BN-pairs

$B N -pairs$

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 27 November 2014

$B N -pairs$

A group with $B N -pair$ is a quadruple $(G, B, N, S)$ where $G$ is a group,
$B$ is a subgroup of $G,$
$N$ is a subgroup of $G,$
$S$ is a subset of the coset space $N / (B \cap N),$ and the following axioms are satisfied:

(T1)	$B \cup N$ generates $G,$ $B \cap N$ is a normal subgroup of $N,$
(T2)	$S$ generates the group $W = B / (B \cap N)$ and each $s \in S$ has order $2,$
(T3)	For each $s \in S$ and $w \in W,$ $(B s B) (B w B) \subseteq B w B \cup B s w B,$
(T4)	For each $s \in S,$ $(B s B) (B s B) = B \cup B s B .$

Every $w \in W$ is of the form $w = n (B \cap N),$ $n \in N .$ By $B w B$ is meant the double coset $B n B .$ Since for $n \in N,$ the double coset $B n B$ depends only on $n$ modulo $B \cap N,$ this is well defined. Similarily we can define $w B,$ $B w,$ $w^{- 1} B w$ by, respectively, $n B,$ $B n,$ $n^{- 1} B n .$

Taking inverses shows axiom (T3) is equivalent to:

For each $s \in S$ and $w \in W,$ $(B w B) (B s B) \subseteq B w B \cup B w s B .$

The elements $s \in S$ are called simple reflections. Every element in $w \in W$ can be written as a product, $w = s_{1} \dots s_{q},$ of simple reflections.

(Length) For $w \in W,$ its length, denoted $l (w),$ is the minimal $q$ such that $w = s_{1} \dots s_{q},$ $s_{i} \in S .$

(Bruhat Decomposition) The group $G$ is a disjoint union over $w \in W$ of the double cosets $B w B .$ $G = ⋃_{w \in W} B w B, (disjoint union) is called the Bruhat decomposition.$

Proof.

$G = ⋃_{w \in W} B w B$ will be proved below.

We show here that $B w_{1} B = B w_{2} B$ implies $w_{1} = w_{2} .$

We prove this by induction on $q,$ assuming $l (w_{1}) \geq l (w_{2}) = q .$

First note $B w B = B$ implies $w = 1 .$ This settles the case $q = 0 .$

Now suppose $B w_{1} B = B w_{2} B$ with $l (w_{1}) \geq l (w_{2}) = q \geq 1 .$ Then we can find an $s \in S$ such that $l (s w_{2}) = q - 1 .$ (If we write $w_{2} = s_{1} \dots s_{q}$ as a product of $q$ simple reflections, $s = s_{1}$ will do.)

So $l (w_{1}) > l (s w_{2}) = q - 1 .$ Hence by induction $B w_{1} B$ and $B s w_{2} B$ are different, so disjoint.

But $(B s w_{2} B) \subseteq (B s B) (B w_{2} B) = (B s B) (B w_{1} B) \subseteq (B w_{1} B) \cup (B s w_{1} B) .$

Hence $B s w_{2} B = B s_{1} w B,$ $l (s w_{1}) \geq l (w_{1}) - 1 \geq l (w_{2}) = q - 1 .$

By induction $s w_{1} = s w_{2},$ and we deduce $w_{1} = w_{2} .$

$□$

Let $s_{1}, s_{2}, \dots, s_{q}$ and $w \in W .$ Then $(B s_{1} s_{2} \dots s_{q} B) (B w B) \subseteq ⋃_{1 \leq i_{1} < \dots < i_{p} \leq q} (B s_{i_{1}} \dots s_{i_{p}} w B)$ Note: The union is over all substrings of $1 < 2 < \dots < q,$ including the empty string. Thus the union includes the double coset $B w B .$

Proof.

The proof is by induction on $q$ using the third $B N$ pair axiom.

The case $q = 0$ is trivial.

