N_{i}\geq 1$. Each rectangle
$\tilde{R}_{i,s}\in{\mathcal{R}}_{S(\underline{w})}$, which is contained
within $R_{i}$ at vertical position $s$, is labeled as
$\tilde{R}:=\tilde{R}_{r}$, where the number $r$ is determined by the formula:

$r=\tilde{r}(i,s):=\sum_{i^{\prime}<i}^{i}N_{i^{\prime}}+s.$

######  Definition 5.30.

The Markov partition ${\mathcal{R}}_{S({\mathcal{W}})}$ with the
geometrization given by Definitions 5.28 and 5.29 is called the $s$-_boundary
refinement_ of ${\mathcal{R}}$ with respect to the family ${\mathcal{W}}$. Its
geometric type is denoted by:

(5.31)
$T_{S({\mathcal{W}})}=\\{N,\\{H_{r},V_{r}\\}_{r=1}^{N},\Phi_{S({\mathcal{W}})}:=(\rho_{S({\mathcal{W}})},\epsilon_{S({\mathcal{W}})})\\}.$

We give the following notation to the rectangles of the partition:

(5.32) ${\mathcal{R}}_{S({\mathcal{W}})}:=\\{\tilde{R}_{r}\\}_{r=1}^{N}$

####  5.2.3. The number of rectangles in ${\mathcal{R}}_{S({\mathcal{W}})}$

Let $\underline{w}^{q}\in{\mathcal{W}}$ be any code. We’ll denote
$\textbf{Per}(\underline{w}^{q})$ as the period of $\underline{w}^{q}$, and
let $w^{q}_{t}$ represent the term in position $t$ of the code
$\underline{w}^{q}$.

######  Definition 5.31.

For every $i\in\\{1,\cdots,n\\}$ let:

(5.33)
${\mathcal{O}}(i,{\mathcal{W}})=\cup_{q=1}^{Q}{\mathcal{O}}(i,\underline{w}^{q})=\\{(t,\underline{w}^{q}):t\in\\{0,\ldots,\textbf{Per}(\underline{w}^{q})-1\\}\text{
and }w^{q}_{t}=i\\}.$

the indexes of the stable intervals determined by ${\mathcal{W}}$ that are
contained into $R_{i}$. Let $O(i,{\mathcal{W}})$ be its cardinality.

The next lemma is immediate.

######  Lemma 5.30.

For all $i\in\\{1,\ldots,n\\}$, the number of rectangles in the Markov
partition ${\mathcal{R}}_{S({\mathcal{W}})}$ that are contained in $R_{i}$ is
equal to $O(i,{\mathcal{W}})+1$

######  Corollary 5.7.

The number $N$ in the geometric type $T_{S({\mathcal{W}})}$ is equal to the
number of rectangles in the $s$-boundary refinement respect to ${\mathcal{W}}$
and is determined by the following formula:

(5.34) $N=\sum_{i=1}^{n}O(i,{\mathcal{W}})+1.$

####  5.2.4. The vertical sub-rectangles in $T_{S({\mathcal{W}})}$.

######  Definition 5.32.

Let $\tilde{R}_{r}$ be a rectangle in the geometric Markov partition
${\mathcal{R}}_{S({\mathcal{W}})}$ and suppose that $r=\tilde{r}(i,s)$. Let
$V^{i}_{l}$ be a vertical sub-rectangle of the Markov partition
$(f,{\mathcal{R}})$ contained in $R_{i}$. Define:

$\tilde{V}^{r}_{l}:=\overline{\overset{o}{\tilde{R}_{r}}\cap\overset{o}{V^{i}_{l}}}.$

as the closure of the intersection between the interior of $\tilde{R}_{r}$ and
the interior of $V^{i}_{l}$.

The proof of the following Lemma is completely analogous to the such presented
in Lemma 5.7.

######  Lemma 5.31.

The set $\tilde{V}^{r}_{l}$ is a unique horizontal sub-rectangle of
$\tilde{R}_{r}$

######  Remark 5.7.

The labeling of the vertical sub-rectangles of $\tilde{R}_{r}$:

$\\{\tilde{V}^{r}_{l}\\}_{l=1}^{v_{i}},$

is coherent with the horizontal orientation of $R_{i}$.

Taking in account the previous Remark 5.7, the argument used to prove Lemma
5.8 can be applied to demonstrate the following generalized result:

######  Lemma 5.32.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$ with
$r=\tilde{r}(i,s)$. Let $\tilde{H}\subset\tilde{R}_{r}$ be a horizontal sub-
rectangle of ${\mathcal{R}}_{S({\mathcal{W}})}$. Then there is a unique pair
$(i,j)\in{\mathcal{H}}(T)$ such that
$\overset{o}{H}\subset\overset{o}{H^{i}_{j}}$.

######  Definition 5.33.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S_{(}{\mathcal{W}})}$ with
$r=\tilde{r}(i,s)$. Let $\tilde{H}\subset\tilde{R}_{r}$ be a horizontal sub-
rectangle of ${\mathcal{R}}_{S_{(}{\mathcal{W}})}$, we call
$(i,j_{\tilde{H}})$ to the only pair that was determine in Lemma 5.32.

######  Lemma 5.33.

The rectangles $\\{\tilde{V}^{r}_{l}\\}_{l=1}^{v_{i}}$ described in Definition
5.9 are the vertical sub-rectangles of the Markov partition
$(f,{\mathcal{R}}_{S({\mathcal{W}})})$ contained in $\tilde{R}_{r}$.

###### Proof.

Consider a single rectangle $\tilde{V}^{r}_{l}$. Following the reasoning from
Lemma 5.9, we can observe that $f^{-1}(\tilde{V}^{r}_{l})$ is a subset of
$f^{-1}(V^{k}_{l})$, and specifically, it is contained in some horizontal sub-
rectangle of ${\mathcal{R}}$, $H^{i}_{j}$. The preimage
$f^{-1}(\tilde{V}^{r}_{l})$ is a horizontal sub-rectangle of $H^{i}_{j}$ whose
interior don’t intersect $\partial^{s}{\mathcal{R}}_{S({\mathcal{W}})}$, then
and it’s exclusively contained within a unique rectangle $\tilde{R}_{r}$. Even
more, the stable boundary of $f^{-1}(\tilde{V}^{r}_{l})$ has image contained
within the stable boundary of the Markov partition
${\mathcal{R}}_{S({\mathcal{W}})}$. Therefore, $f^{-1}(\tilde{V}^{r}_{l})$ is
a horizontal sub-rectangle of ${\mathcal{R}}_{S({\mathcal{W}})}$, and then
$\tilde{V}^{r}_{l}$ is a vertical sub-rectangle of the Markov partition
${\mathcal{R}}_{S({\mathcal{W}})}$.

Conversely, if $\tilde{V}$ is a vertical sub-rectangle of
${\mathcal{R}}_{S({\mathcal{W}})}$, then $f^{-1}(\tilde{V})=\tilde{H}$ is a
horizontal sub-rectangle of ${\mathcal{R}}_{S({\mathcal{W}})}$. Let
$(i,j_{\tilde{H}})$ be the indexes give in Definition 5.33. Thus, $\tilde{V}$
is a horizontal sub-rectangle of $V^{k}_{l}=f(H^{i}_{j_{\tilde{H}}})$, and its
interior does not intersect the stable boundaries of
${\mathcal{R}}_{S({\mathcal{W}})}$. However, its stable boundaries are
contained in $\partial^{s}{\mathcal{R}}_{S({\mathcal{W}})}$. This implies that
$\tilde{V}\subset V^{k}_{l}\cap\tilde{R}_{r}$ for some $r\in\\{1,\cdots,n\\}$,
and it turns out to be one of the rectangles $\tilde{V}^{r}_{l}$ given by
$\tilde{V}^{r}_{l}:=\overline{\overset{o}{\tilde{R}_{r}}\cap\overset{o}{V^{k}_{l}}}$.

∎

This result have the next immediate corollary.

######  Corollary 5.8.

If $r=\tilde{r}(i,s)$ then $V_{r}=v_{i}$.

The following result is a generalization of Proposition 5.1, and it is proven
in a similar manner. The primary element in this proof is the consistency of
the horizontal orientation of $R_{i}$ and the labeling of the vertical sub-
rectangles $\\{\tilde{V}^{r}_{l}\\}_{l=1}^{v_{i}}$ within
$\tilde{R}_{r}\subset R_{i}$.

######  Corollary 5.9.

Let $\tilde{H}\subset\tilde{R}_{r}$ any horizontal sub-rectangle of
${\mathcal{R}}_{S({\mathcal{W}}}$, assume that $\tilde{H}\subset H^{i}_{j}$
and $f(H^{i}_{j})=V^{k}_{l}$. If $f(H)=\tilde{V}^{r^{\prime}}_{l^{\prime}}$
then $l=l^{\prime}$.

####  5.2.5. The stable boundary of a rectangle
$\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$

Almost any result in this sub-section can be proved in the same manner as the
respective result was proved for only one code. For example, the next lemma is
proved with the same strategy developed in Lemma 5.12. The number $M$ is the
minimum natural number such that $w^{1}_{t+1+M}\neq w^{2}_{t_{2}+M}$, and all
the other properties follow from this number.

######  Lemma 5.34.

Suppose that $\underline{w}^{1}$ and $\underline{w}^{2}$ are two codes in
${\mathcal{W}}$ with periods $P_{1}$ and $P_{2}$, respectively. Let $t_{1}$ be
a number in $\\{0,\cdots,P_{1}-1\\}$ and $t_{2}$ be a number in
$\\{0,\cdots,P_{2}-1\\}$ such that
$\sigma^{t_{1}}(\underline{w}^{1})\neq\sigma^{t_{2}}(\underline{w}^{2})$, but
both $w^{1}_{t_{1}}$ and $w^{2}_{t_{2}}$ are equal to $i\in\\{1,\cdots,n\\}$.
Consider the two elements in ${\mathcal{O}}(i,{\mathcal{W}})$ determined by
these iterations: $(t_{1},\underline{w}^{1})$ and $(t_{2},\underline{w}^{2}).$
In this way:

  * i)

There exists an integer $M\in\\{1,\cdots,P-1\\}$ such that
$w^{1}_{t_{1}+M}\neq w^{2}_{t_{2}+M}$, but for all $0\leq m\leq M-1$, the
numbers $w^{1}_{t_{1}+m}$ and $w^{2}_{t_{2}+m}$ are equal.

  * ii)

If $M=1$, the indexes $j_{t_{1},\underline{w}^{1}}$ and
$j_{t_{2},\underline{w}^{2}}$ (Definition 5.10) are distinct. However, when
$M\geq 2$, for every $m$ in the set $\\{0,\cdots,M-2\\}$, the indices
$j_{t_{1}+m,\underline{w}^{1}}$ and $j_{t_{2}+m,\underline{w}^{1}}$ (see
Definition 5.10) are equal. Thus, for all $m\in\\{0,\cdots,M-2\\}$:

$\epsilon_{T}(w^{1}_{t_{1}+m},j_{t_{1}+m,\underline{w}^{1}})=\epsilon_{T}(w^{2}_{t_{2}+m},j_{t_{2}+m,\underline{w}^{2}}).$

  * iii)

The indices $j_{t_{1}+M-1,\underline{w}^{1}}$ and
$j_{t_{2}+M-1,\underline{w}^{2}}$ in
$\\{1,\cdots,h_{w^{1}_{t_{1}+M-1}}=h_{w^{2}_{t_{2}+M-1}}\\}$ are different.

This permits us to define the interchange function as in Definition 5.12 but
this time respect two non necessarily equal codes in ${\mathcal{W}}$.

######  Definition 5.34.

Suppose that $\underline{w}^{1}$ and $\underline{w}^{2}$ are two codes in
${\mathcal{W}}$ with periods $P_{1}$ and $P_{2}$, respectively. Let $t_{1}$ be
a number in $\\{0,\cdots,P_{1}-1\\}$ and $t_{2}$ be a number in
$\\{0,\cdots,P_{2}-1\\}$ such that
$\sigma^{t_{1}}(\underline{w}^{1})\neq\sigma^{t_{2}}(\underline{w}^{2})$ but
$w^{1}_{t_{1}}=w^{2}_{t_{2}}=i$. Let $M$ be the number determined in Lemma
5.34. We define the _interchange order_ between $(t_{1},\underline{w}^{1})$
and $(t_{2},\underline{w}^{2})$ as follows:

  * •

If $M=1$ then $\delta((t_{1},\underline{w}^{1}),(t_{2},\underline{w}^{2}))=1$.

  * •

If $M>1$ then

$\delta((t_{1},\underline{w}^{1}),(t_{2},\underline{w}^{2}))=\prod_{m=0}^{M-2}\epsilon_{T}(w^{1}_{t_{1}+m},j_{t_{1}+m,\underline{w}^{1}})=\prod_{m=0}^{M-2}\epsilon_{T}(w^{2}_{t_{2}+m},j_{t_{2}+m,\underline{w}^{2}}).$

Lemma 5.13 admits the following generalization whose proof is essentially the
same.

######  Lemma 5.35.

Let $I_{t_{1},\underline{w}^{1}}$ and $I_{t_{2},\underline{w}^{2}}$ be two
stable segments such that $w^{1}_{t_{1}}=w^{2}_{t_{2}}=i$. Then,
$I_{t_{1},\underline{w}^{1}}<I_{t_{2},\underline{w}^{2}}$ with respect to the
vertical orientation of $R_{i}$ if and only if one of the following situations
occurs:

  * •

$j_{t_{1}+M-1,\underline{w}^{1}}<j_{t_{2}+M-1,\underline{w}^{2}}$ and
$\delta((t_{1},\underline{w}^{1})(t_{2},\underline{w}^{2}))=1$, or

  * •

$j_{t_{1}+M-1,\underline{w}^{1}}>j_{t_{2}+M-1,\underline{w}^{2}}$ and
$\delta((t_{1},\underline{w}^{1})(t_{2},\underline{w}^{2}))=-1$

Now, we can establish a formal order for the set of stable segments in $R_{i}$
that are part of the stable boundary of the rectangles in
${\mathcal{R}}_{S({\mathcal{W}})}$ contained within $R_{i}$.

######  Definition 5.35.

Let $R_{i}\in{\mathcal{R}}$ be any rectangle, denote
$I_{i,-1}:=\partial^{s}_{-1}R_{i}$ as the lower boundary of $R_{i}$, and
$I_{i,+1}:=\partial^{s}_{+1}R_{i}$ as the upper boundary. Define

$\underline{s}:{\mathcal{O}}(i,{\mathcal{W}})\cup\\{(i,+1),(i,-1)\\}\rightarrow\\{0,1,\cdots,O(i,{\mathcal{W}}),O(i,{\mathcal{W}})+1\\}.$

as the unique function such that:

  * •

$\underline{s}(t_{1},\underline{w}^{1})<\underline{s}(t_{2},\underline{w}^{2})$
if and only if $I_{t_{1},\underline{w}^{1}}<I_{t_{1},\underline{w}^{2}}$.

  * •

It assigns $0$ to $(i,-1)$ and $O(i,{\mathcal{W}})+1$ to $(i,+1)$.

