Differentials and Mod 2 Reductions

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Mike Hill

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Dec 13, 2005, 2:18:30 PM12/13/05
to Algebraic Topology
So I was checking some Adams spectral sequence computations today, and
I noticed the following odd result. It seems like there should be a
general theorem that covers this, but I'm not sure how I would go about
proving it.

Let a be an element in Ext_A(M,F_2) such that
1. h_0^k a=0 for some 0<k<infty
2. h_1 a=h_0^{k-1}b\neq 0 for some b, and for k the smallest value such
that 1 holds.
3. d_r(c)=h_1 a.

Then I think that in the mod 2 Adams spectral sequence, we have that
d_{r+k-2}(c)=\beta(a).

To be completely honest, the examples I have are for k=3, so I might
need other assumptions (ie maybe k>1, etc). There also seems to be a
dual result, but I've not thought of it enough to work it out. Any
ideas how to approach this? Anyone want to help me try?

Mike

Mike Hill

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Dec 13, 2005, 2:31:50 PM12/13/05
to Algebraic Topology
If it makes it easier, here is another condition: there is a
differential taking b to h_0 a, and c has lower filtration than b.

Mark Behrens

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Dec 13, 2005, 7:20:57 PM12/13/05
to algebrai...@googlegroups.com
Hey - what is beta?

-Mark

Mike Hill

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Dec 13, 2005, 8:02:42 PM12/13/05
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Sorry, beta is the bockstein.

And I think I can now prove this. It actually relies not on the eta multiplication (that is a happy coincidence). Here's a sketch of the argument.

d(b)=h_0a, so the class \eta a (in this case) is the Massey product <h_0,a,h_0^{k-1}>. The other condition is that c bounds this class. However, the Bockstein on a is represented by the exact same Massey product. That gives us what we need.

So yeah. Feel free to use this, if you ever need it!

M
--
Mais il avait compté sans une force qui, si elle est nourrie d'abord par vanité, vainc le dégoût, le mépris, l'ennui même : c'est l'habitude.
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