Suppose $q \geq 1 .$ Then using $B x y B \subseteq (B x B) (B y B),$ $(B s_{1} \dots s_{q} B) (B w B) \subseteq (B s_{1} B) (B s_{2} \dots s_{q} B) (B w B) .$ So by induction, $\begin{array}{rcl} (B s_{1} \dots s_{q} B) (B w B) & \subseteq & (B s_{1} B) (⋃_{2 \leq j_{1} < \dots < j_{p} \leq q} (B s_{j_{1}} \dots s_{j_{p}} w B)) \\ = & ⋃_{2 \leq j_{1} < \dots < j_{p} \leq q} (B s_{1} B) (B s_{j_{1}} \dots s_{j_{p}} w B) . \end{array}$

The third $B N -pair$ axiom (T3) says for $s \in S,$ $(B s B) (B w B) \subseteq (B w B) \cup (B s w B)$ and thus we get $\begin{array}{rcl} (B s_{1} \dots s_{q} B) (B w B) & \subseteq & ⋃_{2 \leq j_{1} < \dots < j_{p} \leq q} ((B s_{1} s_{j_{1}} \dots s_{j_{p}} w B) \cup (B s_{j_{1}} \dots s_{j_{p}} w B)) \\ = & ⋃_{1 \leq i_{1} < i_{2} < \dots < i_{p} \leq q} (B s_{i_{1}} \dots s_{i_{p}} B) \end{array}$ which proves the lemma.

$□$

For $I \subseteq S$ set $W_{I} = ⟨ s | s \in I ⟩ .$

$P_{I} = B W_{I} B = ⋃_{w \in W_{I}} B w B$ is a subgroup of $G .$ In particular $G = B W B .$

Proof.

It is clear that if $x \in P_{I}$ then $x^{- 1} \in P_{I},$ since $B$ and $W_{I}$ are both groups.

Suppose $x$ and $y$ are elements of $P_{I} .$ Then for some $s_{1}, \dots, s_{q} \in S$ and $w \in W_{I},$ $x \in B s_{1} \dots s_{q} B and y \in B w B .$ Then $x y \in (B s_{1} \dots s_{q} B) (B w B) .$ So by the lemma $x y \in ⋃_{1 \leq i_{1} < \dots < i_{p} \leq q} (B s_{i_{1}} \dots s_{i_{p}} w B) \subseteq P_{I} .$ Hence $P_{I}$ is a group.

If $I = S$ then $P_{I} = B W B$ is a group containing both $B$ and $N .$ Therefore $P_{S} = G .$

$□$

Let $I, J \subseteq S$ and let $P_{I} = B W_{I} B$ and $P_{J} = B W_{J} B$ be as above. Then, for $w \in W,$ $P_{I} w P_{J} = B W_{I} w W_{J} B .$

Proof.

$B W_{I} w W_{J} B \subseteq P_{I} w P_{J}$ because $B W_{I} \subseteq P_{I}$ and $W_{J} B \subseteq P_{J} .$

We now derive the opposite inclusion. Let $t_{1}, \dots, t_{p} \in I$ and $s_{1}, \dots, s_{q} \in J .$ Then by the lemma (and its analogous right hand version), $(B t_{1} \dots t_{p} B) (B w B) (B s_{1} \dots s_{q})$ is contained in a union of double cosets, $B w_{1} w W_{2}$ with $w_{1} \in P_{I}$ and $w_{2} \in P_{J} .$

Hence $P_{I} w P_{J} \subseteq B W_{I} w W_{J} B .$

$□$

If $w$ has length $q,$ and $w = s_{1}, s_{2}, \dots, s_{q},$ then $B w B = (B s_{1} B) (B s_{2} B) \dots (B s_{q} B) .$

Proof.

By induction on $q .$ The case $q = 1$ is trivial.

For $q > 1,$ $s_{2} \dots s_{q}$ and $s_{1} \dots s_{q - 1}$ each have length $q - 1 .$ By induction $\begin{array}{rcl} B s_{2} \dots s_{q} B & = & (B s_{2} B) \dots (B s_{q} B) \\ B s_{1} \dots s_{q - 1} B & = & (B s_{1} B) \dots (B s_{q - 1} B) \end{array}$ Hence by $B N -pair$ axiom (T3), $\begin{array}{rcl} (B s_{1} B) (B s_{2} B) \dots (B s_{q} B) & = & (B s_{1} B) (B s_{2} \dots s_{q} B) \\ \subseteq & B s_{1} \dots s_{q} B \cup B s_{2} \dots s_{q} B, \\ (B s_{1} B) \dots (B s_{q - 1} B) (B s_{q} B) & = & (B s_{1} \dots s_{q - 1} B) (B s_{q} B) \\ \subseteq & B s_{1} \dots s_{q} B \cup B s_{1} \dots s_{q - 1} B . \end{array}$ If we can show $B s_{2} \dots s_{q} B \cap B s_{1} \dots s_{q - 1} = \emptyset,$ then we deduce $(B s_{1} B) \dots (B s_{q} B) \subseteq B s_{1} \dots s_{q} B,$ and since the opposite inclusion is always true, that $(B s_{1} B) \dots (B s_{q} B) = B s_{1} \dots s_{q} B .$