Lemma 5.36 is a generalization of 5.14 , and Lemma 5.37 is a generalization of
5.15. Their proofs are exactly the same as the results they generalize.

######  Lemma 5.36.

Let $\tilde{R}_{r}$ be a rectangle in the geometric Markov partition
${\mathcal{R}}_{S({\mathcal{W}})}$ with $r=\tilde{r}(i,s)$. Then, the stable
boundary of $\tilde{R}_{r}$ consists of two stable segments of $R_{i}$
determined as follows:

  * i)

If $s=1$, the lower boundary of $\tilde{R}_{r}$ is $I_{i,-1}$ and if
$O(i,{\mathcal{W}})=0$, its upper boundary is $I_{i,+1}$. However, if
$O(i,{\mathcal{W}})>0$, there exists a unique
$(t,\underline{w}^{q})\in{\mathcal{O}}(i,{\mathcal{W}})$ such that
$\underline{s}(t,\underline{w}^{q})=1$, and the upper boundary of
$\tilde{R}_{r}$ is the stable segment $I_{t,\underline{w}^{q}}$.

  * ii)

If $O(i,{\mathcal{W}})>1$, $s\neq 1$ and $s\neq O(i,{\mathcal{W}})+1$, there
are unique
$(t_{1},\underline{w}^{1}),(t_{2},\underline{w}^{2})\in{\mathcal{O}}(i,{\mathcal{W}})$
such that: $\underline{s}(t_{1},\underline{w}^{1})=s-1$,
$\underline{s}(t_{2},\underline{w}^{2})=s$. Furthermore, the lower boundary of
$\tilde{R}_{r}$ is the stable segment $I_{t_{1},\underline{w}^{1}}$, and the
upper boundary of $\tilde{R}_{r}$ is the stable segment
$I_{t_{2},\underline{w}^{2}}$.

  * iii)

If $O(i,{\mathcal{W}})>0$ and $s=O(i,{\mathcal{W}})+1$, then the upper
boundary of $\tilde{R}_{r}$ is the stable segment $I_{i,+1}$. Additionally,
there exists a unique $(t,\underline{w}^{q})\in{\mathcal{O}}(i,{\mathcal{W}})$
such that $\underline{s}(t,\underline{w}^{q})=O(i,{\mathcal{W}})$, and the
inferior boundary of $\tilde{R}_{r}$ is the stable segment
$I_{t,\underline{w}^{q}}$.

######  Lemma 5.37.

Let $\tilde{R}_{r}$ with $r=\tilde{r}(i,s)$ be a rectangle in the geometric
Markov partition ${\mathcal{R}}_{S({\mathcal{W}})}$. Then, the stable boundary
components of $\tilde{R}_{r}$ have images under $f$ determined in the
following manner:

  * i)

If the lower boundary of $\tilde{R}_{r}$ is $I_{i,-1}$ and
$\Phi_{T}(i,1)=(k,l,\epsilon)$, then: if $\epsilon=1$, $f(I_{i,-1})\subset
I_{k,-1}$, but if $\epsilon=-1$, $f(I_{i,-1})\subset I_{k,+1}$.

  * ii)

If one boundary component of $\tilde{R}_{r}$ is the stable segment
$I_{t,\underline{w}}$, then $f(I_{t,\underline{w}})\subset
I_{t+1,\underline{w}}$.

  * iii)

If $I_{i,+1}$ is the upper boundary of $\tilde{R}_{r}$ and
$\Phi_{T}(i,h_{i})=(k,l,\epsilon)$, then: if $\epsilon=1$, $f(I_{i,+1})\subset
I_{k,+1}$, but if $\epsilon=-1$, $f(I_{i,+1})\subset I_{k,-1}$.

####  5.2.6. The relative position of $H^{i}_{j}$ with respect to
$\tilde{R}_{r}$

Like we did before, we are going to determine the set of horizontal sub-
rectangles of ${\mathcal{R}}$ that intersect a certain rectangle $R_{r}$ from
${\mathcal{R}}_{S({\mathcal{W}})}$.

######  Definition 5.36.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$ with
$r=\tilde{r}(i,s)$. We define:

${\mathcal{H}}(r):=\\{(i,j)\in{\mathcal{H}}(T):\overset{o}{H^{i}_{j}}\cap\overset{o}{\tilde{R}_{r}}\neq\emptyset\\}.$

The fundamental tool to prove the next result is Lemma 5.10 , which we
presented earlier and is valid for any element in
${\mathcal{O}}(i,{\mathcal{W}})$. We’ll leave the proof out as the philosophy
is the same as our previous Lemma 5.16.

######  Lemma 5.38.

Let $\tilde{R}_{r}$ be a rectangle in the geometric Markov partition
${\mathcal{R}}_{S(\underline{w})}$ with $r=\tilde{r}(i,s)$. Then the set
${\mathcal{H}}(r)$ is determined by one of the three following formulas:

  * i)

If $O(i,{\mathcal{W}})=0$, then:
${\mathcal{H}}(r)=\\{(i,j)\in{\mathcal{H}}(T)\\}$.

  * ii)

If $O(i,{\mathcal{W}})>0$, $s\neq 1$ and $s\neq O(i,{\mathcal{W}})+1$, let
$I_{t_{1},\underline{w}^{1}}$ be the inferior boundary of $\tilde{R}_{r}$, and
let $I_{t_{2},\underline{w}^{2}}$ be the upper boundary of $\tilde{R}_{r}$.
Let $j_{t_{1},\underline{w}^{1}}$ and $j_{t_{1},\underline{w}^{2}}$ be the
indices determined in Lemma 5.10. In this situation:

${\mathcal{H}}(r)=\\{(i,j)\in{\mathcal{H}}(T):j_{t_{1},\underline{w}^{1}}\leq
j\leq j_{t_{2},\underline{w}^{2}}\\}.$

  * iii)

If $O(i,{\mathcal{W}})>0$ and $s=O(i,{\mathcal{W}})+1$, let
$I_{t,\underline{w}^{1}}$ be the inferior boundary of $\tilde{R}_{r}$ and
$I_{i,+1}$ the upper boundary of $\tilde{R}_{r}$. Let
$j_{t,\underline{w}^{1}}$ be the index determined in 5.10. In this manner,

${\mathcal{H}}(r)=\\{(i,j)\in{\mathcal{H}}(T):j_{t,\underline{w}^{1}}\leq
j\leq h_{i}\\}.$

  * iv)

If $O(i,{\mathcal{W}})>0$ and $s=1$, let $I_{t,\underline{w}^{1}}$ be the
upper boundary of $\tilde{R}_{r}$, let $I_{i,-1}$ be its inferior boundary and
let $j_{t,\underline{w}^{1}}$ be the index determined in 5.10. In this manner:

${\mathcal{H}}(r)=\\{(i,j)\in{\mathcal{H}}(T):1\leq j\leq
j_{t,\underline{w}^{1}}\\}.$

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$ be a rectangle with
$r=\tilde{r}(i,s)$. For each $(i,j)\in{\mathcal{H}}(r)$, there are three
possible situations regarding the placement of $\tilde{R}_{r}$ with respect to
the sub-rectangle $H^{i}_{j}$ of $R_{i}\in{\mathcal{R}}$:

  * i)

$H^{i}_{j}\subset\tilde{R}_{r}$

  * ii)

$\tilde{R}_{r}\subset H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$

  * iii)

$H^{i}_{j}$ contains only one horizontal boundary component of
$\tilde{R}_{r}$.

As done in Proposition 5.2, we need to provide a criterion to determine which
of the three situations previously mentioned we are dealing with, in fact the
criterion is the same, but we reproduce here just for completeness.

######  Proposition 5.3.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$ be a rectangle with
$r=\tilde{r}(i,s)$.

Assume that $\tilde{R}_{r}\neq R_{i}$ and
${\mathcal{H}}(r)=\\{(i,j_{1}),\cdots,(i,j_{K})\\}$ with $j_{k}<j_{k+1}$. Let
$(i,j_{k})\in{\mathcal{H}}(r)$ in this manner:

  * i)

$\tilde{R}_{r}\subset H^{i}_{j_{k}}\setminus\partial^{s}H^{i}_{j_{k}}$ if and
only if ${\mathcal{H}}(r)=\\{(i,j_{k})\\}$, i.e. $K=1$.

  * ii)

$H^{i}_{j_{k}}\subset\tilde{R}_{r}$ if and only if $j_{1}<j_{k}<j_{K}$.

  * iii)

The sub-rectangle $H^{i}_{j_{k}}$ contains only one horizontal boundary
component of $\tilde{R}_{r}$ if and only if $K>1$, i. e. the cardinality of
${\mathcal{H}}(r)$ is bigger than $1$ ( $\sharp({\mathcal{H}}(r))>1$ ). In
this case, $H^{i}_{j_{k}}$ contains the inferior boundary of $\tilde{R}_{r}$
if $k=1$ and the upper boundary of $\tilde{R}_{r}$ if $k=K$.

In case that $\tilde{R}_{r}=R_{i}$, for all $(i,j)\in{\mathcal{H}}(r)$,
$H^{i}_{j}\subset\tilde{R}_{r}$.

####  5.2.7. The number of horizontal sub-rectangles of
${\mathcal{R}}_{S({\mathcal{W}})}$

We are going to count how many rectangles in the Markov partition
${\mathcal{R}}_{S({\mathcal{W}})}$ are contained in the intersection
$\tilde{R}_{r}\cap H^{i}_{j}$ and describe the way that $f$ send such
horizontal sub-rectangles into vertical sub-rectangles. Demonstrations of all
the following Lemmas is a compilation of the arguments developed in the
previous section, and we believe they don’t require further explanation.
However, we enunciate them and give some hits about their proof for
completeness.

##### Fist case: $\tilde{R}_{r}\subset
H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$

######  Lemma 5.39.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S(cW)}$, with $r=\tilde{r}(i,s)$. Suppose
we have a pair of indices $(i,j)\in{\mathcal{H}}(r)$ such that
$\tilde{R}_{r}\subset H^{i}_{j}\setminus\partial^{s}H^{i}_{j}$. In this case
$s\neq 1$ and $s\neq O(i,{\mathcal{W}})+1$ and we take
$I_{t_{1},\underline{w}^{1}}$ as the lower boundary of $\tilde{R}_{r}$ and
$I_{t_{2},\underline{w}^{2}}$ as the upper boundary of $\tilde{R}_{r}$.

Assume that $\rho_{T}(i,j)=(k,l)$. Let
$s^{-}=\underline{s}(t_{1}+1,\underline{w}^{1})$ and
$s^{+}=\underline{s}(t_{2}+1,\underline{w}^{1})$. Then,
$w^{2}_{t_{1}+1}=w^{2}_{t_{2}+1}=k$ and the number of horizontal sub-
rectangles of ${\mathcal{R}}_{S({\mathcal{W}})}$ contained in the intersection
$\tilde{R}_{r}\cap H^{i}_{j}$ is given by the formula:

(5.35) $\underline{h}(r,j)=|s^{-}-s^{+}|.$

######  Definition 5.37.

The horizontal sub-rectangles of $\tilde{R}_{r}$ as described in Lemma 5.39
are labeled from the bottom to the top with respect to the orientation of
$\tilde{R}_{r}$ as:

$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}.$

######  Lemma 5.40.

With the hypothesis of Lemma 5.39, assume that $\Phi_{T}(i,j)=(k,l,\epsilon)$
and $f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$. Then:

  * i)

If $\epsilon=\epsilon(i,j)=1$, then $s^{-}<s^{+}$,
$r^{\prime}=\tilde{r}(k,s^{-}+J)$ and $l=l^{\prime}$.

  * ii)

If $\epsilon=\epsilon(i,j)=-1$, then $s^{-}>s^{+}$,
$r^{\prime}=\tilde{r}(k,s^{-}-J+1)$ and $l=l^{\prime}$.

######  Remark 5.8.

The only modification in the statement of Lemmas 5.39 and 5.40 respect to such
in Lemma 5.17 and 5.18 is that the numbers $s^{-}$ and $s^{+}$ in our previous
section depend on the same code $\underline{w}$ in this case $s^{-}$ depends
on $\underline{w}^{1}$ and $s^{+}$ depends on $\underline{w}^{2}$. But once we
have fixed the values $s^{-}$ and $s^{+}$ all the arguments in the proofs are
the same.

######  Definition 5.38.

With the hypothesis of Lemma 5.39, assume that $\Phi_{T}(i,j)=(k,l,\epsilon)$.
Then:

If $\epsilon(i,j)=1$ we define:

(5.36) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{-}+J),l).$

and if $\epsilon(i,j)=-1$:

(5.37) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{+}-J+1),l).$

##### Second case: $H^{i}_{j}\subset\tilde{R}_{r}$

######  Lemma 5.41.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$, with
$r=\tilde{r}(i,s)$. Let $(i,j)\in{\mathcal{H}}(r)$ be a pair of indices such
that $H^{i}_{j}\subset\tilde{R}_{r}$, assume that $\rho_{T}(i,j)=(k,l)$, then
the number of horizontal sub-rectangles of ${\mathcal{R}}_{S({\mathcal{W}})}$
that are contained in the intersection $\tilde{R}_{r}\cap H^{i}_{j}$ is given
by the formula:

(5.38) $\underline{h}(r,j)=O(k,{\mathcal{W}})+1,$

###### Proof.

The image of $f^{(}H^{i}_{j}\cap\tilde{R}_{r})=f(H^{i}_{j})=V^{k}_{l}$ and
$V^{k}_{l}$ intersect all the rectangles of ${\mathcal{R}}_{S({\mathcal{W}})}$
that are contained in $R_{k}$, the total amount of them is
$O(k,{\mathcal{W}})+1,$ and this is the number of horizontal sub-rectangles of
${\mathcal{R}}_{S({\mathcal{W}})}$ that are contained in $\tilde{R}_{r}\cap
H^{i}_{j}$. ∎

######  Definition 5.39.

The horizontal sub-rectangles of $\tilde{R}_{r}$ described in Lemma 5.41 are
labeled from the bottom to the top with respect to the orientation of
$\tilde{R}_{r}$ as:

$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}.$

Now we can determine the image of any horizontal sub-rectangle.

######  Lemma 5.42.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$, with
$r=\tilde{r}(i,s)$. Let $(i,j)\in{\mathcal{H}}(r)$ be a pair of indices such
that $H^{i}_{j}\subset\tilde{R}_{r}$, and let $\tilde{H}^{r}_{(}j,J)$ be an
horizontal sub-rectangle of $\tilde{R}_{r}\cap H^{i}_{j}$. Assume that
$\rho_{T}(i,j)=(k,l)$, and that
$f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$, then:

  * i)

If $\epsilon(i,j)=1$, then $r^{\prime}=\tilde{r}(k,J)$ and $l=l^{\prime}$.

  * ii)

If $\epsilon_{T}(i,j)=-1$, then
$r^{\prime}=\tilde{r}(k,O(k,{\mathcal{W}})+1-J+1)$ and $l=l^{\prime}$.

