## Chapter I. Hyperplane Systems and Galleries

Last update: 3 May 2013

## §1. Pregalleries

(1.1) Equivalence relations on wordsets. We need terminology used in connection with free semigroups, relations, presentations, etc.

Let Al and Ob be sets, called the alphabet and the objects respectively. Let two maps be given: $\text{beg, end}:\text{Al}\to \text{Ob.}$ A word is a finite sequence ${a}_{1},\dots ,{a}_{n}$ with ${a}_{i}\in \text{Al}$ and $\text{end}\left({a}_{i-1}\right)=\text{beg}\left({a}_{i}\right)$ for $i=2,\dots ,n\text{.}$

Let $W$ be the set of all words. Mostly we add to $W$ for each $C\in \text{Ob}$ an empty word ${\varnothing }_{C}\text{.}$ Define $\text{beg, end}:W\to \text{Ob}$ by $\text{beg}\left({a}_{1},\dots ,{a}_{n}\right)≔\text{beg}\left({a}_{1}\right),$ $\text{end}\left({a}_{1},\dots ,{a}_{n}\right)≔\text{end}\left({a}_{n}\right)$ and $\text{beg}\left({\varnothing }_{C}\right)≔\text{end}\left({\varnothing }_{C}\right)≔C\text{.}$ If $w,w\prime$ are words and $\text{end}\left(w\right)=\text{beg}\left(w\prime \right),$ then by juxtaposition we can define $ww\prime \in W\text{.}$ This operation is associative and has, if empty words are added left and right neutral elements. This way we get a category with the property that each map from Al to the morphisms together with a map from Ob to the objects such that the image of $\text{beg}\left(a\right)$ and $\text{end}\left(a\right)$ are the domain and range respectively of the image of $a\in \text{Al},$ extends uniquely to a functor. Therefore we call this construction the free category generated by the quadruple (Al, Ob; beg, end). If $|\text{Ob}|=1,$ then this is nothing else then the usual free semigroup on (or generated by) Al.

Now let $R$ be any subset of $W×W$ such that each pair $\left(a,b\right)\in R$ satisfies $\text{beg}\left(a\right)=\text{beg}\left(b\right)$ and $\text{end}\left(a\right)=\text{end}\left(b\right)\text{.}$ Define an equivalence relation ${\sim }_{R}$ by: $w{\sim }_{R}w\prime$ iff there exists a sequence $w={w}_{0},\dots ,{w}_{n}=w\prime$ such that ${w}_{i-1}={x}_{i}{a}_{i}{y}_{i}$ and ${w}_{i}={x}_{i}{b}_{i}{y}_{i}$ for some ${x}_{i},{y}_{i},{a}_{i}$ and ${b}_{i}\in W$ with $\left({a}_{i},{b}_{i}\right)\in R$ or $\left({b}_{i},{a}_{i}\right)\in R\text{.}$ If $w{\sim }_{R}w\prime ,$ then we say that $w$ and $w\prime$ are equivalent by means of the operations (or relations) in $R\text{.}$ Remark that in that case $\text{beg}\left(w\right)=\text{beg}\left(w\prime \right)$ and $\text{end}\left(w\right)=\text{end}\left(w\prime \right)\text{.}$ This is the smallest equivalence relation with the property $\left(a,b\right)\in R⇒xay{\sim }_{R}xby$ and hence we call ${\sim }_{R}$ the equivalence relation generated by $R\text{.}$

(1.2) A fundamental topological theorem.

Let $X$ be a topological space and $𝒮$ a cover of $X$ consisting of open and simply connected sets. We choose in each $S\in 𝒮$ a point ${a}_{S}$ and now we want to give a description of the groupoid $\Pi ,$ which has as objects the set $𝒮,$ and ${\Pi }_{ST}$ for $S,T\in 𝒮$ consists of the homotopy classes of paths from ${a}_{S}$ to ${a}_{T}\text{.}$ This subgroupoid of the fundamental groupoid of $X$ essentially describes the latter, because of the simply connectedness of the sets in $𝒮\text{.}$

(1.3) Definition. A pregallery $G$ consists of a sequence of elements of $𝒮:$ ${S}_{0},\dots ,{S}_{k}$ such that ${S}_{i-1}\cap {S}_{i}\ne \varnothing$ for $i=1,\dots ,k,$ together with a sequence ${K}_{1},\dots ,{K}_{k}$ where ${K}_{i}$ is a connected component of ${S}_{i-1}\cap {S}_{i}$ for $i=1,\dots ,k\text{.}$ We set $\text{beg}\left(G\right)≔{S}_{0}$ and $\text{end}\left(G\right)≔{S}_{k}\text{.}$ This pregallery is denoted:

$S0 K1 S1 K2 S2… Sk-1 Kk Sk.$

If $\text{end}\left(G\right)=\text{beg}\left(G\prime \right),$ then obviously we can define the pregallery $G.G\prime \text{.}$ This way we get a category $\text{Pgal}$ with objects $𝒮$ and whose morphisms are the pregalleries.

A pregallery of the form ${S}^{K}T$ is called direct. Remark that the category $\text{Pgal}$ is in fact the free category on the direct pregalleries, so we can use the terminology of (1.1).

$S T K d σ γ γ1 γ′ γ1′ d′ aS aT Figure 1.$

Now we assign to each direct pregallery ${S}^{K}T$ an element of ${\Pi }_{ST}$ in the following way: Choose $d\in K$ and a path $\gamma$ in $S$ from ${a}_{S}$ to $d$ and a path $\gamma \prime$ in $T$ from $d$ to ${a}_{T}\text{.}$ Now the homotopy class of $\gamma \gamma \prime$ doesn't depend on the choice of $d,\gamma$ and $\gamma \prime \text{.}$ For let also ${d}_{1},{\gamma }_{1}$ and ${\gamma }_{1}^{\prime }$ be chosen. Choose also a path $\sigma$ in $K$ from $d$ to $d\prime$ $\text{(}K$ is connected). Because $\gamma \sigma$ and ${\gamma }_{1}$ stay inside the simply connected set $S,$ we have $\gamma \sigma \simeq {\gamma }_{1}\text{.}$ Also, ${\sigma }^{-1}\gamma \prime \simeq {\gamma }_{1}^{\prime }\text{.}$ So $\gamma \gamma \prime \simeq \gamma \sigma {\sigma }^{-1}\gamma \prime \simeq {\gamma }_{1}{\gamma }_{1}^{\prime }\text{.}$ The class defined this way is denoted by $\phi \left({S}^{K}T\right)\text{.}$ Because of freeness $\phi$ extends in a unique way to a functor $\phi :\text{Pgal}\to \Pi \text{.}$

(1.4) Definition. Consider the pairs

$R≔ { ( SKULT, SMT ) ∈Pgal×Pgal | K∩L∩M=∅ } ∪ { ( SSS, ∅C ) ∈Pgal×Pgal | S∈𝒮 }$

and let ${\sim }_{R}$be the equivalence relation on $\text{Pgal}$ generated by $R\text{.}$

The following fundamental theorem says that $\Pi$ can be described as the quotient category $\text{Pgal}/{\sim }_{R}\text{.}$ This observation is the basis for all other presentations of fundamental groupoids and groups in the sequel.

(1.5) Theorem. The functor $\phi :\text{Pgal}\to \Pi$ is surjective and $\phi \left(G\right)=\phi \left(G\prime \right)$ iff $G{\sim }_{R}G\prime \text{.}$

For the proof it is convenient to have the following:

(1.6) Definition. Let $\gamma :\left[0,1\right]\to X$ be a continuous curve. We call a pregallery $G={{S}_{0}}^{{K}_{1}}{S}_{1}\dots {}^{{K}_{k}}{S}_{k}$ linked with $\gamma$ by a division $D:$ $0={t}_{0}\le {t}_{1}\le \dots \le {t}_{k+1}=1,$ notation: $G\in {L}_{\gamma }\left(D\right),$ if:

 (i) $\gamma \left(\left[{t}_{i},{t}_{i+1}\right]\right)\subset {S}_{i}$ for $i=1,\dots ,k$ (ii) $\gamma \left({t}_{i}\right)\in {K}_{i}$ for $i=1,\dots ,k\text{.}$

If there exists a division $D$ such that $G\in {L}_{\gamma }\left(D\right),$ then we say that $G$ is linked with $\gamma ,$ notation $G\in {L}_{\gamma }\text{.}$

(1.7) Remark. If ${S}_{0},\dots ,{S}_{k}$ is a sequence of elements of $𝒮$ and $\gamma$ a curve and $D$ a division of $\left[0,1\right]$ such that condition (i) of (1.6) is satisfied, then it follows that $\gamma \left({t}_{i}\right)\in {S}_{i-1}\cap {S}_{i},$ so $\gamma \left({t}_{i}\right)\in {K}_{i}$ for some component ${K}_{i}$ of ${S}_{i-1}\cap {S}_{i}\text{.}$ Of course then $G\in {L}_{\gamma }\left(D\right),$ where $G={{S}_{0}}^{{K}_{1}}{S}_{1}\dots {}^{{K}_{k}}{S}_{k}\text{.}$

Proof of (1.5).

First the surjectivity of $\phi \text{.}$ Let $\gamma :\left[0,1\right]\to X$ be a continuous curve from ${a}_{S}$ to ${a}_{T}\text{.}$ The collection $\left\{{\gamma }^{-1}\left(S\right) | S\in 𝒮\right\}$ forms an open cover of $\left[0,1\right],$ so by compactness of $\left[0,1\right]$ there is a division $D:0={t}_{0}\le {t}_{1}\le \dots \le {t}_{k+1}=1$ such that $\left[{t}_{i},{t}_{i+1}\right]$ is contained in ${\gamma }^{-1}\left({S}_{i}\right)$ for some ${S}_{i}\in 𝒮\text{.}$ As remarked in (1.7) we now know that there is a pregallery $G={{S}_{0}}^{{K}_{1}}{S}_{1}\dots {}^{{K}_{k}}{S}_{k}$ such that $G\in {L}_{\gamma }\text{.}$ The surjectivity of $\phi$ follows now from the

Claim. If $G\in {L}_{\gamma }$ then $\phi \left(G\right)=\left\{\gamma \right\}\text{.}$

 Proof. Let $G\in {L}_{\gamma }$ and suppose $G,\gamma$ and $D$ as above. Put ${d}_{i}≔\gamma \left({t}_{i}\right)$ for $i=1,\dots ,k+1$ and choose for $i=1,\dots ,k$ a curve ${\gamma }_{i}$ in ${S}_{i-1}$ from ${a}_{{S}_{i-1}}$ to ${d}_{i}$ and a curve ${\gamma }_{1}^{\prime }$ in ${S}_{i}$ from ${d}_{i}$ to ${a}_{{S}_{i}}\text{.}$ Furthermore ${\gamma }_{0}^{\prime }$ is the constant curve on ${a}_{{S}_{0}}≕{d}_{0}$ and ${\gamma }_{k+1}$ is the constant curve on ${a}_{{S}_{k}}≕{d}_{k+1}\text{.}$ Let ${\gamma }_{i}^{\prime \prime }≔\gamma \circ \left[{t}_{i},{t}_{i+1}\right]$ (see notational remark 3). Now for $i=0,\dots ,k,$ ${\gamma }_{i}^{\prime \prime }$ as well as ${\gamma }_{i}^{\prime }\circ {\gamma }_{i+1}$ are curves in ${S}_{i}$ from ${d}_{i}$ to ${d}_{i+1},$ so they are homotopic. So $\gamma \simeq {\gamma }_{0}^{\prime \prime }\circ {\gamma }_{1}^{\prime \prime }\circ \dots \circ {\gamma }_{k}^{\prime \prime }\simeq {\gamma }_{0}^{\prime }\circ {\gamma }_{1}\circ {\gamma }_{1}^{\prime }\circ {\gamma }_{2}\circ \dots \circ {\gamma }_{k}\circ {\gamma }_{k}^{\prime }\circ {\gamma }_{k+1}\simeq \prod _{i=1}^{k}{\gamma }_{i}\circ {\gamma }_{i}^{\prime }\text{.}$ So $\phi \left(G\right)=\left\{\prod _{i=1}^{k}{\gamma }_{i}\circ {\gamma }_{i}^{\prime }\right\}=\left\{\gamma \right\}\text{.}$ $S0 S1 S2 aS0=γ(t0) aS1 aS2=γ(t3) γ(t1) γ(t2) γ1 γ2 γ1′ γ2′ Figure 2.$ $\square$

