Re: Extensions of Z+Z by Z

["Ronald Brown" <ronnie@ll319dg.fsnet.co.uk>]

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



----- Original Message -----
From: "Colin McLarty" <cxm7@po.cwru.edu>
To: <categories@mta.ca>
Sent: Monday, April 26, 2004 3:58 AM
Subject: categories: Extensions of Z+Z by Z


> 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
>
>
>