But by Bruhat decomposition, $B s_{2} \dots s_{q} B \cap B s_{1} \dots s_{q - 1} \neq \emptyset$ if and only if $s_{2} \dots s_{q} = s_{1} s_{2} \dots s_{q - 1}$ which on left multiplication by $s_{1}$ gives $s_{1} s_{2} \dots s_{1} = s_{2} \dots s_{q - 1},$ contradicting $l (w) = q .$

$□$

For $w \in W$ and $s \in S$ there are a priori three possibilities for the length of $s w$ or $w s :$ either $l (w),$ $l (w) + 1$ or $l (w) - 1 .$ In fact the first possibility never occurs, as the following proposition, which strengthens the previous, one shows.

Suppose $w \in W$ and $s \in S .$
Then $\begin{matrix} l (s w) \geq l (w) if and only if (B s B) (B s w B) = B s w B, \\ l (s w) \leq l (w) if and only if (B s B) (B s w B) = B s w B \cup B w B, \end{matrix}$ Similarly for multiplication of $w$ by $s$ on the right.

Proof.

We proceed by induction on $q = l (w),$ the case $q = 0$ being trivial.

Suppose $l (s w) \geq l (w) = q .$

Write $w = w' s',$ $l (w') = q - 1,$ $s' \in S,$

Then by induction $(B s B) (B w' B) = B s w' B$ and $(B w' B) (B s' B) = B w' s' B = B w B .$

Thus we have $\begin{array}{rcl} (B s B) (B w B) & = & (B s B) (B w' B) (B s' B) \\ = & (B s w' B) (B s' B) \\ \subseteq & (B s w' s' B) \cup (B s w' B) (axiom) \\ = & B s w B \cup B s w' B \end{array}$ We cannot have $(B s B) (B w B) = B s w B \cup B w B$ unless $B w B = B s w' B .$ But then, by Bruhat decomposition we deduce, $s w = w',$ giving $l (s w) = q - 1,$ contradicting $l (s w) \geq l (w) = q .$

Hence $(B s B) (B s w B) = B s w B .$

Now consider the case $l (s w) \leq l (w) .$

This is equivalent to $l (s s w) = l (w) \geq l (s w) .$ Then by the result above, $B w B = B s s w B = (B s B) (B s w B) .$ Hence we deduce $\begin{array}{rcl} (B s B) (B w B) & = & (B s B) (B s B) (B w B) \\ = & (B \cup B s B) (B w B) (axiom) \\ = & B w B \cup (B s B) (B w B) \\ \supseteq & B w B \cup B s w B \\ = & B w B \cup B s w B . \end{array}$ This together with (T3) shows $(B s B) (B w B) = B w B \cup B s w$ in this case.

$□$

(Cancellation Properties)

(a)	Suppose $w = s_{1} s_{2} \dots s_{q},$ $l (w) = q,$ and $l (s w) < l (w) .$ Then for some $i,$ $s w = s_{1} \dots {\hat{s}}_{i} \dots s_{q} .$ Similarity if $l (w s) < l (w),$ $w s = s_{1} \dots {\hat{s}}_{i} \dots s_{q}$ for some $i .$
(b)	Suppose we express $w \in W$ as a word in $S :$ $w = s_{1} s_{2} \dots s_{q} .$ Then if $l (w) < q,$ $w = s_{1} \dots {\hat{s}}_{i} \dots {\hat{s}}_{j} \dots s_{q},$ for some $i, j .$

Proof.

(a)

Set

w_{i} = s_{1} \dots s_{i} .

Then

l (w_{i}) = i .

