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]