Since $G_{\alpha }-B_{\alpha }=B_{\alpha }{w}_{\alpha }{B}_{\alpha },$ the second case above really
is more general than the first. We have

Now assume $b{y}_{\alpha }{y}_{\beta }\dots y_{\gamma }{y}_{\delta }=b\prime y_{\alpha }^{\prime }{y}_{\beta }^{\prime }\dots y_{\gamma }^{\prime }{y}_{\delta }^{\prime }$ with $b,b\prime \in B,$ etc. Then $b{y}_{\alpha }\dots y_{\gamma }=b\prime y_{\alpha }^{\prime }\dots y_{\gamma }^{\prime }{y}_{\delta }^{\prime }{y}_{\delta }^{-1}\text{.}$ We have $y_{\delta }^{\prime }{y}_{\delta }^{-1}\in B$ or $B{w}_{\delta }B\text{.}$ The second case can not occur since then the left side would be in $Bw{w}_{\delta }B$ and the right side in $BwB$ (by Lemma 25). From the definition of $Y_{\delta }$ it follows that $y_{\delta }=y_{\delta }^{\prime },$ and then by induction that $y_{\gamma }=y_{\gamma }^{\prime },\dots ,$ whence, the uniqueness in Theorem 15.

$\square$

We have (a) by Theorem 4' applied to $G_{\alpha }\text{.}$ To verify (b) and (c) we may assume that $G_{\alpha }$ is ${SL}_2$ and $B_{\alpha }$ the superdiagonal subgroup $B_2$ since $\text{ker}\hspace{0.17em}{\phi }_{\alpha }\subseteq B_2\text{.}$ Any element of ${SL}_2\left(ℂ\right)$ can be converted to one of ${SU}_2$ by adding a multiple of the second row to the first and normalizing the lengths of the rows. Thus ${SL}_2\left(ℂ\right)=B_2\left(ℂ\right)·{SU}_2\text{.}$ Then $B_2\left(ℂ\right)\backslash {SL}_2\left(ℂ\right)\sim (B_2\left(ℂ\right)\cap {SU}_2)\backslash {SU}_2,$ whence (b). Now assume $\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\in {SL}_2\left(k\right)$ with $k$ as in (c). We choose $a,c$ in $\theta$ relatively prime and such that $pa+qc=0$ (using unique factorization), and then $b,d$ in $\theta$ so that $ad-bc=1\text{.}$ Multiplying the preceding matrix on the right by $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ we get an element of $B_2\left(k\right)\text{.}$ Thus ${SL}_2\left(k\right)=B_2\left(k\right){SL}_2\left(\theta \right),$ and (c) follows.

$\square$

This basic result is proved, e.g., in Jacobson, Lie algebras, p. 147.

$\square$

Let ${\sigma }_1$ be ${\sigma }_0$ in Lemma 44 composed with complex conjugation, and $\rho$ the representation of $ℒ$ used to define $G\text{.}$ Applying Theorem 4', Cor. 5 to the Chevalley groups (both equal to $G\text{)}$ constructed from the representations $\rho$ and ${\rho }_0{\sigma }_1$ of $ℒ,$ we get an automorphism of $G$ which aside from complex conjugation satisfies the equations of (a), hence composed with conjugation satisfies these equations. From Theorem 7 adapted to the present situation (see the remark at the end of §5) it follows that $\sigma$ is analytic, whence (a). We observe that if $G$ is defined by the ajoint representation of $ℒ,$ then $\sigma$ is effected by conjugation by the semiautomorphism ${\sigma }_0$ of Lemma 44.

$\square$

The kernel of ${\phi }_{\alpha }:{SL}_2\to G_{\alpha }$ is contained in $\{\pm 1\},$ and $\sigma$ pulls back to the inverse transpose conjugate, say ${\sigma }_2,$ on ${SL}_2\text{.}$ Since the equation ${\sigma }_{2}x=-x$ has no solutions we get (a).

