Last update: 8 July 2013

Let $G$ be a Chevalley group, $k,B\dots $ as usual. We recall (Theorems 4 and 4'):

(a) | $G={\displaystyle \bigcup _{w\in W}bwB,}$ a disjoint union. |

(b) | For each $w\in W,$ $BwB=Bw{U}_{w},$ with, uniqueness of expression on the right. Our purpose is to present some analogues of (b) with applications. |

For each simple root, $\alpha $ we set ${G}_{\alpha}=\u27e8{\U0001d51b}_{\alpha},{\U0001d51b}_{-\alpha}\u27e9,$ a group of $\text{rank}\hspace{0.17em}1,$ ${B}_{\alpha}=B\cap {G}_{\alpha},$ and assume that the representative of ${w}_{\alpha}$ in $N/H,$ also denoted ${w}_{\alpha},$ is chosen in ${G}_{\alpha}\text{.}$

*Theorem 15:* For each simple root $\alpha $ let ${Y}_{\alpha}$ be a system of representatives
for ${B}_{\alpha}\backslash ({G}_{\alpha}-{B}_{\alpha}),$
or more generally for $B\backslash B{w}_{\alpha}B\text{.}$ For each
$w\in W$ choose a minimal expression $w={w}_{\alpha}{w}_{\beta}\dots {w}_{\delta}$
as a product of reflections relative to simple roots $\alpha ,\beta \dots \text{.}$
Then $BwB=B{Y}_{\alpha}{Y}_{\beta}\dots {Y}_{\delta}$
with uniqueness of expression on the right.

Proof. | |

Since ${G}_{\alpha}-{B}_{\alpha}={B}_{\alpha}{w}_{\alpha}{B}_{\alpha},$ the second case above really is more general than the first. We have $$\begin{array}{cccc}BwB& =& B{w}_{\alpha}B{w}_{\alpha}wB& \text{(by Lemma 25)}\\ & =& B{w}_{\alpha}B{Y}_{\beta}\dots {Y}_{\delta}& \text{(by induction)}\\ & =& B{Y}_{\alpha}{Y}_{\beta}\dots {Y}_{\delta}& \text{(by the choice of}\hspace{0.17em}{Y}_{\alpha}\text{).}\end{array}$$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 $ |

*Lemma 43:* Let ${\phi}_{\alpha}:{SL}_{2}\to {G}_{\alpha}$
be the canonical homomorphism (see Theorem 4', Cor. 6). Then ${Y}_{\alpha}$ satisfies the conditions of Theorem 15 in each of
the following cases.

(a) | ${Y}_{\alpha}={w}_{\alpha}{\U0001d51b}_{\alpha}\text{.}$ |

(b) | $k=\u2102$ (resp. $\mathbb{R}\text{)}$ and ${Y}_{\alpha}$ is the image under ${\phi}_{\alpha}$ of the elements of ${SU}_{2}$ (resp. ${SO}_{2}\text{)}$ (standard compact forms) of the form $\left[\begin{array}{cc}a& -\stackrel{\u203e}{b}\\ b& \stackrel{\u203e}{a}\end{array}\right]$ with $b>0\text{.}$ |

(c) | If $\theta $ is a principal ideal domain (commutative with $1\text{),}$ ${\theta}^{*}$ is the group of units, $k$ is the quotient field, and ${Y}_{\alpha}$ is the image under ${\phi}_{\alpha}$ of the elements of ${SL}_{2}\left(\theta \right)$ of the form $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ with $c$ running through a set of representatives for $(\theta -0)/{\theta}^{*},$ and for each $c,$ a running over a set of representatives for the residue classes of $\theta $ mod $c\text{.}$ |

Proof. | |

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 ${SL}_{2}\left(\u2102\right)$ 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 ${SL}_{2}\left(\u2102\right)={B}_{2}\left(\u2102\right)\xb7{SU}_{2}\text{.}$ Then ${B}_{2}\left(\u2102\right)\backslash {SL}_{2}\left(\u2102\right)\sim ({B}_{2}\left(\u2102\right)\cap {SU}_{2})\backslash {SU}_{2},$ 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 $ |

*Remarks:* (a) The case (a) above is essentially Theorem 4' since
$w{U}_{w}={w}_{\alpha}{\U0001d51b}_{\alpha}\xb7{w}_{\beta}{\U0001d51b}_{\beta}\dots {w}_{\delta}{\U0001d51b}_{\delta}$
in the notation of Theorem 15, by Appendix II 25, or else by induction on the length of the expression. (b) In (c) above the choice can be made precise in the
following cases:

(1) | $\theta =\mathbb{Z}\text{;}$ choose $a,c$ so that $0\le a<c\text{.}$ |

(2) | $\theta =F\left[X\right]$ $\text{(}F$ a field); choose so that $c$ is monic and $\text{dg}\hspace{0.17em}a<\text{dg}\hspace{0.17em}c\text{.}$ |

(3) | $\theta ={\mathbb{Z}}_{p}$ $\text{(}p\text{-adic}$ integers); choose $c$ a power of $p$ and $a$ an integer such that $0\le a<c\text{.}$ |

In what follows we will give separate but parallel developments of the consequences of (b) and (c) above. In (b) we will treat the case $k=\u2102$ for definiteness, the case $k=\mathbb{R}$ being similar.

*Lemma 44:* Let $\mathcal{L}$ and $\{{X}_{\alpha},{H}_{\alpha}\}$
be as in Theorem 1.

(a) | There exists an involutory semiautomorphism ${\sigma}_{0}$ of $\mathcal{L}$ (relative to complex conjugation of $\u2102\text{)}$ such that ${\sigma}_{0}{X}_{\alpha}=-{X}_{-\alpha}$ and ${\sigma}_{0}{H}_{\alpha}=-{H}_{\alpha}$ for every root $\alpha \text{.}$ |

(b) | On $\mathcal{L}$ the form $\{X,Y\}$ defined by $(X,{\sigma}_{0}Y)$ in terms of the Killing form is negative definite. |

Proof. | |

This basic result is proved, e.g., in Jacobson, Lie algebras, p. 147. $\square $ |

*Theorem 16:* Let $G$ be a Chevalley group over $\u2102$ viewed as a Lie group over
$\mathbb{R}\text{.}$

(a) | There exists an analytic automorphism $\sigma $ of $G$ such that $\sigma {x}_{\alpha}\left(t\right)={x}_{-\alpha}(-\stackrel{\u203e}{t})$ and $\sigma {h}_{\alpha}\left(t\right)={h}_{\alpha}\left({\stackrel{\u203e}{t}}^{-1}\right)$ for all $\alpha $ and $t\text{.}$ |

(b) | The group $K={G}_{\sigma}$ of fixed points of $\sigma $ is a maximal compact subgroup of $G$ and the decomposition $G=BK$ holds (Iwasawa decomposition). |

Proof. | |

Let ${\sigma}_{1}$ be ${\sigma}_{0}$ in Lemma 44 composed with complex conjugation, and $\rho $ the representation of $\mathcal{L}$ 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 $\mathcal{L},$ 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 $\mathcal{L},$ then $\sigma $ is effected by conjugation by the semiautomorphism ${\sigma}_{0}$ of Lemma 44. $\square $ |

*Lemma 45:* Let $K={G}_{\sigma},$
${K}_{\alpha}=K\cap {G}_{\alpha}$ for each simple root
$\alpha \text{.}$

(a) | ${K}_{\alpha}={\phi}_{\alpha}{SU}_{2}$ (see Lemma 43(b)), hence ${Y}_{\alpha}\subset {K}_{\alpha}\text{.}$ |

(b) | ${B}_{\sigma}={H}_{\sigma}=\{h\in H\hspace{0.17em}|\hspace{0.17em}\left|\stackrel{\u02c6}{\mu}\left(h\right)\right|=1\hspace{0.17em}\text{for all}\hspace{0.17em}\stackrel{\u02c6}{\mu}\in \stackrel{\u02c6}{L}\hspace{0.17em}\text{(global weights)}\}=\{\prod {h}_{i}\left({t}_{i}\right)\hspace{0.17em}\text{(see Lemma 28)}\hspace{0.17em}|\hspace{0.17em}\left|{t}_{i}\right|=1\}=$ maximal torus in $K\text{.}$ |

Proof. | |

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 $\sigma {h}_{\alpha}\left(t\right)={h}_{\alpha}\left({\stackrel{\u203e}{t}}^{-1}\right),$ $\stackrel{\u02c6}{\mu}\left({h}_{\alpha}\left(t\right)\right)={t}^{\mu \left({H}_{\alpha}\right)}$ (here $\mu $ and $\stackrel{\u02c6}{\mu}$ are corresponding weights on $\mathscr{H}$ and $H\text{),}$ and the ${h}_{\alpha}\left(t\right)$ generate $H,$ we have $\stackrel{\u02c6}{\mu}\left(\sigma h\right)={\stackrel{\u203e}{\stackrel{\u02c6}{\mu}\left(h\right)}}^{-1}$ for all $h\in H,$ so that $\sigma h=h$ if and only if $\left|\stackrel{\u02c6}{\mu}\left(h\right)\right|=1$ for all weights $\stackrel{\u02c6}{\mu}\text{.}$ If $h=\prod {h}_{i}\left({t}_{i}\right),$ then $\stackrel{\u02c6}{\mu}\left(h\right)=\prod {t}_{i}^{\mu \left({H}_{i}\right)}\text{.}$ Since there are $\ell $ linearly independent weights $\mu ,$ we see that if $\left|\stackrel{\u02c6}{\mu}\left(h\right)\right|=1$ for all $\stackrel{\u02c6}{\mu},$ 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 $\left\{{h}_{i}(\xb7)\right\},$ 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 $\stackrel{\u02c6}{\alpha}\left(h\right)$ $(\alpha \in \Sigma )$ 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 $ |

*Exercise:* Check out the existence of $h$ and the property
${G}_{h}=H$ above.

Now we consider part (b) of Theorem 16. By Theorem 15 and Lemmas 43(b) and 45(a) we have $G=BK\text{.}$ By the same results ${\left(BwB\right)}_{\sigma}\subseteq {B}_{\sigma}{K}_{\alpha}\dots {K}_{\delta},$ a compact set since each factor is (the compactness of tori and ${SU}_{2}$ is being used). Thus $K={G}_{\sigma}$ is compact. (This also follows easily from Lemma 44(b)). Let ${K}_{1}$ be a compact subgroup of $G,$ ${K}_{1}\supseteq K\text{.}$ Assume $x\in {K}_{1}\text{.}$ Write $x=by$ with $b\in B,$ $y\in K,$ and then $b=uh$ with $u\in U,$ $h\in H\text{.}$ Since ${K}_{1}$ is compact, all eigenvalues $\stackrel{\u02c6}{\mu}\left({h}^{n}\right)$ $(n=0,\pm 1,\pm 2,\dots )$ are bounded, whence $h\in K$ by Lemma 45(b). Then all coefficients of all ${u}^{n}$ are bounded so that $u=1\text{.}$ Thus $x\in K,$ so that $K$ is maximal compact.

*Remark:* It can be shown also that $K$ is semisimple and that a complete set of semisimple compact Lie groups is got from the above construction.

*Corollary 1:* Let $G\prime $ be of the same type as $G$ with a weight lattice
containing that of $G,K\prime ={G}_{\sigma}^{\prime},$
and $\pi :G\prime \to G$ the natural projection. Then
$\pi K\prime =K\text{.}$

Proof. | |

This follows from the fact proved in Lemma 45 that $K$ is generated by the groups ${\phi}_{\alpha}{SU}_{2}\text{.}$ $\square $ |

*Examples:*

(a) | If $G={SL}_{n}\left(\u2102\right),$ then $K={SU}_{n}\text{.}$ |

(b) | If $G={SO}_{n}\left(\u2102\right),$ then $K$ fixes simultaneously the forms $\Sigma {x}_{i}{x}_{n+1-i}$ and $\Sigma {x}_{i}\stackrel{\u203e}{{x}_{i}},$ hence equals ${SO}_{n}\left(\mathbb{R}\right)$ (compact form) after a change of coordinates. Prove this. |

(c) | If $G={Sp}_{2n}\left(\u2102\right),$ then $K$ fixes the forms $\underset{1}{\overset{n}{\Sigma}}({x}_{i}{y}_{2n+1-i}-{x}_{2n+1-i}{y}_{i})$ and $\Sigma {x}_{i}\stackrel{\u203e}{{x}_{i}},$ and is isomorphic to ${SU}_{n}\left(\mathbb{H}\right)$ (compact form, $\mathbb{H}=$ quaternions). For this see Chevalley, Lie groups, p. 22. |

(d) | We have isomorphisms and central extensions, ${\mathbb{H}}^{*}={SU}_{1}\left(\mathbb{H}\right)\cong {SU}_{2}\left(\u2102\right)\to {SO}_{3}\left(\mathbb{R}\right),$ ${SU}_{2}\left(\mathbb{H}\right)\to {SO}_{5}\left(\mathbb{R}\right),$ ${SU}_{2}{\left(\u2102\right)}^{2}\to {SO}_{4}\left(\mathbb{R}\right),$ ${SU}_{4}\left(\u2102\right)\to {SO}_{6}\left(\mathbb{R}\right)$ (compact forms). |

This follows from (a), (b), (c), Corollary 1 and the equivalences ${C}_{1}={A}_{1}={B}_{1},$ ${C}_{2}={B}_{2},$ ${A}_{1}^{2}={D}_{2},$ ${A}_{3}={D}_{3}\text{.}$

*Corollary 2:* The group $K$ is connected.

Proof. | |

As already remarked, $K$ is generated by the groups ${\phi}_{\alpha}{SU}_{2}\text{.}$ Since ${SU}_{2}$ is connected, so is $K\text{.}$ $\square $ |

*Corollary 3:* If $T$ denotes the maximal torus ${H}_{\sigma},$ then
$T\backslash K$ is homeomorphic to $B\backslash G$ under the natural map.

