User henri johnston - MathOverflow

Character fields and Clifford's theorem

2013-03-17T14:41:01Z

Let $G$ be a finite group with normal subgroup $N$. Let $\chi$ be an irreducible complex character of $G$. Then Clifford's Theorem says that $\mathrm{res}^{G}_{N}\chi = e(\eta_1 + \cdots + \eta_r)$ where $e$ is a positive integer and each $\eta_i$ is a complex irreducible character of $N$ such that the $\eta_i$'s are the distinct conjugates of $\eta := \eta_1$ (i.e. for each $i$ there exists $g_i \in G$ such that $\eta_i(n)=\eta(g_i n g_i^{-1})$ for all $n \in N$.) Let $\mathbb{Q}(\eta) = \mathbb{Q}(\eta(n) : n \in N)$ be the character field of $\eta$ and similarly for $\mathbb{Q}(\mathrm{res}^{G}_{N} \chi)$ . Since $\mathbb{Q}(\eta_i)=\mathbb{Q}(\eta)$ for each $i$, we have $K:=\mathbb{Q}(\mathrm{res}^{G}_{N} \chi) \subseteq \mathbb{Q}(\eta)$.

Question: Is $e$ somehow related to $[\mathbb{Q}(\eta):K]$? For example, do we have $e \geq [\mathbb{Q}(\eta):K]$? Otherwise, can you tell me anything about the relation between $K$ and $\mathbb{Q}(\eta)$? e.g. do we somehow get $\mathbb{Q}(\eta)$ from $K$ by adjoining certain roots of unity? If so, which ones (are they related to say $[G:N]$)? (I am actually interested in the case where we replace $\mathbb{Q}$ by the $p$-adics $\mathbb{Q}_p$ throughout and $[G:N]$ is a power of $p$.)

Vanishing of certain $\mu$-invariants attached to abelian extensions of imaginary quadratic fields

2012-09-03T19:11:03Z

In Fonctions L p-adiques des corps quadratiques imaginaires et de leurs extensions abéliennes, J. Reine Angew. Math. 358 (1985), 76–91, Roland Gillard shows the following result (I mainly follow the MR review here):

Let $K$ be an imaginary quadratic field and let $p > 3$ be a prime number which splits in K into $(p)=\mathfrak{p}\mathfrak{p'}$. Let $K_{\infty}$ be the unique $\mathbb{Z}_{p}$ -extension of $K$ unramified outside $\mathfrak{p}$ (thus noncyclotomic). Let $F$ be a finite abelian extension of $K$ and let $M$ be the maximal abelian $p$-extension of $F$ unramified outside $\mathfrak{p}$. Then Theorem 3.4 states that $\mathrm{Gal}(M/FK_{\infty})$ is $\mathbb{Z}_{p}$-torsion-free; in particular its $\mu$-invariant is 0.

Question: does anyone know if the vanishing of this $\mu$-invariant is also proven somewhere when $p=3$ (even special cases would be of interest)?

Note: with some work, you can get a PDF of the article in question without a subscription by following the link from here: http://www.ams.org/dmr/JournalListJ.html

Representation theory of a finite p-group over a field of characteristic p: dim of invariants =1 => dim of coinvariants = 1?

2012-05-04T15:10:56Z

I am trying to understand the proof of Proposition 4 in S. Ullom, Integral normal bases in Galois extensions of local fields, Nagoya Math. J. Volume 39 (1970), 141-148. The PDF is available here: http://projecteuclid.org/euclid.nmj/1118798052

It appears that the following result is used, but I'm afraid that I don't quite see the proof. Let $k$ be a field of characteristic $p$ and let $G$ be a finite $p$-group. Let $W$ be a left $k[G]$-module such that $\dim_k W = |G|$. Suppose that $\dim_k W^{G} = 1$. Then $\dim_k W_G = 1$. Here $W^G$ denotes invariants and $W_G$ denotes coinvariants.

I have to admit that I know relatively little about representation theory in characteristic $p$. One idea would be to consider the dual representation $\hat{W}$, but I only got so far: $\dim_k W^{G} = 1 \implies W$ is indecomposable $\implies \hat{W}$ is indecomposable. But maybe this is not the right approach. If I could show that the Tate cohomology groups of $W$ vanish, then of course the desired result drops out, but I think this is rather strong medicine.

Is anyone able to give a proof of the above claim? I suspect the solution is fairly easy, but I just don't see it at the moment.

