Multiple factors of the character of a representation

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 $\ell$.

• 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 ... – darij grinberg Nov 18 '10 at 13:40
• ... 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 ... – darij grinberg Nov 18 '10 at 13:43
• ... $\left(ABCD–BACD−ABDC+BADC−ACBD+CABD+ACDB\right.$ $\left. –CADB+ADBC–DABC−ADCB+DACB+CDAB−CDBA−DCAB+DCBA \right.$ $\left.–BDAC+BDCA+DBAC–DBCA+BCAD−BCDA−CBAD+CBDA\right)v=0$ 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 – darij grinberg Nov 18 '10 at 13:45
• ... 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? – darij grinberg Nov 18 '10 at 13:47
• 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. – darij grinberg Nov 18 '10 at 13:53

• 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. – Geoff Robinson Dec 1 '14 at 23:28