Now we prove: $G{\sim }_{R}G\prime ⇒\phi \left(G\right)=\phi \left(G\prime \right)\text{.}$ It is sufficient to show that $\phi \left({S}^{K}{U}^{L}T\right)=\phi \left({S}^{M}T\right)$ for each pair from $R$ i.e. $K\cap L\cap M\ne \varnothing \text{.}$ Choose then a point $d\in K\cap L\cap M\text{.}$ Let $\gamma ,\gamma \prime ,{\gamma }^{\prime \prime }$ be curves from ${a}_{S},$ ${a}_{U}$ and ${a}_{T}$ respectively to $d$ inside $S,U$ and $T$ respectively. Then $\phi \left({S}^{K}{U}^{L}T\right)=\left\{\gamma \circ {\left(\gamma \prime \right)}^{-1}\circ \gamma \prime \circ {\left(\gamma \prime \prime \right)}^{-1}\right\}=\left\{\gamma \circ {\left(\gamma \prime \prime \right)}^{-1}\right\}=\phi \left({S}^{M}T\right)\text{.}$

$d S T U aS aT aU γ γ′ γ′′ Figure 3.$

We also have to remark that $\phi \left({S}^{S}S\right)=\phi \left({\varnothing }_{S}\right)≔1,$ but this is true by the simply connectedness of $S\in 𝒮\text{.}$

We prove the converse: $\phi \left(G\right)=\phi \left(G\prime \right)⇒G{\sim }_{R}G\prime ,$ in four steps:

Step 1. If $G,G\prime \in {\text{Pgal}}_{AB}$ and $G,G\prime \in {L}_{\gamma }\left(D\right),$ then $G{\sim }_{R}G\prime \text{.}$

 Proof. $A=S0 S1 S2 S3 S4=B A=T0 T1 T2 T3 T4=B Figure 4.$ Let $G,\gamma$ and $D$ be as before and let $G\prime$ be the pregallery $G\prime ={{T}_{0}}^{{L}_{1}}{T}_{1}\dots {}^{{L}_{k}}{T}_{k}\text{.}$ We have ${T}_{0}={S}_{0}=A$ and ${T}_{k}={S}_{k}=B\text{.}$ For $i=0,\dots ,k$ we know $\gamma \left(\left[{t}_{i},{t}_{i+1}\right]\right)\subset {S}_{i}\cap {T}_{i}\text{.}$ Let ${W}_{i}$ be the component of ${S}_{i}\cap {T}_{i}$ such that $\gamma \left(\left[{t}_{i},{t}_{i+1}\right]\right)\subset {W}_{i}\text{.}$ Furthermore we know that $\gamma \left({t}_{i+1}\right)\in {S}_{i}\cap {T}_{i+1},$ so let ${W}_{i+1}^{\prime }$ be the component of ${S}_{i}\cap {T}_{i+1}$ such that $\gamma \left({t}_{i+1}\right)\in {W}_{i+1}^{\prime }\text{.}$ Now define for $i=1,\dots ,k:$ $Gi ≔ S0 K1 S1 … Si-1 Wi′ Ti Li+1 Ti+1 … Lk Tk Gi′ ≔ S0 K1 S1 … Si-1 Ki Si Wi Ti Li+1 Ti+1 … Lk Tk So: Gi′ ≔ S0 K1 S1 … Si-1 Ki Si Wi+1′ Ti+1 … Lk Tk$ where you get this last line by substituting $i+1$ for $i$ in the first line for $i=0,\dots ,k-1\text{.}$ Now $\gamma \left({t}_{i}\right)\in {W}_{i}^{\prime }\cap {K}_{i}\cap {W}_{i},$ so ${G}_{i}{\sim }_{R}{G}_{i}^{\prime }$ and also $\gamma \left({t}_{i+1}\right)\in {W}_{i+1}^{\prime }\cap {W}_{i}\cap {L}_{i+1},$ so ${G}_{i}^{\prime }{\sim }_{R}{G}_{i+1},$ so ${G}_{i}{\sim }_{R}{G}_{i+1}$ for $i=1,\dots ,k-1\text{.}$ So ${G}_{1}{\sim }_{R}{G}_{k}\text{.}$ Now ${G}_{1}={{S}_{0}}^{{W}_{1}^{\prime }}{{T}_{1}}^{{L}_{2}}\dots {}^{{L}_{k}}{T}_{k},$ but because ${S}_{0}={T}_{0}$ we have ${W}_{1}^{\prime }={L}_{1},$ so ${G}_{1}={G}^{\prime }\text{.}$ Analogously ${G}_{k}=G\text{.}$ So $G{\sim }_{R}{G}^{\prime }\text{.}$ $\square$

Step 2. If $G\in {L}_{\gamma }\left(D\right)$ and $D\prime$ is a refinement of $D,$ then there exists a $G\prime ,$ such that $G\prime \in {L}_{\gamma }\left(D\prime \right)$ and $G{\sim }_{R}G\prime \text{.}$

 Proof. Let again $G,$ $\gamma$ and $D$ be as before. Of course it is sufficient to prove this for the case that $D\prime$ has one more division point than $D,$ so $D\prime =\left\{0={t}_{0}\le \dots \le {t}_{i}\le s\le {t}_{i+1}\le \dots \le {t}_{k+1}=1\right\}\text{.}$ Now we can take simply $G\prime ={{S}_{0}}^{{K}_{1}}{{S}_{1}}^{}\dots {{S}_{i}}^{{S}_{i}}{{S}_{i}}^{{K}_{i+1}}{{S}_{i+1}}^{}\dots {}^{{K}_{k}}{S}_{k}\text{.}$ $\square$

Step 3. If $G,G\prime \in {L}_{\gamma },$ then $G{\sim }_{R}G\prime \text{.}$

 Proof. Let $G\in {L}_{\gamma }\left(D\right)$ and $G\prime \in {L}_{\gamma }\left(D\prime \right)\text{.}$ Denote by ${D}^{\prime \prime }$ the common refinement of $D$ and $D\prime \text{.}$ According to step 2 there are ${G}_{1}$ and ${G}_{2}$ such that ${G}_{i}\in {L}_{\gamma }\left({D}^{\prime \prime }\right)$ for $i=1,2$ and ${G}_{1}{\sim }_{R}G$ and ${G}_{2}{\sim }_{R}G\prime \text{.}$ But step 1 says that also ${G}_{1}{\sim }_{R}{G}_{2}$ so $G{\sim }_{R}G\prime \text{.}$ $\square$

Step 4. Let $G,G\prime \in {\text{Pgal}}_{AB}\text{.}$ If $G\in {L}_{\gamma },G\prime \in {L}_{\gamma },$ and $\gamma \simeq \gamma \prime ,$ then $G{\sim }_{R}G\prime \text{.}$

 Proof. $aA aB γ γ γ′ γj+1 (i,j) Ii,j Figure 5.$ Since $\gamma \simeq \gamma \prime ,$ there is a homotopy $F:\left[0,1\right]×\left[0,1\right]\to X$ such that $F\left(t,0\right)=\gamma \left(t\right),$ $F\left(t,1\right)=\gamma \prime \left(t\right),$ $F\left(0,s\right)={a}_{A}$ and $F\left(1,s\right)={a}_{B}\text{.}$ The collection $\left\{{F}^{-1}\left(S\right) | S\in 𝒮\right\}$ is an open cover of ${\left[0,1\right]}^{2}$ and so by compactness of the latter there is a division $D=\left\{0={t}_{0}\le {t}_{1}\le \dots \le {t}_{k+1}=1\right\}$ such that for each pair $\left(i,j\right)\in {\left\{0,\dots ,k\right\}}^{2}$ there is a ${S}_{i,j}\in 𝒮$ such that $F\left({I}_{i,j}\right)\subset {S}_{i,j}$ where ${I}_{i,j}≔\left[{t}_{i},{t}_{i+1}\right]×\left[{t}_{j},{t}_{j+1}\right]\text{.}$ For $j=0,\dots ,k+1$ define ${\gamma }_{j}\left(t\right)≔F\left(t,{t}_{j}\right)\text{.}$ Now fix an index $j\text{.}$ $F\circ \left[\left({t}_{i},{t}_{j}\right),\left({t}_{i},{t}_{j+1}\right)\right]$ (see notational remark (0.3)) is a curve in ${S}_{i-1,j}\cap {S}_{i,j}$ for $i=1,\dots ,k,$ so $F\left({t}_{i},{t}_{j}\right)$ and $F\left({t}_{i},{t}_{j+1}\right)$ lie in the same component of ${S}_{i-1,j}\cap {S}_{i,j};$ call this component ${K}_{i}\text{.}$ From $F\left({I}_{i,j}\right)\subset {S}_{i,j}$ it follows that ${\gamma }_{j}\left(\left[{t}_{i},{t}_{i+1}\right]\right)\subset {S}_{i,j}$ for $i=0,\dots ,k$ and also ${\gamma }_{j+1}\left(\left[{t}_{i},{t}_{i+1}\right]\right)\subset {S}_{i,j}\text{.}$ We already saw ${\gamma }_{j}\left({t}_{i}\right),$ ${\gamma }_{j+1}\left({t}_{i}\right)\in {K}_{i}$ for $i=1,\dots ,k\text{.}$ Then it follows from remark (1.7) that ${G}_{j}\in {L}_{{\gamma }_{i}}\cap {L}_{{\gamma }_{j+1}},$ where ${G}_{j}≔{{S}_{0,j}}^{{K}_{1}}{{S}_{1,j}}^{{K}_{2}}{S}_{2,j}\dots {{S}_{k-1,j}}^{{K}_{k}}{S}_{k,j}\text{.}$ So for all $j=0,\dots ,k$ we have ${G}_{j}\in {L}_{{\gamma }_{j+1}}$ and ${G}_{j+1}\in {L}_{{\gamma }_{j+1}}$ so by step 3 we have ${G}_{j}{\sim }_{R}{G}_{j+1}\text{.}$ Moreover ${\gamma }_{0}=\gamma ,G\in {L}_{\gamma }$ and ${G}_{0}\in {L}_{{\gamma }_{0}},$ so $G{\sim }_{R}{G}_{0}$ (step 3). Analogously ${G}_{k+1}{\sim }_{R}G\prime \text{.}$ So $G{\sim }_{R}{G}_{0}{\sim }_{R}{G}_{1}{\sim }_{R}{G}_{2}{\sim }_{R}\dots {\sim }_{R}{G}_{k+1}{\sim }_{R}G\prime \text{.}$ So $G{\sim }_{R}G\prime \text{.}$ $\square$

To finish the proof of theorem (1.5), suppose $\phi \left(G\right)=\phi \left(G\prime \right)\text{.}$ For the defining curves $\gamma$ and $\gamma \prime$ (i.e. $\phi \left(G\right)=\left\{\gamma \right\}$ and $\phi \left(G\prime \right)=\left\{\gamma \prime \right\}\text{)}$ certainly $G\in {L}_{\gamma }$ and $G\prime \in {L}_{\gamma },$ holds. And from $\phi \left(G\right)=\phi \left(G\prime \right)$ it follows that $\gamma \simeq \gamma \prime ,$ so by step 3 (Suggested correction: 4) we have $G{\sim }_{R}G\prime \text{.}$

$\square$

## §2. Hyperplane Systems

Let $𝕍$ be a real finite dimensional vector space.