Proof. | |

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 $ |

*Corollary 4:*

(a) | $G$ is contractible to $K\text{.}$ |

(b) | If $G$ is universal, then $K$ is simply connected. |

Proof. | |

Let $A=\{h\in H\hspace{0.17em}|\hspace{0.17em}\stackrel{\u02c6}{\mu}\left(h\right)>0\hspace{0.17em}\text{for all}\hspace{0.17em}\stackrel{\u02c6}{\mu}\in \stackrel{\u02c6}{L}\}\text{.}$ 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 $ |

*Corollary 5:* For $w\in W$ set
${\left(BwB\right)}_{\sigma}=BwB\cap K={K}_{w},$
and let $\alpha ,\beta ,\dots ,\delta $ be as in
Theorem 15. Then $K=\bigcup _{w}{K}_{w}$ and
${K}_{w}=T{Y}_{\alpha}\dots {Y}_{\delta},$
with uniqueness of expression on the right.

Proof. | |

This follows from Theorem 15 and Lemma 43(b). $\square $ |

*Remark:*
Observe that ${K}_{w}$ is essentially a cell since each ${Y}_{\alpha}$
is homeomorphic to $\u2102$ (consider the values of $a$ in Lemma 43(b)). A true cellular decomposition is obtained by writing
$T$ as a union of cells. Perhaps this decomposition can be used to give an elementary treatment of the cohomology of $K\text{.}$

*Corollary 6:* $B\backslash G$ and $T\backslash K$ have as their Poincaré
polynomials $\underset{w\in W}{\Sigma}{t}^{2N\left(w\right)}\text{.}$
They have no torsion.

Proof. | |

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 $\u2102\text{.}$ $\square $ |

*Remark:* The above series will be summed in the next section, where it arises in connection with the orders of the finite Chevalley groups.

*Corollary 7:* For $w\in W$ let
$w={w}_{\alpha}\dots {w}_{\delta}$ be a minimal expression
as before and let $S$ denote the set of elements of $W$ each of which is a product of some subsequence of the expression for
$w\text{.}$ Then ${K}_{w}$ (topological closure)
$=\bigcup _{w\prime \in S}{K}_{w\prime}\text{.}$

Proof. | |

If ${T}_{\alpha}=T\cap {K}_{\alpha},$ we have ${K}_{\alpha}={T}_{\alpha}{Y}_{\alpha}\cup {T}_{\alpha}$ by Lemma 45(a) and $\stackrel{\u203e}{{T}_{\alpha}{Y}_{\alpha}}={K}_{\alpha}$ by the corresponding result in ${SU}_{2}\text{.}$ Now $BwB=B\xb7{T}_{\alpha}{Y}_{\alpha}\dots {T}_{\delta}{Y}_{\delta}$ by Lemma 43(b). Hence ${K}_{w}=T\xb7{T}_{\alpha}{Y}_{\alpha}\dots {T}_{\delta}{Y}_{\delta},$ so that ${\stackrel{\u203e}{K}}_{w}\supseteq T{K}_{\alpha}\dots {K}_{\delta},$ 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 $ |

*Corollary 8:*

(a) | $T={K}_{1}$ is the closure of every ${K}_{w}\text{.}$ |

(b) | ${K}_{w}$ is closed if and only if $w=1\text{.}$ |

*Corollary 9:* The set $S$ of Cor. 7 depends only on $w,$ not on the minimal expression
chosen, hence may be written $S\left(w\right)\text{.}$

Proof. | |

Because ${\stackrel{\u203e}{K}}_{w}$ doesn't depend on the expression. $\square $ |

*Lemma 46:* Let ${w}_{0}$ be the element of $W$ which makes all positive roots negative. Then
$S\left({w}_{0}\right)=W\text{.}$

Proof. | |

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 $ |

*Corollary 10:* If ${w}_{0}$ is as above and
${w}_{0}={w}_{\alpha}{w}_{\beta}\dots {w}_{\delta}$
is a minimal expression, then

(a) | $K={\stackrel{\u203e}{K}}_{{w}_{0}}\text{.}$ |

(b) | $K={K}_{\alpha}{K}_{\beta}\dots {K}_{\delta}\text{.}$ |

Proof. | |

(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 $ |

*Exercise:* If $G$ is any Chevalley group and
${w}_{0},\alpha ,\beta ,\dots $ are as
above, show that $G=B{G}_{\alpha}{G}_{\beta}\dots {G}_{\delta}\text{.}$

*Remarks:* (a) If $\u2102$ and ${SU}_{2}$ are replaced by
$\mathbb{R}$ and ${SO}_{2}$ in accordance with Lemma 43(b), then everything above goes
through except for Cor. 4, Cor. 6 and the fact that $T$ is no longer a torus. In this case each ${K}_{\alpha}$
is a circle since ${SO}_{2}$ is. The corresponding angles in Cor. 10(b), which we have to restrict
suitably to get uniqueness, may be called the Euler angles in analogy with the classical ease:

(b) If ${K}_{w}$ is replaced by $BwB=B{K}_{w}$ in Cor. 7, the formula for $\stackrel{\u203e}{BwB}$ is obtained. (Prove this.) If $\u2102$ (or $\mathbb{R}\text{)}$ is replaced by any algebraically closed field and the Zariski topology is used, the same formula holds. So as not to interrupt the present development, we give the proof later, at the end of this section.

*Theorem 17:* (Cartan). Again let $G$ be a Chevalley group over $\u2102$ or $\mathbb{R},$
$K={G}_{\sigma}$ as above, and
$A=\{h\in H\hspace{0.17em}|\hspace{0.17em}\stackrel{\u02c6}{\mu}\left(h\right)>0\hspace{0.17em}\text{for all}\hspace{0.17em}\stackrel{\u02c6}{\mu}\in \stackrel{\u02c6}{L}\}\text{.}$

(a) |
$G=KAK$ |

(b) | In (a) the $A\text{-component}$ is determined uniquely up to conjugacy under the Weyl group. |

Proof. | |

(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 $a=\text{exp}\hspace{0.17em}H$ $\text{(}H\in {\mathscr{H}}_{\mathbb{R}},$ uniquely determined by $a\text{),}$ then set $\left|a\right|=\left|H\right|,$ the Killing norm in ${\mathcal{L}}_{\mathbb{R}}\text{.}$ 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 $u=\prod _{\beta >0}{u}_{\beta}$ $({u}_{\beta}\in {\U0001d51b}_{\beta})\text{.}$ 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 $u\prime \in {\U0001d51b}_{P-\left\{\alpha \right\}}$ (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 ${\U0001d51b}_{P-\left\{\alpha \right\}}$ (since ${\U0001d51b}_{\alpha}$ and ${\U0001d51b}_{-\alpha}$ 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 $\stackrel{\u02c6}{\mu}\left(\sigma a\right)={\stackrel{\u203e}{\stackrel{\u02c6}{\mu}\left(a\right)}}^{-1}=\stackrel{\u02c6}{\mu}\left({a}^{-1}\right)$ for all $\stackrel{\u02c6}{\mu}\in \stackrel{\u02c6}{L}\text{.}$ $\square $ |

*Lemma 47:* If elements of $H$ are conjugate in $G$ (any Chevalley group), they are conjugate under the Weyl group.