Modular representation theory: central idempotents in $\mathbb{Z}_p[G]$

2012-08-08T10:19:46Z

Let $G$ be a finite group and let $p$ be a prime dividing the order of $G$. Let $\chi$ be a $\mathbb{C}_p$ -valued irreducible character of $G$ and let $e_{\chi} = |G|^{-1}\chi(1)\sum_{g \in G} \chi(g^{-1})g$ be the associated primitive central idempotent in $\mathbb{C}_p[G]$ . Let $\mathbb{Q}_{p}(\chi)=\mathbb{Q}_p(\chi(g) : g \in G)$ be the character field. Let $H=\mathrm{Gal}(\mathbb {Q}_{p}(\chi)/\mathbb{Q}_p)$ and let $e=\sum_{h \in H} e_{\chi^h}$ ($H$ acts on characters in the usual way.) Then $e$ is a central primitive idempotent of $\mathbb{Q}_p[G]$. Let $v_p$ denote the usual $p$-adic valuation.

Claim: $v_p(|G|)=v_p(\chi(1))$ if and only if $e \in \mathbb{Z}_p[G]$ and $e\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order.

If $v_p(|G|)=v_p(\chi(1))$ then it is clear that $e \in \mathbb{Z}_p[G]$. That $e\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order follows from Jacobinski's formula for the central conductor of $\mathbb{Z}_p[G]$ in a maximal order (see Curtis-Reiner, Methods of representation theory, vol 1 section 27).

For the converse, I can prove the claim for $p$ odd again using Jacobinski's formula and some calculations of the different of the extension $\mathbb{Q}_p(\chi)/\mathbb{Q}_p$.

Question: can anyone provide a proof of counterexample for the missing part for $p=2$?

Here is a related claim that would prove the claim and make everything much simpler if true: If $e \in \mathbb{Z}_p[G]$ then $\mathbb{Q}_p(\chi)/\mathbb{Q}_p$ is unramified (i.e. $\mathbb{Q}_p(\chi) \subseteq \mathbb{Q}_p(\zeta_n)$ for some $n$ relatively prime to $p$).

Also, maybe I can drop the maximal order part of the claim altogether?

I have a reasonable knowledge of ordinary representation theory but have only really started to look at modular representation theory in the past few days. I know that this is related to "blocks of defect zero", but in the books I have looked at (Serre, Curtis-Reiner) it is assumed that the ground field is "sufficiently large", which doesn't really help me. But I suspect this is an easy problem for someone who knows the subject well.

Triviality of SK_0(Lambda) for Lambda an order in a group algebra over a $p$-adic field

2012-07-04T12:52:01Z

CR refers to Methods of Representation Theory by Charles Curtis and Irving Reiner.

Let $F$ be a finite extension of $\mathbb{Q}_p$ with valuation ring $\mathcal{O}_F$. Let $G$ be a finite group and let $A$ be the group algebra $F[G]$. Let $\Lambda$ be an $\mathcal{O}_F$-order in $A$. Let $K_0(\Lambda)$ (resp. $K_0(A)$) denote the Grothendieck group of the category of finitely generated projective left $\Lambda$-modules (resp. $A$-modules).

Let $SK_0(\Lambda)=\ker \varphi$ where $\varphi: K_0(\Lambda) \longrightarrow K_0(A)$ is the map given by $\varphi([M])=[F \otimes_{\mathcal{O}_F} M]$ (this definition is taken from CR vol 2, top of page 222.)

Is $SK_0(\Lambda)$ always trivial? I know that this is true in the following cases:

$\Lambda=\mathcal{O}_{F}[G]$ - see CR vol 1, Theorem 32.1
more generally, when $\Lambda$ satisfies the conditions of CR vol 1, Theorem 32.5
$\Lambda$ is a maximal order - see CR vol 1, Theorem 26.24iii
$\Lambda$ is commutative - see CR vol 1, Proposition 35.7

Can anyone provide (a reference to) a proof of the general case or counterexample?

Answer by Henri Johnston for Looking for criterion for $\mathbb{Z}G$-modules to be projective

2012-03-14T20:47:06Z

As Geoff mentioned, there is a lot of material on this type of question in "Methods of Representation Theory" by Curtis and Reiner, though I think Volume 1 is more relevant here.

