Disconnected diagrams, direct products, and comparison of invariant forms

(Commuting factors and trivial intersections.) If $S=\mathrm{\varnothing }$, all four clauses hold: $W=\{1\}$, $V=\{0\}$, the product and sums are empty, and positive definiteness is vacuous. Hence assume $S\mathrm{\ne }\mathrm{\varnothing }$ for the remaining argument. Let $s\in S_i$, $t\in S_j$ with $i\mathrm{\ne }j$; by [F2] $m(s,t)=2$, so $(st{)}^2$ is a relator and $st=ts$ in $W$ by [F1]; since the $s\in S_i$ generate $W_i$ [F1, F5], the subgroups $W_i,W_j$ commute elementwise. For the intersection, $W_i\cap W_j=W_{S_i}\cap W_{S_j}=W_{S_i\cap S_j}=W_{\mathrm{\varnothing }}=\{1\}$ by the support description of [F1], because $S_i\cap S_j=\mathrm{\varnothing }$ [F2].

(Orthogonal decomposition.) For $s\in S_i$ and $t\in S_j$ with $i\mathrm{\ne }j$, $m(s,t)=2$ [F2] and hence $B(e_s,e_t)=-\cos (\pi \mathrm{/}2)=0$ by [F3]; by bilinearity [F6] this gives $B(V_i,V_j)=0$. Since the $e_s$ form a basis of $V$ and the $S_i$ partition it, grouping the unique basis expansion by its supports $S_i$ gives a unique decomposition into vectors of the $V_i$, so $V=V_1\oplus \cdots \oplus V_k$ is a $B$-orthogonal direct sum and $B=B_1\oplus \cdots \oplus B_k$ with $B_i=B{\mathrm{\mid }}_{V_i\times V_i}$ [F6]. For the action, the reflection formula gives $r_{e_s}{e}_t=e_t-2B(e_t,e_s)e_s$: for $t\in S_i$ this lies in $V_i$, while for $t\in S_j$, $j\mathrm{\ne }i$, it equals $e_t$ because $B(e_t,e_s)=B(e_s,e_t)=0$; hence $\rho (s)$ preserves $V_i$ and fixes each $V_j$ with $j\mathrm{\ne }i$ pointwise, and the same holds for every $\rho (w)$ with $w\in W_i$ since these are products of such generators [F4, F5].

(Proportionality on one generator.) Fix $s\in S$ and put $H=\{v:B(v,e_s)=0\}$; by [F3] $H$ is a hyperplane fixed pointwise by $r_{e_s}=\rho (s)$ and $r_{e_s}{e}_s=-e_s$. For $v\in H$, invariance of $\beta$ under $\rho (s)$ gives $\beta (e_s,v)=\beta (\rho (s)e_s,\rho (s)v)=\beta (-e_s,v)=-\beta (e_s,v)$, so $\beta (e_s,v)=0$; thus the linear functional $\beta (e_s,\cdot )$ vanishes on $H$, as does $B(e_s,\cdot )$, which is nonzero because $B(e_s,e_s)=1$ [F3]. Any linear functional $\psi$ vanishing on $H=\ker \phi$ is a multiple of the nonzero functional $\phi$: if $\phi (v_0)\mathrm{\ne }0$, then $u-(\phi (u)\mathrm{/}\phi (v_0))v_0\in H$ for every $u$, so $\psi (u)=(\psi (v_0)\mathrm{/}\phi (v_0))\phi (u)$. Hence $\beta (e_s,\cdot )={\lambda }_{s}B(e_s,\cdot )$ with ${\lambda }_s:=\beta (e_s,e_s)$, and evaluating at $e_t$ gives $\beta (e_s,e_t)={\lambda }_{s}B(e_s,e_t)$ for all $t$.

(The multiplication map is an isomorphism.) Every relator of $(S,m)$ is mapped to $1$ by the assignment $s\mapsto (1,\dots ,s,\dots ,1)\in W_1\times \cdots \times W_k$ placing $s$ in the factor $W_i$ with $s\in S_i$: the relators $s^2$ and $(st{)}^{m(s,t)}$ with $s,t$ in one component hold in that factor because they hold in $W$, and for $s\in S_i$, $t\in S_j$ with $i\mathrm{\ne }j$ the two images have disjoint supports, hence commute and are involutions, so the image of $(st{)}^2$ is $1$; by the universal property [F1] there is a homomorphism $\phi :W\to W_1\times \cdots \times W_k$ with $\phi (s)=(1,\dots ,s,\dots ,1)$. Conversely the inclusions $W_i\to W$ are homomorphisms [F1] with pairwise commuting images by step 1.1, so $(w_1,\dots ,w_k)\mapsto w_1\cdots w_k$ is a homomorphism $\psi :W_1\times \cdots \times W_k\to W$ [F5]. The two are mutually inverse: $\psi \phi$ and the identity of $W$ are homomorphisms agreeing on the generating set $S$, and $\phi \psi$ and the identity of $W_1\times \cdots \times W_k$ are homomorphisms agreeing on each coordinate generating set $W_i$ (a generator $s\in S_i$ of the $i$-th factor is sent by $\phi \psi$ to $\phi (s)=(1,\dots ,s,\dots ,1)$); hence $\mu =\psi$ is an isomorphism [F5].

(The scalars are constant on components.) For $s\mathrm{\ne }t$ in the same component, step 1.3 gives $\beta (e_s,e_t)={\lambda }_{s}B(e_s,e_t)$ and, by symmetry of $\beta$ and $B$, $\beta (e_s,e_t)=\beta (e_t,e_s)={\lambda }_{t}B(e_t,e_s)={\lambda }_{t}B(e_s,e_t)$; since $s,t$ lie in one component and are joined by a path, it suffices to treat adjacent pairs, where $B(e_s,e_t)\mathrm{\ne }0$: indeed for finite labels the cosine is positive when $m(s,t)\ge 3$, while an infinite label has $B(e_s,e_t)=-1$ [F3]; thus $B(e_s,e_t)=0$ forces $m(s,t)=2$, i.e. no edge [F2]. For such a pair $({\lambda }_s-{\lambda }_t)B(e_s,e_t)=0$ gives ${\lambda }_s={\lambda }_t$, and equality propagates along the edges of the connected component [F2], so there is ${\lambda }_i$ with ${\lambda }_s={\lambda }_i$ for all $s\in S_i$. Evaluating $\beta$ on pairs of basis vectors of $V_i$ step 1.3 then gives $\beta {\mathrm{\mid }}_{V_i\times V_i}={\lambda }_i{B}_i$ for every $i$, that is, $\beta ={\lambda }_1{B}_1\oplus \cdots \oplus {\lambda }_k{B}_k$ [F6]; if $k=1$ this is $\beta =\lambda B$.

(Length additivity.) Let $w_i\in W_i$. Choosing a reduced expression of each $w_i$, concatenation represents $w_1\cdots w_k$ with ${\sum }_i\mathrm{\ell }(w_i)$ letters, so $\mathrm{\ell }(w_1\cdots w_k)\le {\sum }_i\mathrm{\ell }(w_i)$. For the reverse inequality let $w_1\cdots w_k=s_1{s}_2\cdots s_{\mathrm{\ell }}$ be a reduced expression of length $\mathrm{\ell }=\mathrm{\ell }(w_1\cdots w_k)$, with letters $s_j\in S$; letters lying in distinct components commute in $W$ by step 1.1, so we may reorder the $s_j$ within this word so that the letters of each $S_i$ become consecutive (the value in $W$ is unchanged), obtaining $w_1\cdots w_k=w_1^{\mathrm{\prime }}\cdots w_k^{\mathrm{\prime }}$ with $w_i^{\mathrm{\prime }}$ a product of $n_i$ letters from $S_i$ and ${\sum }_i{n}_i=\mathrm{\ell }$. By the isomorphism of step 2.1 the projection $W\to W_i$ is the restriction of the inverse map and is a homomorphism [F5], so it sends $w_1\cdots w_k$ to $w_i$ and $w_1^{\mathrm{\prime }}\cdots w_k^{\mathrm{\prime }}$ to $w_i^{\mathrm{\prime }}$; hence $w_i^{\mathrm{\prime }}=w_i$ and $\mathrm{\ell }(w_i)\le n_i$ for each $i$; summing, ${\sum }_i\mathrm{\ell }(w_i)\le \mathrm{\ell }$.

(Positive definite invariant forms give positive definite $B$.) Assume $\beta$ positive definite. By step 2.2 $\beta ={\lambda }_1{B}_1\oplus \cdots \oplus {\lambda }_k{B}_k$, and each ${\lambda }_i=\beta (e_s,e_s)>0$ for $s\in S_i$ because $e_s\mathrm{\ne }0$ and $\beta$ is positive definite [F7]. Hence $B_i={\lambda }_i^{-1}\beta {\mathrm{\mid }}_{V_i\times V_i}$ is a positive multiple of the restriction of a positive definite form and is positive definite [F7]; a $B$-orthogonal direct sum of positive definite forms is positive definite, since a nonzero vector has some nonzero component $v_i$ and $B(v,v)\ge B_i(v_i,v_i)>0$ [F6, F7]. Thus $B=B_1\oplus \cdots \oplus B_k$ is positive definite, which is (3)(iii).

(Finite $W$ has positive definite $B$.) Assume $W$ finite and define $\beta (u,v):={\sum }_{w\in W}{\beta }_0(\rho (w)u,\rho (w)v)$, a finite sum over the finite set $W$ [F8] of symmetric bilinear terms, hence a symmetric bilinear form. It is $\rho (W)$-invariant: for $g\in W$, substituting $w^{\mathrm{\prime }}=wg$ and using that $w\mapsto wg$ is a bijection of the finite set $W$ [F8] gives $\beta (\rho (g)u,\rho (g)v)={\sum }_w{\beta }_0(\rho (wg)u,\rho (wg)v)={\sum }_{w^{\mathrm{\prime }}}{\beta }_0(\rho (w^{\mathrm{\prime }})u,\rho (w^{\mathrm{\prime }})v)=\beta (u,v)$. It is positive definite: every summand is $\ge 0$ by [F8] and the summand with $w=1$ equals ${\beta }_0(u,u)>0$ for $u\mathrm{\ne }0$, so $\beta (u,u)>0$. Applying step 3.2 to this $\beta$ gives that $B$ is positive definite, which is (4).