Since $\sigma h_{\alpha }\left(t\right)=h_{\alpha }\left({\bar{t}}^{-1}\right),$ $\hat{\mu }\left(h_{\alpha }\left(t\right)\right)=t^{\mu \left(H_{\alpha }\right)}$ (here $\mu$ and $\hat{\mu }$ are corresponding weights on $\mathscr{H}$ and $H\text{),}$ and the $h_{\alpha }\left(t\right)$ generate $H,$ we have $\hat{\mu }\left(\sigma h\right)={\overline{\stackrel{ˆ}{\mu }\left(h\right)}}^{-1}$ for all $h\in H,$ so that $\sigma h=h$ if and only if $\left|\hat{\mu }\left(h\right)\right|=1$ for all weights $\hat{\mu }\text{.}$ If $h=\prod h_i\left(t_i\right),$ then $\hat{\mu }\left(h\right)=\prod t_i^{\mu \left(H_i\right)}\text{.}$ Since there are $\ell$ linearly independent weights $\mu ,$ we see that if $\left|\hat{\mu }\left(h\right)\right|=1$ for all $\hat{\mu },$ then ${\left|t_i\right|}^n=1$ for some $n>0,$ whence $\left|t_i\right|=1,$ for all $i\text{.}$ If $G$ is universal, then $B_{\sigma }$ is the product of the $\ell$ circles $\left\{h_i(·)\right\},$ hence is a torus; if not, we have to take the quotient by a finite group, thus still have a torus. Now if $h\in H_{\sigma }$ is general enough, so that the numbers $\hat{\alpha }\left(h\right)$ $(\alpha \in \Sigma )$ are distinct and different from $1,$ then $G_h,$ the centralizer of $h$ in $G,$ is $H,$ by the uniqueness in Theorem 4', so that $H_{\sigma }$ is in fact a maximal Abelian subgroup of $G_{\sigma },$ which proves the lemma.

$\square$

This follows from the fact proved in Lemma 45 that $K$ is generated by the groups ${\phi }_{\alpha }{SU}_2\text{.}$

$\square$

As already remarked, $K$ is generated by the groups ${\phi }_{\alpha }{SU}_2\text{.}$ Since ${SU}_2$ is connected, so is $K\text{.}$

$\square$

The map $K\to B\backslash G,$ $k\to Bk,$ is continuous and constant on the fibres of $T\backslash K,$ hence leads to a continuous map of $T\backslash K$ into $B\backslash G$ which is $1-1$ and onto since $T=B\cap K$ and $G=BK\text{.}$ Since $T\backslash K$ is compact, the map is a homeomorphism.

$\square$

Let $A=\{h\in H\hspace{0.17em}|\hspace{0.17em}\hat{\mu }\left(h\right)>0\hspace{0.17em}\text{for all}\hspace{0.17em}\hat{\mu }\in \hat{L}\}\text{.}$ Then we have $H=AT,$ so that $G=BK=UAK\text{.}$ On the right there is uniqueness of expression. Since $K$ is compact it easily follows that the natural map $UA\times K\to G$ is a homeomorphism. Since $UA$ is contractible to a point, $G$ is contractible to $K\text{.}$ If also $G$ is universal, then $G$ is simply connected by Theorem 13; hence so is $K\text{.}$

$\square$

This follows from Theorem 15 and Lemma 43(b).

$\square$

We have $B\backslash BwB$ homeomorphic to $w{U}_w,$ a cell of real
dimension $2N\left(w\right)\text{.}$ Since each dimension is even, it follows that the cells represent independent elements of the homology group and that there is no torsion (essentially because the boundary operator lowers dimensions by exactly 1), whence Cor. 6. Alternately one may use the fact that each $Y_{\alpha }$ is homeomorphic to $ℂ\text{.}$

$\square$

If $T_{\alpha }=T\cap K_{\alpha },$ we have $K_{\alpha }=T_{\alpha }{Y}_{\alpha }\cup T_{\alpha }$ by Lemma 45(a) and $\overline{T_{\alpha }{Y}_{\alpha }}=K_{\alpha }$ by the corresponding result in ${SU}_2\text{.}$ Now $BwB=B·T_{\alpha }{Y}_{\alpha }\dots T_{\delta }{Y}_{\delta }$ by Lemma 43(b). Hence $K_w=T·T_{\alpha }{Y}_{\alpha }\dots T_{\delta }{Y}_{\delta },$ so that ${\bar{K}}_w\supseteq T{K}_{\alpha }\dots K_{\delta },$ and we have equality since each factor on the right is compact, so that the right side is compact, hence closed. Since $K_{w^{\prime \prime }}{K}_{\alpha }\subseteq K_{w^{\prime \prime }}\cup K_{w^{\prime \prime }{w}_{\alpha }}$ if $w^{\prime \prime }\in W$ and $\alpha$ is simple, by Lemma 25, Cor. 7 follows.

$\square$

Because ${\bar{K}}_w$ doesn't depend on the expression.