(2.1) If $M$ is a hyperplane and the set $A\subset 𝕍$ is contained in one of the two open halfspaces determined by $M$ (i.e. components of $𝕍-M\text{),}$ then we denote this halfspace by ${D}_{M}\left(A\right)\text{.}$ For singletons $\left\{x\right\}$ we also write ${D}_{M}\left(x\right)\text{.}$

Let $H\subset 𝕍$ be an open set (so "open in $H\text{"}$ and "open in $𝕍\text{"}$ is the same thing) and $ℳ$ a collection of hyperplanes of $𝕍$ (not necessarily containing the origin), which is locally finite on $H\text{.}$ Recall that this means, chat each $x\in H$ has a neighbourhood $U\in H$ such that $U\cap M\ne \varnothing$ for only finitely many $M\in ℳ\text{.}$

(2.2) Proposition. Let $𝕍,H$ and $ℳ$ be as above. Then $H-\bigcup _{M\in ℳ}M$ is open.

 Proof. Let $x\in H-\bigcup _{M\in ℳ}M\text{.}$ There is a neighbourhood $U$ of $x$ such that $ℳ\prime ≔\left\{M\in ℳ | U\cap M\ne \varnothing \right\}$ is finite. Then $U\cap \bigcap _{M\in ℳ\prime }{D}_{M}\left(x\right)$ is a neighbourhood of $x$ inside $H-\bigcup _{M\in ℳ}M\text{.}$ $\square$

(2.3) Corollary. The components of $H-\bigcup _{M\in ℳ}M$ are open.

(2.4) Definition. These components are called chambers in $H$ with respect to $ℳ\text{.}$ We denote the collection of chambers by $K\left(ℳ,H\right)$ or by $K\left(ℳ\right)$ if no confusion is caused by the ambiguity.

If $A\in K\left(ℳ\right),$ then ${D}_{M}\left(A\right)$ is defined for all $M\in ℳ\text{.}$ Let $B$ also be a chamber. If $M\in ℳ,$ then there are two possibilities:

1. ${D}_{M}\left(A\right)\cap {D}_{M}\left(B\right)=\varnothing$ in which case we say that $M$ separates $A$ and $B$ or
2. ${D}_{M}\left(A\right)={D}_{M}\left(B\right)\text{.}$

(2.5) Definition. $ℳ\left(A,B\right)≔\left\{M\in ℳ | M \text{separates} A \text{and} B\right\}\text{.}$

After these preliminaries we come to the main definitions of this paragraph.

(2.6) Definition. Let $𝕍,H$ and $ℳ$ be as in (2.1). If, moreover, $H$ is convex then the triple $\left(𝕍,H,ℳ\right)$ is called a hyperplane configuration.

(2.7) Definition. Let $\left(𝕍,H,ℳ\right)$ be a hyperplane configuration and let $ℳ\ni M\stackrel{p}{\to }{p}_{M}$ be given, where ${p}_{M}$ is a locally finite collection of hyperplanes in $𝕍\text{.}$ Then we call the quadruple $\left(𝕍,H,ℳ,p\right)$ a hyperplane system. Let us summarize the defining properties of a hyperplane system $\left(𝕍,H,ℳ,p\right):$

(2.8)

 (i) $𝕍$ is a finite dimensional real vector space. (ii) $H$ is an open convex subset of $𝕍$ (iii) $ℳ$ is a collection of hyperplanes of $𝕍$ locally finite on $H\text{.}$ (iv) $p$ is an assignment $ℳ\ni M\to {p}_{M},$ where ${p}_{M}$ is a locally finite collection of hyperplanes of $𝕍\text{.}$

We associate a space $Y\left(𝕍,H,ℳ,p\right)$ (or briefly $Y\text{)}$ to a hyperplane system $\left(𝕍,H,ℳ,p\right)$ as follows:

(2.9) Definition. $Y≔\left\{x+\sqrt{-1}y\in {𝕍}_{ℂ} | y\in H \text{and} \left(x\in N\in {p}_{M}⇒y\notin M\right)\right\}\text{.}$

In the sequel the complex structure of ${𝕍}_{ℂ}$ is not relevant. We merely use the notations ${𝕍}_{ℂ}$ for $𝕍×𝕍$ and Re and Im for the projections as it is convenient. Note that we can assume that if $M\in ℳ$ then $M\cap H\ne \varnothing$ and ${p}_{M}\ne \varnothing ,$ for otherwise $M$ doesn't occur in the definition of the space $Y\text{.}$ Remark also that $\left\{N+\sqrt{-1}M | M\in ℳ,N\in {p}_{M}\right\}$ is locally finite on $𝕍+\sqrt{-1}H={\text{Im}}^{-1}\left(H\right),$ hence $Y$ is open in ${𝕍}_{ℂ}\text{.}$

(2.10) Definition. Together with each hyperplane configuration $\left(𝕍,H,ℳ\right)$ there is a canonical simple hyperplane system defined, also denoted by the triple $\left(𝕍,H,ℳ\right),$ by setting ${p}_{M}≔\left\{M\right\}\text{.}$ For instance, if the origin of $𝕍$ is contained in all $M\in ℳ,$ then $Y$ can be considered as a complement of complex hyperplanes: $Y={\text{Im}}^{-1}\left(H\right)-\bigcup _{M\in ℳ}{M}_{ℂ}\text{.}$

The theory we are going to develop has two main applications

1. The case of simple hyperplane systems.
2. Generalized root systems.

In the latter case, the collection ${p}_{M}$ consists of hyperplanes all parallel to $M,$ hence the use of the letter $p\text{.}$

Now we construct a cover of $Y$ such that the result of §1 is applicable.

(2.11) Definition. For $C\in K\left(ℳ\right)$ we define:

$YC≔ { x+-1y | y∈H and x∈N∈pM⇒ y∈DM(C) } .$

Note that ${Y}_{C}\subset Y\text{.}$

(2.12) Proposition. ${Y}_{C}$ is open in ${𝕍}_{ℂ}\text{.}$

 Proof. Let $x+\sqrt{-1}y\in {Y}_{C}\text{.}$ Choose $c\in C$ and put $M\prime =\left\{M\in ℳ | M\cap \left[y,c\right]=\varnothing \right\}\text{.}$ Due to the compactness of $\left[y,c\right]$ and locally finiteness of $ℳ$ on $H,$ $ℳ-ℳ\prime$ is finite. Then $p\prime ≔\bigcup _{M\in ℳ-ℳ\prime }{p}_{M}$ is locally finite on $𝕍,$ since the finite union of a locally finite collection is again locally finite. If $x\in N\in {p}_{M},$ then $y\in {D}_{M}\left(C\right)\text{.}$ Thus $\left[y,c\right]\cap M=\varnothing ,$ so $M\in ℳ\prime ,$ which means $N\notin p\prime \text{.}$ Therefore $x$ lies in a chamber $U\in K\left(p\prime \right)\text{.}$ Choose also a chamber $V\in K\left(ℳ\prime ,H\right)$ such that $\left[y,c\right]\subset V\text{.}$ This is possible, because $\left[y,c\right]$ is connected and for any $M\in ℳ$ we have $\left[y,c\right]\cap M\ne \varnothing ⇒M\in ℳ\prime \text{.}$ Now we have $U+\sqrt{-1}V\subset {Y}_{C},$ for if $U\cap N\ne \varnothing$ and $N\in {p}_{M},$ then $N\notin p\prime ,$ so $M\in M\prime$ which implies that $V\subset {D}_{M}\left(V\right)={D}_{M}\left(c\right)={D}_{M}\left(C\right)\text{.}$ So $U+\sqrt{-1}V$ is a neighbourhood of $x+\sqrt{-1}y$ inside ${Y}_{C}\text{.}$ $\square$

(2.13) Proposition. $Y=\bigcup _{C\in K\left(ℳ\right)}{Y}_{C}\text{.}$

 Proof. Let $x+\sqrt{-1}y\in Y\text{.}$ Choose an open neighbourhood $U$ of $y$ such that $ℳ\prime ≔\left\{M\in ℳ | U\cap M\ne \varnothing \right\}$ is finite. By deleting from $U\bigcup _{y\notin M\in ℳ\prime }M$ (which is closed because of the finiteness of $ℳ\prime \text{)}$ we see that we can suppose that $ℳ\prime =\left\{M\in ℳ | y\in M\right\}\text{.}$ Moreover we may assume that $U$ is connected. Then ${D}_{M}\left(U\right)$ is defined for all $M\notin ℳ\prime$ and ${D}_{M}\left(U\right)={D}_{M}\left(y\right)\text{.}$ Because the interior of $\bigcup _{M\in ℳ}M$ is empty there has to be a $C\in K\left(ℳ\right)$ such that $U\cap C\ne \varnothing \text{.}$ For this $C$ we have ${D}_{M}\left(C\right)={D}_{M}\left(U\right)$ for all $M\in ℳ\prime \text{.}$ Now let $x\in N\in {p}_{M}\text{.}$ Then $y\notin M,$ so $M\notin ℳ\prime \text{.}$ Then $y\in {D}_{M}\left(y\right)={D}_{M}\left(U\right)={D}_{M}\left(C\right)\text{.}$ So we have proved that $x+\sqrt{-1}y\in {𝕍}_{ℂ}\text{.}$ $\square$

(2.14) Corollary (from 2.12) and (2.13)). The space $Y$ is open in ${𝕍}_{ℂ}$ and $\left\{{Y}_{C} | C\in K\left(ℳ\right)\right\}$ is an open cover of $Y\text{.}$

In order to prove that ${Y}_{C}$ is simply connected (even contractible) and to study the intersections we consider now the space $X≔{Y}_{{C}_{1}}\cap {Y}_{{C}_{2}}\cap \dots \cap {Y}_{{C}_{n}}$ for certain ${C}_{1},\dots ,{C}_{n}\in K\left(ℳ\right)\text{.}$ First observe:

$X= { x+-1y | y∈ H and x∈N∈pM ⇒y∈⋂i=1n DM(Ci) } .$

In particular, if $x+\sqrt{-1}y\in X$ and $x\in N\in {p}_{M}$ then $\bigcap _{i=1}^{n}{D}_{M}\left({C}_{i}\right)\ne \varnothing \text{.}$ The latter is equivalent to ${D}_{M}\left({C}_{1}\right)={D}_{M}\left({C}_{2}\right)=\dots ={D}_{M}\left({C}_{n}\right)$ and this means that $M\notin ℳ\left({C}_{i},{C}_{j}\right)$ for $i\ne j\text{.}$ So if we give the following:

(2.15) Definition. Let $A,B\in K\left(ℳ\right)\text{.}$ $p\left(A,B\right)≔\bigcup _{M\in ℳ\left(A,B\right)}{p}_{M}$ and

$p(C1,…,Cn)≔ ⋃i≠jp (Ci,Cj).$

(\$) Then we have $N\notin p\left({C}_{1},\dots ,{C}_{n}\right)\text{.}$ Now note first that $p\left({C}_{1},\dots ,{C}_{n}\right)$ is locally finite on $𝕍,$ for we have the following:

(2.16) Lemma. If $A,B\in K\left(ℳ\right),$ then $|ℳ\left(A,B\right)|<\infty \text{.}$

 Proof. By compactness of $S≔\left[a,b\right],$ where $a\in A$ and $b\in B,$ and locally finiteness of $ℳ$ on $H,$ we see that $ℳ\prime ≔\left\{M\in ℳ | M\cap S\ne \varnothing \right\}$ is finite. But it is easy to see that $ℳ\prime =ℳ\left(A,B\right)\text{.}$ $\square$

From (\$) we see that the projection $\text{Re}:{𝕍}_{ℂ}\to 𝕍$ maps the set $X$ into the union of chambers with respect to $p\left({C}_{1},\dots ,{C}_{n}\right)\text{.}$ In fact we have the following:

(2.17) Proposition. The projection $\text{Re}:{𝕍}_{ℂ}\to 𝕍$ maps each connected component of $X$ onto a chamber w.r.t. $p\left({C}_{1},\dots ,{C}_{n}\right)\text{.}$ This sets up a bijective correspondence between the connected components of $X$ and $K\left(p\left({C}_{1},\dots ,{C}_{n}\right)\right)\text{.}$ Furthermore, these components are contractible.