If we specialise Corollary (25.16) to the case you are interested in, then we get the following:

Let $G$ be a finite group of order $n$. Let $M$ be a finitely generated left $\mathbb{Z}[G]$-module. Suppose that $M$ is free as a $\mathbb{Z}$-module. Then $M$ is $\mathbb{Z}[G]$-projective if and only if $M \otimes_{\mathbb{Z}}\mathbb{Z}_{p}$ is $\mathbb{Z}_{p}[G]$ -projective for each prime $p$ dividing $n$.

"Maximal Orders" by Reiner will probably also be a useful reference.

Commutator subgroups and normal $p$-complements

2012-03-09T14:30:06Z

Let $G$ be a finite group with commutator subgroup $G'$. Let $p$ be a prime number. Then $p \nmid |G'|$ if and only if $G$ has an abelian Sylow $p$-subgroup $P$ and normal $p$-complement $N$ (and in this case $G=N \rtimes P$ ).

This is surely a standard result, but I can't seem to find a proof written down anywhere. I have worked out a straightforward proof below, but if someone could provide a good reference to a proof (so that I can just cite it in a paper I'm writing), that would be great.

Proof: Suppose that $G$ has an abelian Sylow $p$-subgroup $P$ with normal complement $N$. Then $P \simeq G/N$ is abelian so $G' \leq N$. But $p \nmid |N|$ so $p \nmid |G'|$. Suppose conversely that $p \nmid |G'|$. Let $P$ be a Sylow $p$-subgroup of $G$ and let $\theta:G \rightarrow G/G'$ be the natural projection. Then $P \cap G'$ is trivial and so $\theta$ restricted to $P$ is an isomorphism. But $G/G'$ is abelian, so $P$ is abelian. Since $G/G'$ is abelian, there exists a subgroup $M$ such that $G/G' = \theta(P) \times M$. Let $\psi:G/G' \rightarrow \theta(P)$ be the natural projection. Then put $N:=\ker(\psi \circ \theta)$.

Answer by Henri Johnston for Introductory reading on the Scholz reflection principle?

2011-09-10T00:03:58Z

This is covered in Ralph Greenberg's book-in-progress "Topics in Iwasawa theory" http://www.math.washington.edu/~greenber/book.pdf It also contains lots of other interesting stuff on class groups.

Answer by Henri Johnston for number of galois extensions of local fields of fixed degree

2011-06-23T17:08:28Z

If $K$ contains the $p$-th roots of unity, then Kummer theory tells us that the degree $p$ Galois extensions of $K$ are in bijective correspondence with the subgroups of $K^{\times}/(K^{\times})^p$ of order $p$. The structure of $K^{\times}$ is well-known; see http://en.wikipedia.org/wiki/Local_fields or any decent book on local fields. So you can work out the answer in this case.

If $K$ doesn't contain the $p$-th roots of unity, then it becomes harder. Here are some special cases. For every $d \in \mathbb{N}$, $K$ has a unique unramified extension of degree $d$, which is necessarily cyclic - see Corollary 4.4 of these nice notes: http://websites.math.leidenuniv.nl/algebra/localfields.pdf So in particular there is a unique unramified degree p Galois extension of $K$.

If $l \neq p$, then any ramified extension of $K$ must be totally and tamely ramified. But then by 5.3 and 5.4 of the above notes, $\mathbb{Z}/p\mathbb{Z}$ must embed into the unit group of the residue field of $K$. Then by Hensel's Lemma, $K$ must contain $p$-th roots of unity and so we are reduced to the Kummer case above.

So we are left with the case $l=p$ and $K$ not containing $p$-th roots of unity. I'll think about this some more, but you should be able to use class field theory as mentioned above.

Answer by Henri Johnston for How to distinguish division algebras from matrix algebras?

2011-03-10T12:42:39Z

This may be repeating what others have said as it essentially follows the maximal order approach, but have you looked at Nebe, Gabriele; Steel, Allan, Recognition of division algebras, J. Algebra 322 (2009), no. 3, 903–909? http://dx.doi.org/10.1016/j.jalgebra.2009.04.026

Preprint version and magma code available here: http://www.math.rwth-aachen.de/~nebe/pl.html

(I know this should really be a comment, but I'm afraid that I don't have enough reputation yet.)

Answer by Henri Johnston for Taking lecture notes in lectures

2010-10-23T20:15:19Z

Tom Körner's essay "In Praise of Lectures" gives lots of tips on how to get the most out of lectures, and can be found here: http://www.dpmms.cam.ac.uk/~twk/Lecture.pdf (The issue of note taking in touched up on page 5.)