######  Definition 5.40.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$, with
$r=\tilde{r}(i,s)$. Let $(i,j)\in{\mathcal{H}}(r)$ be a pair of indices such
that $H^{i}_{j}\subset\tilde{R}_{r}$, and let $\tilde{H}^{r}_{j,J}$ be an
horizontal sub-rectangle of $\tilde{R}_{r}\cap H^{i}_{j}$. Assume that
$\rho_{T}(i,j)=(k,l)$, then:

If $\epsilon(i,j)=1$ we define:

(5.39) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,J),l),$

and if $\epsilon(i,j)=-1$

(5.40)
$\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,O(k,{\mathcal{W}})+1-J+1),l).$

##### Third case: $H^{i}_{j}$ contains only one horizontal boundary component
of $\tilde{R}_{r}$

######  Lemma 5.43.

Consider $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$, where
$r=\tilde{r}(i,s)$, and assume that $O(i,{\mathcal{W}})>0$. Take a pair of
indexes $(i,j)\in{\mathcal{H}}(r)$ such that $H^{i}_{j}$ exclusively contains
the upper boundary of $\tilde{R}_{r}$. Let $I_{t,\underline{w}^{1}}$ (with
$\underline{w}^{1}\in{\mathcal{W}}$) be the upper boundary of $\tilde{R}_{r}$.
Let $s^{+}=\underline{s}(t+1,\underline{w}^{1})$ and assume that
$\rho_{T}(i,j)=(k,l)$. Then, the number of horizontal sub-rectangles of
${\mathcal{R}}_{S({\mathcal{W}})}$ that are contained in $\tilde{R}_{r}\cap
H^{i}_{j}$ can be computed using one of the following formulas:

  * •

If $\epsilon_{T}(i,j)=1$:

(5.41) $\underline{h}(r,j)=s^{+}$

  * •

If $\epsilon_{T}(i,j)=-1$

(5.42) $\underline{h}(r,j)=O(k,{\mathcal{W}})+1-s^{+}.$

###### Proof.

In fact the proof is exactly the same as Lemma 5.21 the only difference is
that function $\underline{s}$ now take in count all the rectangles determined
by the family ${\mathcal{W}}$, when previously only depended on a single code
$\underline{w}$.

∎

######  Lemma 5.44.

Consider $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$, where
$r=\tilde{r}(i,s)$, and assume that $O(i,{\mathcal{W}})>0$. Take a pair of
indexes $(i,j)\in{\mathcal{H}}(r)$ such that $H^{i}_{j}$ just contains the
lower boundary of $\tilde{R}_{r}$. Let $I_{t,\underline{w}^{1}}$ (with
$\underline{w}^{1}\in{\mathcal{W}}$ ) be the lower boundary of
$\tilde{R}_{r}$, and denote $s^{+}=\underline{s}(t+1,\underline{w}^{1})$. In
this scenario, the number of horizontal sub-rectangles of
${\mathcal{R}}_{S({\mathcal{W}})}$ that are contained in $\tilde{R}_{r}\cap
H^{i}_{j}$ can be determined using one of the following formulas:

  * •

If $\epsilon_{T}(i,j)=1$:

(5.43) $\underline{h}(r,j)=O(k,{\mathcal{W}})+1-s^{+}$

  * •

If $\epsilon_{T}(i,j)=-1$

(5.44) $\underline{h}(r,j)=s^{+}.$

######  Definition 5.41.

The horizontal sub-rectangles of $\tilde{R}_{r}$ described in Lemma 5.43 and
Lemma 5.44 are labeled from the bottom to the top with the vertical
orientation of $\tilde{R}_{r}$ as:

$\\{\tilde{H}^{r}_{j,J}\\}_{J=1}^{\underline{h}(r,j)}.$

######  Lemma 5.45.

With the hypothesis of Lemma 5.43, suppose that $H^{i}_{j}$ just contains the
upper boundary of $\tilde{R}_{r}$ and assume that
$\Phi_{T}(i,j)=(k,l,\epsilon_{T}(i,j))$ and
$f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$. Then we have the
following situations:

  * i)

If $\epsilon_{T}(i,j)=1$, then $r^{\prime}=\tilde{r}(k,J)$ and $l=l^{\prime}$

  * ii)

If $\epsilon_{T}(i,j)=-1$, then
$r^{\prime}=\tilde{r}(k,O(k,{\mathcal{W}})+1-J+1)$ and $l=l^{\prime}$.

######  Lemma 5.46.

With the hypothesis of Lemma 5.44, suppose that $H^{i}_{j}$ just contains the
lower boundary of $\tilde{R}_{r}$ and assume that
$\Phi_{T}(i,j)=(k,l,\epsilon_{T}(i,j))$ and
$f(\tilde{H}^{r}_{j,J})=\tilde{V}^{r^{\prime}}_{l^{\prime}}$. Then we have the
following situations:

  * i)

If $\epsilon_{T}(i,j)=1$, then $r^{\prime}=\tilde{r}(k,s^{+}+J)$ and
$l=l^{\prime}$

  * ii)

If $\epsilon_{T}(i,j)=-1$, then $r^{\prime}=\tilde{r}(k,s^{+}-J+1)$ and
$l=l^{\prime}$.

######  Definition 5.42.

With the hypothesis of Lemma 5.43, suppose that $H^{i}_{j}$ just contains the
upper boundary of $\tilde{R}_{r}$ and assume that
$\Phi_{T}(i,j)=(k,l,\epsilon)$. Then:

  * i)

If $\epsilon_{T}(i,j)=1$, we define:

(5.45) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,J),l).$

  * ii)

if $\epsilon_{T}(i,j)=-1$ we take:

(5.46)
$\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,O(k,{\mathcal{W}})+1-J+1),l).$

######  Definition 5.43.

With the hypothesis of Lemma 5.44, suppose that $H^{i}_{j}$ just contains the
lower boundary of $\tilde{R}_{r}$ and assume that
$\Phi_{T}(i,j)=(k,l,\epsilon)$. Then:

  * i)

If $\epsilon_{T}(i,j)=1$, we define:

(5.47) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{+}+J),l).$

  * ii)

if $\epsilon_{T}(i,j)=-1$ we take:

(5.48) $\underline{\rho}_{(r,j)}(r,J)=(\tilde{r}(k,s^{+}-J+1),l).$

##### Collecting the information

Now, we can collect the information from the three cases to compute the number
of horizontal sub-rectangles in $\tilde{R}_{r}$.

######  Corollary 5.10.

Let $\tilde{R}_{r}\in{\mathcal{R}}_{S({\mathcal{W}})}$ with
$r=\tilde{r}(i,s)$, then the number $H_{r}$ of horizontal sub-rectangles of
the Markov partition $(f,{\mathcal{R}}_{S({\mathcal{W}})})$ that are contained
in $\tilde{R}_{r}$ is given by:

(5.49) $H_{r}=\sum_{\\{j:(i,j)\in{\mathcal{H}}(r)\\}}\underline{h}(r,j).$

####  5.2.8. The function $\rho_{S({\mathcal{W}})}$ in the geometric type
$T_{S({\mathcal{W}})}$

######  Definition 5.44.

The sub-rectangles of the geometric Markov partition
${\mathcal{R}}_{S({\mathcal{W}})}$ contained in $\tilde{R}_{r}$ are labeled
as:

$\\{H^{r}_{\underline{J}}\\}_{\underline{J}=1}^{H_{r}}$

from the bottom to the top with respect to the vertical orientation of
$\tilde{R}_{r}$.

######  Lemma 5.47.

For each $\underline{J}\in\\{1,\cdots,H_{r}\\}$, there exists a unique index
$j_{(\underline{J})}$ such that $(i,j)\in{\mathcal{H}}(r)$ and
$\tilde{H}^{r}_{\underline{J}}\subset H^{i}_{j}\cap\tilde{R}_{r}$. The index
$j_{(\underline{J})}$ is determined by satisfy the next inequalities:

(5.50) $\sum_{\\{j^{\prime}:(i,j^{\prime})\in{\mathcal{H}}(r)\text{ and
}j^{\prime}<j_{(\underline{J})}\\}}\underline{h}(j^{\prime},r)<\underline{J}\leq\sum_{\\{j^{\prime}:(i,j^{\prime})\in{\mathcal{H}}(r)\text{
and }j^{\prime}\leq j_{(\underline{J})}\\}}\underline{h}(j^{\prime},r).$

######  Definition 5.45.

Given $\underline{J}\in\\{1,\cdots,H_{r}\\}$ we call the index
$j_{(\underline{J})}$ that was determined in Lemma 5.47 the _horizontal
parameter_ of $\underline{J}$.

######  Definition 5.46.

Let $r=\tilde{r}(i,s)$, $\underline{J}\in\\{1,\cdots,H_{r}\\}$, and let
$j_{(\underline{J})}$ be the horizontal parameter of $\underline{J}$. The
_relative index_ of $H^{r}_{J}$ inside of $\tilde{R}_{r}\cap
H^{i}_{j_{(\underline{J}})}$ is given by:

(5.51)
$J_{(\underline{J})}:=\underline{J}-\sum_{j^{\prime}<j_{({\underline{J})}}}\underline{h}(r,j^{\prime}).$

######  Definition 5.47.

Let $\underline{J}\in\\{1,\cdots,H_{r}\\}$ the _relative position_ of
$\underline{J}$ inside $\tilde{R}_{r}$ is the pair
$(j_{\underline{J}},J_{(\underline{J})})$ of the horizontal parameter of
$\underline{J}$ and the relative index of $H^{r}_{J}$ inside of
$\tilde{R}_{r}\cap H^{i}_{j_{\underline{J}}}$

######  Lemma 5.48.

The rectangle $\tilde{H}^{r}_{j_{(\underline{J})},J_{(\underline{J})}}$ and
the rectangle $\tilde{H}^{r}_{\underline{J}}$ are the same.

######  Corollary 5.11.

Let $r=\tilde{r}(i,s)$ and $\underline{J}\in\\{1,\cdots,H_{r}\\}$. Let
$(j_{(\underline{J})},J_{(\underline{J})})$ the relative position of
$\underline{J}$ as was given in Definition 5.47. Then the function
$\rho_{S({\mathcal{W}})}$ is given by the formula:

(5.52)
$\rho_{S({\mathcal{W}})}(r,\underline{J}):=\underline{\rho}_{(r,j_{(\underline{J})})}(r,J_{(\underline{J})}).$

####  5.2.9. The orientation $\epsilon_{S({\mathcal{W}})}$

######  Lemma 5.49.

Let $r=\tilde{r}(i,s)$ and $\underline{J}\in\\{1,\cdots,H_{r}\\}$. Let
$j_{(\underline{J})}$ the horizontal parameter of $\underline{J}$ (Definition
5.45). Then:

(5.53)
$\epsilon_{S(\underline{w})}(r,\underline{J}):=\epsilon_{T}(i,j_{(\underline{J})}).$

####  5.2.10. Conclusion

######  Corollary 5.12.

Let $T$ be a geometric type in the pseudo-Anosov class with a binary incidence
matrix denoted as $A:=A(T)$. Let ${\mathcal{W}}$ a family of periodic codes
contained in $\Sigma_{A}$, there is algorithm to compute the geometric type:

$T_{S({\mathcal{W}})}:=\\{N,\\{H_{r},V_{r}\\}_{r=1}^{N},\Phi_{S({\mathcal{W}})}:=(\rho_{S({\mathcal{W}})},\epsilon_{S({\mathcal{W}})})\\}.$

using the given geometric type $T$ and the codes in ${\mathcal{W}}$.

###  5.3. The $u$-boundary refinement

.

In this part, we are going to define the $u$-boundary refinement of a Markov
partition along a family of periodic codes. This procedure involves cutting
${\mathcal{R}}$ along the unstable segments that correspond to the orbit of
any of the codes in ${\mathcal{W}}$. In fact, we can define it using our
previous $s$-boundary refinement applied to $T^{-1}$ and $f^{-1}$.

######  Definition 5.48.

Let $T$ be a geometric type in the pseudo-Anosov class with an incidence
matrix $A:=A(T)$ that is binary. Consider a generalized pseudo-Anosov
homeomorphism $f:S\rightarrow S$ with a geometric Markov partition
${\mathcal{R}}$ of geometric type $T$. Let

${\mathcal{W}}=\\{\underline{w}^{1},\cdots,\underline{w}^{Q}\\},$

be a family of periodic codes that are non-$u$-boundary.

  * •

Let ${\mathcal{R}}_{U({\mathcal{W}})}$ the $s$-boundary refinement of
${\mathcal{R}}$ when ${\mathcal{R}}$ is viewed as a Markov partition of
$f^{-1}$. i.e $({\mathcal{R}},f^{-1})$. Denote by $T^{-1}_{S({\mathcal{W}})}$
the geometric type of $({\mathcal{R}}_{U({\mathcal{W}})},f^{-1})$.

  * •

Let $T_{U({\mathcal{W}})}$ the geometric type of
$({\mathcal{R}}_{U({\mathcal{W}})},f)$.

In this case the geometry Markov partition ${\mathcal{R}}_{U({\mathcal{W}})}$
for $f$ called the $u$-boundary refinement of ${\mathcal{R}}$ respect the
family ${\mathcal{W}}$.

######  Remark 5.9.

We can make the following direct observations:

  1. (1)

If $\underline{w}$ is not a $u$-boundary point for $({\mathcal{R}},f)$, the
$\underline{w}$ it is not a $s$-boundary point for $({\mathcal{R}},f^{-1})$,
which justifies our assumption.

  2. (2)

The geometric Markov partition $({\mathcal{R}}_{U({\mathcal{W}})},f)$ has a
geometric type $T_{U({\mathcal{W}})}$, and this geometric type is the inverse
of the geometric type of $({\mathcal{R}}_{U({\mathcal{W}})},f^{-1})$,
therefore:

$T_{U({\mathcal{W}})}:=(T^{-1}_{S({\mathcal{W}})})^{-1}.$

and we can determine $T_{U({\mathcal{W}})}$ in an algorithmic manner.

  3. (3)

For every $\underline{w}\in{\mathcal{W}}$, the point $\pi_{f}(\underline{w})$
(where $\pi_{f}$ is the projection with respect to $({\mathcal{R}},f)$) is a
$u$-boundary point of ${\mathcal{R}}_{U({\mathcal{W}})}$.

  4. (4)

If $\underline{w}$ is a $u$-boundary point, clearly
${\mathcal{R}}_{U(\underline{w})}={\mathcal{R}}$ and $T_{U(\underline{w})}=T$.
This justifies taking any code in ${\mathcal{W}}$ as a non $u$-boundary code.

