Proof involving retractions onto apartments

Question

Let $\Delta$ be a (thick) building and let $\Sigma$ be an apartment. Let $C$ and $C'$ be adjacent chambers of $\Sigma$. Then $C$ and $C'$ have common wall $B \in \Sigma$. Since $\Delta$ is thick, there is another chamber $C'' \in \Delta$ that contains $B$ and since $\Sigma$ is thin we must have $C'' \not\in \Sigma$. There is an apartment $\Sigma'$ that contains $C$ and $C''$. Let $\phi : \Sigma \rightarrow \Sigma$ be the map $$\rho(\Sigma', C) \rho(\Sigma, C') |_{\Sigma} $$ where we compose from left to right and $\rho(\Sigma', C)$ is the retraction onto apartment $\Sigma'$. Also there must be an apartment $\Sigma''$ which contains $C'$ and $C''$ and we define $\phi'$ similarly $$ \phi' = \rho(\Sigma'', C') \rho(\Sigma, C)|_{\Sigma}. $$

We can show that $\phi(C')=C$ and $\phi'(C')=C'=\phi'(C)$. Let $D$ be any chamber in $\Sigma$ and let $G=\{C_i \}_{i=0}^m$ be a shortest gallery joining $B$ and $D$. We know that all $C_i \in \Sigma$. This implies that $C_0$ is either $C$ or $C'$.

I am trying to show by induction on $m$ that if $C_0=C$ then for $E \subset D$ we have

$\phi(E)=\phi(\phi'(E))=E.$

I can show that $\phi(E)=E$ for any $m$ but I am having trouble showing that $\phi(\phi'(E))=E$. Even for the case when $m=0$ and so $C_0=C=D$ I just cannot see how $\phi(\phi'(E))=E$.

Any help on this?

If you want more of a background where this is from, it is Group theory I by Suzuki and proposition 3.18 on page 322.

Thanks

Answer 1

This follows from the fact that retractions are type-preserving, because they are defined using type-preserving isomorphisms between apartments. (See the argument used in the proof of (3.16) in Suzuki's book.)

In particular, in the case $m=0$ where $E \subset C$, you already know that $\phi(\phi'(C)) = C$, so $\phi(\phi'(E)) \subset C$. The only subset of $C$ of the same type as $E$ is $E$ itself, so you are done.

Edit:

Without using types, this seems to require a bit more detailed work. We can spell out $\phi(\phi'(E))$ as $$ \rho(\Sigma, C') \rho(\Sigma', C) \rho(\Sigma, C) \rho(\Sigma'', C') (E) , $$ and now notice that the product of the middle two retractions $\rho(\Sigma', C) \circ \rho(\Sigma, C)$ acts as the identity on $\Sigma'$. However, $\rho(\Sigma'', C') (E)$ is contained in $C''$ and hence in $\Sigma'$, so we get $$ \phi(\phi'(E)) = \rho(\Sigma, C') \rho(\Sigma'', C') (E) . $$ We now apply the same idea again: the product $\rho(\Sigma, C') \circ \rho(\Sigma'', C')$ is the identity on $\Sigma$, and we conclude that indeed $\phi(\phi'(E)) = E$.