$\square$

Assume $w\in W,$ and let $w=w_1\dots w_m$ be a minimal expression as a product of simple reflections and similarly for $w^{-1}{w}_0=w_{m+1}\dots w_n\text{.}$ Then $w_0=w_1\dots w_m\dots w_n$ is one for $w_0$ since if $N$ is the number of positive roots then m $m=N\left(w\right),$ $n=N-N\left(w\right),$ and $m+n=N=N\left(w_0\right)\text{.}$ Looking at the initial segment of $w_0$ we see that $w\in S\left(w_0\right)\text{.}$

$\square$

(a) By Cor. 7 and Lemma 46.

(b) By (a) $K=T{K}_{\alpha }{K}_{\beta }\dots K_{\delta }\text{.}$ We may write $T=\prod T_{\gamma }$ $\text{(}\gamma$ simple), then absorb the $T_{\gamma }\text{'s}$ in appropriate $K_{\gamma }\text{'s}$ to get (b).

$\square$

(a) Assume $x\in G\text{.}$ By the decompositions $H=AT$ and $G=BK$ (Theorem 16), t here exist elements in $KxK\cap UA\text{.}$ Given such an element $y=ua,$ we write $a=\text{exp}\hspace{0.17em}H$ $\text{(}H\in {\mathscr{H}}_{ℝ},$ uniquely determined by $a\text{),}$ then set $\left|a\right|=\left|H\right|,$ the Killing norm in ${ℒ}_{ℝ}\text{.}$ This norm is invariant under $W\text{.}$ We now choose $y$ to maximize $\left|a\right|$ (recall that $K$ is compact). We must show that $u=1\text{.}$ This follows from: $(*)$ if $u\ne 1,$ then $\left|a\right|$ can be increased. We will reduce $(*)$ to the rank 1 case. Write $u=\prod _{\beta >0}{u}_{\beta }$ $(u_{\beta }\in {𝔛}_{\beta })\text{.}$ We may assume $u_{\alpha }\ne 1$ for some simple $\alpha :$ choose $\alpha$ of minimum height, say $n,$ such that $u_{\alpha }\ne 1,$ then if $n>1,$ choose $\beta$ simple so that $(\alpha ,\beta )>0$ and $\text{ht}\hspace{0.17em}{w}_{\beta }\alpha <n,$ then replace $y$ by $w_{\beta }\left(1\right)y{w}_{\beta }{\left(1\right)}^{-1}$ and proceed by induction on $n\text{.}$ We write $u=u\prime u_{\alpha }$ with $u\prime \in {𝔛}_{P-\left\{\alpha \right\}}$ (here $P$ is the set of positive roots). Then we write $a=\text{exp}\hspace{0.17em}H,$ choose $c$ so that $H\prime =H-c{H}_{\alpha }$ is orthogonal to $H_{\alpha },$ set $a_{\alpha }=\text{exp}\hspace{0.17em}c{H}_{\alpha }\in A\cap G_{\alpha },$ $a\prime =\text{exp}\hspace{0.17em}H\prime \in A,$ $a=a_{\alpha }a\prime \text{.}$ Then $a\prime$ commutes with $G_{\alpha }$ elementwise and is orthogonal to $a_{\alpha }$ relative to the bilinear form corresponding to the norm introduced above. By $(*)$ for groups of rank 1, there exist $y,z\in K_{\alpha }$ such that $y{u}_{\alpha }{a}_{\alpha }z=a_{\alpha }^{\prime }\in A\cap G_{\alpha }$ and $\left|a_{\alpha }^{\prime }\right|>\left|a_{\alpha }\right|\text{.}$ Then $yuaz=yu\prime u_{\alpha }{a}_{\alpha }a\prime z=yu\prime y^{-1}{a}_{\alpha }^{\prime }a\prime \text{.}$ Since $G_{\alpha }$ normalizes ${𝔛}_{P-\left\{\alpha \right\}}$ (since ${𝔛}_{\alpha }$ and ${𝔛}_{-\alpha }$ do), $yu\prime y^{-1}\in U\text{.}$ Since ${\left|a_{\alpha }^{\prime }a\prime \right|}^2={\left|a_{\alpha }^{\prime }\right|}^2+{\left|a\prime \right|}^2>{\left|a_{\alpha }\right|}^2+{\left|a\prime \right|}^2={\left|a_{\alpha }a\prime \right|}^2={\left|a\right|}^2,$ we have $(*),$ modulo the rank 1 case. This case, essentially $G={SL}_2,$ will be left as an exercise.