###  5.4. The corner refinement

Let $T$ be a geometric type within the pseudo-Anosov class with binary
incidence matrix $A(T)$. Consider a generalized pseudo-Anosov homeomorphism
$f:S\to S$ with a geometric Markov partition ${\mathcal{R}}$ of geometric type
$T$. The projection $\pi_{f}:\Sigma_{A(T)}\to S$, as defined in 3.7, depends
on the specific geometric type $T$ of the Markov partition. We will relabel it
as $\pi_{T}:\Sigma_{A(T)}\to S$ to enable easier comparison with another
projection $\pi_{T^{\prime}}$ associated with a different geometric type
$T^{\prime}$

For a given geometric type $T$, we have defined its $s$ and $u$ generating
functions in 3.12 and 3.13, respectively. Iterating these functions generates
codes in $\Sigma_{A(T)}$, which project to the stable and unstable boundary
components of ${\mathcal{R}}$. Furthermore, as described in Section 3.1, using
$T$, it is feasible to determine the sets of periodic $s$-boundary codes of
$T$:

$\textit{Per}(\underline{S(T)}):=\\{\underline{w}\in\Sigma_{A(T)}:\underline{w}\text{
is periodic for $\sigma$ and
}\pi_{T}(\underline{w})\in\partial^{s}{\mathcal{R}}\\}.$

and the sets of periodic $u$-boundary codes:

$\textit{Per}(\underline{(T)}):=\\{\underline{w}\in\Sigma_{A(T)}:\underline{w}\text{
is periodic for $\sigma$ and
}\pi_{T}(\underline{w})\in\partial^{u}{\mathcal{R}}\\}.$

Their union is the set of boundary periodic codes:

$\textit{Per}(\underline{B(T)}):=\textit{Per}(S(T))\cup\textit{Per}(U(T)).$

In order to avoid to heavy notations we relabel these sets:

$\underline{S(T)}:=\textit{Per}(\underline{S(T)}),\,\,\underline{U(T)}:=\textit{Per}(\underline{U(T)})\text{
and }\underline{B(T)}:=\textit{Per}(\underline{B(T)})$

Finally a corner periodic code is any code in the intersection:

$\underline{C(T)}:=\underline{S(T)}\cap\underline{U(T)}.$

Definition 3.16 reveals that $\underline{B(T)}$ can be described in a
combinatorial manner using $T$, avoiding the use of the Markov partition or
the homeomorphism $f$. Based on the theory developed in Section 3.1, we can
draw the following conclusion, which we present in the form of a corollary:

######  Corollary 5.13.

Given a specific geometric type $T$, we can compute the sets:
$\underline{(S(T)})$, $\underline{U(T)}$, $\underline{B(T)}$, and
$\underline{C(T)}$, in terms of the geometric type $T$.

######  Definition 5.49.

A geometric type $T$ have the _corner property_ if every periodic boundary
code is a corner peridic code, i.e
$\underline{B(T)}=\underline{S(T)}\cap\underline{U(T)}$.

We fix the following notation:

  * •

$P_{s}(f)$ is the set of $s$-boundary periodic points of $(f,{\mathcal{R}})$.

  * •

$P_{u}(f)$ is the set of $u$-boundary periodic points of $(f,{\mathcal{R}})$

  * •

$P_{u}(f)=P_{s}(f)\cup P_{u}(f)$ is the set of periodic boundary codes of
$(f,{\mathcal{R}})$, and

  * •

$P_{c}(f)=P_{s}(f)\cap P_{u}(f)$ is the set of corner points of
${\mathcal{R}}$.

######  Definition 5.50.

A geometric Markov partition ${\mathcal{R}}$ of $f$ have the _corner property_
if every rectangle $R\in{\mathcal{R}}$ that contains a boundary point $p\in
P_{b}(f,{\mathcal{R}})$ have $p$ as a corner point.

A corner point $p$ of a rectangle $R\in{\mathcal{R}}$ is on the boundary of
another rectangle $R^{\prime}\in{\mathcal{R}}$ but it is not necessarily a
corner point of $R^{\prime}$; it can be on the interior of the unstable
boundary of $R^{\prime}$, for example. In this situation,
$\pi_{T}^{-1}(p)\subset\Sigma_{A(T)}$ contains a corner code and a
$u$-boundary code that is not an $s$-boundary code. Therefore,
$\underline{C(T)}\neq\underline{B(T)}$ and $T$ don’t have the corner property.
This discrepancy makes it necessary to introduce the following lemma

######  Lemma 5.50.

The geometric Markov partition ${\mathcal{R}}$ has the corner property if and
only if its geometric type $T$ has the corner property.

###### Proof.

If ${\mathcal{R}}$ satisfies the corner property, and we have an $s$-boundary
code $\underline{w}\in\underline{S(T)}$, then by Lemma 3.15,
$\pi_{T}(\underline{w})$ is a stable boundary periodic point contained in the
rectangle $R_{w_{0}}$, where $w_{0}=\underline{w}_{0}$, is $0$ therm of the
code $\underline{w}$. However, by hypothesis, if $p$ is a $s$-boundary point
of $R_{w_{0}}$, it is a corner point of $R_{w_{0}}$, then, $p$ is a
$u$-boundary point. We can use Lemma 3.15 to deduce that the code
$\underline{w}$ is also a $u$-boundary code. So,
$\underline{w}\in\underline{C(T)}$. The case when the code $\underline{w}$ is
a $u$-boundary code, i.e., $\underline{w}\in\underline{U(T)}$, is similarly
proven.

On the other hand, assume $T$ satisfies the corner property. Let $p\in
R_{w_{0}}\in{\mathcal{R}}$ a $s$-boundary point of $R$. There exist a
$s$-boundary code $\underline{w}\in\pi^{-1}_{T}(p)$ such that
$\underline{w}_{0}=w_{0}$. Like $T$ have the corner property, any $s$-boundary
periodic code is $u$-boundary, which in turn means $\underline{w}$ is a
$u$-boundary code. As a consequence of Lemma 3.15,
$p=\pi_{T}(\underline{w})\in R_{w_{0}}$ is a $u$-boundary point contained in
$R_{w_{0}}$, therefore $p$ is a corner point of such rectangle, $R_{w_{0}}$.
The situation when $p$ is $s$-boundary point of $R_{w_{0}}$ is proved in the
same manner.

On the other hand, assume that $T$ satisfies the corner property. Let
$p\in\in\partial^{s}R_{w_{0}}$ be an $s$-boundary point of $R_{w_{0}}$. There
exists an $s$-boundary code $\underline{w}\in\pi^{-1}_{T}(p)$ such that
$\underline{w}0=w_{0}$. Since $T$ has the corner property, any $s$-boundary
periodic code is also a $u$-boundary code, which, in turn, means that
$\underline{w}$ is a $u$-boundary code. As a consequence of 3.15,
$p=\pi_{T}(\underline{w})\in\partial^{s}R{w_{0}}$ is a $u$-boundary point
contained in $R{w_{0}}$; therefore, $p$ is a corner point of such a rectangle,
$R_{w_{0}}$. The situation when $p$ is an $s$-boundary point of $R_{w_{0}}$ is
proved in the same manner.

∎

####  5.4.1. Construction of the corner refinement

The first goal of this section is to provide a method for creating a geometric
Markov partition ${\mathcal{R}}_{C(T)}$ for the homeomorphism $f$, along with
its corresponding geometric type $C(T)$, in such a way that both
${\mathcal{R}}_{C(T)}$ and $T_{C}$ satisfy the corner property.

Let $T$ be a geometric type in the pseudo-Anosov class with a binary incidence
matrix $A(T)$. Consider a generalized pseudo-Anosov homeomorphism $f:S\to S$
with a geometric Markov partition ${\mathcal{R}}$ of geometric type $T$. The
procedure is as describe in the following paragraph:

  1. (1)

Get the $s$-boundary refinement of ${\mathcal{R}}$ along the family of
boundary periodic of codes $\underline{B(T)}$ to obtain:

${\mathcal{R}}_{S(\underline{B(T)})}.$

and its geometric type: $T_{S(\underline{B(T)})}$.

  2. (2)

Compute the set of boundary periodic codes of the geometric type
$T_{S(\underline{B(T)})}$ (Corollary 5.13):

$\underline{S(T_{S(\underline{B(T)})})}\subset\Sigma_{A(T_{S(\underline{B(T)})})}.$

  3. (3)

Obtain the $u$-boundary refinement of ${\mathcal{R}}_{S(\underline{B(T)})}$
along the family of boundary periodic codes
$\underline{S(T_{S(\underline{B(T)})})}$ to obtain the corner refinement of
${\mathcal{R}}$:

(5.54)
${\mathcal{R}}_{C(T)}:=[{\mathcal{R}}_{S(\underline{B(T)})}]_{U(\underline{S(T_{S(\underline{B(T)})})})}.$

whose geometric type is $T_{C}$.

######  Definition 5.51.

The geometric Markov partition for $f$, ${\mathcal{R}}_{C(T)}$, described by
equation (5.54), is the _Corner refinement_ of the geometric Markov partition
$(f,{\mathcal{R}})$, and its geometric type $T_{C}$ is the _Corner refinement_
of $T$.

######  Proposition 5.4.

Let $T$ be a geometric type in the pseudo-Anosov class with a binary incidence
matrix $A(T)$. Consider a generalized pseudo-Anosov homeomorphism $f:S\to S$
with a geometric Markov partition ${\mathcal{R}}$ of geometric type $T$. Let
$(f,{\mathcal{R}}_{C(T)})$ and $T_{C}$ be their corner refinement. Then:

  * i)

The boundary periodic points of $(f,{\mathcal{R}}_{C(T)})$ and the boundary
periodic points of $(f,{\mathcal{R}})$ are exactly the same. In simpler terms:
$P_{b}(f,{\mathcal{R}}_{C(T)})$ equals $P_{b}(f,{\mathcal{R}})$.

  * ii)

The Markov partition $(f,{\mathcal{R}}_{C(T)})$ exhibits the corner property

  * iii)

The new geometric type $T_{C}$ also exhibits the corner property:

$\underline{B(T_{C})}=\underline{U(T_{C})}\cap\underline{S(T_{C})}.$

In other terms every boundary periodic code of $T_{C}$ is a corner code.

###### Proof.

Item $i)$. For the $s$-boundary refinement, the following equality holds:

$P_{b}(f,{\mathcal{R}}_{S(\underline{B(T)}))}):=P_{b}(f,{\mathcal{R}})\cup\pi_{T}(\underline{B(T)})=P_{b}(f,{\mathcal{R}}).$

As a consequence, the boundary points of $(f,{\mathcal{R}}_{C(T)})$ coincide
with the boundary points of $(f,{\mathcal{R}}).$

Item $ii)$. Let $p$ be a boundary point of
${\mathcal{R}}_{S(\underline{B(T)})}$, and let
$R\in{\mathcal{R}}_{S(\underline{B(T)})}$ be a rectangle that contains $p$.
Let $R_{0}\in{\mathcal{R}}$ be a rectangle of ${\mathcal{R}}$ that also
contains $R$. After the $s$-boundary refinement:

  1. (1)

If $p$ belonged to both $\partial^{s}R_{0}$ and $\partial^{u}R_{0}$, then $p$
becomes a corner point of the rectangle $R$.

  2. (2)

If $p$ were in $\partial^{s}R_{0}$ but not in $\partial^{u}R_{0}$, it remains
in $\partial^{s}R$ but not in $\partial^{u}R$. There is no change.

  3. (3)

If $p$ was in $\partial^{u}R_{0}$ but not in $\partial^{s}R_{0}$, then it
becomes a corner point of $R$.

Let $p$ be a periodic boundary point in ${\mathcal{R}}_{C(T)}$. Consider a
rectangle $R$ in

${\mathcal{R}}_{C(T)}=[{\mathcal{R}}_{S(\underline{B(T)}}]_{U(\underline{B(T_{S(\underline{B(T)})})})},$

that contains $p$, along with a rectangle $R_{0}$ in
$R_{0}\in{\mathcal{R}}_{S(\underline{B(T)})}$ that contains $R$. After the
$u$-boundary refinement, we can have two possible outcomes

  1. (1)

If $p\in\partial^{s}R_{0}\cap\partial^{u}R_{0}$ was a corner point of
$R_{w_{0}}$, then the rectangle $R$ has $p$ as one of its corner points.

  2. (2)

If $p\in\partial^{u}R_{0}\setminus\partial^{s}R_{0}$, after the $u$-boundary
refinement, $p$ is in $\partial^{s}R\cap\partial^{u}R$ and is a corner point
of $R$.

In conclusion, the geometric Markov partition $(f,{\mathcal{R}}_{C(T)})$
exhibits the corner property, as asserted in Item $ii$.

Item $iii)$. In light of Lemma 5.50, $T_{C(T)}$ exhibits the corner property,
and thus, we can proceed to prove item $iii$.

∎

####  5.4.2. The corner refinement in a family of periodic codes

Let ${\mathcal{R}}$ be a geometric Markov partition of $f:S\to S$ with
geometric type $T$ that has the corner property. Take a family of periodic
codes ${\mathcal{W}}\subset\Sigma_{A(T)}$ with its corresponding projected
family $P_{{\mathcal{W}}}=\pi_{T}({\mathcal{W}})$ of periodic points for $f$.
The $s$-boundary refinement of ${\mathcal{R}}$ concerning this family yields
${\mathcal{R}}_{S({\mathcal{W}})}$ and its geometric type
$T_{S({\mathcal{W}})}$. The boundary points of this Markov partition are given
by the union of the boundary points of ${\mathcal{R}}$ and the periodic points
determined by the projections of the codes in ${\mathcal{W}}$:

$P_{b}(f,{\mathcal{R}}_{S({\mathcal{W}})})=P_{b}(f,{\mathcal{R}})\cup
P_{{\mathcal{W}}}$

Let ${\mathcal{R}}_{C(T_{S({\mathcal{W}})})}$ be the corner refinement of
${\mathcal{R}}_{S({\mathcal{W}})}$ with geometric type
$[T_{S({\mathcal{W}})}]_{C}$. A corollary of Proposition 5.4, when applied to
the geometric type $T_{S({\mathcal{W}})}$, is as follows:

######  Corollary 5.14.

Let ${\mathcal{R}}_{C(T_{S({\mathcal{W}})})}$ be the corner refinement of
${\mathcal{R}}_{S({\mathcal{W}})}$ with geometric type
$T_{C(T_{S({\mathcal{W}})})}$. Then:

  * i)

The boundary periodic points of $(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}})})})$
and the boundary periodic points of $(f,{\mathcal{R}}_{S({\mathcal{W}})})$ are
exactly the same. In simpler terms:
$P_{b}(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}})})})=P_{b}(f,{\mathcal{R}}_{S({\mathcal{W}})})$.

  * ii)

The Markov partition $(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}})})})$ exhibits
the corner property.

  * iii)

The new geometric type $[T_{S({\mathcal{W}})})]_{C}$ also exhibits the corner
property:

$\underline{B([T_{S({\mathcal{W}})})]_{C})}=\underline{U([T_{S({\mathcal{W}})})]_{C})}\cap\underline{S([T_{S({\mathcal{W}})})]_{C})}.$

In other terms, every boundary periodic code of $[T_{S({\mathcal{W}})})]_{C}$
is a corner code.

####  5.4.3. The corner refinement in codes of bounded period

Finally, let us make a construction that will help us during the discussion of
the Béguin algorithm.

