The Iwahori–Matsumoto basis and its multiplication table

$T_sT_w$ is supported on $IsI\cdot IwI$ and is a combination of the $T_z$ (Hecke algebras, Lemma 2.2(ii)); (F3) lists the double cosets in that product. Since $s^2=e$ in $\Wt$, $\dot s^2\in T(\cO)\subset I$, so $\dot s^{-1}\in\dot sI$ and $(I\dot sI)^{-1}=I\dot s^{-1}I=I\dot sI$.

1. By Step 1, $T_sT_w=c\,T_{sw}$ with $c=c^{\dot s\dot w}_{s,w}=\#\{uI\subseteq IsI:\ u^{-1}\dot s\dot w\in IwI\}$. The coset $u=\dot s$ counts, so $c\ge1$.
2. Let $uI\subseteq IsI$ with $uI\neq\dot sI$; write $u=x\dot s$ with $x\in I$ (every element of $I\dot sI$ lies in $I\dot s\,I$, and the right factor can be dropped from the coset). Then $y:=\dot s^{-1}x^{-1}\dot s\notin I$, since otherwise $uI=x\dot sI=\dot s(\dot s^{-1}x\dot s)I=\dot sI$. But $y\in\dot s^{-1}I\dot s\subseteq IsI\cdot IsI\subseteq I\cup IsI$ by (F3) applied to $w=s$ (where $\len(ss)=0=\len(s)-1$). Hence $y\in IsI$.
3. If this $u$ counted, then $u^{-1}\dot s\dot w=y\dot w\in IwI$. Also $y\dot w\in IsI\cdot IwI=IswI$ by (F3). So $IwI\cap IswI\neq\varnothing$, contradicting (F1) since $sw\neq w$. Therefore only $u=\dot s$ counts: $c=1$.
4. Apply the homomorphism $\ind$ to $T_sT_w=T_{sw}$: $\ind(s)\ind(w)=\ind(sw)$, and $\ind(s)=q$ by (F2).

1. For $w\in\Waff$, induct on $\len(w)$: $\ind(e)=|I/I|=1$; if $\len(w)\ge1$ choose a reduced expression and write $w=sw'$ with $\len(w')=\len(w)-1$; Step 2 gives $\ind(w)=q\,\ind(w')=q\cdot q^{\len(w)-1}$.
2. For $w=w'\sigma$ with $w'\in\Waff$, $\sigma\in\Om$: by (F4), $Iw'\sigma I=Iw'I\sigma$, and $zI\mapsto z\sigma I=zI\sigma$ is a bijection of $\GF/I$ carrying $Iw'I/I$ onto $Iw'\sigma I/I$. So $\ind(w'\sigma)=\ind(w')=q^{\len(w')}=q^{\len(w)}$ by (F5).

1. By Step 1 with $w=s$, $T_s^2=a\,T_e+b\,T_s$ with $a,b\in\N$.
2. $a=c^{e}_{s,s}=\#\{uI\subseteq IsI:\ u^{-1}\in IsI\}=|IsI/I|=q$, because $(IsI)^{-1}=IsI$ (Step 1) and (F2).
3. Apply $\ind$: $q^2=a+bq=q+bq$, so $b=q-1$.

Put $w'=sw$; then $w=sw'$ with $\len(w)=\len(w')+1$, so $T_w=T_sT_{w'}$ by Step 2. Hence, by Step 4,

By (F4), $I\sigma I=\dot\sigma I$ is a single coset. So $(T_\sigma T_w)(z)=\#\{uI\subseteq\dot\sigma I:u^{-1}z\in IwI\}=\one_{\dot\sigma IwI}(z)$, and $\dot\sigma IwI=I\dot\sigma\dot wI$ since $\dot\sigma I=I\dot\sigma$. For the right-hand formula, $(T_wT_\sigma)(z)=\#\{uI\subseteq IwI:\ u^{-1}z\in\dot\sigma I\}=\#\{uI\subseteq IwI:\ uI=z\dot\sigma^{-1}I\}=\one_{IwI\dot\sigma}(z)=\one_{Iw\sigma I}(z)$. The last two formulas follow: $\sigma\sigma'\in\Om$, and $T_\sigma T_s=T_{\sigma s}=T_{(\sigma s\sigma^{-1})\sigma}=T_{\sigma s\sigma^{-1}}T_\sigma$ with $\sigma s\sigma^{-1}\in\Saff$ by (F5).

1. Generation. For $w=s_{i_1}\cdots s_{i_r}\sigma$ reduced, each suffix $s_{i_k}\cdots s_{i_r}$ has length $r-k+1$, so Step 2 applies $r$ times and Step 6 once: $T_w=T_{s_{i_1}}\cdots T_{s_{i_r}}T_\sigma$. This is [IM65, Thm 3.3].
2. Braid relations. For $s\neq t$ in $\Saff$ with $(st)^{m}=1$, $m<\infty$, the two alternating words of length $m$ are reduced expressions of one element $w_{st}\in\Waff$ (F5), so by step 1 both $T_sT_tT_s\cdots$ and $T_tT_sT_t\cdots$ equal $T_{w_{st}}$ [IM65, Prop 3.4(ii)].
3. Presentation. Let $\mathcal A$ be the algebra generated by symbols $\Delta_s$ subject to the quadratic and braid relations; $\Delta_s\mapsto T_s$ is a surjection $\mathcal A\to\Haff$ by step 1. For each $w\in\Waff$ put $\Delta_w=\Delta_{s_{i_1}}\cdots\Delta_{s_{i_r}}$ along a reduced expression; by Matsumoto's theorem (any two reduced expressions are joined by braid moves) this does not depend on the expression [IM65, Prop 1.15]. The relations give $\Delta_s\Delta_w\in\C\Delta_{sw}+\C\Delta_w$ in both length cases, so the $\Delta_w$ span $\mathcal A$; they map onto the basis $T_w$, hence $\mathcal A\simeq\Haff$ [IM65, Thm 3.5]. The decomposition $\cH=\bigoplus_\sigma\Haff T_\sigma$ is (a) and Step 6 [IM65, Prop 3.8].
4. Right-hand versions. $\iota(f)(x)=f(x^{-1})$ is an anti-automorphism of $\cH$ ($\iota(f\ast g)=\iota(g)\ast\iota(f)$, by the substitution $y\mapsto x^{-1}y^{-1}$ in the convolution) with $\iota(T_w)=T_{w^{-1}}$, since $(IwI)^{-1}=Iw^{-1}I$. Applying $\iota$ to the formulas for $T_sT_{w^{-1}}$ and using $\len(w^{-1}s)=\len(sw^{-1})$... precisely $\len((sw^{-1})^{-1})=\len(ws)$ by (F5), gives $T_wT_s=T_{ws}$ or $q\,T_{ws}+(q-1)T_w$ according as $\len(ws)=\len(w)\pm1$.

$K=\bigsqcup_{w\in W}IwI$ (Hecke algebras, Lemma 5.1) gives the first formula; applying $\ind$ and Step 3 gives the second, which agrees with Groups, Lemma 2.2. For left $K$-invariant $\varphi$, $(\one_K\ast\varphi)(x)=\sum_{y\in K/I}\varphi(y^{-1}x)=[K:I]\varphi(x)$; with $\varphi=\one_K$ this is the third formula, and dividing by $[K:I]$ gives the rest. $\HK=\eK\cH\eK$ is the source page, Lemma 4.1. If $W\neq1$ then $\one_K\neq T_e$ since $K\neq I$.

The quadratic relation reads $T_s\big(T_s-(q-1)T_e\big)=q\,T_e$; Step 6 gives $T_\sigma T_{\sigma^{-1}}=T_e$; Step 7 writes $T_w$ as a product.

1. $\ind$ is the integral, an algebra homomorphism by Hecke algebras, Lemma 3.2, with $\ind(T_w)=q^{\len(w)}$ by Step 3. (Directly: $q^2=q+(q-1)q$ and $q^m=q^m$ verify the relations.)
2. $\eps$: by the presentation (Step 7) it suffices to check the relations on the generators. Quadratic: $(-1)^2=1=q+(q-1)(-1)$. Braid: both sides are $(-1)^m$. Twist: $\eps(T_\sigma T_sT_\sigma^{-1})=-1=\eps(T_{\sigma s\sigma^{-1}})$. Then $\eps(T_w)=(-1)^{r}$ for $w=s_{i_1}\cdots s_{i_r}\sigma$ by step 1 of Step 7. This is the remark of [IM65, §3.2].
3. $\eps(\one_K)=\sum_{w\in W}(-1)^{\len(w)}$ by . If $W\neq1$, choose a simple reflection $s\in S$; $w\mapsto ws$ is an involution of $W$ without fixed points and changes $\len$ by $\pm1$ (F5), so it pairs each even-length element with an odd-length one and the sum vanishes. If $W=1$ then $\one_K=T_e$.