This easily follows from the uniqueness in Theorem 4'.

By the lemma $x$ above uniquely determines ${a}^{2}$ up to conjugacy under the Weyl group, hence also $a$ since square-roots in $A$ are unique.

*Remark:* We can get uniqueness in (b) by replacing $A$ by
${A}^{+}=\{a\in A\hspace{0.17em}|\hspace{0.17em}\stackrel{\u02c6}{\alpha}\left(a\right)\ge 1\hspace{0.17em}\text{for all}\hspace{0.17em}\stackrel{\u02c6}{\alpha}>0\}\text{.}$
This follows from Appendix III 33.

*Corollary:* Let $P$ consist of the elements of $G$ which satisfy
$\sigma x={x}^{-1}$ and have all eigenvalues positive.

(a) | $A\subset P\text{.}$ |

(b) | Every $p\in P$ is conjugate under $K$ to some $a\in A,$ uniquely determined up to conjugacy under $W$ (spectral theorem). |

(c) | $G=KP,$ with uniqueness on the right (polar decomposition). |

Proof. | |

(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 $x={k}_{1}{k}_{2}\xb7{k}_{2}^{-1}a{k}_{2}\in KP\text{.}$ 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 $ |

*Example:* If $G={SL}_{n}\left(\u2102\right),$
so that $K={SU}_{n}\left(\u2102\right),$
$A=\{$ positive diagonal matrices $\},$
$P=\{$ positive-definite Hermitean matrices $\},$
then (b) and (c) reduce
to classical results.

We now consider the case (c) of Lemma 43. The development is strikingly parallel to that for case (b) just completed although the results are basically arithmetic in one case, geometric in the other. Throughout we assume that $\theta ,{\theta}^{*},k,{Y}_{\alpha}$ are as in Lemma 43(c) and that the Chevalley group $G$ under discussion is based on $k\text{.}$ We write ${G}_{\theta}$ for the subgroup of elements of $G$ whose coordinates, relative to the original lattice $M,$ all lie in $\theta \text{.}$

*Lemma 48:* If ${\phi}_{\alpha}$ is as in Theorem 4', Cor. 6, then
${\phi}_{\alpha}{SL}_{2}\left(\theta \right)\subseteq {G}_{\theta}\text{.}$

Proof. | |

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 $ |

*Remark:* A version of Lemma 48 is true if $\theta $ is any commutative ring since
${\phi}_{\alpha}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
is generically expressible as a polynomial in $a,b,c,d$
with integral coefficients (proof omitted). The proof just given works if $\theta $ is any integral domain for which
$\theta =\bigcap {\theta}_{p}$
$\text{(}p=$ maximal ideal), which includes most of the interesting cases.

*Lemma 49:* Write $K={G}_{\theta},$
${K}_{\alpha}={G}_{\alpha}\cap K\text{.}$

(a) | $B\cap K=(U\cap K)(H\cap K)\text{.}$ |

(b) | $U\cap K=\left\{\prod _{\alpha >0}{x}_{\alpha}\left({t}_{\alpha}\right)\hspace{0.17em}\right|\hspace{0.17em}{t}_{\alpha}\in \theta \}\text{.}$ |

(c) | $H\cap K=\{h\in H\hspace{0.17em}|\hspace{0.17em}\stackrel{\u02c6}{\mu}\left(h\right)\in {\theta}^{*}\hspace{0.17em}\text{for all}\hspace{0.17em}\stackrel{\u02c6}{\mu}\in \stackrel{\u02c6}{L}\}=\{\prod {h}_{i}\left({t}_{i}\right)\hspace{0.17em}|\hspace{0.17em}\text{all}\hspace{0.17em}{t}_{i}\in {\theta}^{*}\}\text{.}$ |

(d) | ${\phi}_{\alpha}{SL}_{2}\left(\theta \right)={K}_{\alpha}\text{.}$ Hence ${Y}_{\alpha}\subset {K}_{\alpha}\text{.}$ |

Proof. | |

(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 $\stackrel{\u02c6}{\mu}\left(h\right)$ must be in $\theta $ for each weight $\stackrel{\u02c6}{\mu}$ 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 $x\in \u27e8{x}_{\alpha}\left(\theta \right),{x}_{-\alpha}\left(\theta \right)\u27e9={S}_{\alpha},$ whence (d). $\square $ |

*Theorem 18:* Let $\theta ,k,G$ and
$K={G}_{\theta}$ be as above. Then $G=MK$
(Iwasawa decomposition).

Proof. | |

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 $ |

*Corollary 1:* Write ${K}_{w}=BwB\cap K\text{.}$

(a) | $K=\bigcup _{w\in W}{K}_{w}\text{.}$ |

(b) | ${K}_{w}=(B\cap K){Y}_{\alpha}\dots {Y}_{\delta},$ with $B\cap K$ given by Lemma 49, and on the right there is uniqueness of expression. |

*Remark:* This normal form in $K={G}_{\theta}$ has all components in ${G}_{\theta}$
whereas the usual one obtained by imbedding ${G}_{\theta}$ in $G$ doesn't.

*Corollary 2:* $K$ is generated by the groups ${K}_{\alpha}\text{.}$

Proof. | |

By Lemma 49 and Cor. 1. $\square $ |

*Corollary 3:* If $\theta $ is a Euclidean domain, then $K$ is generated by
$\left\{{x}_{\alpha}\left(t\right)\hspace{0.17em}\right|\hspace{0.17em}\alpha \in \Sigma ,t\in \theta \}\text{.}$

Proof. | |

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

*Example:* Assume $\theta =\mathbb{Z},$
$k=\mathbb{Q}\text{.}$ We get that ${G}_{\mathbb{Z}}$ is generated by
$\left\{{x}_{\alpha}\left(1\right)\right\}\text{.}$
The normal form in Cor. 1 can be used to extend Nielsen's theorem (see (1) on p. 96) from ${SL}_{3}\left(\mathbb{Z}\right)$
to ${G}_{\mathbb{Z}}$ whenever $\Sigma $ has
$\text{rank}\ge 2,$ is indecomposable, and has all roots of equal length (W. Wardlaw, Thesis, U.C.L.A. 1966).
It would be nice if the form could be used to handle ${SL}_{3}\left(\mathbb{Z}\right)$
itself since Nielsen's proof is quite involved. The case of unequal root lengths is at present in poor shape. In analogy with the fact that in the earlier
development $K$ is a simple compact group if $\Sigma $ is indecomposable, we have here:
Every normal subgroup of ${G}_{\mathbb{Z}}$ is finite or of finite index if $\Sigma $
is indecomposable and has $\text{rank}\ge 2\text{.}$ The proof isn't easy.

