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Let $G$ be a finite group and let $(K,R,k)$ be a $p$-modular system (large enough for $G$ etc.) and consider a block algebra $B \subseteq RG$ with cyclic defect group.

My question is about the following statement found for example in Feit's 'Representation Theory of Finite Groups' Chapter 7 Theorem 2.24

If $\chi \in \text{Irr}(B)$ is real-valued, then $\widehat{\chi} = \chi\mid_{G_{p'}}$ is the sum of at most two irreducible Brauer characters.

My goal is to prove a version of this statement where we replace the invariance under complex conjugation by other more general operations. I believe the proof by Feit can easily be adapted to show what I have in mind but there are some doubts which I would like to have clarified.

The proof given in Feit's book is roughly as follows (if I understand it correctly):

  1. Let $\varphi_1, \dots, \varphi_s$ denote the distinct irreducible Brauer characters which are constituents of $\widehat{\chi}$.
  2. For every $1 \leq i \leq s$ there exists an $RG$-lattice $X_i$ such that $K \otimes X_i$ affords $\chi$, $k \otimes X_i$ is indecomposable and serial and its socle affords $\varphi_i$.
  3. There exists an ordering $\varphi^{(1)}, \dots, \varphi^{(s)}$ of the $\varphi_1, \dots, \varphi_s$ such that if $\varphi^{(t)} = \varphi_i$, then the composition factors of $k \otimes X_i$ are in ascending order: $\varphi^{(t)}, \varphi^{(t + 1)}, \dots $ with superscripts read modulo $s$. This is Theorem (2.23).

This is the preparation needed. Now starts the real proof:

  1. If $\chi$ is real valued, then $(k \otimes X_i^*) \cong k \otimes X_j$ for some $j$.

Feit claims to use Theorem 2.20 here, I am not sure if he wants to say that every $RG$-lattice $X$ for which $K \otimes X$ affords $\chi$ and $k \otimes X$ is indecomposable satisfies $k \otimes X \cong k \otimes X_i$ for some $i$. But this is not the main problem I have.

  1. Without loss of generality we may assume $\varphi^{(1)} = \varphi_j$. Let $n$ be such that ${\varphi^{(n)}}^* = \varphi^{(1)}$. It follows that ${\varphi^{(n)}}^*,{\varphi^{(n-1)}}^* \dots$ and ${\varphi^{(1)}}, {\varphi^{(2)}} \dots $ both describe the composition factors of $k \otimes X_j$ in ascending order. We thus have ${\varphi^{(n - k)}}^* = \varphi^{(k +1)}$

For the next part I will cite from Feit's book:

  1. Hence $\varphi^{(n - t)} = {\varphi^{(n-t)}}^*$ if and only if $n - t = t + 1$ (mod $s$) or $2t = n - 1$ (mod $s$).There are at most two solutions to this congruence which implies the result.

According to Feit, this finishes the proof.

To make use of what he proved in 6. we would have to have $\varphi^{(n -t)} = {\varphi^{(n-t)}}^*$ for all $t$ and it is not clear to me where this identity comes from and I do not see any reason why it should be true. Wouldn't this imply that every module affording $\varphi_i$ would be isomorphic to its dual? Does this follow from something I have missed? Any clarification is appreciated.

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  • $\begingroup$ I'm not clear what you don't understand. If you're happy with (5) then that proves that $\varphi^{(n-k)^*}=\varphi^{(k+1)}$. And then all (6) is saying is that if $\varphi^{(n-k)^*}=\varphi^{(n-k)}$ then $n-k$ must be the same (mod $s$) as $k+1$. $\endgroup$ – Jeremy Rickard Jun 11 '16 at 10:34
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    $\begingroup$ @JeremyRickard : I do not understand why ${\varphi^{(n−t)}}^*= {\varphi^{(n−t)}}$ should hold. The equivalence of the statements in 6. is clear to me. My problem is that it seems to me that Feit wants to say that we actually have ${\varphi^{(n−t)}}^*= {\varphi^{(n−t)}}$ for all $t$. Otherwise he would not have to write down the statement in 6. which according to him ends the proof. $\endgroup$ – Matthias Klupsch Jun 11 '16 at 10:55
  • $\begingroup$ Ah, sorry! Before the edit I (mis)understood that you were questioning the proof of (6). $\endgroup$ – Jeremy Rickard Jun 11 '16 at 13:37
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    $\begingroup$ @JeremyRickard : Your comment made me realize that what I wrote was not to the point. Thank you for helping me improving my question. $\endgroup$ – Matthias Klupsch Jun 11 '16 at 14:16
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The problem here does not lie in the mathematics but in my reading capapabilities. In fact, the correct Theorem 2.24 says

If $\chi \in \text{Irr}(B)$ is real-valued, then there are at most two real-valued irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi\mid_{G_{p'}}$.

For such a character $\varphi$ we have ${\varphi}^* = \varphi$ and this is exactly what is used in 6.

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  • $\begingroup$ I think your formulation is still ambiguous and can be mistaken to say that $\widehat{\chi}$ is the sum of at most two irreducible Brauer characters, and these are real valued. But the theorem says: At most two of the irreducible Brauer constituents of $\widehat{\chi}$ are real valued (but there may be others). $\endgroup$ – Frieder Ladisch Jun 13 '16 at 13:22
  • $\begingroup$ @FriederLadisch You are right, I tried to fix it. Do you think it is Ok now? $\endgroup$ – Matthias Klupsch Jun 13 '16 at 14:12
  • $\begingroup$ Maybe better, but I'm not completely sure... I still find my formulation, or Feit's original formulation, less ambiguous. Of course it is a question of english language, and it's not my native language, either. $\endgroup$ – Frieder Ladisch Jun 13 '16 at 14:42
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    $\begingroup$ @MatthiasKlupsch I think from an English perspective I'd slightly modify your formulation as follows: "If $\chi \in \mathrm{Irr}(B)$ is real valued then there are at most two real-valued irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi|_{G_{p'}}$." This, for me, is then quite clear. $\endgroup$ – Jay Taylor Jun 13 '16 at 15:00
  • $\begingroup$ @JayTaylor : You are right, this looks better. I edited it. $\endgroup$ – Matthias Klupsch Jun 15 '16 at 7:45

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