Let

i

be minimal such that

l (s w_{i}) < l (w_{i}) .

Such

i

exist as this is the case for

i = q .

Hence

l (s w_{i - 1}) = i > i - 1 = l (w_{i - 1})

and

l (s w_{i}) < l (w_{i}) .

By the proposition above,

(B s B) (B w_{i - 1} B) = B s w_{i - 1} B and (B s B) (B w_{i}) B = B s w_{i} B \cup B w_{i} B .

We calculate

(B s B) (B w_{i - 1} B) (B s_{i} B) = {\begin{cases} (B s w_{i - 1} B) (B s_{i} B) \subseteq B s w_{i} B \cup B s w_{i - 1} B, \\ (B s B) (B w_{i} B) = B s w_{i} B \cup B w_{i} B . \end{cases}

Thus we must have equality in the first case and

B s w_{i - 1} B = B w_{i} B .

By Bruhat decomposition we deduce,

s w_{i - 1} = w_{i} .

This gives

s w = s s_{1} \dots s_{i - 1} s_{i} \dots s_{1} = s_{1} \dots s_{i - 1} s_{i} s_{i} \dots s_{1} = s_{1} \dots {\hat{s}}_{i} \dots s_{q} .

(b)

Put

w_{j} = s_{1} s_{2} \dots s_{j} .

Let

j \geq 1

be minimal such that

l (w_{j}) < j .

(There is at least one such

j,

j = q,

l (w) = l (w_{q}) < q).

Then

l (s_{j} w_{j - 1}) = l (w_{j}) \leq j - 1 = l (w_{j - 1}) .

By the cancellation lemma there is an

i

such that

s_{1} \dots s_{j - 1} s_{j} = s_{1} \dots {\hat{s}}_{i} \dots s_{j - 1} .

Postmultiplication by

s_{j + 1} \dots s_{q},

gives

w = s_{1} \dots {\hat{s}}_{i} \dots {\hat{s}}_{j} \dots s_{q} .

$□$

We deduce immediately from the cancellation results:

If $w = s_{1} \dots s_{p}$ and $l (w) = q,$ then $w = s_{i_{1}} \dots s_{i_{q}}$ for some sequence, $1 \leq i_{1} < \dots < i_{q} \leq p .$

Since the elements of any $W_{I}$ are products of $s_{i} \in I$ we deduce:

If $w \in W_{I},$ $I \subseteq S,$ has length $l (w) = q$ then $w$ can be expressed $w = s_{1} \dots s_{q},$ all $s_{i} \in I .$

This means that the length of any $w \in W_{I},$ defined relative to $W_{I},$ i.e. with respect to the generators $s \in I$ of $W_{I},$ is the same as its length defined with respect to the set of generators $S$ of $W .$

In particular as $l (w) = 1$ only if $w = s \in S,$ we deduce:

We have $s \in W_{I}$ if and only if $s \in I .$

This generalizes.

Suppose $l (w) = q$ and $w = s_{1} \dots s_{q} .$ Then $w \in W_{I},$ $I \subseteq S,$ if and only if each $s_{i} \in I .$

Proof.

The if part follows from the definition of $W_{I} .$

For the converse we work by induction on $q = l (w) .$ The case $q = 1$ is the corollary above.

For $q > 1,$ $s_{2} \dots s_{q}$ has length $q - 1 .$

By the previous corollaries we can write $w = t_{1} \dots t_{1},$ all $t_{i} \in I .$

Then using the cancellation results: for some $i,$ $s_{2} \dots s_{q} = s_{1} t_{1} \dots t_{q} = t_{1} \dots {\hat{t}}_{j} \dots t_{q} \in W_{I} .$

Hence by induction $s_{2}, \dots, s_{q} \in I .$ From this we deduce $s_{1} \in W_{I},$ which we know implies $s_{1} \in I,$ also.

$□$

From the above, given $w = s_{1} \dots s_{q}$ of length $q,$ we have $w \in W_{I} \cup W_{J}$ if and only if each $s_{i} \in I \cap J .$ This together with the definition $P_{K} = B W_{K} B,$ $K \subseteq S$ proves:

(a)	$W_{I} \cap W_{J} = W_{I \cap J} .$
(b)	$P_{I} \cap P_{J} = P_{I \cap J} .$

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