*Exercise:* Prove that ${G}_{\mathbb{Z}}/\mathcal{D}{G}_{\mathbb{Z}}$
is finite, and is trivial if $\Sigma $ is indecomposable and not of type
${A}_{1},{B}_{2}$ or ${G}_{2}\text{.}$

Returning to the general set up, if $p$ is a prime in $\theta ,$ we write ${\left|\hspace{0.17em}\right|}_{p}$ for the $p\text{-adic}$ norm defined by ${\left|0\right|}_{p}=0$ and ${\left|x\right|}_{p}={2}^{-r}$ if $x={p}^{r}a/b$ with $a$ and $b$ prime to $p\text{.}$

*Theorem 19:* (Approximation theorem): Let $\theta $ and $k$ be as above, a principal ideal domain and its quotient
field, $S$ a finite set of inequivalent primes in $\theta ,$ and for each
$p\in S,$ ${t}_{p}\in k\text{.}$
Then for any $\epsilon >0$ there exists $t\in k$ such that
${|t-{t}_{p}|}_{p}<\epsilon $
for all $p\in S$ and ${\left|t\right|}_{q}\le 1$
for all primes $q\notin S\text{.}$

Proof. | |

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 $ |

Now given a matrix $x=\left({a}_{ij}\right)$ over $k,$ we define ${\left|x\right|}_{p}=\text{max}{\left|{a}_{ij}\right|}_{p}\text{.}$ The following properties are easily verified.

(1) | ${|x+y|}_{p}\le \text{max}{\left|x\right|}_{p},$ {\left|y\right|}_{p}\text{.}$$ |

(2) | ${\left|xy\right|}_{p}\le {\left|x\right|}_{p}{\left|y\right|}_{p}\text{.}$ |

(3) | If ${\left|{x}_{i}\right|}_{p}={\left|{y}_{i}\right|}_{p}$ for $i=1,2,\dots ,n$ then ${|\prod {x}_{i}-\prod {y}_{i}|}_{p}\le {\text{max}}_{i}{\left|{y}_{1}\right|}_{p}\dots {\left|{\stackrel{\u02c6}{y}}_{i}\right|}_{p}\dots {\left|{y}_{n}\right|}_{p}{|{x}_{i}-{y}_{i}|}_{p}\text{.}$ |

*Theorem 20:* (Approximation theorem for split groups): Let $\theta ,k,S,\epsilon $
be as in Theorem 19, $G$ a Chevalley group over $k,$ and
${x}_{p}\in G$ for each $p\in S\text{.}$
Then there exists $x\in G$ so that
${|x-{x}_{p}|}_{p}<\epsilon $
for all $p\in S$ and ${\left|x\right|}_{q}\le 1$
for all $q\notin S\text{.}$

Proof. | |

Assume first that all ${x}_{p}$ are contained in some ${\U0001d51b}_{\alpha},{x}_{p}={x}_{\alpha}\left({t}_{p}\right)$ 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 ${x}_{pi}\in {\U0001d51b}_{{\alpha}_{i}}$ for all $p\in S\text{.}$ By the first case there exists ${x}_{i}\in {\U0001d51b}_{{\alpha}_{i}}$ so that $${|{x}_{i}-{x}_{pi}|}_{p}<{\left|{x}_{pi}\right|}_{p}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\epsilon {\left|{x}_{pi}\right|}_{p}/{\left|{x}_{p1}\right|}_{p}{\left|{x}_{p2}\right|}_{p}\dots $$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 $ |

With Theorem 20 available we can now prove:

*Theorem 21:* (Elementary divisor theorem): Assume $\theta ,k,G,K={G}_{\theta}$
are as before. Let ${A}^{+}$ be the subset of $H$ defined by:
$\stackrel{\u02c6}{\alpha}\left(h\right)\in \theta $ for all positive roots
$\stackrel{\u02c6}{\alpha}\text{.}$

(a) | $G=K{A}^{+}K$ (Cartan decomposition). |

(b) | The ${A}^{+}$ component in (a) is uniquely determined $\text{mod}\hspace{0.17em}H\cap K,$ i.e. mod units (see Lemma 49); in other words, the set of numbers $\left\{\stackrel{\u02c6}{\mu}\left(h\right)\hspace{0.17em}\right|\hspace{0.17em}\stackrel{\u02c6}{\mu}$ weight of the representation defining $G\}$ is. |

*Example:* The classical case occurs when $G={SL}_{n}\left(k\right),$
$K={SL}_{n}\left(\theta \right),$
and ${A}^{+}$ consists of the diagonal elements
$\text{diag}({a}_{1},{a}_{2},\dots ,{a}_{n})$
such that ${a}_{i}$ is a multiple of ${a}_{i+1}$ for
$i=1,2,\dots \text{.}$

Proof of theorem. | |

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 $\stackrel{\u02c6}{\mu}\left({a}_{p}\right)$ 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 $ |

We now consider the local case, $p$ being the unique prime in $\theta \text{.}$ The proof to follow is quite close to that of Theorem 17. Let $A$ be the subgroup of all $h\in H$ such that all $\stackrel{\u02c6}{\mu}\left(h\right)$ are powers of $p,$ and redefine ${A}^{+},$ casting out units, so that in addition all $\stackrel{\u02c6}{\alpha}\left(h\right)$ $(\stackrel{\u02c6}{\alpha}>0)$ are nonnegative powers of $p\text{.}$

*Lemma 50:* For each $a\in A$ there exists a unique
$H\in {\mathscr{H}}_{\mathbb{Z}},$ the
$\mathbb{Z}\text{-module}$ generated by the elements ${H}_{\alpha}$
of the Lie algebra $\mathcal{L},$ such that
$\stackrel{\u02c6}{\mu}\left(a\right)={p}^{\mu \left(H\right)}$
for all weights $\mu \text{.}$

Proof. | |

Write $a=\prod {h}_{\alpha}\left({c}_{\alpha}{p}^{{n}_{\alpha}}\right)$ with ${c}_{\alpha}\in {\theta}^{*},{n}_{\alpha}\in \mathbb{Z}\text{.}$ Then $\stackrel{\u02c6}{\mu}\left(a\right)=\prod {\left({c}_{\alpha}{p}^{{n}_{\alpha}}\right)}^{\mu \left({H}_{\alpha}\right)}\text{.}$ Since $\stackrel{\u02c6}{\mu}\left(a\right)$ is a power of $p$ the ${c}_{\alpha},$ being units, may be omitted, so that $\stackrel{\u02c6}{\mu}\left(a\right)={P}^{\mu \left(H\right)}$ 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 $ |