Answer by Henri Johnston for Where are mathematics jobs advertised if not on mathjobs (e.g. in Europe and elsewhere)?

2010-08-27T11:25:34Z

Another site for jobs in the UK & Ireland is http://www.lms.ac.uk/jobs/index.html

Jobs at Oxford and Cambridge are often advertised in the Oxford Gazette http://www.ox.ac.uk/gazette/ and the Cambridge reporter http://www.admin.cam.ac.uk/reporter/

Comment by Henri Johnston

2013-03-18T13:05:45Z

Thanks for the counterexample! As you say, it looks like the relationship is not clear in general.

Comment by Henri Johnston

2012-10-18T19:47:57Z

The discussion in section 4.5 of my joint paper with Werner Bley may be relevant: arxiv.org/pdf/1006.4381v2.pdf

Comment by Henri Johnston

2012-10-06T10:55:02Z

"I do not believe these $\mu$-invariants are relevant in a trivial way, so the "answer" in the link was not really correct."

Comment by Henri Johnston

2012-10-06T10:54:34Z

Ming-Lun Hsieh said (with minor edits) "In the works of Hida and I, we consider the $\mu$-invariant of anticyclotomic $\mathbb{Z}_{p}$-extension or the $\mu$-invariant of the $\mathbb{Z}_{p}^{2}$-extension of an imaginary quadratic fields. The latter is shown to be always zero in the compositio paper of Hida and my IMRN paper with Burungale. However, what you want to know is the $\mu$-invariant of the $\mathbb{Z}_{p}$-extension unramified outside $\mathfrak{p}$, $p=\mathfrak{p}\mathfrak{p}^c$. [ctd]

Comment by Henri Johnston

2012-10-05T11:20:37Z

I emailed both Hida and Ming-Lun Hsieh to ask them about this, and they both said that actually the result I am looking for does not follow from the work of either of them. I am sorry about this - I should have contacted them before placing the bounty on this question.

Comment by Henri Johnston

2012-09-24T11:18:01Z

It's possible that the result can be easily derived from the work of Hida, Ming-Lun Hsieh, or others. However, the problem is that I really know very little about the techniques and language they use, and so I just don't see it.

Comment by Henri Johnston

2012-09-04T10:39:46Z

Thanks for pointing out the typos - now corrected. I'll look at the Compositio paper later today.

Comment by Henri Johnston

2012-08-09T13:48:10Z

@F. Landish: Ah, yes; I see that this is a bit simpler than using the full power of Proposition 5.

Comment by Henri Johnston

2012-08-09T13:21:10Z

It looks like Proposition 5 of Külshammer's paper will give us that $e$ is supported on $p$-regular elements.

Comment by Henri Johnston

2012-08-09T12:54:10Z

@F. Landish: many thanks for this - it is indeed much simpler :) The only point that I don't yet understand is how exactly we know that $e$ is supported on $p$-regular elements. In the paper of M. Leitz that you mention, he assumes that the defect is zero, which is exactly what we want to show, and I don't see how to modify this argument. But maybe I am missing something easy.

Comment by Henri Johnston

2012-08-08T20:51:10Z

@F. Landish - Thanks! I guess I first need to read a lot more on modular representation theory before I fully understand this; I'll probably ask some more questions on specific points tomorrow.

Comment by Henri Johnston

2012-07-04T18:54:26Z

Thanks, Florian!

Comment by Henri Johnston

2012-05-05T17:32:45Z

A very nice answer - thanks! If I understand correctly: under the given hypotheses, by combining your argument with its `dual' we in fact have $\dim_k W^G=1 \Leftrightarrow W≅k[G] \Leftrightarrow dim_k W_G = 1$.

Comment by Henri Johnston

2012-05-04T15:14:16Z

@Alireza: good point - I was being a bit sloppy!

Comment by Henri Johnston

2012-03-15T10:54:20Z

A slightly different way of looking at Jef's comment is as follows. If $p$ does not divide the order of $G$, then $Λ:=\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order and hence is a hereditary ring. This implies that every submodule of a free (left) $\Lambda$-module is projective. Of course, if M is a $\Lambda$-module, then M is a submodule of a free Λ module if and only if M is free as a $\mathbb{Z}_p$ module.