######  Definition 5.52.

Let $T$ be a geometric type in the pseudo-Anosov class with a binary incidence
matrix $A(T)$, and let $f:S\to S$ be a generalized pseudo-Anosov homeomorphism
with a geometric Markov partition ${\mathcal{R}}$ of geometric type $T$.
Assume that $(f,{\mathcal{R}})$ (and $T$) has the corner property. Now,
consider the following definitions:

  * •

Let $P_{\underline{B(T)}}$ be the maximum period among the boundary periodic
codes of $T$:

$P_{\underline{B(T)}}:=\max\\{\textbf{Per}(\underline{w},\sigma_{T}):\underline{w}\in\underline{B(T)}\\}$

  * •

For any integer $P\geq P_{\underline{B(T)}}$, define ${\mathcal{W}}_{P}$ as
the set of admissible codes with periods less than or equal to $P$:

${\mathcal{W}}_{P}:=\\{\underline{w}\in\Sigma_{A(T)}:\textbf{Per}(\underline{w},\sigma_{T})\leq
P\\}.$

  * •

The corner refinement of ${\mathcal{R}}_{S({\mathcal{W}}_{P})}$ along the
family ${\mathcal{W}}_{P}$ is given by
${\mathcal{R}}_{C(T_{S({\mathcal{W}}_{P})})}$, with its corresponding
geometric type referred to as $T_{C(P)}$.

######  Lemma 5.51.

Let $P\geq P_{\underline{B(T)}}$. Suppose $p\in S$ is a periodic point of $f$
with a period less than or equal to $P$; i.e., $\textbf{Per}(p,f)\leq P$.
Consider $\underline{w}\in\Sigma_{A(T)}$ as a code that projects to $p$:
$\pi_{T}(\underline{w})=p$. Then, the period of $\underline{w}$ is less than
or equal to $P$, and $\underline{w}\in{\mathcal{W}}_{P}$.

Conversely, if $\underline{w}\in{\mathcal{W}}_{P}$, then the point
$p:=\pi_{T}(\underline{w})\in S$ is a periodic point of $f$ with a period
$\textbf{Per}(p,f)$ less than or equal to $P$. In this manner:

(5.55) $\pi_{T}({\mathcal{W}}_{P})=\\{p\in S:\textit{Per}(p,f)\leq P\\}\text{
and }\pi_{T}^{-1}(\\{p\in S:\textit{Per}(p,f)\leq P\\})={\mathcal{W}}_{P}$

###### Proof.

Let $p\in S$ be a periodic point of $f$ with period less or equal than $P$. We
have two possible situations:

  * •

The periodic point $p$ is a boundary point. In this case,
$\underline{w}\in\underline{B(T)}\subset{\mathcal{W}}_{P}$ just by definition
of ${\mathcal{W}}_{P}$. Let’s recall that any code in $\underline{B(T)}$ has a
period less than or equal to $P_{\underline{B(T)}}\leq P$.

  * •

The periodic point $p$ is interior. In this case, $\pi^{-1}(p)$ contains a
unique code $\underline{w}$ whose period coincides with the period of $p$,
i.e., $\textbf{Per}(\underline{w},\sigma_{T})=\textbf{Per}(p,f)\leq P$.
Therefore, $\underline{w}\in{\mathcal{W}}_{P}$.

This implies that:

(5.56) $\pi_{T}^{-1}(\\{p\in S:\textit{Per}(p,f)\leq
P\\})\subset{\mathcal{W}}_{P}$

and

(5.57) $\\{p\in S:\textit{Per}(p,f)\leq P\\}=\pi_{T}(\pi_{T}^{-1}(\\{p\in
S:\textit{Per}(p,f)\leq P\\}))\subset\pi_{T}({\mathcal{W}}_{P}).$

For any code $\underline{w}\in{\mathcal{W}}_{P}$ with a period $Q$ no greater
than $P$, we can observe the following relationship:

$p=\pi_{T}(\underline{w})=\pi_{T}(\sigma^{Q}(\underline{w}))=f^{Q}(\pi_{T}(\underline{w}))=f^{Q}(p)$

This implies that $P\geq Q\geq\textbf{Per}(p,f)$, i.e., the period of $p$ with
respect to $f$ is less than or equal to $P$. Then:

(5.58) $\pi_{T}({\mathcal{W}}_{P})\subset\\{p\in S:\textit{Per}(p,f)\leq
P\\}.$

This inclusion, along with the information provided in 5.57, implies that:

$\pi_{T}({\mathcal{W}}_{P})=\\{p\in S:\textit{Per}(p,f)\leq P\\}.$

Furthermore, if $\underline{w}\in{\mathcal{W}}_{P}$, trivially
$\underline{w}\in\pi_{T}^{-1}(\pi_{T}(\underline{w}))$ and
$p:=\pi_{T}(\underline{w})$ has a period less than or equal to $P$. In this
manner:

(5.59) ${\mathcal{W}}_{P}\subset\pi_{T}^{-1}(\\{p\in S:\textit{Per}(p,f)\leq
P\\}).$

Using this information and the contention provided in 5.56, we establish the
second equality of our lemma:

$\pi_{T}^{-1}(\\{p\in S:\textit{Per}(p,f)\leq P\\})={\mathcal{W}}_{P}.$

This ends our proof. ∎

######  Corollary 5.15.

When $P\geq P_{B(T)}$, the periodic boundary points of the corner refinement
$(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}}_{P})})})$ coincides with the set of
all periodic points of $f$ having periods less than or equal to $P$ i.e.,

$P_{b}(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}}_{P})})})=\\{p\in
S:\textit{Per}(p,f)\leq P\\}.$

Furthermore, $(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}}_{P})})})$ also possesses
the corner property.

###### Proof.

From Corollary 5.14:

$P_{b}(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}}_{P})})})=P_{b}(f,{\mathcal{R}}_{S({\mathcal{W}}_{P})}).$

But, we recall that:

$P_{b}(f,{\mathcal{R}}_{S({\mathcal{W}}_{P})})=P_{b}(f,{\mathcal{R}})\cup\pi_{T}({\mathcal{W}}_{P}).$

In view of Lemma 5.51

$\pi_{T}({\mathcal{W}}_{P})=\\{p\in S:\textit{Per}(p,f)\leq P\\}$

and clearly $P_{b}(f,{\mathcal{R}})\subset\\{p\in S:\textit{Per}(p,f)\leq
P\\}$. Therefore:

$P_{b}(f,{\mathcal{R}}_{C(T_{S({\mathcal{W}}_{P})})})=P_{b}(f,{\mathcal{R}}_{S({\mathcal{W}}_{P})})=P_{b}(f,{\mathcal{R}})\cup\pi_{T}({\mathcal{W}}_{P})=\\{p\in
S:\textit{Per}(p,f)\leq P\\}.$

∎

###  5.5. The boundary refinement respect to ${\mathcal{W}}$

In this section we deal with next objects:

  1. (1)

A geometric type in the pseudo-Anosov class, $T$ whose incidence matrix $A(T)$
is binary.

  2. (2)

A generalized pseudo-Anosov homeomorphism $f:S\rightarrow S$. With a geometric
Markov partition $(f,{\mathcal{R}})$ of geometric type $T$.

  3. (3)

Finite family of periodic points of $f$,
${\mathcal{P}}:={p_{\alpha}}_{\alpha=1}^{A}$

Given a geometric type ${\mathcal{G}}$ we must fix the following notation,
whenever it have seance:

  1. (1)

$A({\mathcal{G}})$ its it incidence matrix,
$(\Sigma_{A({\mathcal{G}})},\sigma_{A({\mathcal{G}})})$ the sub-shift of
finite associate to ${\mathcal{G}}$ (??????)and
$\pi_{{\mathcal{G}}}:\Sigma_{A({\mathcal{G}})}\to S$ the natural projection
defined in Def ????.

  2. (2)

${\mathcal{W}}_{{\mathcal{G}}}({\mathcal{P}})=\pi^{-1}_{\pi_{{\mathcal{G}}}}({\mathcal{P}})$.

######  Definition 5.53.

The boundary refinement of ${\mathcal{R}}$ and $T$ with respect to the family
of periodic points ${\mathcal{P}}$ is:

${\mathcal{R}}_{B({\mathcal{P}})}:=({\mathcal{R}}_{S({\mathcal{W}}_{T})})_{U({\mathcal{W}}_{T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}}({\mathcal{P}}))}.$

To clarify the ideas:

  * •

We obtain the $s$-boundary refinement of ${\mathcal{R}}$ along the periodic
codes ${\mathcal{W}}_{T}$, to get another Markov partition
${\mathcal{R}}_{S({\mathcal{W}}_{T})}$ with geometric type
$T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}$.

  * •

${\mathcal{W}}_{T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}}({\mathcal{P}}))$ is a
family of periodic codes in
$\Sigma_{A(T_{S({\mathcal{W}}_{T}({\mathcal{P}}))})}$. We obtain the
$u$-boundary refinament of ${\mathcal{R}}_{S({\mathcal{W}}_{T})}$ along this
family of peridoic codes. To obtain:

$({\mathcal{R}}_{S({\mathcal{W}}_{T})})_{U({\mathcal{W}}_{T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}}({\mathcal{P}}))}.$

as was defined.

This definition have total seance from a formal point but we is necessarily
that $({\mathcal{R}}_{S({\mathcal{W}}_{T})})$ be commutable in therms of the
geometric type $T$ and the family of periodic codes ${\mathcal{W}}_{T}$ to be
interesting. We have develop a effective approach to compute the $s$ and $u$
boundary refinements of a Markov partition and their respective geometric
type, by applied such process to the respective Markov partitions we should
obtain the boundary refinement an its geometric type. The only difficulty is
loved a effective computation of the family codes of
${\mathcal{W}}_{T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}}({\mathcal{P}})$ in
therms of $T$.

Lets to recall that $T\neq T_{S({\mathcal{W}}_{T})}$ at leas the family just
contains $s$-boundary codes. Therefore the shift space $\Sigma_{A(T)}$ is
different that $\Sigma_{A(T_{S({\mathcal{W}}_{T})})}$. The objective of this
section is solve the following problem:

######  Problem 5.1.

Let $T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}$ be the geometric type of the
$s$-boundary refinement of $T$ a long the family
${\mathcal{W}}_{T}({\mathcal{P}})$ of periodic codes in $\Sigma_{A(T)}$.
Compute every code in the family
${\mathcal{W}}_{T_{S({\mathcal{W}}_{T}({\mathcal{P}}))}}({\mathcal{P}})$, in
therms of the geometric type $T$ and ${\mathcal{W}}_{T}({\mathcal{P}})$.

####  5.5.1. From a code in $\Sigma_{A(T)}$ to a pair of codes in
$\Sigma_{A(T_{S({\mathcal{W}})})}$

We are going to consider the next simplified problem:

######  Problem 5.2.

Let $\\{p_{m}\\}_{m=1}^{m}$ a family of periodic points of $f$ that are no $s$
or $u$-boundary codes. Let ${\mathcal{W}}:=\cup_{m=1}^{m}\pi_{f}^{-1}(p_{i})$
and let ${\mathcal{W}}_{s}:=\cup_{m=1}^{m}\pi_{f_{s}}^{-1}(p_{i})$. Construct
any element of ${\mathcal{W}}_{s}$ in therms of elements in ${\mathcal{W}}$.

Let $\underline{w}\in{\mathcal{W}}$ lets see how the boundary refinement split
it. Lets to take:

  1. (1)

The period of $\underline{w}$ is $P$.

  2. (2)

We can determine for all $t\in\\{0,\cdots,P-1\\}$ the following parameters:
$s_{t}:=\underline{s}(t,\underline{w})$,
$\tilde{r}(w_{t},\underline{s}(t,\underline{w}))$ and
$\epsilon_{t}:=\epsilon_{t}(w_{t},j_{t,\underline{w}})$.

Now we are going to construct the two codes in
$\Sigma_{A(T_{S({\mathcal{W}})})}$ that project to $p:=\pi_{f}(\underline{w})$
using a recursive process:

  * •

The point $p$ is in the stable boundary of two adjacent rectangles of
${\mathcal{R}}_{S({\mathcal{W}})}$
$\tilde{R}_{\tilde{r}(w_{0},\underline{s}(0,\underline{w}))}$ and
$\tilde{R}_{\tilde{r}(w_{0},\underline{s}(0,\underline{w})+1)}$. Clearly
$\tilde{R}_{\tilde{r}(w_{0},\underline{s}(0,\underline{w}))}$ is below
$\tilde{R}_{\tilde{r}(w_{0},\underline{s}(0,\underline{w})+1)}$. Depending on
this choose we are going to construct the different codes. In this
construction we are going to take the inferior rectangle

$v_{0}:=\tilde{R}_{\tilde{r}(w_{0},\underline{s}(0,\underline{w}))}$

which means that we need to look for the orbit of the any sector $e$ of $p$
that point toward the bottom of $R_{w_{0}}$.

  * •

Now the stable segment $f(I_{0,\underline{w}})\subset I_{1,\underline{w}}$ and
$I_{1,\underline{w}}$ is the stable boundary of two adjacent rectangles of
${\mathcal{R}}_{S({\mathcal{W}})}$ they are
$\tilde{R}_{\tilde{r}(w_{1},\underline{s}(1,\underline{w}))}$ and
$\tilde{R}_{\tilde{r}(w_{1},\underline{s}(1,\underline{w})+1)}$. To determine
the index $v_{1}$ we need to know if $f$ change or not the vertical direction
of our sector $e$, i.e. we need to look at $\epsilon_{0}$, in this manner:

    * –

If $\epsilon_{0}=1$ we take $v_{1}=\tilde{r}(w_{1},\underline{s}))$ because
$f(e)$ point towards the bottom of $R_{w_{1}}$.

    * –

If $\epsilon_{0}=-1$ we take $v_{1}=\tilde{r}(w_{1},\underline{s}+1))$ because
$f(e)$ point towards the top of $R_{w_{1}}$.

In any case we know the therm $v_{1}$ and the direction towards $f(e)$ points,
and this position is given by the sing og $-\epsilon_{0}$.

  * •

Now it comes our recursive law. Imagine we have determined the therm $v_{T}$
and we know the direction towards $f^{T}(e)$ points towards by use the
formula:

$e_{T}:=-\prod_{0}^{T-1}\epsilon_{t}$

then:

    * i)

If $e_{T}=-\prod_{t=0}^{T-1}\epsilon_{t}=-1$, $f^{T}(e)$ points toward the
bottom of $R_{w_{T}}$ and :

      * a)

If $\epsilon_{T}=1$,
$v_{t+1}=\tilde{r}(w_{T+1},\underline{s}(T+1,\underline{w}))$ and
$-\prod_{t=0}^{T-1}\epsilon_{t}=-1$.

      * b)

If $\epsilon_{T}=1$,
$v_{t+1}=\tilde{r}(w_{T+1},\underline{s}(T+1,\underline{w})+1)$ and
$-\prod_{t=0}^{T-1}\epsilon_{t}=1$.

    * i)

If $e_{T}=-\prod_{t=0}^{T-1}\epsilon_{t}=1$, $f^{T}(e)$ points toward the top
of $R_{w_{T}}$ and :

      * a)

If $\epsilon_{T}=1$,
$v_{t+1}=\tilde{r}(w_{T+1},\underline{s}(T+1,\underline{w})+1)$ and
$-\prod_{t=0}^{T-1}\epsilon_{t}=1$.

      * b)

If $\epsilon_{T}=1$,
$v_{t+1}=\tilde{r}(w_{T+1},\underline{s}(T+1,\underline{w}))$ and
$-\prod_{t=0}^{T-1}\epsilon_{t}=-1$.