If $a$ and $H$ are as above, we write $H={\text{log}}_{p}a,a={p}^{H},$ and introduce a norm: $\left|a\right|=\left|H\right|,$ the Killing norm. This norm is invariant under the Weyl group. Now assume $x\in G\text{.}$ We want to show $x\in K{A}^{+}K\text{.}$ From the definitions if $T=H\cap K$ then $H=AT\text{.}$ Thus by Theorem 18 there exists $y=ua\in KxK$ with $u\in U,a\in A\text{.}$ There is only a finite number of possibilities for $a\text{:}$ if $a={p}^{H},$ then $\left\{\mu \left(H\right)\hspace{0.17em}\right|\hspace{0.17em}\mu $ a weight in the given $\text{representation}\}$ is bounded below (by $-n$ if $n$ is chosen so that the matrix of ${p}^{n}x$ is integral, because $\left\{{p}^{\mu \left(H\right)}\right\}$ are the diagonal entries of $y\text{),}$ and also above since the sum of the weights is $0,$ so that $H$ is confined to a bounded region of the lattice ${\mathscr{H}}_{\mathbb{Z}}\text{.}$ We choose $y=ua$ above so as to maximize $\left|a\right|\text{.}$ If $u=\prod {u}_{\alpha}$ $({u}_{\alpha}\in {\U0001d51b}_{\alpha}),$ we set $\text{supp}\hspace{0.17em}u=\left\{\alpha \hspace{0.17em}\right|\hspace{0.17em}{u}_{\alpha}\ne 1\}$ and then minimize $\text{supp}\hspace{0.17em}u$ subject to a lexicographic ordering of the supports based on an ordering of the roots consistent with addition (thus $\text{supp}\hspace{0.17em}u<\text{supp}\hspace{0.17em}u\prime $ means that the first $\alpha $ in one but not in the other lies in the second). We claim $u=1\text{.}$ Suppose not. We claim $(*)$ ${u}_{\alpha}\notin K$ and ${a}^{-1}{u}_{\alpha}a\notin K$ for $\alpha \in \text{supp}\hspace{0.17em}u\text{.}$ If ${u}_{\alpha}$ were not in $K,$ we could move it to the extreme left in the expression for $y$ and then remove it. The new terms introduced by this shift would, by the relations (B), correspond to roots higher than $\alpha ,$ so that $\text{supp}\hspace{0.17em}u$ would be diminished, a contradiction. Similarly a shift to the right yields the second part of $(*)\text{.}$ Now as in the proof of Theorem 17 we may conjugate $y$ by a product of ${w}_{\beta}\left(1\right)\text{'s}$ (all in $K\text{)}$ to get ${u}_{\alpha}\ne 1$ for some simple $\alpha ,$ as well as $(*)\text{.}$ We write $a={p}^{H},$ choose $c$ so that $H\prime =H-c{H}_{\alpha}$ is orthogonal to ${H}_{\alpha},$ set ${a}_{\alpha}={p}^{c{H}_{\alpha}},$ $a\prime ={p}^{H\prime},$ $a={a}_{\alpha}a\prime \text{.}$ We only know that $2c=\u27e8H,{H}_{\alpha}\u27e9\in \mathbb{Z},$ so that this may involve an adjunction of ${p}^{1/2}$ which must eventually be removed. If we bear this in mind, then after reducing $(*)$ to the rank 1 case, exactly as in the proof of Theorem 17, what remains to be proved is this:

*Lemma 51:* Assume
$y=ua=\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]\left[\begin{array}{c}{p}^{c}\\ & {p}^{-c}\end{array}\right]$
with $2c\in \mathbb{Z},$ $t\in k,$
$t\notin \theta $ and $t{p}^{-2c}\notin \theta \text{.}$
Then $c$ can be increased by an integer by multiplications by elements of $K\text{.}$

Proof. | |

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 $\mathcal{L}\text{-module}$ 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 ${\stackrel{\u02c6}{\mu}}_{i}\left(a\right)={p}^{-{n}_{i}}\text{.}$ 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. $\square $ |

*Corollary 1:* If $\theta $ is not a field, the group $K$ is maximal in its commensurability class.

Proof. | |

Assume $K\prime $ is a subgroup of $G$ containing $K$ properly. By the theorem there exists $a\in {A}^{+}\cap K\prime ,$ $a\notin K\text{.}$ Some entry of the diagonal matrix $a$ is nonintegral so that by unique factorization $|K\prime /K|$ is infinite. $\square $ |

*Remark:* The case $\theta =\mathbb{Z}$ is of some importance here.

*Corollary 2:* If $\theta ={\mathbb{Z}}_{p}$ and
$k={\mathbb{Q}}_{p}$ $\text{(}p\text{-adic}$ integers and
numbers) and the $p\text{-adic}$ topology is used, then $K$ is a maximal compact subgroup of
$G\text{.}$

Proof. | |

We will use the fact that ${\mathbb{Z}}_{p}$ is compact. (The proof is a good exercise.) We may assume that $G$ is universal. Let $\stackrel{\u203e}{k}$ be the algebraic closure of $k$ and $\stackrel{\u203e}{G}$ the corresponding Chevalley group. Then $G=\stackrel{\u203e}{G}\cap SL(V,k)$ (Theorem 7, Cor. 3), so that $K=\stackrel{\u203e}{G}\cap SL(V,\theta )\text{.}$ Since $\theta $ is compact, so is $\text{End}(V,\theta ),$ hence also is $K,$ the set of solutions of a system of polynomial equations since $\stackrel{\u203e}{G}$ is an algebraic group, by Theorem 6. If $K\prime $ is a subgroup of $G$ containing $K$ properly, there exists $a\in {A}^{+}\cap K\prime ,$ $a\notin K,$ by the theorem. Then $\left\{{\left|{a}^{n}\right|}_{p}\hspace{0.17em}\right|\hspace{0.17em}n\in \mathbb{Z}]\}$ is not bounded so that $K\prime $ is not compact. $\square $ |

*Remark:* We observe that in this case the decompositions $G=BK$ and
$G=K{A}^{+}K$ are relative to a maximal compact subgroup just
as in Theorems 16 and 17. Also in this case the closure formula of Theorem 16, Cor. 7 holds.

*Exercise* (optional): Assume that $G$ is a Chevalley group over $\u2102,\mathbb{R}$ or
${\mathbb{Q}}_{p}$ and that $K$ is the corresponding maximal compact subgroup discussed above. Prove the
commutativity under convolution of the algebra of functions on $G$ which are complex-valued, continuous, with compact support, and invariant
under left and right multiplications by elements of $K\text{.}$ (Such functions are sometimes called zonal functions and
are of importance in the harmonic analysis of $G\text{.)}$ Hint: prove that there exists an antiautomorphism
$\phi $ of $G$ such that
$\phi {x}_{\alpha}\left(t\right)={x}_{-\alpha}\left(t\right)$
for all $\alpha $ and $t,$ that $\phi $ preserves every double coset relative to
$K,$ and that $\phi $ preserves Haar measure. A much harder exercise is to determine the exact structure
of the algebra.

Next we consider a double coset decomposition of $K={G}_{\theta}$ itself in the local case. We will use the following result, the first step in the proof of Theorem 7.

