   Date: Wed, 29 Dec 93 17:43:28 -0500 (EST)   From: Dan Hoey <hoey@aic.nrl.navy.mil >
~~~ ~~~ Subject: Correction Re: Some Additional Distances in the Edge Group

A couple of days ago, I said that proofs are a good idea. I'll say it
again today with a redder face.

Yesterday I discussed the edge group positions

I = Solved,
P = Pons Asinorum (or Mirror),
E = All edges flipped, and
PE = P E = Pons Asinorum with all edges flipped

and the function from the edge group to 4-tuples of distances

```f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).
```

I wrote:

```?? If f(X)=(a,b,c,d), then conjugation shows us that              ??
?? f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a).     ??
?? So the set of quadruples has the symmetries of the rectangle.  ??
```

The first sentence is incorrect, though the argument as a whole is
reparable.

First, I'll do what I should have done yesterday, and define the
distance function d(X,Y). We want the minimum length process Z such
that X Z = Y. But premultiplying both sides by X', we have Z = X' Y.
So I define d(X,Y)=Length(X' Y). From the properties of the length
function (Length(I)=0, Length(X)=Length(X'), and
Length(X Y)<=Length(X) + Length(Y)) we can conclude that d(X,Y) is a
metric.

```Suppose f(X)=(a,b,c,d).  I claim f(E X)=(b,a,d,c), f(P X)=(c,d,a,b),
and F(PE X)=(d,c,b,a).
```

Proof: To show f(E X)=(b,a,d,c), first observe that I=I', E=E', and
P E = E P.

```d(I,E X) = Length(I' E X) = Length(E' X) = d(E,X),
so d(E,E X) = d(I, E E X) = d(I,X);
d(P,E X) = Length(P' E X) = Length((PE)' X) = d(PE,X)
so d(PE,E X)=d(P,E E X)=d(P,X).
```

To show that f(P X)=(c,d,a,b), exchange P and E in the above argument.
To show that f(PE X)=(d,c,b,a), use both occurrences of the argument.
QED.

So the idea of yesterday's message is correct, but I had X E, X P, and
X PE instead of E X, P X, and PE X, respectively. I would show you a
counterexample to yesterday's formulation, but it turns out there is
none. I claim that f(X,E)=f(E,X), f(X,P)=f(P,X), and f(X,PE)=f(PE,X).

Proof: Recall that E commutes with every element of the Rubik cube
group, so f(X E)=f(E X). It turns out that ``up to M-conjugacy'', P
commutes with every element of the edge group as well. For P performs
a mirror-reflection of the edges, and so can be regarded as an element
of M acting on the edge group. So P' X P = Xbar is an M-conjugate of
X, and X P = P Xbar. Since Length(X) agrees on M-conjugates, so does
d(X,Y), and so f(X), so f(X P)=f(P Xbar) = f(P X). Finally,
f(X PE) = f(X P E) = f(E X P) = f(P E X) = f(PE X). QED.

So it turns out it that the statement about f was true. But I am no
less embarrassed for asserting it, for I had no reason to think it
would be true. It's only rescued by the surprising commutativity of
the Pons Asinorum.

Finally, I would like to note something that I nearly included in
yesterday's message, but yanked when I decided it was false:
f(X')=f(X). Now I'll prove it:

Proof: For W among {I,E,P,PE}, we have X W = W Xbar, for Xbar an
M-conjugate of X. So
d(X,W)=Length(X'W)=Length(W'Xbar')=Length(W'X')=d(W,X').
QED.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil     