Use the counit from part b for the first map and the restriction map $\mathcal{G}\to i_{\ast }{i}^{-1}\mathcal{G}$ for the second. Exactness can be checked on stalks. At $P\in U$ the sequence iswhereas at $P\in Z$ it isThusis a short exact sequence of sheaves.

On the defining presheaf, send a section over $V\subseteq U$ by the identityand use the unique zero map when $V⊈U$. These maps commute with restrictions and therefore sheafify to the natural counit

After restricting $j^p\mathcal{F}$ back to $U$, every open set lies in $U$, so one recovers the original sheaf $\mathcal{F}$. Equivalently, the natural map $\mathcal{F}\to j^{-1}{j}_{!}\mathcal{F}$ is an isomorphism on every stalk and hence an isomorphism of sheaves.

Take $X=\mathbb{R}$, $U=(0,1)$, and let $\mathcal{F}$ be the skyscraper sheaf with value a nonzero abelian group $A$ at $p=3/4$. Cover $V=(-1,1)$ by $V_1=(-1,2/3)$ and $V_2=(1/2,1)$. The presheaf gives $(j^p\mathcal{F})(V)=0=(j^p\mathcal{F})(V_1)$ and $(j^p\mathcal{F})(V_2)=A$. A nonzero section on $V_2$ and the zero section on $V_1$ agree on the overlap, whose skyscraper sections vanish, but cannot be glued on $V$. Thus $j^p\mathcal{F}$ need not be a sheaf.

Sheafification preserves stalks. If $P\in U$, neighborhoods contained in $U$ are cofinal, so $(j^p\mathcal{F}{)}_P={\mathcal{F}}_P$. If $P\notin U$, no neighborhood of $P$ lies in $U$, so every term defining the presheaf stalk is zero. Therefore

First suppose $(W,S)$ is a Coxeter system. The usual exchange condition implies that multiplication by a simple generator changes length by exactly one. Let $w=t_1\cdots t_n$ be reduced and suppose both $s_{i}w$ and $w{s}_j$ have length $n+1$. The length of $s_{i}w{s}_j$ is therefore either $n+2$ or $n$. In the latter case, apply exchange to the reduced word $s_i{t}_1\cdots t_n$ followed by $s_j$. If exchange deleted one of the $t_k$, multiplying the resulting equality on the left by $s_i$ would express the length-$n+1$ element $w{s}_j$ using only $n-1$ generators. Hence exchange must delete the initial $s_i$, giving $s_{i}w{s}_j=w$. This is exactly the folding condition.

Conversely, suppose the folding condition holds, and let $W_M$ be the abstract Coxeter group with generators $S$ and matrix $(m_{ij})$. The defining relations hold in $W$, so there is a surjective homomorphismIt remains to prove injectivity. Take any word in the kernel. If its image word in $W$ is not reduced, choose its shortest nonreduced prefix $ut$, where $u$ is reduced and $t$ is its last generator. Part b gives $ut=v$, where $v$ is obtained by deleting one letter from $u$. Hence $u=vt$, and $u$ and $vt$ are two reduced expressions for the same element. By the assumed braid-equivalence theorem they are related by braid moves. Those moves are defining relations in $W_M$, after which the end of the prefix becomes $vtt$ and $t^2=1$ shortens the original word by two.

Repeating this process turns the kernel word, using only Coxeter relations, into a word that is reduced in $W$. Since its image is the identity, that reduced word is empty. The original word is therefore already the identity in $W_M$, so $\ker \pi =1$. Hence $W\cong W_M$ is the Coxeter group with generators $S$.

If $m_{ij}<\infty$, the braid relation in a Coxeter group isFor $m_{ij}=2$ this is commutation; no relation is imposed when $m_{ij}=\infty$. Two reduced expressions are braid equivalent when one can be transformed into the other by finitely many replacements of one side of such a relation by the other.

We prove the assertion by induction on $n$. Write $w=av$, where $a=s_{i_1}$ and $v=s_{i_2}\cdots s_{i_n}$; both displayed words are reduced. The first clause of the folding condition and part a imply that $\ell (vs)$ is either $n-2$ or $n$.