*Lemma 52:* Let $\mathcal{L}$ be the Lie algebra of $G$ (the original Lie algebra of §1 with its coefficients
transferred to $k\text{),}$ $N$ the number of positive roots, and
$\{{Y}_{1},{Y}_{2},\dots ,{Y}_{r}\}$
a basis of ${\bigwedge}^{N}\mathcal{L}$ made up of products of
${X}_{\alpha}\text{'s}$ and ${H}_{i}\text{'s}$ with
${Y}_{1}=\underset{\alpha >0}{\bigwedge}{X}_{\alpha}\text{.}$
For $x\in G$ write $x{Y}_{1}=\Sigma {c}_{j}\left(x\right){Y}_{j}\text{.}$
Then $x\in {U}^{-}HU$ if and only if
${c}_{1}\left(x\right)\ne 0\text{.}$

*Theorem 22:* Assume that $\theta $ is a local principal ideal domain, that $p$ is its unique prime, and that
$k$ and $G$ are as before.

(a) | ${B}_{I}={U}_{p}^{-}{H}_{\theta}{U}_{\theta}$ is a subgroup of ${G}_{\theta}\text{.}$ |

(b) | ${G}_{\theta}=\bigcup _{w\in W}{B}_{I}w{B}_{I}$ (disjoint), if the representatives for $W$ in $G$ are chosen in ${G}_{\theta}$ |

(c) | ${B}_{I}w{B}_{I}={B}_{I}w{U}_{w,\theta}$ with the last component of the right uniquely determined $\text{mod}\hspace{0.17em}{U}_{w,p}\text{.}$ |

Proof. | |

Let $\stackrel{\u203e}{\theta}$ denote the residue class field $\theta /p\theta ,$ ${G}_{\stackrel{\u203e}{\theta}}$ the Chevalley group of the same type as $G$ over $\stackrel{\u203e}{\theta},$ and ${B}_{\stackrel{\u203e}{\theta}},{H}_{\stackrel{\u203e}{\theta}},\dots $ the usual subgroups. By Theorem 18, Cor. 3 reduction mod $p$ yields a homomorphism $\pi $ of ${G}_{\theta}$ onto ${G}_{\stackrel{\u203e}{\theta}}\text{.}$ (1) ${\pi}^{-1}\left({U}_{\stackrel{\u203e}{\theta}}^{-}{H}_{\stackrel{\u203e}{\theta}}{U}_{\stackrel{\u203e}{\theta}}\right)\subseteq {U}^{-}HU\text{.}$ We consider $G$ acting on ${\bigwedge}^{N}\mathcal{L}$ as in Lemma 52. As is easily seen ${G}_{\theta}$ acts integrally relative to the basis of $Y\text{'s.}$ Now assume $\pi x\in {U}_{\stackrel{\u203e}{\theta}}^{-}{H}_{\stackrel{\u203e}{\theta}}{U}_{\stackrel{\u203e}{\theta}}\text{.}$ Then ${c}_{1}\left(\pi x\right)\ne 0$ by the lemma applied to ${G}_{\stackrel{\u203e}{\theta}},$ whence ${c}_{1}\left(x\right)\ne 0$ and $x\in {U}^{-}HU$ again by the lemma.
(2) (3) ${B}_{I}={\pi}^{-1}{B}_{\stackrel{\u203e}{\theta}}\text{.}$ Assume $x\in {\pi}^{-1}{B}_{\stackrel{\u203e}{\theta}}\text{.}$ Then $x\in {U}^{-}BU$ by (1). From this and $x\in {G}_{\theta}$ it follows as in the proof of Theorem 7(b) that $x\in {U}_{\theta}^{-}{H}_{\theta}{U}_{\theta},$ and then that $x\in {B}_{I}\text{.}$
(4) $\square $ |

*Remark:* The subgroup ${B}_{I}$ above is called an Iwahori subgroup. It was introduced in an interesting paper by
Iwahori and Matsumoto (Pabl. Math. I.H.E.S. No. 25 (1965)). There a decomposition which combines those of Theorems 21(a) and 22(b) can be found. The present
development is completely different from theirs.

There is an interesting connection between the decomposition ${G}_{\theta}=\bigcup {B}_{I}w{B}_{I}$ above and the one, ${G}_{\theta}=\bigcup {\left(BwB\right)}_{\theta},$ that ${G}_{\theta}$ inherits as a subgroup of $G,$ namely:

*Corollary:* Assume $w\in W,$ that $S\left(w\right)$
is as in Theorem 16, Cor. 9, and that $\pi :{G}_{\theta}\to {G}_{\stackrel{\u203e}{\theta}}$
is, as above, the natural projection. Then $\pi \left({B}_{I}w{B}_{I}\right)={B}_{\stackrel{\u203e}{\theta}}w{B}_{\stackrel{\u203e}{\theta}},$
and $\pi {\left(BwB\right)}_{\theta}=\bigcup _{w\prime \in S\left(w\right)}{B}_{\stackrel{\u203e}{\theta}}w\prime {B}_{\stackrel{\u203e}{\theta}}\text{.}$
Hence if $\stackrel{\u203e}{\theta}$ is a topological field, e.g.
$\u2102,\mathbb{R}$ or ${\mathbb{Q}}_{p},$ then
$\pi {\left(BwB\right)}_{\theta}$ is the topological closure of
$\pi \left({B}_{I}w{B}_{I}\right)\text{.}$

Proof. | |

The first equation follows from ${\pi}^{-1}{B}_{\stackrel{\u203e}{\theta}}={B}_{I},$ proved above. Write $w={w}_{\alpha}{w}_{\beta}\dots $ as in Lemma 25, Cor. Then $(*)$ ${\left(BwB\right)}_{\theta}={\left(B{w}_{\alpha}B\right)}_{\theta}{\left(B{w}_{\beta}B\right)}_{\theta}\dots $ by Theorem 18, Cor. 1. Now ${\left(B{w}_{\alpha}B\right)}_{\theta}\supseteq {x}_{-\alpha}\left(p\right)$ and ${w}_{\alpha}\left(1\right)$ and is a union of ${B}_{\theta}$ double cosets. Thus $\pi {\left(B{w}_{\alpha}B\right)}_{\theta}\supseteq {B}_{\stackrel{\u203e}{\theta}}\cup {B}_{\stackrel{\u203e}{\theta}}{w}_{\alpha}{B}_{\stackrel{\u203e}{\theta}}={B}_{\theta}{G}_{\alpha ,\stackrel{\u203e}{\theta}}\text{.}$ The reverse inequality also holds since ${\left(B{w}_{\alpha}B\right)}_{\theta}\subseteq {B}_{\theta}{G}_{\alpha ,\theta}$ by Theorem 18, Cor. 1. From this, $(*),$ the definition of $S\left(w\right),$ and Lemma 25, the required expression for $\pi {\left(BwB\right)}_{\theta}$ now follows. $\square $ |

*Appendix.* Our purpose is to prove Theorem 23 below which gives the closure of $BwB$ under very general
conditions. We will write $w\prime \le w$ if
$w\prime \in S\left(w\right)$ with
$S\left(w\right)$ as in Theorem 16, Cor. 9, i.e. if $w\prime $
is a subexpression (i.e. the product of a subsequence) of some minimal expression of $w$ as a product of simple reflections.

