well, the moon is just about full. If you'd like I could just show you. It would be much easier.

Owoooooooo!

The sad part is...Fuzzy writes that out longhand every time...and does it from memory....

Wow so that's fuzzy logic, I am impressed and feel so boring as mine is just a combination of my dog's name Shaza and mother's maiden name hence shazray voila!

Jafo: The sad part is...Fuzzy writes that out longhand every time...and does it from memory.

No - the sad part is - I didn't understand a word of it..

PS: I didnt see anyone ask what JAFO stands for..tell the " new peeps "

J.A.F.O. ....it's an acronym.

Stands for  'Jolly And Felicitous Ocker'....

Maybe...

Googling will elucidate...

J.A.F.O. ....it's an acronym. Stands for 'Jolly And Felicitous Ocker'....

uh huh

Guess I use an Aussie dictionary

Kupotek comes from 2 interests of mine about 5 years ago, Final Fantasy and Egyptology.

Jester is an anagram derived from the first three letters of my two sons' names.

Of course every variation of JESTER was already taken (too many old Top Gun fans, I suppose) so I stuck my initials, T and B on the ends.
I think it looks kind of silly now, but I'm too lazy to change it. And it's never taken on any site I register on.

The sad part is...Fuzzy writes that out longhand every time...and does it from memory...

The other sad part is, I tried to read the whole damn thing again.

starkers began as a spoonerism of letters in my name - the St being the first two letters of my surname and ark being the last three of my Christian name. The ers was added after an unfortunate incident with a cat by my first wife and others who thought it not only hilarious but appropriate....so starkers it has been for the last 35 years or so.

Funny thing is, I was chatting with another starkers on another site who uses the nick for similar reasons, an anagram of his real name....though he had added the year of his birth because there were two others registered including myself.

whitedragon comes from Anne McCaffrey's Dragonriders of Pern series. Unique & easily underestimated. 33 was my age when I started using the handle...

Since algebra is used to expain handles here is my bit...

First a small comment which relates to the subject:
while it's true that Brown initially includes S (the set of
involutions generating the Weyl group) as part of the data
of a BN-pair, he proves later in the chapter that S is
uniquely determined (as the set of w in W such that
B union BwB is a subgroup of G). So you were right:
S is described by *properties* of a BN-pair.

So for us the data will be (G, B, N) where B and N are
subgroups which jointly generate G, and T = B intersect N
is normal in N. Define W := N/T. Before stating the Hecke
algebra-based axioms, I'll make some general remarks:

(1) There is a sup-preserving map between power sets

[ ]: 2[W] --> 2[B\G/B]

sending w to BwB (where BwB := Bw'B for any w' in N which
maps to w under the projection N --> W). I claim that even under
the minimal hypotheses given above, this is a lax homomorphism
from the Boolean group algebra to the Boolean Hecke algebra.

To see this more clearly, let's express the Boolean Hecke algebra
structure directly in terms of 2[B\G/B], starting from the formula

J_[k] (gB) = Sum_(g' in gB) g'kB

(the normalizing factor 1/|B| being absorbed in the Boolean case).
Composing, we obtain

J_[k'] J_[k] (gB) = J_[k'] Sum_(g' in gB) g'kB

= Sum_(g' in gB) J_[k'] (g'kB)

= Sum_(g' in gB) Sum_(g'' in g'kB) g''k'B

= Sum_(g' in gB) [ Sum_(b in g'kbk'B]

which makes it clear that

J_[k'] J_[k] = Sum_(b in J_[kbk'].

Let me abbreviate the right side to J_[kBk'], and the last equation
to

[k].[k'] = [kBk'] := Sum_(b in [kbk']

where [k] denotes the orientation BkB for any element k in G,
and [k].[k'] denotes multiplication in 2[B\G/B], here taken
as the multiplication which is *opposite* to composition of the
random jump operators J_[k]. We have

[ww'] <= [wBw'] = [w].[w']

so that [ ]: 2[W] --> 2[B\G/B] is a (normal) lax homomorphism.

