Subject: categories: Extensions of Z+Z by Z

From: Colin McLarty <[email-masked]>
Date: Sun, 25 Apr 2004 22:58:28 -0400

After calculating the group extensions of Z+Z by Z, with constant action, I
am curious whether the groups have any more natural form than I found.  I
mean extension of Z+Z by Z in this sense, as a sequence of groups where E
need not be commutative:

    0 --> Z --> E --> Z+Z --> 0

and the kernel is in the center of E.

The form I found is parametrized by the integers this way:  For any integer
c, the group E_c has triples of integers (i,,j,k) as elements and the
multiplication rule is coordinate-wise addition plus an extra bit in the
first coordinate.

(i,,j,k).(q,r,s) =  ( (i+j+c.(kr)), j+r, k+s)

When c=0 this is commutative and is just the coproduct Z+Z+Z.  In any group
E_c, the element (c,0,0) is the commutator of (0,0,1) and (0,1,0).  The
Baer sum of extensions corresponds to addition of the parameters c as
integers.  So I understand the group of extensions.  Of course I understood
it before I calculated it, since it is the second cohomology group of the
torus.  That is why I tried the algebraic calculation.

But is there a natural way to think about each group E_c, for non-zero
values of c?  Do these groups appear in any other natural way?

thanks, colin

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From: "Ronald Brown" <[email-masked]>
Date: Wed, 28 Apr 2004 07:27:54 +0100

Try the Heisenberg group of upper triangular matrices with 1's on the
diagonal and the integers i,j,k in the upper non diagonal entries.

1   k   i
0   1   j
0   0   1

This should give the case c=1, but your formula is not quite correct as the
RHS does not involve q.

Should it be q + j + c.(kr)?

Ronnie Brown

http://www.bangor.ac.uk/~mas010