 Proof. The projection Re maps each component of $X$ into a component of the union of the chambers w.r.t. $p\left({C}_{1},\dots ,{C}_{n}\right)$ i.e. into a chamber. Choose $b\in {C}_{1},$ then ${D}_{M}\left(b\right)={D}_{M}\left({C}_{1}\right)=\dots ={D}_{M}\left({C}_{n}\right)$ for each $M\notin ℳ\left({C}_{1},\dots ,{C}_{n}\right)≔\bigcup _{i\ne j}ℳ\left({C}_{i},{C}_{j}\right)\text{.}$ Then for each $U\in K\left(p\left({C}_{1},\dots ,{C}_{n}\right)\right),$ $U+\sqrt{-1}b\subset X$ holds. So (\$\$) $U+\sqrt{-1}b\subset X\cap {\text{Re}}^{-1}\left(U\right)$ and the latter is nonempty. We want to show that it is contractible. If $u+\sqrt{-1}y\in X\cap {\text{Re}}^{-1}\left(U\right);$ then $\left[u+\sqrt{-1}y,u+\sqrt{-1}b\right]\subset X\cap {\text{Re}}^{-1}\left(U\right)\text{.}$ For if $u\in N\in {p}_{M},$ then $y\in {D}_{M}\left({C}_{i}\right)$ for $i=1,\dots ,n$ and so also $y\in {D}_{M}\left(b\right)$ for $M\notin ℳ\left({C}_{1},\dots ,{C}_{n}\right)\text{.}$ So we have $\left[y,b\right]\subset {D}_{M}\left(b\right),$ i.e. $u+\sqrt{-1}\left[y,b\right]\subset X\cap {\text{Re}}^{-1}\left(U\right)\text{.}$ From this we can see that $U+\sqrt{-1}b$ is a deformation retract of $X\cap {\text{Re}}^{-1}\left(U\right)\text{.}$ $U+\sqrt{-1}b$ itself is contractible (for so is $U\text{),}$ so $X\cap {\text{Re}}^{-1}\left(U\right)$ is indeed contractible. Hence for each $U\in K\left(p\left({C}_{1},\dots ,{C}_{n}\right)\right)$ there is one and only one connected, even contractible, component of $X,$ namely $X\cap {\text{Re}}^{-1}\left(U\right),$ that is mapped onto (this follows from (\$\$)) $U$ by the projection Re. $\square$

(2.18) Corollary $\left(n=1\right)\text{.}$ If $C\in K\left(ℳ\right)$ then ${Y}_{C}$ is contractible, in particular simply connected.

(2.19) Corollary $\left(n=2\right)\text{.}$ If $A,B\in K\left(ℳ\right)$ then the components of ${Y}_{A}\cap {Y}_{B}$ are in a bijective correspondence with the elements of $K\left(p\left(A,B\right)\right)\text{.}$ This correspondence is given by $K\left(p\left(A,B\right)\right)\ni U\to {Y}_{A}\cap {Y}_{B}\cap {\text{Re}}^{-1}\left(U\right)\text{.}$

Denote such a component ${Y}_{A}\cap {Y}_{B}\cap {\text{Re}}^{-1}\left(U\right)$ for the moment by $\text{Co}\left(U\right)\text{.}$ From (2.18) it follows that we have the situation of (1.2). A pregallery is now given by a sequence ${C}_{0},{C}_{1},\dots ,{C}_{n},{C}_{i}\in K\left(ℳ\right),$ together with (by (2.19)) a sequence ${U}_{1},\dots ,{U}_{n},$ where ${U}_{i}\in K\left(p\left({C}_{i-1},{C}_{i}\right)\right)\text{.}$ This pregallery ${{Y}_{{C}_{0}}}^{\text{Co}\left({U}_{1}\right)}{Y}_{{C}_{1}}\dots {}^{\text{Co}\left({U}_{n}\right)}{Y}_{{C}_{n}}$ is denoted by ${{C}_{0}}^{{U}_{1}}{C}_{1}\dots {}^{{U}_{n}}{C}_{n}$ and from now on all the pregalleries we are considering, are of this type. We need one more corollary of (2.17):

(2.20) Corollary $\left(n=3\right)\text{.}$ Let $A,L,B\in K\left(ℳ\right)$ and $U\in K\left(p\left(A,L\right)\right),$ $V\in K\left(p\left(L,B\right)\right)$ and $W\in K\left(p\left(A,B\right)\right)\text{.}$ Then $U\cap V\cap W\ne \varnothing ⇔\text{Co}\left(U\right)\cap \text{Co}\left(V\right)\cap \text{Co}\left(W\right)\ne \varnothing \text{.}$

 Proof. $\left(⇐\right)$ is easy. For $\left(⇒\right)$ choose $x\in U\cap V\cap W\text{.}$ Then for all hyperplanes $N\in p\left(A,L,B\right)$ we have $x\notin N,$ so there is a $O\in K\left(p\left(A,L,B\right)\right)$ with $x\in O\text{.}$ According to proposition (2.17) the component ${Y}_{A}\cap {Y}_{B}\cap {Y}_{L}\cap {\text{Re}}^{-1}\left(O\right)$ is mapped onto $O$ by Re, so it contains a $d$ with $\text{Re}\left(d\right)=x\text{.}$ Now we have $d\in \text{Co}\left(U\right)\cap \text{Co}\left(V\right)\cap \text{Co}\left(W\right)\text{.}$ $\square$

Combining the results of §1 and §2 we now get the main result of this section:

(2.21) Theorem. Let $\left(𝕍,H,ℳ,p\right)$ be a hyperplane system. Let $\text{Pgal}$ be the free category on the direct pregalleries ${A}^{U}B$ where $A,B\in K\left(ℳ\right),$ $U\in K\left(p\left(A,B\right)\right),$ $\text{beg}\left({A}^{U}B\right)≔A$ and $\text{end}\left({A}^{U}B\right)≔B\text{.}$ Then there is a surjective functor $\phi :\text{Pgal}\to \Pi ,$ where $\Pi$ is the subgroupoid of the fundamental groupoid of $Y\left(𝕍,H,ℳ,p\right),$ consisting of all homotopy classes of paths from and to points of the form ${a}_{C}$ where in each ${Y}_{C}$ there is chosen a point ${a}_{C}\text{.}$

Two pregalleries have the same $\phi \text{-image}$ if and only if they are equivalent by means of the relations

$R≔ { ( AU LV B, AW B ) ∈Pgal×Pgal | U∩V∩W ≠∅ } ∪ { ( A𝕍A, ∅A ) | A∈K(ℳ) } .$

 Proof. This is a restatement of theorem (1.5) for the case of hyperplane systems. By (2.20) the equivalence relations are indeed the same. $\square$

## §3. Tits Galleries

Given a path in the space $Y$ (defined in §2), then by a small homotopy we can make the imaginary part avoid intersections of distinct hyperplanes of $ℳ\text{.}$ For the set of such points has codimension bigger then one. This leads to the idea, that it is always possible to choose the sequence ${C}_{1},\dots ,{C}_{n}$ of chambers of a pregallerv in such a way, that ${C}_{i-1}$ and ${C}_{i}$ are separated by only one hyperplane. Such a sequence we will call a Tits gallery. Tits himself and Deligne call this a gallery, but we prefer to use this word for the pregalleries that arise this way. Much of this paragraph can be found in Bourbaki V §1 [Bou1968] and in Deiigne [Del1972]. But our situation is slighty more general, because we consider only an open part $H$ of $𝕍\text{.}$ Let $\left(𝕍,H,ℳ\right)$ be a hyperplane configuration as defined in (2.6).

(3.1) Definition. Two points of $H$ are said to be on the same facet, if for each $M\in ℳ$ there are only two possibilities: (i) Both points are contained in $M\text{.}$ (ii) The points are not contained in $M$ and they are moreover on the same side of $M$ i.e. ${D}_{M}\left(x\right)={D}_{M}\left(y\right),$ if $x$ and $y$ are these two points. This is an equivalence relation and the classes are called facets of $H$ with respect to $ℳ\text{.}$ The collection of facets is denoted by $F\left(ℳ,H\right)$ or for short $F\left(ℳ\right)\text{.}$

For each $F\subset H$ (in particular for $F\in F\left(ℳ\right)\text{)}$ we define: ${ℳ}_{F}≔\left\{M\in ℳ | M\supset F\right\}$ and ${L}_{F}≔\bigcap _{M\in {ℳ}_{F}}$ the support of $F\text{.}$ Furthermore $\text{(co)dim}\left(F\right)≔\text{(co)dim}\left({L}_{F}\right)\text{.}$

From the definition it is clear that for $F\in F\left(ℳ\right)$ we have

(3.2) $F={L}_{F}\cap {D}_{ℳ-{ℳ}_{F}}\left(F\right)\cap H$

(In general, if $A\subset 𝕍$ is on one side of $M,$ for instance if $A$ is convex and $A\cap M=\varnothing ,$ for all $M\in ℳ\prime \subset ℳ,$ then ${D}_{ℳ\prime }\left(A\right)$ $\bigcap _{M\in ℳ\prime }{D}_{M}\left(A\right)\text{).}$

(3.3) Proposition. $\stackrel{‾}{F}\cap H={L}_{F}\cap \bigcap _{M\in ℳ-{ℳ}_{F}}\left(\stackrel{‾}{{D}_{M}\left(F\right)}\cap H\right)\text{.}$

Remark. $\stackrel{‾}{F}\cap H$ is just the closure of $F$ as subset of $H\text{.}$ We denote this in the sequel by $\stackrel{‾}{F}\text{.}$

 Proof of (3.3). From (3.2) it follows at once that the left hand side is contained in the right hand side. Conversely, let $y$ be contained in the right hand side and choose $x\in F,$ then $\left[x,y\right]\subset {L}_{F}$ and $\left[x,y\right)\subset {D}_{M}\left(F\right)$ for all $M\in ℳ-{ℳ}_{F}\text{.}$ So $\left[x,y\right)\subset F\text{.}$ But that means that $y\in \stackrel{‾}{F}\text{.}$ $\square$

(3.4) Corollary. The closure (in $H\text{)}$ of a facet is a (disjoint) union of facets.

 Proof. For ${L}_{F}\cap H$ and each $\stackrel{‾}{{D}_{M}\left(F\right)}\cap H$ are so. $\square$

We also have a useful criterion to see if a facet is part of this union:

(3.5) Criterion. Let $F,G\in F\left(ℳ\right)\text{.}$ Then $F\subset \stackrel{‾}{F}$ if and only if for all $M\in ℳ$ $F\not\subset M⇒G\subset {D}_{M}\left(F\right)$ holds.