(b) Assume $x\in G,$ $x=k_{1}a{k}_2$ as in (a). Then $\sigma x=k_1{a}^{-1}{k}_2,$ so that $x\sigma x^{-1}=k_1{a}^2{k}_1^{-1}\text{.}$ Here $\sigma a=a^{-1}$ since $\hat{\mu }\left(\sigma a\right)={\overline{\stackrel{ˆ}{\mu }\left(a\right)}}^{-1}=\hat{\mu }\left(a^{-1}\right)$ for all $\hat{\mu }\in \hat{L}\text{.}$

$\square$

(a) This has been noted in (b) above.

(b) We can assume $p=ka\in KA,$ by the theorem. Apply ${\sigma }^{-1}\text{:}$ $p=a{k}^{-1}\text{.}$ Thus $k$ commutes with $a^2,$ hence also with $a\text{.}$ (Since $a$ is diagonal (relative to a basis of weight vectors) and positive, the matrices commuting with $a$ have a certain block structure which does not change when it is replaced by $a^2\text{.)}$ Then $k^2=1$ and $k=a^{-\frac{1}{2}}p{a}^{-\frac{1}{2}}\in P,$ so that $k$ is unipotent by the definition of $P\text{.}$ Since $K$ is compact, $k=1\text{.}$ Thus $p=a\text{.}$ The uniqueness in (b) follows as before.

(c) If $x\in G,$ then $x=k_{1}a{k}_2$ as in the theorem, so that $x=k_1{k}_2·k_2^{-1}a{k}_2\in KP\text{.}$ Thus $G=KP\text{.}$ Assume $k_1{p}_1=k_2{p}_2$ with $k_i\in K$ and $p_i\in P\text{.}$ By (b) we can assume that $p_2\in A\text{.}$ Then $p_1=k_1^{-1}{k}_2{p}_2\text{.}$ As in (b) we conclude that $k_1^{-1}{k}_2=1,$ whence the uniqueness in (c).

$\square$

If $\theta$ is a Euclidean domain then ${SL}_2\left(\theta \right)$ is generated by its unipotent superdiagonal and subdiagonal elements, so that the lemma follows from the fact that $x_{\alpha }\left(t\right)$ acts on $M$ as an integral polynomial in $t\text{.}$ In the general case it follows that if $p$ is a prime in $\theta$ and ${\theta }_p$ is the localization of $\theta$ at $p$ (all $a/b\in k$ such that $a,b\in \theta$ with $b$ prime to $p\text{)}$ then ${\phi }_{\alpha }{SL}_2\left(\theta \right)\subseteq G_{{\theta }_p}\text{.}$ Since $\bigcap _p{\theta }_p=\theta ,$ e.g. by unique factorization, we have our result.

$\square$

(a) If $b=uh\in B\cap K,$ then its diagonal $h,$ relative to a basis of $M$ made up of weight vectors (see Lemma 18, Cor. 3), must be in $K,$ hence $u$ must also.

(b) If $u=\prod x_{\alpha }\left(t_{\alpha }\right)\in U\cap K,$ then by induction on heights, the equation $x_{\alpha }\left(t\right)=1+t{X}_{\alpha }+\dots$ and the primitivity of $X_{\alpha }$ in $\text{End}\hspace{0.17em}\left(M\right)$ (Theorem 2, Cor. 2) we get all $t_{\alpha }\in \theta \text{.}$

(c) If $h\in H\cap K,$ in diagonal form as above, then $\hat{\mu }\left(h\right)$ must be in $\theta$ for each weight $\hat{\mu }$ of the representation defining $G,$ in fact in ${\theta }^{*}$ since the sum of these weights is $0$ (the sum is invariant under $W\text{).}$ If we write $h=\prod h_i\left(t_i\right)$ and use what has just been proved, we get $t_i^n\in {\theta }^{*}$ for some $n>0,$ whence $t_i\in {\theta }^{*}$ by unique factorization.