If $\ell (vs)=n-2$, the induction hypothesis deletes one unique letter from the reduced word for $v$. Prefixing $a$ gives the required deletion from $w$. If another deletion were possible, induction rules out another internal position, while deletion of $a$ would give $ws=v$ and hence $vs=av$, contradicting the lengths $n-2$ and $n$.

If $\ell (vs)=n$, both $av$ and $vs$ increase the length of $v$. Since $\ell (avs)=\ell (ws)=n-1$, the other alternative in the folding condition must hold:Thus $ws=avs=v$, which deletes the first letter. An additional internal deletion would give $v=a{v}^{\prime }$ for a word $v^{\prime }$ of length $n-2$; multiplying by $a$ would make the length-$n$ element $av=w$ equal to $v^{\prime }$, impossible. The deletion position is therefore unique in every case.

Reversing a word for $w$ gives a word of the same length for $w^{-1}$ because every generator is an involution. Applying the same argument to $w^{-1}$ provesA shortest word for $w$ followed by $s_i$ gives ${\ell }_S(w{s}_i)\le {\ell }_S(w)+1$. Conversely, $w=(w{s}_i)s_i$ gives ${\ell }_S(w)\le {\ell }_S(w{s}_i)+1$. Hence

If $W$ is finite, its finitely many reflecting hyperplanes divide $V^{\ast }$ into chambers, and the closures of these chambers are exactly the translates ${\sigma }^{\ast }(w)(D)$. Hence the Tits coneequals $V^{\ast }$.

Conversely, suppose $W$ is infinite and choose $f\in -C$, so $f(e_i)<0$ for every $i$. If $f\in U$, then $f\in {\sigma }^{\ast }(w)(D)$ for some $w$, and thereforefor every $i$. Since $f$ is strictly negative on every positive root and strictly positive on every negative root, each $\sigma (w)e_i$ must be negative. The length criterion givesfor every $i$. This contradicts the stated fact that in an infinite Coxeter group the length of every element can be increased by multiplication on the right by some simple generator. Thus $-C$ is not contained in $U$, and $U\ne V^{\ast }$.

Suppose ${\sigma }^{\ast }(w)(C)$ meets ${\sigma }^{\ast }(v)(C)$. Applying the homeomorphism ${\sigma }^{\ast }(v^{-1})$ shows that $C$ meets ${\sigma }^{\ast }(v^{-1}w)(C)$. If $v^{-1}w\ne 1$, choose one of its left descents. Part b places its image of $C$ in the corresponding negative half-space, while $C$ lies in the positive half-space. This is impossible, so $v=w$. The unionis disjoint.

When $W$ is finite, the closures ${\sigma }^{\ast }(w)(D)$ are the simplicial chambers cut out by the reflecting hyperplanes. Intersecting them with a sphere centred at the origin gives simplices. A face of type $J\subseteq I$ has stabilizer the standard parabolic subgroup $W_J$, so its translates are indexed by cosets $w{W}_J$, with reverse inclusion of cosets encoding incidence. This is precisely the Coxeter complex of $(W,I)$.

For $f\in C$, duality givesThe positive-root criterion for Coxeter length saysis a positive root. Its basis coefficients are nonnegative and not all zero, so $f(\sigma (w^{-1})e_i)>0$ and ${\sigma }^{\ast }(w)(C)\subseteq A_i$. If the length decreases, that root is negative and the same calculation gives ${\sigma }^{\ast }(w)(C)\subseteq A_i^{-}$.

If ${\sigma }^{\ast }(w)$ is the identity and $w\ne 1$, choose a left descent $x_i$ from the first letter of a reduced expression for $w$. The preceding result givesalthough $C\subseteq A_i$. The two open half-spaces are disjoint, a contradiction. Thus the Dual geometric representation of a Coxeter group is faithful.

Use the finite-dimensional vector-space topology on $V^{\ast }$: relative to the dual basis, it is the ordinary Euclidean topology on ${\mathbb{R}}^{|I|}$. Each coordinate map $f\mapsto f(e_i)$ is continuous, sois closed, while $A_i$, $A_i^{-}$, and the finite intersection $C={\bigcap }_i{A}_i$ are open. Their closures are

Every ${\sigma }^{\ast }(w)$ is an invertible linear map with inverse ${\sigma }^{\ast }(w^{-1})$. Linear maps between finite-dimensional topological vector spaces are continuous, so both it and its inverse are continuous. Hence every ${\sigma }^{\ast }(w)$ is a homeomorphism.