 Proof. Let $F\subset \stackrel{‾}{G}$ and $F\not\subset M,$ then $F\not\subset M,$ for if $G\subset M,$ then $\stackrel{‾}{G}\subset M,$ and so $F\subset M,$ a contradiction. So G is contained in one of both open half spaces of $M\text{.}$ Now we have $F\subset \stackrel{‾}{G}\subset \stackrel{‾}{{D}_{M}\left(G\right)},$ so $F\subset {D}_{M}\left(G\right),$ which also means $G\subset {D}_{M}\left(F\right)\text{.}$ Conversely, suppose that the condition holds. Then $F$ is contained in ${L}_{G}\cap \bigcap _{M\in ℳ-{ℳ}_{G}}\left(\stackrel{‾}{{D}_{M}\left(G\right)}\cap H\right)$ and so by (3.3) also in $\stackrel{‾}{G},$ which is to be proved). For if $G\subset M,$ then $F\subset M,$ by contraposition of the condition, so $F\subset {L}_{G}\text{.}$ If $G\not\subset M$ then two possibilities arise: (1) $F\subset M$ and then certainly $F\subset \stackrel{‾}{{D}_{M}\left(G\right)}\text{.}$ (2) $F\not\subset M$ and then by the condition $G\subset {D}_{M}\left(F\right),$ so $F\subset {D}_{M}\left(G\right)\subset \stackrel{‾}{{D}_{M}\left(G\right)}\text{.}$ $\square$

(3.6) Remark. $K\left(ℳ\right)=\left\{F\in F\left(ℳ\right) | \text{codim}\left(F\right)=0\right\}\text{.}$

(3.7) Definition. Let $C\in K\left(ℳ\right)$ and $F\in F\left(ℳ\right)$ with $F\subset \stackrel{‾}{C}\text{.}$
If $\text{codim}\left(F\right)=1,$ then $F$ is called a face of $C\text{.}$
If $\text{codim}\left(F\right)=2,$ then $F$ is called a plinth of $C\text{.}$
If $M\in ℳ$ is support of a face of $C,$ $M$ is called a wall of $C\text{.}$
${M}^{C}≔\left\{M\in ℳ | M \text{is a wall of} C\right\}\text{.}$

Let also a configuration $\left(𝕍,H\prime ,ℳ\prime \right)$ be given and let $ℳ\prime \subset ℳ$ and $H\prime \supset H\text{.}$ If $F\subset F\left(ℳ,H\right),$ then there is one and only one $F\prime \subset F\left(ℳ\prime ,H\prime \right)$ such that $F\subset F\prime \text{.}$ For if $x\in F,$ then $x\in F\prime$ for some $F\prime \in F\left(ℳ\prime ,H\prime \right)$ and if $y$ is on the same facet w.r.t. $ℳ$ and $H$ (definition(3.1)), then $x$ and $y$ are also on the same facet w.r.t. $ℳ\prime$ and $H\prime$ so then $F\subset F\prime \text{.}$ Uniqueness follows from the fact that $F\left(ℳ\prime ,H\prime \right)$ is a partition of $H\prime \text{.}$

(3.8) Definition. The map $F\left(ℳ,H\right)\to F\left(ℳ\prime ,H\prime \right)$ obtained above is denoted by $\text{pr}\left(ℳ\prime \subset ℳ,H\prime \supset H\right)$ or if $H\prime =H,$ then $\text{pr}\left(ℳ\prime \subset ℳ\right)\text{.}$ For $F\in F\left(ℳ\right)$ we define: ${\text{pr}}_{F}≔\text{pr}\left({ℳ}_{F}\subset ℳ,𝕍\supset H\right):F\left(ℳ,H\right)\to F\left({ℳ}_{F},𝕍\right)\text{.}$

(3.9) Theorem. (cf. [Del1972] Deligne 1.5(ii)).
The map ${\text{pr}}_{F}$ restricted to $\left\{E\in F\left(ℳ\right) | F\subset \stackrel{‾}{E}\right\}$ is bijective.

 Proof. Injectivity. Suppose $F\subset {\stackrel{‾}{E}}_{i}$ $\left(i=1,2\right)$ and ${\text{pr}}_{F}\left({E}_{1}\right)={\text{pr}}_{F}\left({E}_{2}\right)\text{.}$ For $M\in {ℳ}_{F}$ we have ${E}_{i}\subset M$ iff ${\text{pr}}_{F}\left({E}_{i}\right)\subset M;$ So ${E}_{1}\subset M$ iff ${E}_{2}\subset M\text{.}$ And if ${E}_{1}\not\subset M,$ then ${E}_{1}\subset {D}_{M}\left({\text{pr}}_{F}\left({E}_{1}\right)\right)={D}_{M}\left({\text{pr}}_{F}\left({E}_{2}\right)\right)={D}_{M}\left({E}_{2}\right),$ so then ${E}_{1}$ and ${E}_{2}$ on the same side of $M\text{.}$ If $M\ne {ℳ}_{F},$ then $F\not\subset M$ and by criterion (3.5) and the fact that $F\subset {\stackrel{‾}{E}}_{i}$ for $i=1,2$ we see that ${E}_{i}\subset {D}_{M}\left(F\right)\text{.}$ So also in this case ${E}_{1}$ and ${E}_{2}$ on the same side of $M\text{.}$ We have seen that for all $M\in ℳ,$ ${E}_{i}\subset M$ for $i=1,2$ or ${E}_{1}$ and ${E}_{2}$ on the same side of $M,$ holds, so ${E}_{1}={E}_{2}\text{.}$ Surjectivity. Let $G\in F\left({ℳ}_{F},V\right),$ Then $\stackrel{‾}{G}={L}_{G}\cap \bigcap _{G\not\subset M\in {ℳ}_{F}}\stackrel{‾}{{D}_{M}\left(G\right)}\text{.}$ It follows that ${L}_{F}\subset \stackrel{‾}{G},$ so each neighbourhood of a point $f\in F$ has points in common with $G\text{.}$ Choose a neighbourhood $U$ of $f$ such that $U\subset {D}_{M}\left(F\right)\cap H$ for all $M\notin {ℳ}_{F}$ and choose a point $u\in U\cap G\text{.}$ Let $E\subset F\left(ℳ,H\right)$ be the facet such that $u\in E\text{.}$ Then $F\subset \stackrel{‾}{E}$ and ${\text{pr}}_{F}\left(E\right)=G$ (what was to be proved). The latter is clear from $E\cap G\ne \varnothing$ $\left(u\in E\cap G\right)\text{.}$ We prove the fact that $F\subset \stackrel{‾}{E}$ holds, by criterion (3.5): Suppose $F\not\subset M\in ℳ,$ because $U\subset {D}_{M}\left(F\right)$ we have $u\in {D}_{M}\left(F\right),$ so $E\subset {D}_{M}\left(F\right)\text{.}$ $G G G U F u f Figure 6.$ $\square$

(3.10) Remark. $\text{pr}\left(ℳ\prime \subset ℳ,H\prime \supset H\right)$ $\left(K\left(ℳ,H\right)\right)\subset K\left(ℳ\prime ,H\prime \right)$ and in particular (by (3.9)) ${\text{pr}}_{F}\left(K\left(ℳ\right)\right)=K\left({ℳ}_{F}\right)\text{.}$

If $H\prime =𝕍$ and $\bigcap _{M\in ℳ\prime }M\ne \varnothing ,$ then it doesn't harm to suppose that the origin $0\in 𝕍$ is contained in this intersection. This way we can define in this case for each $E\in F\left(ℳ\prime \right):$ $-E≔\left\{x\in 𝕍 | -x\in E\right\}\in F\left(ℳ\prime \right)\text{.}$

(3.11) Definition. Let $E,F\in F\left(ℳ,H\right)$ and $\stackrel{‾}{E}\supset F\text{.}$ Then we define the reflection of the facet $E$ in the facet $F$ by: ${\text{sp}}_{F}\left(E\right)≔{\text{pr}}_{F}^{-1}\left(-{\text{pr}}_{F}\left(E\right)\right)\text{.}$ Remark again ${\text{sp}}_{F}\left(E\right)\in K\left(ℳ\right)$ if $E\in K\left(ℳ\right)\text{.}$

(3.12) Lemma. If $C\in K\left(ℳ\right),$ then $M\in {ℳ}^{C}$ iff $C\ne {D}_{ℳ-\left\{M\right\}}\left(C\right)\text{.}$

 Proof. See Bourbaki [Bou1968] V §1.4. $\square$

(3.13) Theorem. If $C\in K\left(ℳ\right),$ then $C={D}_{{ℳ}^{C}}\left(C\right)\text{.}$

 Proof. See Bourbaki [Bou1968] V §1.4 prop.9. $\square$

(These proofs also work in our (more general) situation)

(3.14) Corollary. If $A,B\in K\left(ℳ\right)$ and $ℳ\left(A,B\right)\ne \varnothing ,$ then ${ℳ}^{A}\cap ℳ\left(A,B\right)\ne \varnothing \text{.}$

 Proof. If ${ℳ}^{A}\cap ℳ\left(A,B\right)=\varnothing ,$ then $B\subset {D}_{M}\left(A\right)$ for all $M\in {ℳ}^{A},$ so $B\subset \bigcap _{M\in {ℳ}^{A}}{D}_{M}\left(A\right)={D}_{{ℳ}^{A}}\left(A\right)=A,$ so $A=B$ conflicting $ℳ\left(A,B\right)\ne \varnothing \text{.}$ $\square$

We need some propositions about separating hvperplanes.

(3.15) Proposition. If $A,L,B\in K\left(ℳ\right),$ then

 (i) $ℳ\left(A,B\right)=ℳ\left(A,L\right)\Delta ℳ\left(L,B\right)\phantom{\rule{2em}{0ex}}\text{(}\Delta :$ symmetric difference of sets) (ii) $|ℳ\left(A,B\right)|=|ℳ\left(A,L\right)|+|ℳ\left(L,B\right)|-2\mid ℳ\left(A,L\right)\cap ℳ\left(L,B\right)\mid$

 Proof. (i) For any $M\in ℳ$ consider ${D}_{M}\left(A\right),$ ${D}_{M}\left(B\right)$ and ${D}_{M}\left(L\right)\text{.}$ Three cases: 1. ${D}_{M}\left(A\right)={D}_{M}\left(B\right)={D}_{M}\left(L\right)$ then $M$ an element of neither side of (i) 2. ${D}_{M}\left(A\right)={D}_{M}\left(B\right)\ne {D}_{M}\left(L\right)$ then also $M$ an element of neither side of (i) 3. ${D}_{M}\left(A\right)={D}_{M}\left(L\right)\ne {D}_{M}\left(B\right)$ then $M$ an element of both sides of (i) (ii) follows from (i). $\square$

(3.16) Proposition. If $A,B,L\in K\left(ℳ\right),$ then the following are equivalent.

 (i) $ℳ\left(A,L\right)\cap ℳ\left(L,B\right)=\varnothing$ (ii) $ℳ\left(A,B\right)=ℳ\left(A,L\right)\cup ℳ\left(L,B\right)$ (iii) $|ℳ\left(A,B\right)|=|ℳ\left(A,L\right)|+|ℳ\left(L,B\right)|$

 Proof. (i) $⇔$ (ii) follows from (3.15)(i). (i) and (ii) $⇒$ (iii) is obvious. (iii) $⇒$ (i) follows from (3.15)(ii). $\square$

(3.17) Proposition. If ${C}_{0},\dots ,{C}_{n}$ is a sequence of elements of $K\left(ℳ\right)$ then

 (i) $ℳ(C0,Cn)⊂ ⋃i=1n ℳ(Ci-1,Ci)$ (ii) Equivalent are: $ℳ\left({C}_{i-1},{C}_{i}\right)\cap ℳ\left({C}_{j-1},{C}_{j}\right)=\varnothing$ for $i\ne j$ $|ℳ\left({C}_{0},{C}_{n}\right)|=\sum _{i=1}^{n}|ℳ\left({C}_{i,1},{C}_{i}\right)|$ and in this case we have equality in (i) and if we have a hyperplane system: $p\left({C}_{0},{C}_{n}\right)=\bigcup _{i=1}^{n}p\left({C}_{i-1},{C}_{i}\right)\text{.}$

Remark. For $n\ge 3$ the implication $ℳ\left({C}_{0},{C}_{n}\right)=\bigcup _{i=1}^{n}ℳ\left({C}_{i-1},{C}_{i}\right)$ $⇒$ (ii)a. need not hold (it does hold for $n=2$ by proposition (3.16) (ii) $⇒$ (i)), consider for instance the configuration of figure 7.

