... --> Z/pn --> Z/pn-1 --> ... --> Z/p2 --> Z/pwhere each map is reduction mod p; that is, it's the map that sends 1 to 1. (Since these are ring maps, the inverse limit has a ring structure, which might play a role in your answer.)
Also, with p still a prime, describe the direct limit (a.k.a. the "colimit") of the sequence
Z/p --> Z/p2 --> ... --> Z/pn-1 --> Z/pn --> ...where each map sends the generator 1 to the element p. (These are group maps but not ring maps, so the direct limit need not have a ring structure.)
When I say "describe", I mean that, ideally, you should relate these to familiar groups and/or rings (for some value of the word "familiar").
References on direct and inverse limits: McCleary's book, page 67. Any good book on homological algebra should work, too; you could try Weibel's book An introduction to homological algebra, or Mac Lane, Categories for the working mathematician. Also Hatcher, Algebraic topology, page 243 (direct limits) and page 312 (inverse).
xi xj = (i+j choose i) xi+j. When n is even, it's a tensor product of an exterior algebra on a class in dimension n-1 with a divided power algebra on a class in dimension 2n-2.
i | Hi(![]() |
0 | Z |
1 | Z |
2 | 0 |
3 | Z/2 |
4 | Z/2 |
5 | Z/2 + Z/3 |
6 | Z/2 + Z/3 |
7 | Z/2 + Z/2 + Z/3 |
From a different point of view: if you do everything
with rational coefficients, then the cohomology of
the base space S3 is
isomorphic to a polynomial algebra on a class in
dimension 2, so the spectral sequence looks just
like the previous part. So with rational
coefficients, the cohomology of
2S3
is an exterior algebra on a class in dimension 1.
With mod p coefficients, the cohomology of
the base space is isomorphic to the algebra
Fp[x2,x2p,x2p2, ...]/(xip).
One can use this description to determine the mod
p cohomology of 2S3
in as large a range of dimensions as you have
patience for.
Back to Math 583GA home page.
Go to John Palmieri's home page.
Last modified: Fri May 24 15:41:44 PDT 2002