  * •

This processes end after $P$ or $2P$ iterations depending on the change of the
vertical orientation of $f^{P}$ in the unstable manifold of $p$. If end after
$2P$ iterations we obtain the two different codes of
$\Sigma_{A(T_{S({\mathcal{W}})})}$ that project to $p$ at the same time. If
not we need to incite the process assuming that the sector $e$ point towards
the top of $R_{w_{0}}$ and take the upper rectangle

$v_{0}:=\tilde{R}_{\tilde{r}(w_{0},\underline{s}(0,\underline{w})+1)}.$

###  5.6. The genus and the geometric type

Along this section, let
$T=\\{n,\\{h_{i},v_{i}\\}_{i=1}^{n},\Phi_{T}:=\\{\rho_{T},\epsilon_{T}\\}\\}$
be a geometric type in the pseudo-Anosov class, assume that $f:S\to S$ is a
generalized pseudo-Anosov homeomorphism with a geometric Markov partition
${\mathcal{R}}$ of geometric type $T$. We must determine the genus of the
surface $S$, gen$(S)$, using the _Euler-Poincaré formula_ ([FM11, Proposition
11.4]):

######  Proposition 5.5.

Let $S$ be a closed and oriented surface with a singular foliation
${\mathcal{F}}$. Let $P_{s}$ be the number of prongs at a singular point
$p_{s}$. Then

$2-2\text{gen}(S)=\frac{1}{2}\sum_{p_{s}\in\text{Sing}({\mathcal{F}})}(2-P_{s}).$

If we are able to compute the number of singularities of $\mathcal{F}^{s,u}$
and the number of prongs in each of them, we have essentially computed the
genus, as we can take the sum over all of the singularities and apply the
Euler-Poincaré formula.

The set of $s$-_boundary periodic codes_ of $T$,
Per$(\underline{{\mathcal{S}}(T)}$, was introduced in 3.16 and Proposition 3.8
establishes that each such periodic code, $\underline{w}(i,\epsilon)$,
projects into $\partial^{s}_{\epsilon}R_{i}$ and the positive part is given by
$I^{+}(i,\epsilon)$ (Definition 3.14). Similarly, the $u$-generating funtion
$\Upsilon(T)$ (3.13) lets us determine the negative codes $J^{-}(i,\epsilon)$
for each unstable boundary component of ${\mathcal{R}}$.

After a refinement in the geometric type, we can assume that the incidence
matrix of $T$, $A(T)$, is binary. In Chapter3 , we use the $s$-_generating
function_ $\Gamma(T)$ (Definition 3.12) to produce the $s$-_boundary codes_ in
the sub-shift of finite type, $(\Sigma_{A(T)},\sigma_{A(T)})$, associated with
$T$ (Definition 3.6). These codes are projected by
$\pi_{(f,{\mathcal{R}})}:\Sigma_{A(T)}\to S$ (Definition 3.7) into the stable
boundary components of the Markov partition $\mathcal{R}$.

If we think for a while, any $k$-prong of the stable foliation
${\mathcal{F}}^{s}$ is always in the boundary of ${\mathcal{R}}$ and if $T$
have the corner property, the number $k$ of prong in each singularity is equal
to the number of the different periodic boundary codes in $\Sigma_{T}$ that
project into $p$. This discussion suggest the following proposition.

######  Proposition 5.6.

Let $T$ be a geometric type in the pseudo-Anosov class whose incidence matrix
$A(T)$ is binary and has the corner property. Let $f:S\to S$ be a pseudo-
Anosov homeomorphism with a geometric Markov partition $\mathcal{R}$ of
geometric type $T$. Let $(\Sigma_{A(T)},\sigma_{A(T)})$ be the sub-shift
associated with $T$, and let $\pi=\pi_{(f,\mathcal{R})}:\Sigma_{A(T)}\to S$ be
the natural projection, then the following statements holds:

  * i)

For each boundary label (3.11) of $T$ it is possible to determine by iterating
$\Gamma(T)$ at most $2n$ times if the corresponding stable boundary
$\partial_{\epsilon}^{s}R_{i}$ of $\mathcal{R}$ is either periodic or not. In
the periodic case, the positive periodic expression is given by the positive
code, $I^{+}(i,\epsilon)$, and has a period less than or equal to $2n$.
Therefore the set of _periodic and positive_ $s$-boundary codes of $T$ is
given by:

$\underline{{\mathcal{S}}}^{+}(T)=\\{I^{+}(i_{\iota},\epsilon_{\iota})\\}_{\iota=1}^{\textbf{P}(T)},$

Where $1\leq\textbf{P}(T)\leq 2n$ is equal to the number of periodic stable
boundary positive codes of $T$, the index $i_{\iota}\in\\{1,\cdots,n\\}$ may
appear in at most $2$ pairs but if $i_{\iota}=i_{\iota^{\prime}}$ then
$\epsilon_{\iota}\neq\epsilon_{\iota^{\prime}}$.

  * ii)

A similar statement holds for the set of _periodic and negative_ $u$-boundary
codes of $T$ by applying the $u$-generating funtion $\Upsilon(T)$ at most $2n$
times. Moreover, the set of _periodic and negative_ $u$-boundary codes of $T$
is given by:

$\underline{{\mathcal{U}}}^{-}(T)=\\{J^{-}(i_{\iota},\epsilon_{\iota}^{\prime}\\}_{\iota=1}^{\textbf{P}(T)},$

where the number $\textbf{P}(T)$ and the indexes
$i_{\iota}\in\\{1,\cdots,n\\}$ are the same than those given in item $i)$ for
the set $\underline{{\mathcal{S}}}^{+}(T)$, but in this case
$\epsilon_{\iota}^{\prime}$ may be different from $\epsilon_{\iota}$.

  * iii)

The set of _periodic boundary codes_ of $T$, denoted by
Per$(\underline{{\mathcal{B}}(T)})$, is given by such periodic codes
$\underline{p}\in\text{Per}(\Sigma_{A(T)})$ for whose there exist
$\iota\in\\{1,\cdots,\textbf{P}(T)\\}$ and such that:

(5.60) $\underline{p}_{+}=I^{+}(i_{\iota},\epsilon_{\iota})\text{ and
}\underline{p}_{-}=J^{-}(i_{\iota},\epsilon_{\iota}^{\prime})$

In this case we label such periodic periodic code as
$\underline{p}_{(}\iota,\epsilon 8_{\iota})$ keeping the label of its positive
part. In this manner:

$\text{ Per
}(\underline{{\mathcal{B}}(T)})=\\{\underline{p}(\iota,\epsilon_{\iota})\\}_{\iota}^{\textbf{P}(T)}$

###### Proof.

Item $i)$ : For each _boundary label_ $(i,\epsilon)$, the procedure to
construct its positive boundary code $\underline{I}_{+}(i,\epsilon)$ as given
in Definition 3.14, determine the image by $f$ of the horizontal boundary,
$\partial^{s}_{\epsilon}R_{i}$ and the image of $\partial^{s}_{\epsilon}R_{i}$
can be in at most $2n$ different boundary components of ${\mathcal{R}}$ before
come back to itself, in this manner after at most $2n$ iteration of
$\Gamma(T)$ we can determine if $\partial^{s}_{\epsilon}R_{i}$ is or not
periodic, while we can recover the periodic expression of
$\underline{I}^{+}(i,\epsilon)$ by look the first $2n$ iteration of
$\Gamma(T)$.

The number of periodic boundary codes of $T$, P$(T)$, is well determined and
is less or equal than $2n$ that is the number of different of boundary codes
of $T$.

If the code $I^{+}(i,\epsilon)$ is periodic the rectangle $R_{i}$ have a
periodic point $p$ in the stable boundary component
$\partial^{s}_{\epsilon}R_{i}$, but for the corner property of
${\mathcal{R}}$, $p$ is a corner point of $R_{i}$. The rectangle $R_{i}$ cant
have two different periodic corner points, but $R_{i}$ can have $p$ as corner
point whit two different quadrants contained in $R_{i}$, is this situation
which force that the periodic codes generated by the $s$-boundary labels,
$I^{+}(i,1)$ and $I^{+}(i,-1)$ be different.

Items $ii)$ : The arguments are the same than in item $ii)$, but we recall
that, by the corner property every periodic boundary point $p$ is in the
corner of each rectangle that contain it. In this manner if
$I^{+}(i_{\iota},\epsilon_{\iota})$ is periodic we can deduce that the
periodic point $p$ in $\partial^{s}_{\epsilon_{\iota}}R_{\iota}$ it is in
$\partial^{s}_{\epsilon_{\iota}^{\prime}}R_{\iota}$ where
$\epsilon_{\iota}^{\prime}$ is such that
$J^{-}(i_{\iota},\epsilon_{\iota}^{\prime})$ is periodic with the same
periodic expression than $I^{+}(i_{\iota},\epsilon_{\iota})$.

Items $iii)$ : As the code $\underline{p}$ is periodic the positive part and
the negative part $\underline{w}$ must have the same periodic expression, and
this determine the $(i_{\iota},\epsilon_{\iota})$ and
$(i_{\iota},\epsilon_{\iota}^{\prime})$ in the proposition. The rest of the
notation is coherent and natural.

∎

The equivalence relation $\sim_{T}$ was introduced in 3.23, when is applied to
the set of periodic boundary codes, each equivalence class,
$[\underline{p}(i_{\iota},\epsilon_{\iota}]_{T}$, is in correspondence with a
unique periodic boundary point by the natural projection
$\pi_{(f,{\mathcal{R}})}$, there fore we can determine the number of periodic
boundary points by cont the number of such equivalence classes:

$\text{ Per
}(\underline{{\mathcal{B}}(T)})/\sim_{T}=\\{[\underline{p}_{s}]\\}_{s=1}^{\textbf{(}S)(T)}$

where $\textbf{(}S)(T)\in{\mathbb{N}}$ is the number of $\sim_{T}$-equivalence
classes, the construction of $\sim_{T}$ permits to determine in finite time
such number and the elements of each conjugacy class, in this manner we pose
the following notation:

$\underline{P}_{s}:=\\#[\underline{p}_{s}]_{T}$

be the number of periodic codes in the $\sim_{T}$ equivalence class of
$\underline{p}_{s}$.

######  Lemma 5.52.

Assume that $T$ has the corner property and the incidence matrix of $T$ is
binary. Let $\underline{w}$ be a code in $[\underline{p}_{s}]_{T}$ and let
$p_{s}:=\pi_{f,{\mathcal{R}}}(\underline{w})\in S$, be the corresponding
periodic boundary point of $(f,{\mathcal{R}})$. The point $p_{s}$ is a
$P_{s}$-prongs where, $P_{s}=\frac{1}{2}\underline{P}_{s}$.

###### Proof.

Like $T$ have the corner property, then the number $\underline{P}_{s}$ is
equal to the number of sectors of the point $p_{s}$. Now, consider that the
number of prongs of $p_{s}$ is the half of the number of its sectors to obtain
the equality. ∎

######  Theorem 5.1.

Let $T$ a geometric type in the pseudo-Anosov class and let $f:S\to S$ a
generalized pseudo-Anosov homeomorphism that realize $T$. If the $T$ have the
corner property and its incidence matrix is binary, the Euler-Characteristic
of $S$ is given by the formula:

$\chi(S)=\frac{1}{2}\sum_{s\in\text{Sing}({\mathcal{F}})}(2-P_{s})=\sum_{s=1}^{\textbf{S}}(2-\frac{1}{2}\underline{P}_{s}.$

###### Proof.

The only thing to observe is that, ff $p_{s}$ is a regular (boundary) point
$2-P_{s}=0$ and $2-\frac{1}{2}\underline{P}_{s}=0$. Therefore we don’t add
anything to the sum in the left side, by consider regular points. ∎

######  Corollary 5.16.

Let $T$ be a geometric type of the pseudo-Anosov class, and let
$f:S\rightarrow S$ be a generalized pseudo-Anosov homeomorphism with a
geometric Markov partition of type $T$. There exists a finite algorithm that,
given $T$, determines the number of singularities of $f$ and the number of
stable and unstable separatrices at each singularity and subsequently the
genus of $S$.

##  Chapter 6 The formal derived from Anosov of a geometric type

###  6.1. The formal DA of a geometric type.

Let $T$ be a geometric type in the pseudo-Anosov class. Our previous results
showed the existence of a closed surface $S$ and a Smale diffeomorphism
$\phi:S\rightarrow S$, that have a mixing saddle-type basic piece $K(T)$ which
has a Markov partition of geometric type $T$. Therefore, we can define
$\Delta(T):=\Delta(K(T))\subset S$ as the domain of $K(T)$. Moreover, the
diffeomorphism $\phi$ uniquely determines (up to conjugation) a diffeomorphism
in the domain $\phi_{T}:\Delta(T)\rightarrow\Delta(T)$. This justify the
following definition.

######  Definition 6.1.

Let $T$ be a geometric type of the pseudo-Anosov class. Let $S$ be a closed
surface and $\phi:S\rightarrow S$ be a Smale surface diffeomorphism having a
mixing saddle-type basic piece $K(T)$ with a geometric Markov partition of
geometric type $T$, let $\Delta(T)\subset S$ be the domain of the basic piece
$K(T)$. We define the _formal derived from Anosov_ of $T$ to be the triplet

$\textbf{DA}(T):=(\Delta(T),K(T),\phi_{T}).$

where $\phi_{T}:\Delta(T)\to\Delta(T)$ is the restriction of the
diffeomorphism $\phi$ restricted to the domain of $K(T)$.

The following proposition is consequence of the Bonatti-Langevin results that
were presented in the Chapter 1.5.1 of Preliminaries.

######  Proposition 6.1.

Let $T$ be a geometric type in the pseudo-Anosov class. Up to a topological
conjugacy in their respective domains, that is the identity in the non-
boundary periodic points of $K(T)$, there exists a unique DA$(T)$.

Let $T$ be a geometric type in the pseudo-Anosov class. Consider a generalized
pseudo-Anosov homeomorphism $f$ with a geometric Markov partition
${\mathcal{R}}$ whose geometric type is $T$. The concept of formal derived
from Anosov is rooted in the classical idea of considering all the periodic
points in the boundary ${\mathcal{R}}$ and open the stable and unstable
manifolds of these periodic points. Then a saddle-type basic piece without
singularities obtained. In this process, each periodic point on the boundary
of the Markov ${\mathcal{R}}$ partition becomes a corner point of the basic
piece. However, with our definition of DA$(T)$, what happens is that, the
periodic points on the stable boundary of ${\mathcal{R}}$ become $s$-boundary
points of the basic piece $K(T)$ while the periodic points on the unstable
boundary of ${\mathcal{R}}$ become $u$-boundary points of $K(T)$. The need to
open all boundary points of ${\mathcal{R}}$ in the stable and unstable
direction is what has motivated us to introduce the corner refinement of $T$.