$C0 C1 C2 C3 Figure. 7$

 Proof of (3.17). Trivial for $n=1\text{.}$ Suppose proved for $n-1\text{.}$ Then by (3.15) (i) we have $ℳ\left({C}_{0},{C}_{n}\right)\subset ℳ\left({C}_{0},{C}_{n-1}\right)\cup ℳ\left({C}_{n-1},{C}_{n}\right)$ and by induction we know that the latter is contained in $\bigcup _{i=1}^{n-1}ℳ\left({C}_{i-1},{C}_{i}\right)\cup ℳ\left({C}_{n-1},{C}_{n}\right)\text{.}$ This proves (i) for $n$ also. We always have: (\$) $|M\left({C}_{0},{C}_{n}\right)|\le \mid \bigcup _{i=1}^{n}ℳ\left({C}_{i-1},{C}_{i}\right)\mid \le \sum _{i=1}^{n}|ℳ\left({C}_{i-1},{C}_{i}\right)|$ The first inequality follows from the already proved fact (i) and the second is obvious. Furthermore the last inequality is an equality iff (ii)a. holds. This proves the implication b. $⇒$ a. Suppose now that a. holds. If we can show now that (\$\$) $ℳ\left({C}_{0},{C}_{n}\right)=\bigcup _{i=1}^{n}ℳ\left({C}_{i-1},{C}_{i}\right)$ then we are done, for then (besides the second) the first inequality of (\$) is also an equality, so b. is verified then. Now by induction we have: (\$\$\$) $ℳ\left({C}_{0},{C}_{n-1}\right)=\bigcup _{i=1}^{n-1}ℳ\left({C}_{i-1},{C}_{i}\right)$ so also $ℳ\left({C}_{0},{C}_{n-1}\right)\cap ℳ\left({C}_{n-1},{C}_{n}\right)=\left(\bigcup _{i=1}^{n-1}ℳ\left({C}_{i-1},{C}_{i}\right)\right)\cap ℳ\left({C}_{n-1},{C}_{n}\right)$ and by a. the latter is empty, so it follows from (3.16) (i) $⇒$ (ii) that $ℳ\left({C}_{0},{C}_{n}\right)=ℳ\left({C}_{0},{C}_{n-1}\right)\cup ℳ\left({C}_{n-1},{C}_{n}\right)$ and this together with (\$\$\$) gives the desired result (\$\$). (3.18) Definition. We call a sequence ${C}_{0},\dots ,{C}_{n}$ of chambers a Tits gallery from ${C}_{0}$ to ${C}_{n}$ if $|ℳ\left({C}_{i-1},{C}_{i}\right)|=1$ for $i=1,\dots ,n\text{.}$ The number $n$ is called the length of the Tits gallery. The only element of $ℳ\left({C}_{i-1},{C}_{i}\right)$ is often denoted by ${M}_{i}\text{.}$ $\square$

(3.19) Remark. This Tits gallery is completely determined by its first chamber ${C}_{0}$ and the sequence ${M}_{1},\dots ,{M}_{n}\text{.}$

 Proof. Let $A,B\in K\left(ℳ\right)$ and $ℳ\left(A,B\right)=\left\{M\right\}\text{.}$ Then $A\cup B\subset {D}_{ℳ-\left\{M\right\}}\left(A\right)$ and so $A$ and $B$ are both unequal to the latter. By lemma (3.12) this means that $M\in {ℳ}^{A}\cap {ℳ}^{B}\text{.}$ Then $M$ is the support of a common face $F$ of $A$ and $B\text{.}$ So $B$ is determined by $A$ and $M$ by the rule $B={\text{sp}}_{F}\left(A\right)\text{.}$ The remark follows now. $\square$

(3.20) Corollary of (3.17). For Tits gallery ${C}_{0},\dots ,{C}_{n}$ are equivalent:

1. ${M}_{i}\ne {M}_{j}$ for $i\ne j$
2. $|ℳ\left({C}_{0},{C}_{n}\right)|=n$

and in that case we have $ℳ\left({C}_{0},{C}_{n}\right)=\left\{{M}_{1},\dots ,{M}_{n}\right\}\text{.}$

(3.21) Definition If the conditions a. and b. of (3.20) are fulfilled then we call such a Tits gallery a direct Tits gallery. (Deligne uses the word minimal but we want to use that word for other things)

(3.22) Theorem (Deligne [Del1972 1.3]). If $A,B\in K\left(ℳ\right),$ then there exists a direct Tits gallery from $A$ to $B\text{.}$

 Proof. Induction on $|ℳ\left(A,B\right)|\text{.}$ If $ℳ\left(A,B\right)=\varnothing ,$ then $A=B,$ then then trivial. Now let $|ℳ\left(A,B\right)|>0$ and let the theorem be proved in cases where the number of separating hyperplanes is less. By corollary (3.14) we know that $ℳ\left(A,B\right)\cap {ℳ}^{A}\ne \varnothing \text{.}$ Let $M\in ℳ\left(A,B\right)\cap {ℳ}^{A}$ and let $F$ be the face of $A$ supported by $M\text{.}$ Set $C≔{\text{sp}}_{F}\left(A\right)\text{.}$ Then we have by (3.15)(i) $ℳ\left(C,B\right)=ℳ\left(C,A\right)\Delta ℳ\left(A,B\right)=ℳ\left(A,B\right)-\left\{M\right\},$ since $ℳ\left(C,A\right)=\left\{M\right\}\subset ℳ\left(A,B\right)\text{.}$ By induction there is a direct Tits gallery $C={C}_{1},\dots ,{C}_{n}=B$ from $C$ to $B\text{.}$ Then $A,{C}_{1},\dots ,{C}_{n}$ is a direct Tits gallery from $A$ to $B\text{.}$ $\square$

(3.23) Theorem. Let $A,B\in K\left(ℳ\right)\text{.}$ If $B={\text{sp}}_{F}\left(A\right)$ for some plinth $F$ of $A$ then there are exactly two direct Tits galleries from $A$ to $B\text{.}$

Proof.

First we remark that, if $C$ and $C\prime$ are chambers with a common plinth $P,$ then $ℳ\left(C,C\prime \right)\subset {ℳ}_{P}\text{.}$ For let $M\notin {ℳ}_{P},$ so $P\not\subset M\text{.}$ Also $P\subset \stackrel{‾}{C}\cap \stackrel{‾}{C\prime },$ so by criterion (3.5) both $C$ and $C\prime$ are contained in ${D}_{M}\left(P\right),$ hence $M\notin ℳ\left(C,C\prime \right)\text{.}$ It follows that ${C}_{0},\dots ,{C}_{n}$ where $\stackrel{‾}{{C}_{i}}\supset P$ is a (direct) Tits gallery iff ${\text{pr}}_{p}\left({C}_{0}\right),\dots ,{\text{pr}}_{p}\left({C}_{n}\right)$ is a (direct) Tits gallery. The theorem follows from

(3.24) Lemma. If $0\in L≔\bigcap _{M\in ℳ}M$ and $\text{codim}\left(L\right)=2,$ then there are exactly two direct Tits galleries from a chamber $A$ to the chamber $-A\text{.}$

 Proof. The quotient space $𝕍/L$ has dimension two, so essentially the situation is now two-dimensional and the lemma is clear from a picture (Although an exact proof is rather length; it uses for instance remark (3.19)). $A -A Figure 8.$ $\square$

$\square$

(3.25) Definition. The equivalence relation in the set (or better: the free category) of all Tits galleries, generated by the pairs $\text{PL}≔\left\{\left(G,G\prime \right) | G$ and $G\prime$ are the two direct Tits galleries from $A$ to ${\text{sp}}_{P}\left(A\right),$ where $P$ is a plinth of some chamber $A\right\}$ is called plinth equivalence and denoted ${\sim }_{\text{PL}}\text{.}$

(3.26) Theorem. Let $A,B\in K\left(ℳ\right)$ and suppose that $G$ and $G\prime$ are the two direct Tits galleries from $A$ to $B\text{.}$ Then $G{\sim }_{\text{PL}}G\prime \text{.}$

Remark. Both theorem and proof are essentially due to Deligne [Del1972] 1.12. We don't want to use his condition that the chambers are simplicial cones. Therefore we need the simple but technically somewhat complicated lemma (3.28). For the proof we first give a definition:

(3.27) Definition. If ${F}_{1}$ and ${F}_{2}$ are two faces of a chamber $A,$ then we call ${F}_{1}$ and ${F}_{2}$ plinthed if there exists a plinth $P$ of $A$ such that $P\subset \stackrel{‾}{{F}_{1}}\cap \stackrel{‾}{{F}_{2}}\text{.}$ The supporting walls are also said to be plinthed w.r.t. $A\text{.}$

Proof of (3.26).

If $|ℳ\left(A,B\right)|=0,$ then there is nothing to prove. Else, let $G$ be the direct Tits gallery $A={C}_{0},{C}_{1},\dots ,{C}_{k}=B$ and $G\prime :A={D}_{0},{D}_{1},\dots ,{D}_{k}=B\text{.}$ Suppose $ℳ\left({C}_{0},{C}_{1}\right)=\left\{M\right\}$ and $ℳ\left({D}_{0},{D}_{1}\right)=\left\{M\prime \right\}\text{.}$

Case 1. If $M=M\prime ,$ then ${C}_{1}={D}_{1}$ (see remark (3.19)) and then ${C}_{1},\dots ,{C}_{k}$ and ${D}_{1},\dots ,{D}_{k}$ are two direct Tits galleries from ${D}_{1}={C}_{1}$ to $B={C}_{k}={D}_{k}\text{.}$ By induction they are plinth equivalent. But then also $G{\sim }_{\text{PL}}G\prime \text{.}$

Case 2. Let $M\ne M\prime$ and $M$ and $M\prime$ are plinthed walls of $A$ by a plinth $P\text{.}$ Let ${G}_{1}$ be the unique direct Tits gallery from $A$ to ${\text{sp}}_{P}\left(A\right)$ that starts with ${C}_{0},{C}_{1}$ and ${G}_{2}$ the one that starts with ${D}_{0},{D}_{1}\text{.}$ Let $E$ be a direct Tits gallery from ${\text{sp}}_{P}\left(A\right)$ to $B\text{.}$ By (3.15)(i) $ℳ\left({\text{sp}}_{P}\left(A\right),B\right)=ℳ\left(A,B\right)\Delta {ℳ}_{P}$ and the latter equals $ℳ\left(A,B\right)={ℳ}_{P\prime }$ since ${ℳ}_{P}\subset ℳ\left(A,B\right)\text{.}$ So by (3.20) ${G}_{1}E$ and ${G}_{2}E$ are direct Tits galleries from $A$ to $B\text{.}$ Because ${G}_{1}E$ and $G$ (respectively ${G}_{2}E$ and $G\prime \text{)}$ start with ${C}_{0},{C}_{1}$ (respectively ${D}_{0},{D}_{1}\text{)}$ by step 1 ${G}_{1}E{\sim }_{\text{PL}}G$ (respectively ${G}_{2}E{\sim }_{\text{PL}}G\prime \text{)}$ holds. It is clear that ${G}_{1}E{\sim }_{\text{PL}}{G}_{2}E,$ so $G{\sim }_{\text{PL}}G\prime \text{.}$

Case 3. General case. By lemma (3.28) (proved in the sequel) there is a sequence $M={M}_{0},{M}_{1},\dots ,{M}_{m}=M\prime$ such that ${M}_{i-1}$ and ${M}_{i}$ are plinthed walls of $A$ for $i=1,\dots ,m\text{.}$ For $i=1,\dots ,m-1$ there exists a direct Tits gallery ${G}_{i}$ from $A$ to $B$ starting with $A,{\text{sp}}_{{M}_{i}}\left(A\right)\text{.}$ Let ${G}_{0}≔G$ and ${G}_{m}≔G\prime \text{.}$ By step 2 we have ${G}_{i-1}{\sim }_{\text{PL}}{G}_{i}$ for $i=1,\dots ,m\text{.}$ So $G{\sim }_{\text{PL}}G\prime \text{.}$ There remains:

(3.28) Lemma. If $M,M\prime \in {ℳ}^{A}\cap ℳ\left(A,B\right),$ where $A,B\in K\left(ℳ\right),$ then there is a sequence $M={M}_{0},\dots {M}_{m}=M\prime$ such that ${M}_{i-1}$ and ${M}_{i}$ are plinthed walls of $A$ for $i=1,\dots ,m$ and ${M}_{i}\in ℳ\left(A,B\right)\text{.}$

$A B M M′ a b x a′ Figure 9.$

 Proof. Let $F$ and $F\prime$ be the faces of $A$ with $M$ and $M\prime$ as support respectively and choose points $a\in F$ and $a\prime \in F\prime \text{.}$ Choose also $b\in B\text{.}$ By choosing $a,a\prime$ and $b$ in general position we may assume that the affine hyperplane spanned by them intersects transversaly each intersection of hyperplanes of $ℳ\left(A,B\right)$ inside the compact triangle $\Delta \left(a,a\prime ,b\right)$ By compactness and locally finiteness $\Delta \left(a,a\prime ,b\right)$ meets only a finite number of facets and these have codimension less than or equal to 2. Furthermore for each $x\in \left[a,a\prime \right]$ $\left[b,x\right]\cap \stackrel{‾}{A}$ is of the form $\left[{a}_{x},x\right]$ for some unique ${a}_{x}\in \partial A$ $\text{(}\stackrel{‾}{A}$ is convex). It follows that $\left[a,a\prime \right]\ni x\to {a}_{x}\subset \partial A\cap \Delta \left(a,a\prime ,b\right)$ is a piecewise linear arc passing only faces and plinths of $A\text{.}$ $\square$

Theorem (3.26) is proved.