(If you don't like passing to the opposite of the algebra of random
jump operators, another option is to consider [ ] as a lax
homomorphism from 2[W^{op}] and use the inversion isomorphism
2[W] ~ 2[W^{op}]. But I'll stick to how I'm doing it here.)

(2) We will define the subset S of W to be the set of elements w of
W for which

[w]^2 = [w] + 1

is a minimal polynomial equation satisfied in the Boolean Hecke
algebra. (This means [w] + 1 is a nontrivial idempotent which,
translated into Brown's language, means that BwB union B is a
submonoid of G. Under the BN-pair axiom that each such w is
an involution in W, this is actually a subgroup of G.)


BN-pair axioms
----------------------

(1) The subset S defined above is a set of involutions which
generates W.

(2) For s in S and w in W, the following holds in the Boolean Hecke
algebra:
[s].[w] <= [w] + [sw].

I hope these conditions seem motivated and intuitively reasonable.
Axiom (2) says that when we start with a flag f in the orientation
class described by [w] (oriented with respect to a flag stabilized
by and apply a random jump which changes a single feature of
f corresponding to [s], either we stay in the same orientation class,
or we jump into a Bruhat cell or chamber which is s-adjacent to the
one described by [w].

Remark: These axioms are not quite the same as those given by
Brown/Tits. In particular, the assumption that the elements of S
are involutions, and the assumption that their minimal polynomial
equations in the Hecke algebra are given as above, may be dropped,
just by adding the assumption that there exists a set S of generators
of W such that axiom (2) is satisfied, and also that sBs^{-1} is not
contained in B for any S. The equations s^2 = 1 in the group
algebra and s^2 = s + 1 in the Hecke algebra are derivable from
these assumptions (and S is uniquely determined as the set of
elements obeying these equations).

As usual in axiomatics, the degree to which one applies Occam's
razor is a matter of personal aesthetics -- at the expense of some
extra whittling down, I have chosen a formulation which emphasizes
the interactions between the group and Hecke algebras. In practice
it is probably no harder to verify the axioms as given here than it is
to verify the Brown/Tits axioms, so the difference is probably harmless.


Consequences
----------------------

(a) One consequence is a complete description of parabolic
subgroups (i.e. subgroups intermediate between B and G --
Brown actually reserves the word 'parabolic' for a subgroup
conjugate to one between B and G) on the basis of the axioms.
First we show that unions of double cosets BwB, where w
ranges over words generated by a subset S' of S, are parabolic
subgroups. Or, translated into the language of Hecke algebras,
that

Sum_(w in ) [w]

is idempotent for each subset S' of S. (The idempotency says
that the union of such double cosets [w] is the submonoid
generated by B and {s: s in S'}, but since elements s in S are
involutions, it's actually a subgroup.)

It suffices that

[w].[w'] <= Sum_(w in ) [w]

whenever w, w' belong to . This is shown by induction on
the length of a word s1...sd used to write w. The case d = 0
is obvious. The inductive step is an easy consequence of the
lax homomorphism and Axiom (2):

[s1(s2...sd)].[w'] <= [s1].[s2...sd].[w'] (lax homomorphism)

<= [s1]. Sum_(w in ) [w] (inductive hypothesis)

<= Sum_(w in ) [w] + [(s1)w] (axiom (2))

= Sum_(w in ) [w]. QED

In particular, the union of the double cosets [w] over all w in W
(which Brown writes as BWB) is a subgroup of G, and since
B and N jointly generate G, this subgroup must be G.

I will save for later the proof of

Proposition 1. All intermediate subgroups between B and G are
described by idempotents in the Hecke algebra of this form. (Hence
they are in one-one correspondence with subsets S' of S.)

( Our next task will be to establish the Bruhat decomposition
of G. We had just observed that the union of the double cosets
BwB over all w in W is G, i.e. that [ ]: 2[W] --> 2[B\G/B] is
surjective. Now we prove this map is injective, i.e. that [w] = [w']
implies w = w'.

