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In algebra, various objects admit a unique decomposition into irreducible elements. For instance integers $n\ge1$, univariate polynomials $p\in k[X]$ (even multivariate ones), or characters in representation theory of finite groups. In each situation, an irreducible occurs with a multiplicity. It is interesting, from a theoretical point of view, to have a reduction to the situation where every multiplicity is $1$ (or $0$ if you insist to write the product/sum with all irreducibles of the structure). This can be done explicitly in the case of polynomials, by dividing $p$ by the g.c.d. of $p$ and $p'$, the latter being calculated with the help of the Euclid algorithm.

Is there something similar for characters in representation theory of finite groups ? Suppose we know only the cardinals of conjugacy classes of $G$, together with the table of multiplication of these classes. But we don't know the table of characters. Given a character $\chi$, is it possible for instance to split it as a sum $\chi_1+\cdots+\chi_r$, where $\chi_\ell$ gathers the irreducible characters entering in $\chi$ with multiplicity $\ell$ ?

Perhaps the question should be restricted to complex characters; who knows ? Even a weaker property could be interesting, provided it is associated with a finite-time algorithm.

Of course, I have in mind to apply such a property to the regular representation. Then $\chi_\ell$ would be $\ell$ times the sum of irreducible characters of degree $\ell$.

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  • $\begingroup$ At least we can do this with REPRESENTATIONS. In fact, if $V$ is a representation of $G$, then $\mathrm{End}_G\left(V\right)$ is a direct product of matrix rings $M_{n_i}\left(\mathbb C\right)$ (with $n_i$ being the multiplicities of the irreps in $V$), and $V$ can be seen as a direct sum of their standard representations $\mathbb C^{n_i}$. Now, while it is hard (if possible at all) to actually (algorithmically) decompose the ring $\mathrm{End}_G\left(V\right)$ into the direct product $M_{n_i}\left(\mathbb C\right)$, we can still find our which part of $V$ is the direct sum of all ... $\endgroup$
    – darij grinberg
    Commented Nov 18, 2010 at 13:40
  • $\begingroup$ ... irreps of multiplicity $1$ - namely, this is the subset $V_1$ of $V$ consisting of all $v\in V$ such that $\left(AB-BA\right)v=0$ for all $A,B\in\mathrm{End}_G\left(V\right)$. Why? Because it is the subset of $V$ on which all $M_{n_i}\left(\mathbb C\right)$ with $n_i\geq 2$ act as zero. Similarly we can obtain the direct sum of all irreps of multiplicity $\leq 2$ in $V$ - it is the subset $V_2$ of $V$ consisting of all $v\in V$ such that ... $\endgroup$
    – darij grinberg
    Commented Nov 18, 2010 at 13:43
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    $\begingroup$ ... $\left(ABCD–BACD−ABDC+BADC−ACBD+CABD+ACDB\right.$ $\left. –CADB+ADBC–DABC−ADCB+DACB+CDAB−CDBA−DCAB+DCBA \right.$ $\left.–BDAC+BDCA+DBAC–DBCA+BCAD−BCDA−CBAD+CBDA\right)v=0$ for all $A,B,C,D\in \mathrm{End}_G\left(V\right)$. What I used here is the Amitsur-Levitzki identity, or, more precisely, the fact that it is an identity for $M_2$ but not for $M_3$. Similarly we can find, for each $k\in \mathbb N$, the direct sum of all irreps which occur with multiplicity $\leq k$ in $V$. This is not exactly the direct sum of all irreps with multiplicity $= k$ in $V$, but now you can $\endgroup$
    – darij grinberg
    Commented Nov 18, 2010 at 13:45
  • $\begingroup$ ... split $V_{k-1}$ away from $V_k$ by using Maschke's theorem (totally constructive) and get it. Now, of course, this is not what you are asking for because you want to do it with characters rather than representations. Is there a way to construct a representation from its character constructively, without decomposing it into irreducibles? $\endgroup$
    – darij grinberg
    Commented Nov 18, 2010 at 13:47
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    $\begingroup$ Oh, and I'm working over an algebraically closed field of characteristic $0$ all the time. Over a non-algebraically closed one, I don't think we can do it algorithmically. In fact, consider a cyclic group. Whether some irreps of degree $1$ over $\overline K$ are actually defined over $K$ or not depends on whether $K$ has some roots of unity, which we cannot know per se. But if we allow a roots-of-unity oracle, then I suspect that we can do pretty much all of representation theory constructively, including computing all the irreps. $\endgroup$
    – darij grinberg
    Commented Nov 18, 2010 at 13:53

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It's important here to work over a field of characteristic 0, since the traditional notion of character gets too complicated otherwise (and is better replaced by the notion of Brauer character). Moreover, the field should at least initially be a splitting field for the group, say algebraically closed.

In the highlighted part of the question, you say: Suppose we know only the cardinals of conjugacy classes of G, together with the table of multiplication of these classes. But we don't know the table of characters. But knowing how the sums over classes multiply would allow you to compute the character table, following ideas which go back to Burnside. See for example Section 33 in the old Curtis-Reiner book Representation Theory of Finite Groups and Associative Algebras (1962). It's rare to know so much in advance anyway when the group is at all complicated, which makes the question look a bit artificial to me. On the other hand, one can certainly study the isotypic components of the regular representation in the spirit of Wedderburn structure theory. For an arbitrary given representation, there is a similar isotypic decomposition. But for character theory the most natural question is how to obtain the full decomposition into irreducibles.

By now there are sophisticated computational methods available for working with characters of fairly large finite groups, but this viewpoint gets farther from the theoretical question here.

ADDED: Though I'm still uncertain what the actual question here is, it's worth pointing to the discussion in Sections 2.6-2.7 of Linear Representations of Finite Groups by the other Serre. There one finds a "canonical" decomposition of a given representation space corresponding to the way its character would decompose into multiples of irreducible characters. How effectively one can find such a decomposition depends of course on precisely what is already known about the finite group in question.

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  • $\begingroup$ I know the book by the other Serre. I even know Jean-Pierre, because we are related (not a secret). But this "canonical" decomposition is not what I look for, for several reasons. First, the elements of the decomposition are multiple of a single irrep. Second, it concerns representations, whereas I should like to work with characters, and we don't know in general to pass from a char to its rep. Third, this canonical decomp is an abstract result, with no algorithm. $\endgroup$
    – Denis Serre
    Commented Nov 18, 2010 at 21:02
  • $\begingroup$ But algorithms in character theory depend on the starting point: what do you already know about the group and its characters? What you wrote in the highlighted part of your question is still mysterious to me, since it is usually impossible to know how the sums over classes multiply. $\endgroup$
    – Jim Humphreys
    Commented Nov 18, 2010 at 23:01
  • $\begingroup$ If you wish to speak in terms of algorithms, then as Jim says, the multiplication on $A = Z(\mathbb{C}G)$ determines the character table. Use the basis of class sums, and for each class sum $C$, write down the matrix (in terms of the standard basis) for the linear transformation of $A$ given by multiplication by $C$, say $M(C)$. Simultaneously diagonalizing the $M(C)$ (which theory tells us can be done) leads to a basis of mutually orthogonal idempotents of $A$. This easily yields the character table, which allows decomposition of all characters. $\endgroup$
    – Geoff Robinson
    Commented Dec 1, 2014 at 23:28

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