$\square$

(3.29) Corollary of (3.22) and (3.26). Let $C\in K\left(ℳ\right)$ be fixed. Then $G\to \text{end}\left(G\right)$ induces a bijection between the plinth equivalence classes of direct Tits galleries $G$ with $\text{beg}\left(G\right)=C$ and $K\left(ℳ\right)\text{.}$

## §4. Galleries

In this section we combine the results of §2 and §3 to get the main result of this chapter. Let $\left(𝕍,H,ℳ,p\right)$ be again a hyperplane system as in §2.

(4.1) Definition. We call a pregallery ${{C}_{0}}^{{U}_{1}}{C}_{1}\dots {}^{{U}_{n}}{C}_{n}$ a gallery if ${C}_{0},\dots ,{C}_{n}$ is a Tits gallery. The number $n$ is called the length. The subcategory of $\text{Pgal}$ we obtain is denoted $\text{Gal}\text{.}$ For $G\in \text{Gal}$ we denote the length by $|G|\text{.}$

(4.2) Proposition. Let $A={C}_{0},\dots ,{C}_{n}=B$ be a direct Tits gallery. Given $W\in K\left(p\left(A,B\right)\right)$ denote by ${U}_{i}$ the unique element of $K\left(p\left({C}_{i-1},{C}_{i}\right)\right)$ which contains $W\text{.}$ Then ${A}^{W}B{\sim }_{R}{{C}_{0}}^{{U}_{1}}{C}_{1}\dots {}^{{U}_{n}}{C}_{n}\text{.}$ Conversely, if ${U}_{i}\in K\left(p\left({C}_{i-1},{C}_{i}\right)\right)$ for $i=1,\dots ,n$ and $W≔\bigcap _{i=1}^{n}{U}_{i}\ne \varnothing ,$ Then $W\in K\left(p\left(A,B\right)\right)$ and ${A}^{W}B{\sim }_{R}{{C}_{0}}^{{U}_{1}}\dots {}^{{U}_{n}}{C}_{n}\text{.}$

 Proof. (In fact the condition that ${C}_{0},\dots ,{C}_{n}$ is a direct Tits gallery, is too strong. It is enough to require that condition (ii)a. of proposition (3.17) holds: $ℳ\left({C}_{i-1},{C}_{i}\right)\cap ℳ\left({C}_{j-1},{C}_{j}\right)=\varnothing ,$ if $i\ne j\text{).}$ Let $W\in K\left(p\left(A,B\right)\right),$ define ${W}_{i}≔\text{pr}\left(p\left({C}_{i},{C}_{n}\right)\subset p\left(A,B\right)\right)\left(W\right)$ and let ${U}_{i}$ be as in the proposition i.e. ${U}_{i}≔\text{pr}\left(p\left({C}_{i-1},{C}_{i}\right)\subset p\left(A,B\right)\right)\left(W\right)\text{.}$ Then $W\subset {W}_{i}$ and $W\subset {U}_{i}$ so we have ${W}_{i-1}\cap {U}_{i}\cap {W}_{i}\ne \varnothing \text{.}$ So $AWS= C0WCn ∼RC0U1 C1W1 Cn∼R C0U1 C1U2 C2W2 Cn∼R…∼R C0U1… UnCn.$ Second part: $\bigcap _{i=1}^{n}{U}_{i}$ is convex, so $\bigcap _{i=1}^{n}{U}_{i}\subset W$ for some $W\in K\left(p\left(A,B\right)\right)\text{.}$ if $x\in \bigcap _{i=1}^{n}{U}_{i}$ and $x\in N\in \bigcup _{M\in ℳ}{p}_{M\prime }$ then $N\notin p\left({C}_{i-1},{C}_{i}\right)$ for $i=1,\dots ,n\text{.}$ Then also $N\notin p\left(A,B\right)$ by proposition (3.17). On the other hand we have $W\subset {U}_{i}$ for $i=1,\dots ,n,$ because $W$ is convex and if $x\in W$ and $x\in N,$ then $N\notin p\left({C}_{i-1},{C}_{i}\right)\text{.}$ So $W=\bigcap _{i=1}^{n}{U}_{i}\text{.}$ Moreover ${U}_{i}=\text{pr}\left(p\left({C}_{i-1},{C}_{i}\right)\subset p\left(A,B\right)\right)\left(W\right)$ and so as in the first part we conclude the equivalence of the pregalleries. $\square$

(4.3) Definition. Such a gallery ${{C}_{0}}^{{U}_{1}}{C}_{1}\dots {}^{{U}_{n}}{C}_{n}$ with ${U}_{1}\cap {U}_{2}\cap \dots \cap {U}_{n}\ne \varnothing$ and ${C}_{0},\dots ,{C}_{n}$ a direct Tits gallery we call a direct gallery.

(4.4) Corollary. Every direct gallery is pregallery-equivalent to a unique direct pregallery. Each direct pregallery is pregallery-equivalent to a direct gallery (which is in general not unique).

 Proof. Clear from proposition (4.2). For the second statement we also need theorem (3.22). $\square$

In this case we call the gallery a direct repacing gallery of the given direct pregallery.

(4.5) Definition. Let $A\in K\left(ℳ\right)$ and $B={\text{sp}}_{P}\left(A\right)$ for some plinth $P$ of $A\text{.}$ Suppose $W\in K\left(p\left(A,B\right)\right)\text{.}$ According to theorem (3.23) there are exactly two direct Tits galleries from $A$ to $B\text{.}$ Let $G$ and $G\prime$ be the two direct replacing galleries belonging to ${A}^{W}B$ by (4.2). Then we call the pair $\left(G,G\prime \right)$ a flip relation. The equivalence relation in $\text{Gal}$ generated by these flip relations is called flip equivalence and denoted by $\approx \text{.}$

(4.6) Corollary of proposition (4.2) and theorem (3.26): Let $G$ and $G\prime$ be two direct replacing galleries of a direct pregallery. Then $G\approx G\prime \text{.}$

(4.7) Definition. Pairs of type $\left({A}^{U}{A}^{B}A,{\varnothing }_{A}\right)\in \text{Gal}×\text{Gal}$ we call cancel relations. The equivalence relation in $\text{Gal}$ generated by the flip relations and the cancel relations is called gallery equivalence or simply equivalence and is denoted by $\sim$ (note that $\sim \ne \approx ,$ see (5.5)).

(4.8) Proposition. If $G,G\prime \in {\text{Gal}}_{AB},$ then $G{\sim }_{R}G\prime ⇔G\sim G\prime \text{.}$

For the proof we need:

(4.9) Lemma. Let $R\prime ≔\left\{\left({A}^{U}{L}^{V}B,{A}^{W}B\right)\in R | |ℳ\left(L,B\right)|\le 1\right\}\cup \left\{\left({A}^{𝕍}A,{\varnothing }_{A}\right) | A\in K\left(ℳ\right)\right\}\text{.}$ Then for $G,G\prime \in \text{Pgal}$ $G{\sim }_{R}G\prime ⇔G{\sim }_{R\prime }G\prime \text{.}$ (This means that if two pregalleries are pregallery-equivalent, then this equivalence can be realised by a sharpened sort of pregallery-equivalence relations, i.e. one can suppose that $|ℳ\left(L,B\right)|\le 1\text{).}$

 Proof. The $\left(⇐\right)$ part is trivial. For the $\left(⇒\right)$ part it is sufficient to show that ${A}^{U}{L}^{V}B{\sim }_{R\prime }{A}^{W}B$ if $U\cap B\cap W\ne \varnothing \text{.}$ We proceed by induction on $|ℳ\left(L,B\right)|\text{.}$ If $|ℳ\left(L,B\right)|\le 1,$ then we are done. Let $|ℳ\left(L,B\right)|>1\text{.}$ Choose a $M\in {ℳ}^{B}\cap ℳ\left(L,B\right)$ (exists by (3.14)) and define $L\prime ≔{\text{sp}}_{M}\left(B\right)\text{.}$ Then $|ℳ\left(L\prime ,B\right)|=|\left\{M\right\}|=1\text{.}$ Put: $V′ ≔ pr ( p(L,L′)⊃ p(L,B) ) (V)⊃V V′′ ≔ pr ( p(L′,B)⊃ p(L,B) ) (V)⊃V$ Let $x\in U\cap V\cap W$ and suppose $x\in N\in \bigcup _{M\prime \in ℳ}{p}_{M\prime }\text{.}$ Then $N\notin p\left(A,L\right)$ (for $x\in U\in K\left(p\left(A,L\right)\right)\text{)}$ and $N\notin p\left(L,B\right)$ (for $x\in V\in K\left(p\left(L,B\right)\right)\text{),}$ so $N\notin p\left(L,L\prime \right)$ $\text{(}\subset p\left(L,B\right)\text{).}$ It follows that $N\notin p\left(A,L\prime \right)=p\left(A,L\right)\Delta p\left(L,L\prime \right)\text{.}$ So $x\in W\prime$ for some $W\prime \in K\left(p\left(A,L\prime \right)\right)\text{.}$ So now we have ${A}^{U}{L}^{V}B{\sim }_{R\prime }{A}^{U}{L}^{V\prime }{L\prime }^{{V}^{\prime \prime }}B$ for $V\cap V\prime \cap {V}^{\prime \prime }\ne \varnothing$ and $|ℳ\left(L\prime ,B\right)|=1\text{.}$ Since $|ℳ\left(L,L\prime \right)|=|ℳ\left(L,B\right)-\left\{M\right\}|=|ℳ\left(L,B\right)|-1$ the induction hypothesis says that ${A}^{U}{L}^{V\prime }L\prime {\sim }_{R\prime }{A}^{W\prime }L\prime ,$ for $x\in U\cap V\prime \cap W\prime \text{.}$ Finally ${A}^{W\prime }{L\prime }^{{V}^{\prime \prime }}B{\sim }_{R\prime }{A}^{W}B$ for $x\in W\prime \cap {V}^{\prime \prime }\cap W$ and $|ℳ\left(L\prime ,B\right)|=1\text{.}$ $\square$