(d) Set $S_{\alpha }={\phi }_{\alpha }{SL}_2\left(\theta \right)\text{.}$ By Lemma 48, $S_{\alpha }\subseteq K_{\alpha }\text{.}$ Since
 $G_{\alpha }=B_{\alpha }\cup B_{\alpha }{Y}_{\alpha }$ by Lemma 43(c) and $Y_{\alpha }\subset S_{\alpha },$ the reverse inclusion follows from: $B_{\alpha }\cap K\subset S_{\alpha }\text{.}$ Now if $x=x_{\alpha }\left(t\right)h_{\alpha }\left(t^{*}\right)\in B_{\alpha }\cap K,$
 then $t\in \theta$ and $t^{*}\in {\theta }^{*}$ by (a), (b), (c) applied to $G_{\alpha },$ so that $x\in ⟨x_{\alpha }\left(\theta \right),x_{-\alpha }\left(\theta \right)⟩=S_{\alpha },$ whence (d).

$\square$

By Lemmas 43(c) and 49(d), $BwB=B{Y}_{\alpha }\dots Y_{\delta }\subseteq BK$ for every $w\in W,$ so that $G=BK\text{.}$

$\square$

By Lemma 49 and Cor. 1.

$\square$

Since the corresponding result holds for ${SL}_2\left(\theta \right),$ this follows from Lemma 49(d) and Cor. 2.

$\square$

We may assume every $t_p\in \theta \text{.}$ To see this write $t_p=p^{r}a/b$ as above. By choosing $s\ge -r$ and $c$ and $d$ so that $a=c{p}^s+db$ and replacing $a/b$ by $d,$ we may assume $b=1\text{.}$ If we then multiply by a sufficiently high power of the product of the elements of $S,$ we achieve $r\ge 0,$ for all $p\in S\text{.}$ If we now choose $n$ so that $2^{-n}<\epsilon ,$ $e=\prod _{p\in S}{p}^n,$ $e_p=e/p^n,$ then $f_p,g_p$ so that $f_p{p}^n+g_p{e}_p=1,$ and finally $t=\Sigma g_p{e}_p{t}_p,$ we achieve the requirements of the theorem.

$\square$

Assume first that all $x_p$ are contained in some ${𝔛}_{\alpha },x_p=x_{\alpha }\left(t_p\right)$ with $t_p\in k\text{.}$ If $x=x_{\alpha }\left(t\right),t\in k,$ then ${\left|x\right|}_q\le \text{max}\hspace{0.17em}{\left|t\right|}_q,1$ because $x_{\alpha }\left(t\right)$ is an integral polynomial in $t$ and similarly ${|x{x}_p^{-1}-1|}_p\le {|t-t_p|}_p,$ so that ${|x-x_p|}_p\le {\left|x_p\right|}_p{|t-t_p|}_p$ by (1) and (2) above. Thus our result follows from Theorem 19 in this case. In the general case we choose a sequence of roots ${\alpha }_1,{\alpha }_2,\dots$ so that $x_p=x_{p1}{x}_{p2}\dots$ with $x_{pi}\in {𝔛}_{{\alpha }_i}$ for all $p\in S\text{.}$ By the first case there exists $x_i\in {𝔛}_{{\alpha }_i}$ so that

if $p\in S$ and ${\left|x_i\right|}_q\le 1$ if $q\notin S\text{.}$ We set $x=x_1{x}_2\dots \text{.}$ Then the conclusion of the theorem holds by (3) above.

$\square$

First we reduce the theorem to the local case, in which $\theta$ has a single prime, modulo units. Assume the result ture in this case. Assume $x\in G\text{.}$ Let $S$ be the finite set of primes at which $x$ fails to be integral. For $p\in S,$ we write ${\theta }_p$ for the local ring at $p$ in $\theta ,$ and define $K_p$ and
 $A_p^{+}$ in terms of ${\theta }_p$ as $K$ and $A^{+}$ are defined for $\theta \text{.}$ By the local case of the theorem we may write $x=c_p{a}_p{c}_p^{\prime }$ with $c_p,c_p^{\prime }\in K_p$ and $a\in A_p^{+},$ for all $p\in S\text{.}$ Since we may choose $a_p$ so that $\hat{\mu }\left(a_p\right)$ is always a power of $p$ and then replace all $a_p$ by their product, adjusting the $c\text{'s}$ accordingly, we may assume that $a_p$ is independent of $p,$ is in $A^{+},$ and is integral outside of $S\text{.}$ We have $c_{p}a{c}_p^{\prime }{x}^{-1}=1$ with $a=a_p$ for $p\in S\text{.}$ By Theorem 20 there exist $c,c\prime \in G$ so that ${|c-c_p|}_p<{\left|c_p\right|}_p$ for $p\in S$ and ${\left|c\right|}_q\le 1$ for $q\notin S,$ the same equations hold for $c\prime$ and $c_p^{\prime },$ and ${|cac\prime x^{-1}-1|}_p\le 1$ for all $p\in S\text{.}$ By properties (1), (2), (3) of ${\left|\hspace{0.17em}\right|}_p,$ it is now easily verified that ${\left|c\right|}_p\le 1,$ $\left|c_p^{\prime }\right|\le 1$ and ${|cac\prime x^{-1}-1|}_p\le 1,$ whether $p$ is in $S$ or not. Thus $c\in K,c\prime \in K$ and $cac\prime x^{-1}\in K,$ so that $x\in K{A}^{+}K$ as required. The uniqueness in Theorem 21 clearly also follows from that in the local case.

