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.


EDIT: There is a MUCH simpler proof than my first one, which I found after looking up the proof of (90.4) in Curtis-Reiner (Rep. Th. of finite groups...) mentioned by Florian Eisele in the comments:
The central idempotent $e\in \mathbb{Z}_p[G]$ is supported on $p$-regular elements. (By the way, a quite elementary proof of this fact can be given using the ideas in a paper of M. Leitz (Proc. Amer. Math. Soc. 128 (2000), no. 11, 3149–3152, MR1676316), see also Külshammers paper cited at the end.) Since $$ e = \frac{\chi(1)}{|G|}\sum_{h\in H}\sum_{g\in G} \chi(g^{-1})^h g, $$ it follows that the character $\beta= \sum_{h\in H} \chi^h$ vanishes on $p$-singular elements. Let $P$ be a Sylow $p$-subgroup of $G$. Then $\beta$ vanishes on $P\setminus 1 $, so the multiplicity of the trivial character of $P$ in the restriction $\beta_P$ is $$ (\beta_P, 1_P) = \frac{\beta(1)}{|P|}= \frac{|H|\chi(1)}{|P|}. $$ On the other hand, we have $$ (\beta_P,1_P) = \sum_{h\in H} (\chi^h_P,1_P)= |H|(\chi_P,1_P). $$ Thus $|P|=|G|_p$ divides $\chi(1)= |P|(\chi_P,1_P)$, q.e.d.

  • $\begingroup$ You reduce the claim of the OP to the claim that whenever you have a block $B$ such that all characters in $\textrm{Irr}(B)$ are Galois conjugates of one another, then $B$ has defect zero. In literature (specifically: Curtis-Reiner: "Rep. Theory of fin. grps. and ass. alg." Theorem (90.4)) it is claimed that a block $B$ such that all characters in $B$ are $p$-conjugate is necessarily of defect zero. Here $p$-conjugate means "conjugate by an element of $Gal(\mathbb Q (\zeta_{|G|})/\mathbb Q(\zeta_{|G|_{p'}}))$". $\endgroup$ Aug 8 '12 at 20:40
  • 1
    $\begingroup$ (continued) Is it maybe possible to reduce your claim to this theorem? I believe it would suffice to show that (in your notation) $L \subseteq K\cdot \mathbb Q_p(\zeta_{|G|_p})$. This seems plausible, but I don't quite see how to show it at the moment. $\endgroup$ Aug 8 '12 at 20:40
  • $\begingroup$ @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. $\endgroup$ Aug 8 '12 at 20:51
  • 1
    $\begingroup$ There is a purely character theoretic proof that if chi is an irreducible character of G and with defect zero (which means that chi(1) is divisible by the full p-part of |G|), then chi(x) = 0 whenever x has order divisible by p. (This proof depends on Brauer's characterization of characters.) By an amazing result of Knorr, there is a very strong converse. If chi vanishes on every element of order p exactly, then chi has defect zero. $\endgroup$ Aug 17 '12 at 23:15
  • 1
    $\begingroup$ @Marty Isaacs: Actually, Knörr's result is even stronger: if $\sum \chi(x) =0$, where the sum runs over elements of order $p$ exactly, then $\chi$ has $p$-defect zero. Even more, it is enough to assume that the $p$-part of $\chi(1)$ is strictly smaller than the $p$-part of the above sum. $\endgroup$ Aug 20 '12 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.