When comparing two saddle-type basic pieces obtained as the formal DA of two
different geometric types in the pseudo-Anosov class, the first condition for
these basic pieces to be topologically conjugate is that they must possess an
equal number of periodic $s$ and $u$-boundary points. To achieve this, we
introduce the concept of a _joint refinement_ of two geometric Markov
partition and its associated geometric type. This process allows us to
transform two basic pieces originating from different geometric types into
ones with the same number of corner points. We will delve into the details of
this construction in the upcoming section.

###  6.2. The joint refinement of two geometric types.

Let to recall that
${\mathcal{W}}_{T,P}:=\\{\underline{w}\in\Sigma_{A(T)}:\textbf{Per}(\sigma_{T},\underline{w})\leq
P\\}$.

######  Definition 6.2.

Let $f$ and $g$ be pseudo-Anosov homeomorphisms with geometric Markov
partitions ${\mathcal{R}}_{f}$ and ${\mathcal{R}}_{g}$ whose geometric types
$T_{f}$ and $T_{g}$ have binary incidence matrices. Let
$P_{\underline{B(T_{f})}}$ be the maximum period among the boundary periodic
codes of $T_{f}$:

$P_{\underline{B(T_{f})}}:=\max\\{\textbf{Per}(\underline{w},\sigma_{T_{f}}):\underline{w}\in\underline{B(T_{f})}\\}$

and similarly $P_{\underline{B(T_{g})}}$ be the maximum period among the
boundary periodic codes of $T_{g}$. Lets take define
$N(f,g):=\max\\{P_{\underline{B(T_{f})}},P_{\underline{B(T_{g})}}\\}$.
Consider:

  1. (1)

The _joint refinement_ (Definition 5.52) of ${\mathcal{R}}_{f}$ with respect
to ${\mathcal{R}}_{g}$ is the corner refinement of ${\mathcal{R}}_{f}$ with
respect to the family of periodic codes
${\mathcal{W}}_{T_{f},N(f,g)}\subset\Sigma_{A(T_{f})}$:

${\mathcal{R}}_{(R_{f},R_{g})}:=[{\mathcal{R}}_{f}]_{C({\mathcal{W}}_{T_{f},N(f,g)})}.$

Its geometric type is denoted by $T_{(T_{f},T_{g})}$.

  2. (2)

The _joint refinement_ of ${\mathcal{R}}_{g}$ with respect to
${\mathcal{R}}_{f}$ is the corner refinement of ${\mathcal{R}}_{g}$ with
respect to the family of boundary periodic codes
${\mathcal{W}}(g,N(f,g))\subset\Sigma_{A(T_{g})}$:

${\mathcal{R}}_{({\mathcal{R}}_{g},{\mathcal{R}}_{f})}:=[{\mathcal{R}}_{g}]_{C({\mathcal{W}}_{g,N(f,g)})}.$

Its geometric type is denoted $T_{(T_{g},T_{f})}$.

######  Remark 6.1.

By construction, the geometric Markov partition
${\mathcal{R}}_{(R_{f},R_{g})}$ has the corner property and all its boundary
points are corners of any rectangle that contains them. The number $N(f,g)$ is
equal to $N(g,f)$, and the corner points of ${\mathcal{R}}_{(R_{f},R_{g})}$
coincide with the set of periodic points of $f$ with periods less than or
equal to $N(f,g)$. Similarly, for
${\mathcal{R}}_{({\mathcal{R}}_{g},{\mathcal{R}}_{f})}$,

###  6.3. The lifting of a geometric Markov partition of $f$ to the formal
DA$(T)$.

The important property of the joint refinement it that every periodic boundary
point is a corner point of any rectangle that contain such point. We must use
this property to prove the following Proposition.

Notation: Once we have fixed ${\mathcal{R}}_{f}$ and ${\mathcal{R}}_{g}$ to
avoid an overwhelming notation, we make the following conventions:

(6.1)
$\displaystyle{\mathcal{R}}_{f,g}:={\mathcal{R}}_{({\mathcal{R}}_{f},{\mathcal{R}}_{g})}\text{
and
}{\mathcal{R}}_{g,f}:={\mathcal{R}}_{({\mathcal{R}}_{g},{\mathcal{R}}_{f})},$
(6.2) $\displaystyle T_{f,g}:=T_{(T_{f},T_{g})}\text{ and
}T_{g,f}:=T_{(T_{g},T_{f})}.$

The formal derived from pseudo-Anosov of such geometric types are:

(6.3)
$\displaystyle\textbf{DA}(T_{f,g}):=\\{\Delta(T_{f,g}),K(T_{f,g}),\Phi_{T_{f,g}})\\}\text{
and }$ (6.4)
$\displaystyle\textbf{DA}(T_{g,f}):=\\{\Delta(T_{g,f}),K(T_{g,f}),\Phi_{T_{g,f}})\\}$

######  Proposition 6.2.

Let $f:S_{f}\rightarrow S_{f}$ and $g:S_{g}\rightarrow S_{g}$ be two
generalized pseudo-Anosov homeomorphisms with geometric Markov partitions,
${\mathcal{R}}_{f}$ and ${\mathcal{R}}_{g}$, whose geometric types, $T_{f}$
and $T_{g}$, have binary incidence matrix. Suppose that $f$ and $g$ are
topologically conjugate through an orientation preserving homeomorphism
$h:S_{f}\rightarrow S_{g}$ and let:

$\pi:\Delta(T_{g,f})\rightarrow S_{g},$

be the projection given by Proposition 4.2. In this situation:

  * (1)

The induced geometric partition, $h({\mathcal{R}}_{f,g})$ is a geometric
Markov partition of $g$ whose geometric type $T_{f,g}$ have the corner
property.

  * 2)

A point $p\in S_{f}$ is a periodic boundary point of the Markov partition
$(f,{\mathcal{R}}_{f,g})$ if and only if $h(p)$ is a periodic boundary point
of $(g,{\mathcal{R}}_{g,f})$. In particular, the corner periodic points of
$h({\mathcal{R}}_{f,g})$ are the corner periodic points of
${\mathcal{R}}_{{\mathcal{R}}_{g,f}}$.

  * (3)

There exit a unique Markov partition
$\underline{h}({\mathcal{R}}_{f,g})=\\{\underline{h}(R_{i})\\}_{i=1}^{n}$ of
$K(T_{g,f})\subset\Delta(T_{(g,f)}))$ such that:

$\pi\left(\overset{o}{\underline{h}}(R_{i})\right)=\overset{o}{(h(R_{i}))}.$

  * (4)

Keep the previous label of the rectangles in
$\underline{h}({\mathcal{R}}_{f,g})$ and give to the rectangle
$\underline{h}({\mathcal{R}}_{f,g})$ the unique vertical and horizontal
orientation such that $\pi$ is increasing along both the vertical and
horizontal leaves within $\underline{h}(R_{i})$. With this geometrization, the
geometric type of $\underline{h}({\mathcal{R}})$ is $T_{f,g}$.

###### Proof.

Item $(1)$. In Definition 2.10, we introduced the induced geometric Markov
partition $h({\mathcal{R}}_{f,g})$. Theorem2.2 establishes that
$h({\mathcal{R}}_{f,g})$ is a geometric Markov partition with the same
geometric type as ${\mathcal{R}}_{f,g}$, specifically $T_{f,g}$. The joint
refinement was obtained as the corner refinement of another geometric type and
Proposition 5.4 implies that $T_{f,g}$ has the corner property.

Item $(2)$. By definition of the joint refinement, a periodic point $p$ of $f$
is a corner point of ${\mathcal{R}}_{f,g}$ if and only if it is a period is
less than or equal to $N(f,g)$. But $h(p)$ has the same period that $p$,
therefore $h(p)$ is a periodic point of $g$ with period less than or equal to
$N(g,f)=N(f,g)$, then $h(p)$ is corner point of ${\mathcal{R}}_{g,f}$, by
definition it is a boundary point. This proves the second item of the
proposition.

Item $(3)$. We are going to prove the second statement through a series of
lemmas. The reader can refer to Figure 6.1 for a visual representation of our
notation.

Figure 6.1. The lift of a rectangle $R$.

######  Lemma 6.1.

Let $R$ be a rectangle in the Markov partition $h({\mathcal{R}}_{f,g})$. Then
$\pi^{-1}(R)$ contains a unique rectangle
$\underline{R}\subset\Delta(T_{g,f})$ with the following properties:

  * •

The interior of $\underline{R}$ is mapped under $\pi$ to the interior of $R$,
i.e., $\pi(\overset{o}{\underline{R}})=\overset{o}{R}$, and the image of
$\underline{R}$ is the rectangle $R$, i.e., $\pi(\underline{R})=R$.

  * •

The horizontal boundary of $\underline{R}$ is contained in a stable
separatrice of an $s$-boundary periodic point of $K(T_{g,f})$, and its
vertical boundary is in an unstable separatrice of an $u$-boundary point.

  * •

The rectangle $\underline{R}$ is isolated from the basic piece $K(T_{g,f})$
from its exterior, i.e., there is an open neighborhood $U$ of $\underline{R}$
such that $K(T_{g,f})\cap U=K(T_{g,f})\cap\underline{R}$.

###### Proof.

Denote the left side of $R$ as $J^{-}$, its right side as $J^{+}$, its lower
boundary as $I^{-}$, and its upper boundary as $I^{+}$. All these intervals
are contained within the invariant manifold of a corner periodic point of
$h({\mathcal{R}}_{f,g})$. Moreover, the boundary points of
$h({\mathcal{R}}_{f,g})$ are all corner points and as proven in Item $(2)$,
they coincide with the corner points of ${\mathcal{R}}_{g,f}$. This implies
that neither $J^{\pm}$ nor $I^{\pm}$ contain a periodic point in their
interior; therefore, they are contained within a unique separatrice of a
corner point. Now, let’s analyze the set $\pi^{-1}(I^{-})$ to get some
conclusions that can be applied to the remaining intervals $J^{+},J^{-},I^{+}$
in the boundary of $R$.

Assume that $I^{-}$ is contained in the stable separatrice $F^{s}(p)$ of the
corner point $p$. Inside the domain $\textbf{DA}(T_{g,f})$ there are two
$s$-boundary periodic points of $K(T_{g,f})$, $\underline{p}_{1}$ and
$\underline{p}_{2}$ with different stable separatrices
$\underline{F}^{s}(\underline{p}_{1})$ and
$\underline{F}^{s}(\underline{p}_{2})$, such that:
$\pi^{-1}(\overset{o}{I^{-}})$ has two connected components that are are
contained in such separatrices:

$\underline{\overset{o}{I^{\prime}}}_{1}\subset F^{s}(\underline{p}_{1})\text{
and }\underline{\overset{o}{I^{\prime}}}_{2}\subset F^{s}(\underline{p}_{2}).$

and
$\pi(\underline{\overset{o}{I^{\prime}}}_{1})=\pi(\underline{\overset{o}{I^{\prime}}}_{2})=\overset{o}{I^{-}}$.
Such stable intervals are contained in the stable boundary of $s$-boundary
periodic point, therefore only one of these intervals have points that can be
approximated by singletons (Item $i)$ in Definition 4.4 ) of the form
$\pi^{-1}(x_{n})\in K(T_{g,f})$, where $x_{n}\in\overset{o}{R}$ and $x_{n}$
converge to a point in $I^{-}$. Assume that
$\overset{o}{\underline{I^{\prime}}_{1}}$ is such an interval.

Let’s understated the closure of $\overset{o}{\underline{I^{\prime}}_{1}}$
inside the stable separatrice $\underline{F}^{s}(p_{1})$. We need to consider
two different of extreme points in $I^{-}$:

  * i)

The interval $I^{-}$ has a periodic point $p$ as one of its endpoints. In this
case, $\pi^{-1}(p)$ is contained in a cycle. The endpoint of
$\underline{I^{\prime}}^{-}=\overline{\overset{o}{\underline{I^{\prime}}_{1}}}$
which projects to $p$ needs to be contained in such a cycle. However, at the
same time, such a point can be approximated by points in
$K(T_{g,f})\cap\overset{o}{\underline{I^{\prime}}_{1}}$ and is, therefore, an
element of $K(T_{g,f})$. The unique hyperbolic points in the cycle are
periodic points, so the endpoint of $\underline{I^{\prime}}^{-}$ that projects
to $p$ is the periodic point $\underline{p}_{1}$, which is approximated by
points in $\underline{F}^{s}(\underline{p}_{1})$.

  * ii)

The endpoint $O\in I^{-}$ is not periodic. In this case, $\pi^{-1}(O)$ is a
minimal rectangle. Similar to before, the endpoint of
$\underline{I^{\prime}}^{-}=\overline{\overset{o}{\underline{I^{\prime}}_{1}}}$
which projects to $O$ needs to be contained in such a minimal rectangle and be
an element of the basic piece. Therefore, it is the unique corner point
$\underline{O}$, in the minimal rectangle that is saturated by points in
$K(T_{g,f})\cap\pi^{-1}(\overset{o}{\underline{I^{\prime}}_{1}})$.

In this setting: If $I^{-}$ have a periodic end point:

$\underline{I}^{-}:=[\underline{O},\underline{p}_{1}]^{s}\subset
F^{s}(\underline{p}_{1}).$

If $I^{-}=[O_{1},O_{2}]^{s}$ have no periodic end points, we can applied the
analysis in Item $i)$ to construct point $\underline{O_{1}}$ and
$\underline{O_{2}}$ such that:

$\underline{I}^{-}:=[\underline{O_{1}},\underline{O_{1}}]^{s}\subset
F^{s}(\underline{p}_{1}).$

######  Remark 6.2.

The extreme points of $\underline{I}^{-}$ are, in any case, $u$-boundary
points. Therefore, $\underline{I}^{-}$ is contained within an open interval
$\underline{I}_{1}^{-}\subset\underline{F}^{s}(\underline{p}_{1})$ such that
$K(T_{g,f})\cap\underline{I}_{1}^{-}=K(T_{g,f})\cap\underline{I}^{-}$ and
$\pi(\underline{I^{\prime}}_{1}^{-})=I^{-}$. Furthermore, $\underline{I}^{-}$
is the smallest interval contained in $\underline{F}^{s}(\underline{p}_{1})$
that projects to $I^{\prime}$. These two properties determine it uniquely.

The same construction applies to the other boundary components. That is, there
are unique periodic corner points $\underline{p}_{2}$, $\underline{q}_{1}$,
and $\underline{q}_{2}$ of the basic piece $K(T_{g,f})$ and unique
separatrices of such points that contain minimal intervals:

$\displaystyle\underline{I}^{+}\subset\underline{F}^{s}(\underline{p}_{2})$
$\displaystyle\underline{J}^{-}\subset\underline{F}^{s}(\underline{q}_{1})$
$\displaystyle\underline{J}^{+}\subset\underline{F}^{s}(\underline{q}_{2})$

These intervals project into the respective boundary components of $R$. The
stable separatrice $\underline{F}^{s}(\underline{p}_{2})$ is the only stable
separatrice that is saturated by stable leaves of non $s$ boundary periodic
points that project to $R$. Clearly, these intervals are unique, as they are
minimal with these properties.

