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Date: Sat, 8 Jan 1994 10:52:22 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Calculating |G\M|
Armed with Dan Hoey's note of 4 January "Combining conjugacy classes",
I wish to propose once again a procedure for calculating |G\M|, the
number of M-conjugate classes of G, which I think in some sense is the
"real" size of G. My proposal draws *very* heavily on Dan's note.
My first (incorrect) proposal was based on the following idea. By
computer search, we already have a database of GC\B and GE\B, the
corners and edges of G, respectively, reduced by B-conjugacy. Hence,
we know |GC\B| and |GE\B|. For each X in GC\B and each Y in GE\B,
we calculate BClass(X) and BClass(Y). Then, we can combine
BClass(X) and BClass(Y) in all legal ways (minding parity considerations).
Call the combinations BClass(X) * BClass(Y). For any fixed X and Y,
BClass(X) * BCLass(Y) is a set of cubes in G. Hence we can calculate
(BClass(X) * BClass(Y))\M and |(BClass(X) * BClass(Y))\M|. My idea
then was just to sum |(BClass(X) * BClass(Y))\M| over all values
of X and Y to calculate |G\M|. And we know how many X's and Y's there are!
But, alas, |(BClass(X) * BClass(Y))\M| is not the same across all X's and
Y's because, well because of symmetry. All X's and Y's are not equally
symmetrical. I was assuming that |(BClass(X) * BClass(Y))\M| was
constant, and of course it is not.
My next (incorrect) proposal was never posted to the list. It was
a slight improvement on the first idea. We have a data base of all X's
in GC\B and of all Y's in GE\B. For each X in GC\B and for each
Y in GE\B, we know |BClass(X)| and |BClass(Y)|. (Actually, we don't.
I have to calculate it. I have done so, and I have posted
summaries of those calculations. However, I did not store
the order of the equivalence classes in the data base. I kick myself
for not doing so, but this is a minor problem, so let us continue).
There are only 10 distinct values for |BClass(X)| and for |BClass(Y)|,
namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and 1152. (By the
way, I have never figured out why it is *exactly* the same set of
values for both the corners and for the edges. It is easy to see
why it is approximately the same set of values, but the structure
of the corners is enough different from the structure of the edges
that I see no obvious reason the set of values should be exactly
the same in both cases.)
Let GC[m] be the set of all X for which |BClass(X)| = m and let
GE[n] be the set of all Y for which |BClass(Y)| = n. Hence,
GC\B is partitioned into GC[m]\B for m=24,48..., and GE\B is
partitioned into GE[n]\B for n=24,48,... Now, we form the sets
BClass(X)[m] * BClass(Y)[n] for all X in GC[m]\B and for all
Y in GE[n]\B, and for all legal values of m and n. There will
be 100 such sets.
For any fixed X, Y, m and n, BClass(X)[m] * BCLass(Y)[n] is a
set of cubes in G. Hence we can calculate
(BClass(X)[m] * BClass(Y)[n])\M and |(BClass(X)[m] * BClass(Y)[n])\M|.
My idea then was just to sum |(BClass(X)[m] * BClass(Y)[n])\M|
over all X in GC[m]\B and over all Y in GE[n]\B to calculate
|(BClass(X)[m] * BClass(Y)[n])\M|. We know how many X's there are
in GC[m]\B and we know how many Y's there are in GE[n]\B, so the
calculation seemed possible. I then intended to sum again over all
m and over all n, and I would be done.
But, alas, in order to perform the sum over all X and all Y, I
needed a theorem which I couldn't prove and which I now believe
is not true anyway. I needed to be able to prove that for a fixed
m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same value
for all X in GC[m]\B and all GE[n]\B. For a while I thought it was
true, but right now I can't think of any reason why it should be.
But perhaps Dan Hoey comes to the rescue with his CSymm function.
I still need a theorem which I cannot (yet) prove, but I believe it
is true. If it can be proven, my basic overall scheme can be
rescued.
In my second proposal, I used the values of all possible BClass
sizes as indexes -- 24, 48, 72... It would perhaps be more
convenient to make these sizes a set {24, 48, 72, ...},
and to think of the indexes m and n taking on the values from
1 to 10, where the values from 1 to 10 index the set
{24, 48, 72, ....}. With this understanding, all the above results
are valid, and the indexing is more convenient.
We can now say that GC\B is partitioned into GC[1]\B, GC[2]\B, ...
through GC\B[10] and similarly for GE\B. Unfortunately, using
the B-equivalence class sizes to partition GC\B and GE\B did not
permit us perform the calculations we wanted to perform. However,
suppose we partition GC\B and GE\B a different way, namely using
CSymm. Suppose, for each X in GC\B and for each Y in GE\B,
we calculate CSymm(X) and CSymm(Y). (We would have to do this
by computer). CSymm(X) and CSymm(Y) are sets, but there are only
a (relatively) small number of such sets. Let each distinct value
CSymm(X) and CSymm(Y) be mapped to an index. We can call such a
mapping function CSind, and we can calculate CSind(CSymm(X)) and
CSind(CSymm(Y)). Actually, there is no reason not to define
CSind in such a way that the domain is the set of X's and Y's, so
that we can calculate CSind(X) and CSind(Y).
Now, we use m=CSind(X) and n=CSind(Y) to form a partition of
GC\B and GE\B. All our results from before are valid. The only
issue is, can we now perform the sum? In order to perform the sum,
we need the following to be true:
For a fixed m and n, |BClass(X)[m] * BClass(Y)[n]| is constant
for all X in GC[m]\B and all Y in GE\[n]\B, where GC[m] is
the set of all X in GC such that CSind(X)=m and GE[n] is the set
of all Y in GE such that CSind(Y)=n.
It really seems true to me, and I shall strive to prove it.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 293-5192
Associate Director, WVNET (304) 293-5540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?