Timeline for Adams operation on Q-construction of fields
Current License: CC BY-SA 4.0
9 events
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| Jul 16, 2019 at 12:19 | comment | added | user127776 | So the spheres constructed by a choice of upper triangular basis gives you all the basis elements. Now is there a nice way to express a sphere constructed from non upper triangular basis using the upper triangular ones? Is it like sum of distinct spheres or do higher multiplicities also appear? Could you also please give a reference for the Suslin's conjecture I googled it nothing showed up. Thanks | |
| Jul 16, 2019 at 9:07 | comment | added | Matthias Wendt | Another thing, more conceptual, but I'm not sure if I could make this precise. I would believe that the rank spectral sequence has something to do with the rank filtration on K-theory. An identification of Adams operators would then seem to imply Suslin's rank conjecture (that the rank filtration splits the $\gamma$-filtration on K-theory) which is unknown in general (but known for number fields). | |
| Jul 16, 2019 at 8:32 | comment | added | Matthias Wendt | In your comments you suggested that the action of $GL_n$ on the Steinberg representation is given essentially by permuting the basis elements of the representation (or multiplying them by $-1$), but you can see from the above description that the action is more complicated than that. | |
| Jul 16, 2019 at 8:29 | comment | added | Matthias Wendt | If we write the Steinberg representation as homology of the Tits building, then generators can be described as follows: any apartment of the building gives a sphere. Any basis of the vector space gives an apartment by considering all the possible flags built from the basis. But to get a basis of the representation, we restrict to strictly upper triangular matrices and the bases given by their column vectors. So for $GL_n(\mathbb{F}_q)$ the Steinberg representation has dimension $q^m$ with $m=n(n-1)/2$ the number of positive roots. | |
| Jul 15, 2019 at 17:57 | comment | added | user127776 | @MatthiasWendt The more I read the paper you have linked the more confused I get. They claim that $H_0(GL_p(K^p),st(K^p))$ for a field $K$ is zero for $p\geq 2$. But the zeroth homology is the co-invariants since $GL_p(K^p)$ is acting transitively on the spheres and sends each sphere to another sphere or fixes it or multiplies it by -1 the co-invariants will be always $\mathbb{Z}/2$. I'm definitely missing something ... | |
| Jul 15, 2019 at 16:12 | comment | added | user127776 | If what I wrote above is correct then using the Shapiro lemma you can deduce that $H_q(GL_p(F^p),st(F^p))$ is isomorphic to the group homology of monomial matrices i.e. the product of permutation matrices and $(F^\times)^p$ with coefficients in $\mathbb{Z}$ such that action on $\mathbb{Z}$ is defined by the parity of the permutation. If it is odd it multiplies by -1 otherwise it is identity. | |
| Jul 15, 2019 at 15:34 | comment | added | user127776 | My idea was that the generators of $st(F^p)$ are given by a choice of $p$ linearly independent lines. $GL_p(F^p)$ is acting transitively on these and I assumed that the stabilizer of $p$ linearly independent lines are just the monomial matrices which their rational homology coincides with $H_q((F^{\times})^p)$. But I think the problem is that the stabilizer is not quite the monomials i.e. those with even permutation is fixing but odd permutations are just multiplying by -1 and that prevents us to use Shapiro. (If you think this was incorrect as well please let me know!) | |
| Jul 15, 2019 at 12:57 | comment | added | Matthias Wendt | I'm not sure about the identification $H_q(GL_p(F^p),st(F^p))=H_q((F^\times)^p)$. This doesn't seem to be in the linked paper of Kahn. It also seems to contradict the homological vanishing of Ash-Putnam-Sam arxiv.org/abs/1704.08344 An identification as claimed would follow via Shapiro's lemma if the Steinberg representation was induced from the trivial rep of the Borel, but the character of the Steinberg is the alternating sum of induced reps from all parabolics. | |
| May 9, 2019 at 18:14 | history | asked | user127776 | CC BY-SA 4.0 |