* Extensions of Z+Z by Z
@ 2004-04-26 2:58 Colin McLarty
2004-04-28 6:27 ` Ronald Brown
0 siblings, 1 reply; 2+ messages in thread
From: Colin McLarty @ 2004-04-26 2:58 UTC (permalink / raw)
To: categories
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
^ permalink raw reply [flat|nested] 2+ messages in thread
* Re: Extensions of Z+Z by Z
2004-04-26 2:58 Extensions of Z+Z by Z Colin McLarty
@ 2004-04-28 6:27 ` Ronald Brown
0 siblings, 0 replies; 2+ messages in thread
From: Ronald Brown @ 2004-04-28 6:27 UTC (permalink / raw)
To: categories
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
>
>
>
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