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Let $k$ be an algebraically closed field of characteristic $p$, and $G$ be a finite group whose order is not divisible by $p$. I would like to prove the following: if $V$ is an irreducible representation of $kG$, then $\dim V \neq 0$ in $k$, i.e. $p$ does not divide the dimension of $V$.

Here is one explanation of why it is true: for representations over $\mathbb{C}$, the dimension of a representation must divide the order of the group. Then, there is a claim that the dimensions of the irreducible representations of $G$ over $k$ are equal to the dimensions of the irreducible representations of $G$ over $\mathbb{C}$. (I do not know why this claim is true). This then implies that over $kG$, the dimension of any irreducible representation $V$ divides $|G|$, and hence is coprime to $p$.

Is there a shorter argument, which avoids the character theory over $\mathbb{C}$?

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  • $\begingroup$ In groupprops.subwiki.org/wiki/… holds why $|c|\chi(c)/\chi(1)$ is an algebraic integer ? $\endgroup$
    – reuns
    Jun 29, 2019 at 1:43
  • $\begingroup$ @reuns I don't understand your comment. Also, that proof is for fields of characteristic zero. $\endgroup$
    – Joppy
    Jun 29, 2019 at 4:03
  • $\begingroup$ For the claim, see mathoverflow.net/q/118382/58001 $\endgroup$
    – Asvin
    Jun 29, 2019 at 6:25
  • $\begingroup$ There's a model theoretic argument for why it works for $p$ large enough (you don't get a result as precise as "for $p\nmid |G|$" though - and it actually uses the same argument as you do, but proves it model-theoretically; so if you know some model theory you may be happier with that than with your argument) $\endgroup$
    – Maxime Ramzi
    Jul 1, 2019 at 21:29
  • $\begingroup$ @Max Unfortunately, I have no idea what model theory is. I was also after a much sharper result than "for $p$ large enough". $\endgroup$
    – Joppy
    Jul 2, 2019 at 23:58

1 Answer 1

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The claim that the dimensions of irreducible representations over $k$ are the same as over $\mathbb{C}$ is not particularly hard. It follows from these three general facts about representations in characteristic p:

  1. If an irreducible representation over $k$ lifts to characteristic zero, the lift is irreducible too.
  2. Any representation of a finite group $G$ which is a projective object in the category of $G$ representations over $k$ can be lifted to characteristic zero.
  3. If $p \nmid |G|$ representations of $G$ are semisimple, and therefore every object is projective.

If you insist on not using that fact you could use the general theory of blocks and defect groups in characteristic p, although I'm not sure that's an easier theory. Here are some general facts from that theory that imply what you want:

  1. Each irreducible representation lies in a unique block.
  2. Each block has an associated defect group, a p-subgroup of $G$.
  3. A block contains a representation of dimension prime to $p$ iff the defect group is a Sylow subgroup.
  4. Since $p \nmid |G|$, each irreducible representation forms its own block.

Probably neither of these explanations is quite as simple as you'd like, but given that the characteristic zero result is fairly nontrivial I'm somewhat doubtful a slick proof of this will be found.

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  • $\begingroup$ Thanks for the response. Are you able to elaborate on what you mean by "lifts" to characteristic zero, and why projective objects lift? I've always understood statements about lifting to require something like a $\mathbb{Z}$-lattice which can be into characteristics 0 or $p$, but I don't think every representation of a group can be defined over $\mathbb{Z}$. $\endgroup$
    – Joppy
    Nov 17, 2020 at 22:14
  • $\begingroup$ You are right that representations in characteristic zero can't always be defined over $\mathbb{Z}$, however they can always be defined over the ring of integers $\mathcal{O}$ in some finite extension of $\mathbb{Q}$. If you pick such a form over $\mathcal{O}$ and then reduce it modulo a prime ideal $\mathfrak{p}$ lying above an ordinary prime $p$, you get a representation over the finite field $\mathcal{O}/\mathfrak{p}$. This and any extension of scalars thereof, I would call a reduction mod $p$ of the representation - note though it can depend on the choice of form over $\mathcal{O}$. $\endgroup$
    – Nate
    Nov 17, 2020 at 23:43
  • $\begingroup$ As for lifting, it is a bit easier (although ultimately not necessary) to instead work with representations over a finite unramified extension of $\mathbb{Q}_p$ and reduce forms defined over a ring of $p$-adic integers $\mathcal{O}_p$. The key lemma is about lifting idempotents: Starting a representation $\hat{V}$ over $\mathcal{O}_p$ you can reduce it to a representation $V$ over the residue field. Given an idempotent in the endomorphism algebra of $V$, you can always lift it to $\hat{V}$ -the proof is to just solve for one power of $p$ at a time. $\endgroup$
    – Nate
    Nov 17, 2020 at 23:53
  • $\begingroup$ Applying that idempotent lifting lemma to a free module (e.g. the regular representation) gives the result for projective objects. $\endgroup$
    – Nate
    Nov 17, 2020 at 23:54
  • $\begingroup$ This isn't a complete answer that I was looking for, but could you add a reference to where I can find the first three points explained? $\endgroup$
    – Joppy
    Nov 23, 2020 at 22:33

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