From: gtf@math.rochester.edu (Geoffrey T. Falk)
Newsgroups: sci.math.research
Subject: question about regular sequences
Date: Tue, 2 May 95 16:57:03 GMT
Let Q be a quadratic form on an m-dimensional F_2-vector space V.
Then let B(x,y)=Q(x+y)-Q(x)-Q(y) be the associated bilinear form.
Let R_0 = Q \in (V*)^2, and let R_i(x) = B(x,x^2^i), so that
R_i \in (V*)^(2^i + 1). [Here x^2^i means we take the 2^i'th power
of each coordinate in x.]
For example, let x_i be the coordinate functions. If Q = x1 x2,
then R_1 = x1^2 x2 + x1 x2^2; R_2 = x1^4 x2 + x1 x2^4, etc.
Quillen [1] uses some algebraic geometry to show that there is some
integer r for which {R_0, ..., R_(r-1)} is a regular sequence in
the tensor algebra, and such that R_r = 0 mod (R_0, ..., R_(r-1)).
I don't understand the proof of this.
Now I want to do something slightly different. Suppose we take away
R_0. Then {R_1, R_2, ..., R_(r-1)} is a regular sequence, but now it
is not necessarily true that R_r = 0 mod (R_1, R_2, ..., R_(r-1)).
Is there still an integer s >= r such that {R_1, ..., R_(s-1)} is a
regular sequence, and R_s = 0 mod (R_1, R_2, ..., R_(s-1))?
This question arises in the calculation of Morava K-theory of the
classifying space of extraspecial 2-groups.
Thanks
g.
==============================================================================
Date: Thu, 11 May 95 16:04:49 CDT
From: rusin (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: question about regular sequences
In article <1995May2.165703.28342@galileo.cc.rochester.edu>,
Geoffrey T. Falk wrote:
>Let Q be a quadratic form on an m-dimensional F_2-vector space V.
>Then let B(x,y)=Q(x+y)-Q(x)-Q(y) be the associated bilinear form.
>Let R_0 = Q \in (V*)^2, and let R_i(x) = B(x,x^2^i), so that
>R_i \in (V*)^(2^i + 1). [Here x^2^i means we take the 2^i'th power
>of each coordinate in x.]
>
>Quillen [1] uses some algebraic geometry to show that there is some
>integer r for which {R_0, ..., R_(r-1)} is a regular sequence in
>the tensor algebra, and such that R_r = 0 mod (R_0, ..., R_(r-1)).
>
>Now I want to do something slightly different. Suppose we take away
>R_0. Then {R_1, R_2, ..., R_(r-1)} is a regular sequence, but now it
>is not necessarily true that R_r = 0 mod (R_1, R_2, ..., R_(r-1)).
>Is there still an integer s >= r such that {R_1, ..., R_(s-1)} is a
>regular sequence, and R_s = 0 mod (R_1, R_2, ..., R_(s-1))?
Yes. This is a common pattern in the cohomology of groups: your situation
arises if you wish to compute H^*(G, F_2) from the Serre spectral
sequence of the group extension Z/2Z -> G -> (Z/2Z)^n x (Z/NZ)
corresponding to the cocycle Q + q2 where q2 generates H^2(Z/NZ, F_2).
At the first pass, one mods out by Q+q2, so that at the E3 term
one has just the polynomial ring H^*((Z/2)^n) x H^*(Z/2), and one
starts with the differential d3(t^2) = Sq^1(Q) (the q2 is now gone).
And this is the trick you should use. If you're going to throw away R_0
anyway and begin with R_1=Sq^1(Q), then it doesn't matter which type of
quadratic form Q is -- we might as well assume it's real,
Q = x1 x2 + x3 x4 + ... + x_(2m-1) x_(2m)
for some m (possibly less than dim(V)/2).
Now let V' = V + F_2 and use x_0 to designate the additional coordinate.
Define a new quadratic form Q' = Q + x0^2. This is of complex type;
in Quillen's notation, the codimension h of any maximal Q'-isotropic
subspace of V' is h=m+1, and then Q' and your R_1, R_2, ..., R_m
form a regular sequence. (Note that the difference between Q and Q'
does not affect the bilinear form, and so the R_i are unchanged; in
particular, they do not involve the variable x_0.) Quillen further
shows that R_h = R_(m+1) is in the ideal generated by Q' and the other
R_i, but since everything but Q' is free of x0, R_h must in fact
lie in the ideal generated by the others.
dave
PS -- Quillen's paper inexplicably refers to the primary decomposition of
an ideal as a _union_ of ideals, when of course it is the _intersection_
of primary ideals. Perhaps that will make it easier to follow?
==============================================================================
Date: Thu, 11 May 1995 17:45:18 -0400
From: Geoffrey Falk
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: question about regular sequences
If I understand correctly, you are saying that we need only go to at most the
next element R_h in the sequence, and then {R_1...R_h} is a regular sequence
(one which would have arisen from a modification of the original Q), and
R_{h+1} = 0 mod (R_1,...R_h). This essentially answers the question I asked.
Thank you!
Now, what I am actually trying to do is to calculate the Morava K-theory
K(n)^*(BG) for some extra-special 2-groups G. In the Atiyah-Hirzebruch-Serre
SS of the fibration Z/2-->G-->V, the stuff in the fibre is even-dimensional,
so we don't hit R_0. This sounds similar to the calculation you describe.
Thanks to you I can now say that each page of the SS is a tensor product after
each of these Serre-type differentials. h depends only on m (the rank of V
over F_2). In the case of K(n)^*, there is also an Atiyah-Hirzebruch type
differential (a Milnor Q_n) which is a d_{2^{n+1}-1}. I can arrange it so that
this differential comes after all of the others by taking n large enough,
which is good enough for my purposes. I want to prove that E_infinity is
even-dimensional, which means that I now have to try and understand the
structure of the ideal (R_1, ... R_h).
[deletia - djr]
==============================================================================
Date: Thu, 11 May 95 19:39:20 CDT
From: rusin (Dave Rusin)
To: gtf@theorem.math.rochester.edu
Subject: Re: question about regular sequences
>Now, what I am actually trying to do is to calculate the Morava K-theory
>K(n)^*(BG) for some extra-special 2-groups G. In the Atiyah-Hirzebruch-Serre
I will confess that I don't really understand Morava K-theory. However, let
me point out that in the case of ordinary cohomology, the extension
Z/2 -> G -> V is actually a little more work to describe with the
ordinary Serre SS than with the Eilenberg-Moore spectral sequence. Perhaps
Ravenel can comment on whether something of this sort exists in your
setting.
(Briefly, the fibration BZ/2 -> BG -> BV is turned around to
give BG -> BV -> K(Z/2,2). The EMSS computes the fibre BG starting
with E_2=Tor_{H^*(BV)}(H^*(Z/2), F_2). The initial differential d2 carries
various elements of this Tor to all of the Sq_i(q) at once (q in H^2(V, Z/2)
describes the extension). Thus, Quillen's result that the sequence is
regular implies that there is nothing left in E3 to provide a non-trivial
differential, and so E3 = E_infty. )
[deletia - djr]