The proof is by induction on d = min{d(w), d(w')} where *d(w)* is
the minimum length of a word in S which evaluates to w. We may
assume d = d(w').

If d = 0, then w' = 1, and if [w] = 1, i.e. if Bw'B = B for some w'
which maps to w under N --> W, then w' is in B and hence w = 1
by definition of W.

If d > 0, write w' in the form sw'' where s is in S and d(w'') = d-1.
By assumption, [sw''] = [w] and hence

[w''] = [s^2 w''] <= [s].[sw''] = [s].[w]

<= [w] + [sw] by axiom (2)

and since distinct elements of the form [w] are disjoint, either
[w''] = [w] or [w''] = [sw]. By induction, w'' = w or w'' = sw.
The first case is impossible since d(w'') < d(w). Hence w'' = sw,
whence w' = sw'' = w (using the fact that s is an involution). This
completes the proof.

(c) Now we prove a result which is used in the proof that the
Hecke algebra presentation (recalled below) is correct:

Lemma 1: If d(sw) >= d(w), then [s].[w] = [sw].

Proof: By induction on d = d(w). The case d = 0 is clear. For
d > 0, write w = w't where t is in S and d(w') = d-1. We will show
that
([s].[w]) /\ [w] = 0

so that [s].[w] <= [w] + [sw] (axiom (2)) implies [s].[w] = [sw].

From

d(w') + 1 = d(w) <= d(sw) = d(sw't) <= d(sw') + 1

(where the first inequality is the hypothesis of the lemma), we
get d(w') <= d(sw'). Therefore, by inductive hypothesis,

[sw'] = [s].[w'].

Now we calculate

([s].[w]) /\ [w] = ([s].[w't]) /\ [w]

<= ([s].[w'].[t]) /\ [w] (lax homomorphism)

= ([sw'].[t]) /\ [w] (equation above)

<= ([sw'] + [sw't]) /\ [w] (variant of axiom (2))

= ([sw'] + [sw]) /\ [w]

= ([sw'] /\ [w]) + ([sw] /\ [w])

The proof is complete if both summands in the last line are 0.
Because [ ] is injective (from ( above), it suffices that sw != w
(clear) and that sw' != w -- this follows from w' != sw, which
we know from d(w') < d(w) <= d(sw). QED

(d) Now we prove our principal theorem on Hecke algebra
presentations:

Theorem: If s1...sd is a minimal word in S which evaluates to w,
then [s1].[s2].(...).[sd] = [w] in the Boolean Hecke algebra.

Proof: By induction on d. The case d = 1 is trivial. Now s1...sd
is a minimal word only if s2...sd is a minimal word, so

[s2].(...).[sd] = [s2...sd]

by inductive hypothesis. Since d(s2...sd) = d-1 < d(s1s2...sd) by
definition of minimality, the hypothesis of Lemma 1 is satisfied,
so that

[s1].[s2...sd] = [s1...sd] = [w]

and the inductive step goes through. QED

Corollary: Under the hypothesis of the preceding theorem,
[s1].(...).[sd] = [w] in the Hecke algebra taken over any
Q_(+)-algebra as base rig.

Proof: It suffices to consider the case where the base rig is Q_(+).
Writing out [s1].(...).[sd] as a linear combination of basis elements
[w] in B\G/B,

[s1].(...).[sd] = Sum_(v in W) a_v [v] (a_v in Q_(+)),

we see by applying the change-of-base-rig Q_(+) --> 2 that the
only nonzero a_v occurs when v = w (using the theorem). This
coefficient a_v = 1 by conservation of probability. QED

(e) In serious applications of the theorem and corollary of (d), we
need to take advantage of the fact that (W, S) is a Coxeter system.
To this end we first prove a companion to Lemma 1:

Lemma 2: If d(w) >= d(sw), then [s].[w] = [w] + [sw].

