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