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I am studying character theory from the book "Character Theory of Finite Groups" by Martin Isaac. (I am not too familiar with valuations and algebraic number theory.) In the last chapter on modular representation theory, Brauer characters, blocks, and defect groups are introduced.

My question is this: How do we find the irreducible Brauer characters and blocks, given a group and a prime?

For instance, let's say we have $p=3$ and the group $G = S_5$, the symmetric group. An example of the precise calculation or method used to determine the characters would be very helpful. Thanks.

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  • $\begingroup$ What have you done so far? Do you know how many irreps there are? Have you determined which ones come from reduction (mod 3)? $\endgroup$ – Daniel McLaury Aug 4 '13 at 5:04
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    $\begingroup$ There are 5 conjugacy classes with element orders prime to 3, so there are 5 irreducible Brauer characters. But I am unsure how to proceed, given that we do not have the F-representations to begin with. $\endgroup$ – BharatRam Aug 4 '13 at 8:25
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    $\begingroup$ If you want to learn to compute specific examples, then there is an excellent book that explains how GAP works: math.rwth-aachen.de/~RepresentationsOfGroups $\endgroup$ – Jack Schmidt Aug 4 '13 at 15:31
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This is a difficult question and you would probably need to learn more theory in order to understand the different methods available.

But one method that is often used in practice is to calculate the representations and then just find the Brauer character directly from the matrices of the representations. Of course, you have to express the traces of the matrices as sums of roots of unity over the finite field, and then lift this sum to the complex numbers to get the Brauer character, but that is not particularly difficult. (That is not completely true - since the lifting is not uniquely defined, you may have to work hard if you want to make it consistent.)

With ordinary (complex) character tables, it is generally much easier to calculate the characters than the matrices that define the representations, but that is not always the case with modular representations. There are fast algorithms for computing representations over finite fields, using the so-called MeatAxe algorithm.

I am more familiar with Magma than with GAP, and I expect there are similar commands in GAP, but in Magma I can just type

> G := Sym(5);
> I := AbsolutelyIrreducibleModules(G, GF(3));

and I get the five absolutely irreducible representations in characteristic three as group homomorphisms, and so I can just look at the images of elements from the different conjugacy classes. There is a Magma command that does this for you, giving the Brauer character table:

> [BrauerCharacter(i): i in I];
[
    ( 1, 1, 1, 0, 1, 1, 0 ),
    ( 1, -1, 1, 0, -1, 1, 0 ),
    ( 4, 2, 0, 0, 0, -1, 0 ),
    ( 4, -2, 0, 0, 0, -1, 0 ),
    ( 6, 0, -2, 0, 0, 1, 0 )
]
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  • $\begingroup$ Do you happen to know, whether these Brauer character values of the simple $kG$-modules are in MAGMA computed with respect to the standard $p$-modular system described in the book of Lux and Pahlings (chapter 4 in Representations of Groups: A computational Approach)? $\endgroup$ – Bernhard Boehmler May 25 at 21:00
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In your case, $G = S_5$ and $p = 3$, all irreducible Brauer characters are easily derived by the ordinary character table of $G$.

The ordinary character table of $G$ is given by:

$\begin{array}{c|c|c|c|c|c|c|c} S_5 & () & (12) & (12)(34) & (1234) & (12345) & (123) & (123)(45)\\ \small{\#} & \small{1} & \small{10} & \small{15} & \small{30} & \small{24} & \small{20} & \small{20}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \chi_2 & 1 & -1 & 1 & -1 & 1 & 1 & -1\\ \chi_3 & 4 & 2 & 0 & 0 & -1 & 1 & -1\\ \chi_4 & 4 & -2 & 0 & 0 & -1 & 1 & 1\\ \chi_5 & 5 & -1 & 1 & 1 & 0 & -1 & -1\\ \chi_6 & 5 & 1 & 1 & -1 & 0 & -1 & 1\\ \chi_7 & 6 & 0 & -2 & 0 & 1 & 0 & 0\\ \end{array}$

By dropping the 3-irregular conjugacy classes, we immediately get the following Brauer characters:

$\begin{array}{c|c|c|c|c|c} S_5 & () & (12) & (12)(34) & (1234) & (12345)\\ \hline \beta_1 & 1 & 1 & 1 & 1 & 1\\ \beta_2 & 1 & -1 & 1 & -1 & 1\\ \beta_3 & 4 & 2 & 0 & 0 & -1\\ \beta_4 & 4 & -2 & 0 & 0 & -1\\ \beta_5 & 5 & -1 & 1 & 1 & 0\\ \beta_6 & 5 & 1 & 1 & -1 & 0\\ \beta_7 & 6 & 0 & -2 & 0 & 1\\ \end{array}$

Of course $\beta_1$ and $\beta_2$ are (being linear characters) irreducible. Moreover, since $S_5 / A_5 \cong C_2$, these are all linear Brauer characters. Since the integer $|G|/\chi_7(1)$ is not divisible by 3, we get that $\beta_7$ is irreducible by a general theorem. On the other hand, we have $\beta_5 = \beta_4 + \beta_2$ and $\beta_6 = \beta_3 + \beta_1$. So both $\beta_5$ and $\beta_6$ are reducible. So we get to the following table, which already has the right size of a Brauer character table:

$\begin{array}{c|c|c|c|c|c} S_5 & () & (12) & (12)(34) & (1234) & (12345)\\ \hline \beta_1 & 1 & 1 & 1 & 1 & 1\\ \beta_2 & 1 & -1 & 1 & -1 & 1\\ \beta_3 & 4 & 2 & 0 & 0 & -1\\ \beta_4 & 4 & -2 & 0 & 0 & -1\\ \beta_7 & 6 & 0 & -2 & 0 & 1\\ \end{array}$

Indeed, the remaining Brauer characters $\beta_3$, $\beta_4$ are irreducible. To begin with, we show that these characters have no linear constituents. Let $U = \langle (1,2,3,4,5) \rangle$ be the subgroup of $S_5$ generated by a 5-cycle. Since $|U|$ is not divisible by 3, the restriction of any Brauer character of $S_5$ to $U$ is an ordinary character, so we can use the usual inner product on characters. In particular, we may compute

$\langle (\beta_3)_U, (\beta_1)_U \rangle = \langle (\beta_3)_U, (\beta_2)_U \rangle = \langle (\beta_4)_U, (\beta_1)_U \rangle = \langle (\beta_4)_U, (\beta_2)_U \rangle = 0$,

which shows that neither $\beta_3$ nor $\beta_4$ has a linear constituent. It remains to show that neither $\beta_3$ nor $\beta_4$ is a sum of two irreducible Brauer characters of degree 2. Suppose this is the case, say $\beta_3 = \gamma + \delta$ (which we can assume without loss of generality since $\beta_4 = \beta_3 \cdot \beta_2$). Since $\beta_3$ attains the value $-1$ (and since values of Brauer characters are algebraic integers), we see that $\gamma \neq \delta$. Since $\beta_4 = \beta_2 \cdot \beta_3 \neq \beta_3$, we get that $\beta_4 = \beta_2 \cdot \gamma + \beta_2 \cdot \delta$ decomposes into further irreducible Brauer characters, which we haven't considered so far. But this is a contradiction since $S_5$ has exactly 5 irreducible Brauer characters in characteristics 3.


Note that I am asking the same question in the case $p=2$. The case $p=5$ is also fun to consider. In that case, the Brauer character table can be obtained by similar thoughts as here. But the case $p=5$ is more interesting, since then there are irreducible Brauer characters which are not obtained by ordinary characters.

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