 Proof of proposition (4.8). Sufficient for the $\left(⇐\right)$ part is: 1. ${A}^{U}{B}^{U}A$ and ${\varnothing }_{A}$ are pregallery-equivalent. Well, $U\cap U\cap 𝕍\ne \varnothing ,$ so ${A}^{U}{B}^{U}A{\sim }_{R}{A}^{𝕍}A{\sim }_{R}{\varnothing }_{A}\text{.}$ 2. If $G$ and $G\prime$ are as in the definition of flip relation (4.5), then $G{\sim }_{R}G\prime \text{.}$ The $\left(⇒\right)$ part: Let $G,G\prime \in {\text{Gal}}_{AB}$ and $G{\sim }_{R}G\prime \text{.}$ To prove: $G\sim G\prime \text{.}$ Let $G={G}_{0},{G}_{1},\dots ,{G}_{n}=G\prime$ be a sequence of pregalleries such that we obtain ${G}_{i}$ from ${G}_{i-1}$ by a pregallery-equivalence operation. By lemma (4.9) we may assume that these operations are in the there defined sharpened form. Choose now each ${G}_{i}$ a gallery ${G}_{i}^{\prime }$ in the following way: substitute for each direct piece a replacing direct gallery (4.4). Then automatically ${G}_{0}^{\prime }={G}_{0}=G$ and ${G}_{n}^{\prime }={G}_{n}=G\prime ,$ because these were already galleries. Claim. ${G}_{i-1}^{\prime }\sim {G}_{i}^{\prime }\text{.}$ Sufficient for this is: Let $\left({A}^{U}{L}^{V}B,{A}^{W}B\right)\in R\prime$ as in (4.9). If $G$ (respectively $G\prime \text{)}$ is a replacing direct gallery of ${A}^{U}L$ (respectively ${A}^{W}B\text{),}$ then (£) $G.{L}^{V}B\sim G\prime \text{.}$ This suffices, because all other replacing direct galleries are (even flip) equivalent by (4.6). For the proof of (£) let $ℳ\left(L,B\right)=\left\{M\right\}$ and consider two cases: (1) $M\notin ℳ\left(A,L\right)\text{.}$ In this case $G.{L}^{V}B$ is direct because $U\cap V\ne \varnothing$ and so we have even $G.{L}^{V}B\approx G\prime \text{.}$ (2) $M\in ℳ\left(A,L\right)\text{.}$ Then $G\prime .{B}^{V}L$ is a replacing direct gallery of ${A}^{U}L\text{.}$ So $G\approx G\prime .{B}^{V}L$ and we have $G.{L}^{V}B\approx G\prime .{B}^{V}{L}^{V}B\sim G\prime \text{.}$ $\square$

Summarising we have the main result of this chapter:

(4.10) Theorem. Let $\left(𝕍,H,ℳ,P\right)$ be a hyperplane system. Let $\text{Gal}$ be the category of all galleries i.e. the free category generated by the so-called one step galleries ${A}^{U}B$ where $A$ and $B$ are chambers separated by one common wall $M,$ $U\in K\left({p}_{M}\right),$ $\text{beg}\left({A}^{U}B\right)≔A$ and $\text{end}\left({A}^{U}B\right)≔B\text{.}$ Let in each $C\in K\left(ℳ\right)$ a point ${a}_{C}$ be chosen and let $\Pi$ be the subgroupoid of the fundamental groupoid of this space $Y\left(𝕍,H,ℳ,p\right)$ (2.9), where ${\Pi }_{AB}$ for $A,B\in K\left(ℳ\right)$ consists of the homotopy classes of paths from $\sqrt{-1}{a}_{A}$ to $\sqrt{-1}{a}_{B}\text{.}$

Then there exists a surjective functor $\phi :\text{Gal}\to \Pi \text{.}$ Moreover the equivalence relation on $\text{Gal}$ "having the same $\phi \text{-image"}$ is generated by the

$cancel relations: ( AU BUA, ∅A ) ∈Gal×Gal (4.7) flip relations: ( G,G′ ) ∈Gal×Gal (4.5)$

where $G$ and $G\prime$ are the two direct replacing galleries of a direct pregallery ${A}^{W}$ ${\text{sp}}_{P}\left(A\right),$ $A\in K\left(ℳ\right),$ $P$ plinth of $A$ and $W\in K\left(p\left(A,{\text{sp}}_{P}\left(A\right)\right)\right)\text{.}$

 Proof. The required functor is the restriction to $\text{Gal}$ of the functor of theorem (2.21). This restriction remains surjective, because each pregallery is pregallery-equivalent to some gallery by (4.4). The assertion about the relations follows from the corresponding one in (2.21) and proposition (4.8). $\square$

This theorem describes the fundamental groupoid of the space $Y\left(𝕍,H,ℳ,p\right)$ as a combinatorial object, namely the quotient category $\text{Gal}/\sim$ which is even more related to the structure of the hyperplane system than $\text{Pgal}/{\sim }_{R}$ (cf. (1.5) and (2.21)).

## §5. Simple Hyperplane Systems and Signed Galleries

In the case of a simple hyperplane system we have ${p}_{M}=\left\{M\right\}$ for all $M\in K\left(ℳ\right)\text{.}$ So if ${A}^{U}B$ is a one-step gallery with $ℳ\left(A,B\right)=\left\{M\right\},$ then we have two possibilities: $U={D}_{{M}_{i}}\left(A\right)$ or $U={D}_{M}\left(B\right)\text{.}$ In the first case the one-step gallery is called positive and we denote it by ${A}^{+}B,$ otherwise it is called negative and denoted ${A}^{-}B\text{.}$ Therefore an arbitrary gallery is now of the form

$C0t1 C1t2 …tk Ck$

where ${C}_{0},{C}_{1},\dots ,{C}_{k}$ is a Tits gallery and ${t}_{i}\in \left\{+,-\right\}\text{.}$ For this reason we will call them sometimes signed galleries. Such a signed gallery is called positive (respectively negative), if ${t}_{i}=+$ (respectively ${t}_{i}=-\text{)}$ for $i=1,\dots ,k\text{.}$ There is a bijective correspondence between the Tits galleries and the positive galleries.

(5.1) Proposition. A positive gallery is direct if and only if the underlying Tits gallery is so.

 Proof. To see the if-part of this statement, note that if ${C}_{0},{C}_{1},\dots ,{C}_{k}$ is a direct Tits gallery and $ℳ\left({C}_{i-1},{C}_{i}\right)=\left\{{M}_{i}\right\},$ then ${M}_{i}\notin ℳ\left({C}_{0},{C}_{i-1}\right)=\left\{{M}_{1},\dots ,{M}_{i-1}\right\},$ for all the ${M}_{i}$ are different (def.(3.21)). So ${C}_{0}\subset {D}_{{M}_{i}}\left({C}_{i-1}\right)$ and thus ${D}_{{M}_{1}}\left({C}_{0}\right)\cap {D}_{{M}_{2}}\left({C}_{2}\right)\cap \dots \cap {D}_{{M}_{k}}\left({C}_{k-1}\right)\ne \varnothing \text{.}$ Hence ${{C}_{0}}^{+}{{C}_{1}}^{+}\dots {{C}_{k}}^{+}$ is direct as a gallery (def. (4.3)). $\square$

Corresponding to the two possibilities of $U\in K\left(ℳ\left(A,B\right)\right)$ the cancel relations are now the pairs of the type:

$(5.2) ( A+ B-A, ∅A ) or ( A- B+A, ∅A ) forA,B∈ K(ℳ), |ℳ(A,B)|=1.$

The flip relations are the pairs:

$(5.3) ( C0t1 C1… tk Ck, D0tk D1… t1 Dk )$

where ${C}_{0},{C}_{1},\dots ,{C}_{k}$ and ${D}_{0},{D}_{1},\dots ,{D}_{k}$ are the two direct Tits galleries from ${C}_{0}={D}_{0}$ to ${C}_{k}={D}_{k}={\text{sp}}_{P}\left({C}_{0}\right)$ for some plinth $P$ of ${C}_{0}$ and in the sequence of signs ${t}_{1},\dots ,{t}_{k}$ there is at most one change of sign.

 Proof of this statement. If (5.3) is a plinth relation, then there is a $U\in K\left(ℳ\left({C}_{0},{C}_{k}\right)\right)=K\left({ℳ}_{P}\right)=\left\{{C}_{0}^{\prime },\dots ,{C}_{k}^{\prime },{D}_{1}^{\prime },\dots ,{D}_{k-1}^{\prime }\right\},$ where ${C}_{i}^{\prime }={\text{pr}}_{P}\left({C}_{i}\right)$ and ${D}_{i}^{\prime }={\text{pr}}_{P}\left({D}_{i}\right),$ that "prescribes signs" i.e. such that both galleries are replacing direct galleries of the pregallery ${{C}_{0}}^{U}{C}_{k}\text{.}$ Suppose for instance that $U={C}_{i}^{\prime }\text{.}$ Then ${t}_{1}={t}_{2}=\dots ={t}_{i}=-$ and ${t}_{i+1}=\dots ={t}_{k}=+,$ hence at most one change of sign (no change if $i=0$ or $i=k\text{).}$ Conversely, if there is at most one change in sign, for instance ${t}_{i}=-$ and ${t}_{i+1}=+,$ then $U={C}_{i}^{\prime }$ is as above. $\square$

Example. To give examples we recall (3.19) that a Tits gallery is given by its first chamber and the sequence of separating walls. A signed gallery can be denoted by this sequence, underlining the elements of $ℳ$ corresponding to the negative steps. For instance, if we number the elements of $ℳ$ in Figure 10 from 1 to 4 then the galleries of the plinth relation $\left({{C}_{0}}^{-}{{C}_{1}}^{+}{{C}_{2}}^{+}{{C}_{3}}^{+}{C}_{4},{{D}_{0}}^{+}{{D}_{1}}^{+}{{D}_{2}}^{+}{{D}_{3}}^{-}{D}_{4}\right)$ can be denoted as $\underset{_}{1}$ 2 3 4 respectively 4 3 2 $\underset{_}{1}$ (assuming that one knows what the first chamber is).

$1 1 2 3 3 4 4 P C0=D0 C4=D4 U=C1 C2 C3 D1 D2 D3 Figure 10.$

We depict a signed gallery by drawing an arrow $\begin{array}{c} \end{array}$ for positive one step galleries and an arrow $\begin{array}{c} \end{array}$ for negative one step galleries. The following sequence illustrates the next remark

$1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1_ 2 3 4 1_ 2 3 4 1 1_ 1_ 1 4 3 2 1_ 4 3 2 1_ Figure 11.$

(5.4) Remark. Two signed galleries are equivalent iff they are equivalent by means of the cancel relations and the positive (i.e. with ${t}_{1}=\dots ={t}_{k}=+$ in (5.3)) flip relations.

 Proof. Consider an arbitrary flip relatlon (5.3) with for instance ${t}_{1}={t}_{2}=\dots {t}_{i}=-$ and ${t}_{i+1}=\dots =+\text{.}$ Then $C0- C1…- Ci+ Ci+1… +Ck$ is by means of cancel relations equivalent to $C0- C1…- Ci+ Ci+1… + Ck+ Dk-1+ …+ Dk-i- Dk-i+1- …-Dk$ and this is by means of a positive flip relation equivalent to $C0- C1…- Ci+ Ci-1… + C0+ D1+ …+ Dk-i- Dk-i+1- …-Dk$ and this is by means of cancel relations equivalent to $C0+ D1+ …+ Dk-i- Dk-i+1- …-Dk.$ $\square$

(5.5) We conclude this paragraph with an example of two equivalent galleries which are not flip equivalent:

$1 2 3 1 2 3 1 2 2 1_ 3_ 2 2 3 Figure 12.$

They are equivalent for

$1 2 2 1_∼ 1 2 3_ 3 2 1_≈ 3_ 2 1 3 2 1_≈ 3_ 2 1 1_ 2 3∼ 3_ 2 2 3.$

## Notes and References

This is an excerpt of a thesis entitled The Homotopy Type of Complex Hyperplane Complements, written by Harm van der Lek in 1983.