Proof: Since d(ssw) >= d(sw), Lemma 1 gives

[w] = [ssw] = [s].[sw]

so that
[s].[w] = [s]^2 .[sw]

= ([s] + 1).[sw] (defining property of S)

= [s].[sw] + [sw]

= [w] + [sw]. QED

To prove that (W, S) is a Coxeter system, we verify a somewhat
technical necessary and sufficient condition called the "folding
condition" (which I don't understand yet, but which intuitively has
to do with "folding" a Coxeter complex along a wall, i.e. onto the
half-space on the side of the wall which contains the "preferred"
chamber stabilized by :

Folding Condition: Given w in W and s, t in S such that

d(sw) = d(w) + 1 = d(wt),

either d(swt) = d(w) + 2 or swt = w.

Verification: Suppose d(swt) < d(w) + 2. Then by Lemmas 1 and 2,

[s].[w].[t] = [s].[wt] = [wt] + [swt]

[s].[w].[t] = [sw].[t] = [sw] + [swt].

Using the injectivity of [ ]: 2[W] --> 2[B\G/B], the summands in
each of these equations are disjoint, and it follows that [wt] = [sw],
whence (again by injectivity) wt = sw. Therefore swt = w. QED

Having proven that (W, S) is a Coxeter system, we may give a
presentation of the Hecke algebra (over any Q_(+)-algebra), at
least in the case where G is finite. Here the minimal polynomial
of the orientation [s] for s in S is

[s]^2 = (1/q)((q-1)[s] + [1])

where q is the cardinality of the Bruhat cell which is s-adjacent to
the 1-point identity cell B in the "flag manifold" G/B. Aside from
these, the remaining equations of the presentation are of the form

[s].[t].[s]... = [t].[s].[t]...

where sts... and tst... are alternating words of length m(s, t) in
letters s, t in S, and where m(s, t) is the order of st in W. (To
apply the theorem above, we need to know these alternating words
are minimal; this follows from Tits's solution of the word problem for
the presentation.)

I am not quite sure what modifications might be needed when G
is non-finite, particularly with analogues of the quadratic equations
for [s].

(f) Finally we return to Proposition 1, on the characterization of
parabolic subgroups.

Lemma 3: If s1...sd is a minimal word which evaluates to w, then
the subgroup of G generated by w and B contains the elements
s1, ..., sd. Therefore (Proposition 1) every subgroup intermediate
between B and G is of the form

Sum_(w' in ) [w'] = Sum_(w' in ) Bw'B

for some subset S' of S.

Proof: By induction on d. The case d = 0 is trivial. For d > 0,
s2...sd is a minimal word for (s1)w, so by inductive hypothesis
{s2, ..., sd} is contained in <(s1)w, B>, and also we have
d(s1 w) < d(w). Hence Lemma 2 applies:

[s1].[w] = [w] + [s1 w]

so
[s1].[w] /\ [w] != 0.

In other words, B(s1)BwB and BwB have a common element.
It follows that s1 belongs to and also (as noted above)

{s2, ..., sd} is contained in <(s1)w, B>

is contained in =

which completes the inductive step.

For the second statement of the lemma, let P be a subgroup
between B and G. P is a union of double cosets, i.e. there is a
uniquely determined subset (in fact a subgroup) T of W such that

P = Sum_(w in T) [w]

and if we let S' be the set of s in S which occur in a minimal word
of any w in T, then the first statement shows S' is contained in P,
and therefore T = . QED

I think I'll stop here for now.

PS 01: Just kiddin ya all...  Not that good at math, just passed a GCE A Level Pure Maths graded A, but surely not good enough for advanced algebra. As for the handle is my old school nickname. I hope you didn't spent much time reading this stuff.

PS 02: For those interested these are some notes on BN-Pairs and Boolean Hecke Algebras by Todd Trimble and can be found at: http://math.ucr.edu/home/baez/trimble/BN-pairs.html

J.A.F.O. ....it's an acronym.

It's also a programming language if i remember correctly, right?  

cyberworld is this really necessary?