For the same graph, let $e_0$ correspond to the central vertex and $e_1,\dots ,e_4$ to the leaves. The associated simply-laced Coxeter Gram matrix hasfor distinct leaves. The nonzero vectorsatisfies $(\delta ,e_i)=0$ for every basis vector. Thus the form is degenerate; in fact it is positive semidefinite with one-dimensional radical, as expected for the affine graph ${\tilde{D}}_4$.

The geometric form of a finite Coxeter group is positive definite. Since this form is degenerate, the group cannot be finite.

The exchange condition for a Coxeter group says that if $w=s_1\cdots s_n$ is reduced and $s$ is simple with $\ell (ws)<\ell (w)$, thenfor some $j$. The Matsumoto theorem says that any two reduced expressions for the same element are connected by braid moves. Together they imply the Tits word reduction theorem: a nonreduced word can be transformed by braid moves until two equal adjacent generators can be cancelled.

For the displayed four-armed graph, call the central generator $s$ and the leaves $a,b,c,d$. Different leaves commute, while each leaf $t$ satisfies $sts=tst$. Consider the wordBetween successive occurrences of $s$, the intervening leaf sets alternate between $\{a,b\}$ and $\{c,d\}$. Commuting the two leaves in one block never puts the same leaf on both sides of an $s$, so no length-three braid $tst\leftrightarrow sts$ is ever available. The only possible braid moves are those leaf commutations, and they cannot create adjacent equal letters. Tits reduction therefore shows that $u_N$ is reduced. Since $\ell (u_N)=6N$ is unbounded, the group is infinite.

If a finitely generated Coxeter group $W$ is finite, its integer-valued Coxeter length has a maximum. Conversely, if some $w_0$ has globally maximal length $N$, every group element has a word of length at most $N$. There are only finitely many words of bounded length in the finite set of simple generators, so $W$ is finite.

Realize the finite group as the reflection group of a root system with fundamental system $\Delta$ and positive system $\Pi$. Maximality and the fact that multiplication by a simple generator changes Coxeter length by one giveThe positive-root criterion for Coxeter length therefore gives $w_0\Delta \subseteq -\Pi$. Since $w_0\Delta$ is itself fundamental, it must be the simple system $-\Delta$ of the positive system $-\Pi$.

If $v_0$ is another maximal-length element, the same argument gives $v_0\Delta =-\Delta =w_0\Delta$. Hence $v_0^{-1}{w}_0$ stabilizes $\Delta$, and part c gives $v_0^{-1}{w}_0=1$. The longest element of a finite Coxeter group is therefore unique.

Use the positive-root criterion for Coxeter length: for a simple root $\alpha \in \Delta$,The hypothesis $w\Delta \subseteq \Pi$ therefore says that every simple generator is a right ascent of $w$. If $w\ne 1$, a reduced expression in a Coxeter group for $w$ has a final simple generator $s_{\alpha }$, and deleting it givesa contradiction. Hence $w=1$.

If $w$ stabilizes $\Delta$ setwise, then $w\Delta =\Delta \subseteq \Pi$, so the result just proved gives $w=1$. Thus the stabilizer of every fundamental system is trivial.

The reflection group of a root system isEvery generating reflection permutes $\Phi$, so every $w\in W(\Phi )$ does too. If $\Delta$ is fundamental, then $w\Delta$ is a basis contained in $\Phi$. Writingshows thatthe coefficients are unchanged and therefore still have one sign. Thus $w\Delta$ is another fundamental system, and $W(\Phi )$ acts on the set of all fundamental systems.

A root system in the real inner product space $V$ is a finite spanning set $\Phi \subset V\setminus \{0\}$ such thatwhere the orthogonal reflectionFor a crystallographic root system one additionally requires $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$; that condition is not needed for general finite reflection groups.

A fundamental system of a root system is a basis $\Delta \subset \Phi$ such that every root has either all nonnegative or all nonpositive coordinates in this basis. Its associated positive system of a root system isand $\Phi =\Pi \bigsqcup (-\Pi )$.