$\square$

Write $a=\prod h_{\alpha }\left(c_{\alpha }{p}^{n_{\alpha }}\right)$ with $c_{\alpha }\in {\theta }^{*},n_{\alpha }\in \mathbb{Z}\text{.}$ Then $\hat{\mu }\left(a\right)=\prod {\left(c_{\alpha }{p}^{n_{\alpha }}\right)}^{\mu \left(H_{\alpha }\right)}\text{.}$ Since $\hat{\mu }\left(a\right)$ is a power of $p$ the $c_{\alpha },$ being units, may be omitted, so that $\hat{\mu }\left(a\right)=P^{\mu \left(H\right)}$ with $H=\Sigma n_{\alpha }{H}_{\alpha }\text{.}$ If $H\prime$ is a second possibility for $H,$ then $\mu \left(H\prime \right)=\mu \left(H\right)$ for all $\mu ,$ so that $H\prime =H\text{.}$

$\square$

Let $t=e{p}^{-n}$ with $e\in {\theta }^{*}\text{.}$ Then $n\in \mathbb{Z},$ $n>0$ and $n+2c>0$ by the assumptions, so that $c+n>c,$ $c+n>-c$ and $|c+n|>\left|c\right|\text{.}$ If we multiply $y$ on the left by $\left[\begin{array}{cc}{p}^n& e\\ -e^{-1}& 0\end{array}\right],$ on the right by $\left[\begin{array}{cc}1& 0\\ -e^{-1}{p}^{n+2c}& 1\end{array}\right],$ both in $K,$ we get $\left[\begin{array}{cc}{p}^{c+n}& 0\\ 0& p^{-c-n}\end{array}\right],$ which proves the lemma, hence that $u=1\text{.}$ Thus $y=a\in A,$ so that $x\in KAK\text{.}$ Thus $G=KAK\text{.}$ Finally every element of $A$ is conjugate to an element of $A$ under the Weyl group, which is fully represented in $K$ (every $w_{\alpha }\left(1\right)\in K\text{).}$ Thus $G=K{A}^{+}K\text{.}$ It remains to prove the uniqueness of the $A^{+}$ component. If $G\prime$ is the universal group of the same type as $G$ and $\pi$ is the natural homomorphism, it follows from Lemma 49(d) and Theorem 19, Cor. 2 that $\pi K\prime =K$ and from Lemma 49 that $\pi$ maps ${A\prime }^{+}$ isomorphically onto $A^{+}\text{.}$ Thus we may assume that $G$ is universal. Then $G$ is a direct product of its indecomposable factors so that we may also assume that $G$ is indecomposable. Let ${\lambda }_i$ be the $i^{\text{th}}$ fundamental weight, $V_i$ an $ℒ\text{-module}$ with ${\lambda }_i$ as highest weight, $G_i$ the corresponding Chevalley group, ${\pi }_i:G\to G_i$ the corresponding homomorphism, and ${\mu }_i$ the corresponding lowest weight. Assume now that $x=cac\prime \in G,$ with $c,c\prime \in K$ and $a\in A^{+}\text{.}$ Set ${\hat{\mu }}_i\left(a\right)=p^{-n_i}\text{.}$ Each weight on $V_i$ is ${\mu }_i$ increased by a sum of positive roots, Thus $n_i$ is the smallest integer such that $p^{n_i}{\pi }_{i}a$ is integral, i.e. such that $p^{n_i}{\pi }_{i}x$ is since ${\pi }_{i}c$ and ${\pi }_{i}c\prime$ are integral, thus is uniquely determined by $x\text{.}$ Since $\left\{{\mu }_i\right\}$ is a basis of the lattice of weights $({\mu }_i=w_0{\lambda }_i),$ this yields the uniqueness in the local case and completes the proof of Theorem 21.