Simple and reflection lengths of a long transposition in $S_5$

Put $c:=\cos (\pi \mathrm{/}3)$. By [F7], $0<\pi \mathrm{/}3<\gamma <2$ gives $c>\cos \gamma =0$, while $2c^2-1=\cos (2\pi \mathrm{/}3)=-c$, so $(2c-1)(c+1)=0$ and $c=\frac{1}{2}$; also $\cos (\pi \mathrm{/}2)=0$ by [F7]. Put $v_i:=\frac{1}{\sqrt{2}}(e_i-e_{i+1})\in \mathbf{H}$ for $1\le i\le 4$. Then $⟨v_i,v_i⟩=1$, $⟨v_i,v_{i+1}⟩=-\frac{1}{2}$ and $⟨v_i,v_j⟩=0$ for $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$, matching $B(e_{s_i},e_{s_j})=-\cos (\pi \mathrm{/}m(s_i,s_j))$ by [F2] and [F7]; since the $v_i$ are linearly independent (their coordinates in the order $(e_1,\dots ,e_5)$ are $(a,-a+b,-b+c,-c+d,-d)\mathrm{/}\sqrt{2}$ for $a{v}_1+b{v}_2+c{v}_3+d{v}_4$) and every $x\in \mathbf{H}$ equals $\sqrt{2}{\sum }_{i=1}^4(x_1+\cdots +x_i)v_i$, they form a basis of $\mathbf{H}$, so $\dim \mathbf{H}=4$ and the map $\mathrm{\Theta }:V\to \mathbf{H}$ with $\mathrm{\Theta }(e_{s_i})=v_i$ is a linear isometry onto $\mathbf{H}$. For each $i$ the transposition $(ii+1)$ preserves $\mathbf{H}$, reverses $v_i$ and fixes ${v_i}^{\perp }\cap \mathbf{H}$ pointwise, so it is the reflection with normal line $\mathbb{R}{v}_i$; the image $\mathrm{\Theta }\rho (s_i){\mathrm{\Theta }}^{-1}$ is by [F2] and [F6] likewise an inner-product-preserving involution of $\mathbf{H}$ that reverses $v_i$ and fixes ${v_i}^{\perp }\cap \mathbf{H}$ pointwise, so the two agree on all of $\mathbf{H}$. Since $\phi$ is an isomorphism and $S$ generates $W$ [F1], the homomorphisms $\mathrm{\Theta }\rho {\mathrm{\Theta }}^{-1}$ and $\sigma \mapsto \sigma {\mathrm{\mid }}_{\mathbf{H}}$ agree on $S$, hence everywhere: $\mathrm{\Theta }\rho (w){\mathrm{\Theta }}^{-1}=\phi (w){\mathrm{\mid }}_{\mathbf{H}}$ for all $w\in W$. The permutation action on $\mathbf{H}$ is faithful, because if $\sigma$ acts trivially then $\sigma (e_i-e_k)=e_i-e_k$ for all $i\mathrm{\ne }k$, and choosing $k\mathrm{\notin }\{i,\sigma (i)\}$ forces $\sigma (i)=i$. Therefore for $t=ws{w}^{-1}\in T$ the permutation $\phi (t)$ has moved line $\mathrm{\Theta }\mathbb{R}\rho (w)e_s=\mathbb{R}(e_p-e_q)$ for some $p\mathrm{\ne }q$ and fixes the orthogonal hyperplane pointwise, so $\phi (t)(pq{)}^{-1}$ acts trivially on $\mathbf{H}$ and $\phi (t)=(pq)$; conversely every transposition $(pq)=\sigma (12){\sigma }^{-1}$ equals $\phi (u{s}_1{u}^{-1})$ for $u:={\phi }^{-1}(\sigma )$, so $T=\{{\phi }^{-1}(\tau ):\tau \text{ a transposition}\}$.

For a permutation $\sigma \in S_5$ the inversion number of the transposition $(ij)$, $i<j$, is $2(j-i)-1$: the pairs $a<b$ with $(ij)a>(ij)b$ are exactly the $j-i-1$ pairs $(i,k)$ with $i<k<j$, the single pair $(i,j)$, and the $j-i-1$ pairs $(k,j)$ with $i<k<j$. In particular the transposition $(15)$ has $2\cdot 4-1=7$ inversions, and the reversal $(15)(24)$ inverts every one of the $\left(\genfrac{}{}{0px}{}{5}{2}\right)=10$ pairs, so it has $10$ inversions, the maximal value; in $S_5$ this is the longest element.

By step 1.1 the fixed space of $\rho (w)$ corresponds to $\{x\in \mathbf{H}:\sigma x=x\}$ for $\sigma =\phi (w)$, and ${\mathrm{\ell }}_T(w)=4-\dim \{x\in \mathbf{H}:\sigma x=x\}$ by [F3]. The fixed space of $\sigma$ in ${\mathbb{R}}^5$ is spanned by the incidence vectors of its cycles, so it has dimension $c(\sigma )$ and its intersection with $\mathbf{H}$ is defined by the single equation ${\sum }_C\mathrm{\mid }C\mathrm{\mid }{a}_C=0$ on the cycle coefficients $a_C$. Every $\mathrm{\mid }C\mathrm{\mid }$ is positive, so fixing one cycle lets its coefficient be solved uniquely from the other $c(\sigma )-1$ coefficients; the intersection therefore has dimension $c(\sigma )-1$; hence ${\mathrm{\ell }}_T(w)=4-(c(\sigma )-1)=5-c(\sigma )$, with $c(\sigma )$ the number of cycles of $\sigma$ (fixed points included). In particular a transposition has $c=4$ and ${\mathrm{\ell }}_T=1$, the long transposition $(15)$ has $c=4$ and ${\mathrm{\ell }}_T=1$, the reversal $(15)(24)$ has $c=3$ and ${\mathrm{\ell }}_T=2$, and a $5$-cycle has $c=1$ and ${\mathrm{\ell }}_T=4=\dim V$.

Collecting the results: by step 1.1 the reflections of $W$ are exactly the ${\phi }^{-1}(\tau )$ with $\tau$ a transposition, and each has ${\mathrm{\ell }}_T=1$ by step 2.1, which is claim (i)'s first part, while claim (i)'s second part is the inversion count of step 1.2. For the long transposition, $\mathrm{\ell }({\phi }^{-1}(15))=7$ by steps 1.2 and [F1] and ${\mathrm{\ell }}_T({\phi }^{-1}(15))=1$ by step 2.1, and ${\phi }^{-1}(15)\in T$ because $(15)=(25)(12)(25{)}^{-1}$ exhibits $(15)$ as $\phi (u{s}_1{u}^{-1})$ for the element $u:={\phi }^{-1}(25)$; this is claim (ii). Claim (iii)'s dimension formula, the values $\mathrm{\ell }(w_0)=10$ and ${\mathrm{\ell }}_T(w_0)=2$, and ${\mathrm{\ell }}_T=4$ for every $5$-cycle are steps 1.2, 2.1 and [F1].