*Lemma 53:* The following are true.

(a) | If $w\prime $ is a subexpression of some minimal expression for $w,$ it is a subexpression of all of them. |

(b) | In (a) the subexpressions for $w\prime $ can all be taken to be minimal. |

(c) | The relation $\le $ is transitive. |

(d) | If $w\in W$ and $\alpha $ is a simple root such that $w\alpha >0$ (resp. ${w}^{-1}\alpha >0\text{),}$ then $w{w}_{\alpha}>w$ (resp. ${w}_{\alpha}w>w\text{).}$ |

(e) | ${w}_{0}\ge w$ for all $w\in W\text{.}$ |

Proof. | |

(a) This was proved in Theorem 16, Cor. 7 and 9 in a rather roundabout way. It is a direct consequence of the following fact, which will be proved in a later section: the equality of two minimal expressions for $w$ (as a product of simple reflections) is a consequence of the relations ${w}_{1}{w}_{2}\dots ={w}_{2}{w}_{1}\dots $ $\text{(}{w}_{1},{w}_{2}$ distinct simple reflections, $n$ terms on each side, $n=$ order ${w}_{1}{w}_{2}\text{).}$ (b) If $w\prime ={w}_{1}{w}_{2}\dots {w}_{r}$ is an expression as in (b) and it is not minimal, then two of the terms on the right can be cancelled by Appendix II 21. (c) By (a) and (b). (d) If $w\alpha >0$ and ${w}_{1}{w}_{2}\dots {w}_{s}$ is a minimal expression for $w,$ then ${w}_{1}\dots {w}_{s}{w}_{\alpha}$ is one for $w{w}_{\alpha}$ by Appendix II 19, so that $w{w}_{\alpha}>w,$ and similarly for the other case. (e) This is proved in Lemma 46. $\square $ |

Now we come to our main result.

*Theorem 23:* Let $G$ be a Chevalley group. Assume that $k$ is a nondiscrete topological field and that the
topology inherited by $G$ as a matric group over $k$ is used. Then the following conditions on
$w,w\prime $ are equivalent.

(a) | $Bw\prime B\subseteq \stackrel{\u203e}{BwB}\text{.}$ |

(b) | $w\prime \le w\text{.}$ |

Proof. | |||

Let ${Y}_{1}$ be as in Lemma 52 and more generally ${Y}_{w}=\underset{\alpha >0}{\bigwedge}{X}_{w\alpha}$ for $w\in W\text{.}$ For $x\in G$ let ${c}_{w}\left(x\right)$ denote the coefficient of ${Y}_{w}$ in $x{Y}_{1}\text{.}$ We will show that (a) and (b) are equivalent to:
(a) $\Rightarrow $ (c). We have ${x}_{\beta}\left(t\right){X}_{\alpha}={X}_{\alpha}+\Sigma {t}^{j}{X}_{j}$ with ${X}_{j}$ of weight $\text{(}0$ or a root) $\alpha +j\beta ,$ and ${n}_{w}{X}_{\alpha}=c{X}_{w\alpha}$ $(c\ne 0)$ if ${n}_{w}$ represents $w$ in $W$ in $N/H\text{.}$ Thus $(*)$ $BwB{Y}_{1}\subseteq {k}^{*}{Y}_{w}+$ higher terms in the ordering given by sums of positive roots. Thus ${c}_{w\prime}$ is not identically $0$ on $Bw\prime B,$ hence also not on $BwB,$ by (a). (c) $\Rightarrow $ (b). We use downward induction on $N\left(w\prime \right)\text{.}$ If this is maximal then $w\prime ={w}_{0},$ the element of $W$ making all positive roots negative, and then $w={w}_{0}$ by (c) and $(*)$ above. Assume $w\prime \ne {w}_{0}\text{.}$ Choose $\alpha $ simple so that ${w\prime}^{-1}>0,$ hence $N\left({w}_{\alpha}w\prime \right)>N\left(w\prime \right)\text{.}$ Since ${c}_{w\prime}\left(BwB\right)\ne 0$ and $B{w}_{\alpha}bwB\subseteq BwB\cup B{w}_{\alpha}wB,$ we see that ${c}_{{w}_{\alpha}w\prime}\left(BwB\right)\ne 0$ or ${c}_{{w}_{\alpha}w\prime}\left(B{w}_{\alpha}wB\right)\ne 0,$ so that ${w}_{\alpha}w\prime \le w$ or ${w}_{\alpha}w\prime \le {w}_{\alpha}w\text{.}$ In the first case $w\prime <w$ by Lemma 53(c) and (d). In the second case if ${w}^{-1}\alpha <0$ then ${w}_{\alpha}w<w$ by Lemma 53(d) which puts us back in the first case, while if not we may choose a minimal expression for $w$ starting with ${w}_{\alpha}$ and conclude that $w\prime \le w\text{.}$ (b) $\Rightarrow $ (a). By the definitions and the usual calculus of double cosets, this is equivalent to: if $\alpha $ is simple, then $\stackrel{\u203e}{B{w}_{\alpha}B}=B\cup B{w}_{\alpha}B\text{.}$ The left side is contained in the right, an algebraic group, hence a closed subset of $G\text{.}$ Since $B{w}_{\alpha}B$ contains ${\U0001d51b}_{\alpha}-1$ and the topology on $k$ is not discrete, its closure contains $1,$ hence also $B,$ proving the reverse inequality and completing the proof of the theorem. $\square $ |

*Remark:* In case $k$ above is $\u2102,\mathbb{R}$ or
${\mathbb{Q}}_{p},$ the theorem reduces to results obtained earlier. In case $k$
is infinite and the Zariski topology on $k$ and $G$ are used it becomes a result of Chevalley (unpublished).
Our proof is quite different from his.

*Exercise:* (a) If $w\in W$ and $\alpha $ is a *positive* root such that
$w\alpha >0,$ prove that
$w{w}_{\alpha}>w$ (compare this with Lemma 53(d)), and conversely if
$w\prime \le w$ then $(*)$
there exists a sequence of positive roots ${\alpha}_{1},{\alpha}_{2},\dots ,{\alpha}_{r}$
such that if ${w}_{i}={w}_{{\alpha}_{i}}$ then
$w\prime {w}_{1}\dots {w}_{i-1}{\alpha}_{i}>0$
for all $i$ and
$w\prime {w}_{1}\dots {w}_{r}=w\text{.}$
Thus $w\prime \le w$ and $(*)$ are equivalent.

(b) It seems to us likely that $w\prime \le w$ is also equivalent to: there exists a permutation $\pi $ of the positive roots such that $w\prime \pi \alpha -w\alpha $ is a sum of positive roots for every $\alpha >0\text{;}$ or even to: $\underset{\alpha >0}{\Sigma}(w\prime \alpha -w\alpha )$ is a sum of positive roots.

This is a typed excerpt of *Lectures on Chevalley groups* by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson.
This work was partially supported by Contract ARO-D-336-8230-31-43033.