We claim that $|\underline{I}^{-}\cap\underline{J}^{-}|=1$. Let $O$ be the
corner of $R$ determined by $I^{-}$ and $J^{-}$, and let $e$ be the sector of
$O$ that is determined by the left-inferior corner of $R$. Then $\pi^{-1}(O)$
could be:

  * •

Part of a cycle if $O$ is a periodic point. In this cycle, all the boundary
points are corner points and has only one sector saturated by points in
$K(T_{g,f})$. Let $\\{x_{n}\\}\subset\overset{o}{R}$ be a sequence on non-
boundary periodic codes that converges to $0$ in the sector $e$, then
$\pi^{-1}(x_{n})$ is a singleton and then: $\lim_{n\to\infty}\pi^{-1}(x_{n})$
is simultaneously the periodic endpoint of $\underline{I}^{-}$ and the
periodic endpoint of $\underline{J}^{-}$, i.e
$\underline{p}_{1}=\underline{q}_{1}$.

  * •

A minimal rectangle if $O$ is not periodic. In this case, there is only one
corner of the minimal rectangle where $\underline{I}_{-}$ and
$\underline{J}^{-}$ intersect. Such is the only one accumulated by singletons
in $\pi^{-1}(\overset{o}{R})$ that converge to $\underline{O}$.

The same property follows for the other vertices of $R$. Furthermore, since
the basic piece $K(T_{g,s})$ doesn’t contain impasses, we deduce that
$\underline{I}^{-}$ is in a different stable separatrice than
$\underline{I}^{+}$, and then
$\underline{I}^{-}\cap\underline{I}^{+}=\emptyset$. In conclusion,
$\underline{I}^{-}\cup\underline{I}^{+}\cup\underline{J}^{-}\cup\underline{J}^{+}$
forms a closed curve. We shall see what happens with the interior of
$\tilde{R}$.

Take any point $x\in\overset{o}{R}$. The set $\pi^{-1}(x)$ is not part of a
cycle because the only $s$ and $u$ boundary points of $K(T_{g,f})$ are
projected to corner points of the original Markov partition
$h({\mathcal{R}}_{f,g})$. Therefore, $\pi^{-1}(x)$ is a rectangle and is
trivially bi-foliated. We can deduce that if
$\overset{o}{J}\subset\overset{o}{R}$ is an unstable segment, its preimage
$\pi^{-1}(\overset{o}{J})$ is a rectangle without its horizontal boundary.
Clearly, the horizontal boundary of $\pi^{-1}(\overset{o}{J})$ has one
connected component intersecting $\underline{I}^{-}$ and the other
intersecting $\underline{I}^{+}$. The same holds for any horizontal segment
$\overset{o}{I}\subset\overset{o}{R}$.

In conclusion,
$\underline{I}^{-}\cup\underline{I}^{+}\cup\underline{J}^{-}\cup\underline{J}^{+}$
bounds a region that is trivially bi-foliated, and therefore, it is a
rectangle denoted as $\underline{R}$. It’s evident that its projection is $R$,
and its interior projects to the interior of $R$.

The stable boundary of ${\mathcal{R}}$ consists of two disjoint intervals
contained in the $s$ and $u$ boundary leaves, as we observed in Remark 6.2. By
taking slightly larger intervals that contain
$\underline{I}^{-},\underline{I}^{+},\underline{J}^{-}$, and
$\underline{J}^{+}$, we can construct a small neighborhood $U$ of
$\underline{R}$ such that $K(T_{g,f})\cap U=K(T_{g,f})\cap\underline{R}$. ∎

As a consequence, for all rectangles $h(R_{i})\in h({\mathcal{R}}_{f,g})$,
there is a unique rectangle $\underline{h(R_{i})}\subset\textbf{DA}(T_{g,f})$
as described in Lemma 6.1 that projects to $h(R_{i})$. Furthermore, these
rectangles have their boundaries isolated of the basic piece from its
exterior. In particular, we have the following corollary:

######  Corollary 6.1.

If $i\neq j$, then $\underline{h(R_{i})}\cap\underline{h(R_{j})}=\emptyset$.

Let

$\underline{h}({\mathcal{R}}_{f,g}):=\\{\underline{h(R_{i})}\\}_{i=1}^{n}$

be the family of rectangles constructed in Lemma 6.1 for each rectangle in the
Markov partition $h({\mathcal{R}}_{f,g})$.

######  Lemma 6.2.

$K(T_{g,f})\subset\cup_{i=1}^{n}\underline{h(R_{i})}$

###### Proof.

The set of periodic points of $f$ that are not $(s,u)$-boundary is dense
inside every rectangle $h(R_{i})$ and is contained in its interior. Denote the
set of these periodic points by $P(i)$. Then $\pi^{-1}(P(i))$ consists of all
the periodic points of $\phi_{T_{g,f}}$ that are contained in the interior of
$\pi^{-1}(h(R_{i}))$ and there fore that are not $(s,u)$-boundary. Such points
are dense in $K(T_{g,f})\cap R_{i}$, and then:

$K(T_{g,f})\subset\cup_{i=1}^{n}\overline{\pi^{-1}(P(i))}\subset\cup_{i=1}^{n}\overline{\overset{o}{\underline{h(R_{i})}}}=\cup_{i=1}^{n}\underline{h(R_{i})}$

∎

######  Lemma 6.3.

The stable boundary of the partition $\underline{h}({\mathcal{R}}_{f,g})$ is
$\Phi_{g,f}$-invariant.

Figure 6.2. The stable boundary of the partition
$\underline{h}({\mathcal{R}}_{f,g})$

###### Proof.

Let $\underline{h(R_{i})}\in\underline{h}({\mathcal{R}}_{f,g})$ be a
rectangle. Take $\underline{I}$ as the lower boundary of such a rectangle, and
let $\underline{O_{1}}$ and $\underline{O_{2}}$ its end points, the analysis
for the upper boundary is the same.

There is a sequence
${\underline{x}_{n}}\subset\overset{o}{\underline{h(R_{i})}}$ consisting of
non $s,u$-boundary periodic points that converges to $\underline{O_{1}}$ as
shown in Figure 6.2. Similarly, there exists a sequence
$\\{\underline{y}_{n}\\}\subset\overset{o}{\underline{h(R_{i})}}$ of non
$s,u$-boundary periodic points that converges to $\underline{O_{2}}$. In these
points, $\pi$ is a one-to-one map and serves as a conjugation restricted to
their image. This implies that:

$\Phi_{g,f}(\underline{O_{1}})=\lim_{n\rightarrow\infty}\Phi_{g,f}(\underline{x}_{n})=\lim_{n\rightarrow\infty}\pi^{-1}\circ
g\circ\pi(\underline{x_{n}}).$

and

$\Phi_{g,f}(\underline{O_{2}})=\lim_{n\rightarrow\infty}\Phi_{g,f}(\underline{y}_{n})=\lim_{n\rightarrow\infty}\pi^{-1}\circ
g\circ\pi(\underline{y_{n}}).$

By taking sub sequences of $\\{\underline{x}_{n}\\}$ and
$\\{\underline{y}_{n}\\}$, we can assume that $x_{n}:=\pi(\underline{x_{n}})$
and $y_{n}:=\pi(\underline{y_{n}})$ are always contained in the interior of
the rectangle $h(R_{i}).$ Furthermore, since $h({\mathcal{R}}_{f,g})$ is a
Markov partition of $g$, we can assume that the lower horizontal sub-rectangle
$H:=H^{i}_{1}$ contained in $h(R_{i})$ of the Markov partition
$(g,{\mathcal{R}}_{f,g})$ contains to $\\{x_{n}\\}$ and $\\{y_{n}\\}$ within
its interior. Therefore, if $g(H)=V^{k}_{l}$, then $\\{g(x_{n})\\}$ and
$\\{g(y_{n})\\}$ are contained in
$\overset{o}{V^{k}_{l}}\subset\overset{o}{h(R_{k})}$ and they converge to
$g(O_{1})$ and $g(O_{2})$, which are in the stable boundary of
$\underline{h(R_{k})}$.

We claim that $\Phi_{g,f}(\underline{I})$ is a stable interval contained in a
single stable boundary component of $\underline{h(R_{k})}$. Suppose this is
not the case, then $\Phi_{g,f}(\underline{I})$ contains an $s$-arc,
$\underline{\alpha}^{\prime}=[\underline{a}^{\prime},\underline{b}^{\prime}]^{s}$
joining two corners of the rectangle $\underline{h(R_{k})}$, where
$\underline{a}^{\prime}$ is on the lower boundary of $\underline{h(R_{i})}$
and $\underline{b}^{\prime}$ is on the upper boundary of
$\underline{h(R_{i})}$ , as shown in Figure 6.3. This implies that
$\underline{I}$ contains in its interior the $s$-arc
$\Phi_{g,f}^{-1}(\underline{\alpha}^{\prime}):=\underline{\alpha}=[\underline{a},\underline{b}]^{s}$,
that we recall it is contained in the inferior boundary of
$\underline{h(R_{i})}$

Figure 6.3. $\Phi_{g,f}(\underline{I})$ is in single stable boundary component

Just as we did before for the endpoints of $\underline{I}$ we can take
decreasing successions
$\\{\underline{a}_{n}\\},\\{\underline{b}_{n}\\}\subset\overset{o}{\underline{h(R_{i})}}$
of periodic non-corner points converging to $\underline{a}$ and
$\underline{b}$ in two distinct sectors (the green and purple semicircles in
Figure 6.3). Let us define, $a_{n}:=\pi(\underline{a}_{n})$ and
$b_{n}:=\pi(\underline{b}_{n})$. They are decreasing sequences (for the
vertical order of $h(R_{i})$) that converge to the same point
$c=\pi(\underline{a})0\pi(\underline{b})\in I$. Clearly $\\{g(a_{n})\\}$ is a
decreasing (or increasing) sequence for the vertical order of $h(R_{k})$ if
and only if $\\{g(b_{n})\\}$ is a decreasing (resp. increasing) sequence, and
both converge to $g(c)$, without loss of generality suppose they are
decreasing. Thus $\pi^{-1}(g(a_{n}))$ and $\pi^{-1}(g(b_{n}))$ are decreasing
for the vertical order of $\underline{h(R_{i})}$. Moreover

$\lim\pi^{-1}(g(a_{n}))=\lim\Phi_{g,f}(\underline{a_{n}})=\phi_{g,f}(\underline{a})=\underline{a}^{\prime}$

and

$\lim\pi^{-1}(g(b_{n}))=\lim\phi_{g,f}(\underline{b_{n}})=\phi_{g,f}(\underline{b})=\underline{b}^{\prime}$

This implies that $\underline{a}^{\prime}$ and $\underline{b}^{\prime}$ are on
the lower boundary of $\underline{h(R_{i})}$ which is a contradiction. Then
$\Phi_{g,f}(\underline{I})$ is contained in a single stable boundary component
of $\underline{h(R_{k})}$ and we conclude that the stable boundary of
$\pi^{-1}(h({\mathcal{R}}_{f,g}))$ is $f$-invariant.

∎

The $\Phi^{-1}{g,f}$-invariance of
$\partial^{u}\underline{h}({\mathcal{R}}_{f,g})$ is similarly proved.
Consequently, $\underline{h}({\mathcal{R}}_{f,g})$ is a Markov partition of
the basic piece $K(T_{g,f})$.

Item $(4)$. Considering the geometrization of
$\underline{h}({\mathcal{R}}_{f,g})$ that was indicated in our last item, let
$T^{\prime}$ be its geometric type. It was proven in Proposition 4.2 that:

$\pi(\underline{h}({\mathcal{R}}_{f,g}))=h({\mathcal{R}}_{f,g})$

is a geometric Markov partition of $g$ with geometric type $T^{\prime}$. Since
the geometric type of $h({\mathcal{R}}_{f,g})$ is $T_{f,g}$, we can conclude
that $T=T_{f,g}$, which confirms the last statement of our proposition.

∎

##  Chapter 7 Equivalent geometric types: The Béguin’s algorithm

### The equivalence problem.

Let $T_{f}$ and $T_{g}$ be two geometric types in the pseudo-Anosov class, and
let $(f,{\mathcal{R}})$ and $(g,{\mathcal{R}}_{g})$ be realizations of such
geometric types by generalized pseudo-Anosov homeomorphisms. Their joint
refinement was defined in 6.2 and denoted as $T_{f,g}$ and $T_{g,f)}$. Their
formal derived form Anosov, denoted as DA$(T_{f,g})$ and DA$(T_{g,f})$, were
introduced in 6.1. Two geometric types, $T_{1}$ and $T_{2}$ are _strongly
equivalents_ if DA$(T_{1})$ has a Markov partition of geometric type $T_{2}$
and DA$(T_{2})$ has a Markov partition of geometric type $T_{1}$. Proposition
6.2 yields to the following corollary.

######  Corollary 7.1.

Let $f$ and $g$ be generalized pseudo-Anosov homeomorphisms with geometric
Markov partitions ${\mathcal{R}}_{f}$ and ${\mathcal{R}}_{g}$ of geometric
types $T_{f}$ and $T_{g}$, respectively. Let ${\mathcal{R}}_{f,g}$ be the
joint refinement of ${\mathcal{R}}_{f}$ with respect to ${\mathcal{R}}_{g}$,
and let ${\mathcal{R}}_{g,f}$ be the joint refinement of ${\mathcal{R}}_{g}$
with respect to ${\mathcal{R}}_{f}$, whose geometric types are $T_{f,g}$ and
$T_{g,f}$, respectively. Under these hypotheses: $f$ and $g$ are topologically
conjugated through an orientation preserving homeomorphism if and only if
$T_{f,g}$ and $T_{g,f}$ are strongly equivalent.

###### Proof.

Let’s assume that $f$ and $g$ are topologically conjugated through an
orientation-preserving homeomorphism. Proposition 6.2 establishes that, in
this situation, the saddle-type basic piece $K_{f,g}$ in DA$(T_{f,g})$ has a
geometric Markov partition with geometric type $T_{g,f}$; therefore, $T_{f,g}$
and $T_{g,f}$ are strongly equivalent.

If $T_{f,g}$ and $T_{g,f}$ are strongly equivalent, then the saddle-type basic
piece $K_{f,g}$ in DA$(T_{f,g})$ has a geometric Markov partition
${\mathcal{R}}$ with geometric type $T_{g,f}$. Let

$\pi:\textbf{DA}((T_{f,g})\to S_{f}$

be the projection introduced in Theorem 4.2. Proposition 4.2 states that
$\pi({\mathcal{R}})$ is a geometric Markov partition of $f$ with type
$T_{g,f}$. Theorem 3.1 implies that $f$ and $g$ are topologically conjugated
through an orientation-preserving homeomorphism.

∎

Part of the classification of saddle-type basic pieces for surface Smale
diffeomorphisms was carried out in 2004 by François Béguin in [Bég04]. In that
work, he developed an algorithm that determines in finite time when two
geometric types associated with non-trivial saddle-type basic pieces of
surface Smale diffeomorphisms are strongly equivalent.

######  Theorem 7.1 ( Béguin [Bég04]).

Let $T_{1}$ and $T_{2}$ be geometric types realizable as basic pieces $K_{1}$
and $K_{2}$ of Smale surface diffeomorphisms. There exists a finite algorithm
that determines if $T_{1}$ and $T_{2}$ are strongly equivalent.