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I need to decompose the regular representation of $S_3$ into irreducible ones. What I know so far is this: $S_3$ is generated by $\tau = (12)$ and $\sigma = (123)$. If $v$ is an eigenvector of $\sigma$ with an eigenvalue $\omega$, then $\tau(v)$ is also an eigenvector with a corresponding eigenvalue of $\omega^2$. Putting these together we get that an irreducible representation of $S_3$ has dimension at most 2. I also know both the one-dimensional ones.

I've found one two-dimensional subrepresentation so far by using the isomorphism $S_3\simeq SL_2(\mathbb F_2)$. Then the regular representation is the left multiplication of formal linear combinations of matrixes of $SL_2(\mathbb F_2)$ by matrixes from the same group. The sum of all transpositions turns out to be the eigenvector of $\sigma$ and applying $\tau$ to it gives the sum of both 3-cycles and an identity permutation.

How does one systematically look for the other subrepresentations? I can not use anything even remotely fancy like charachters at this point.

Thank you.

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IANAGT, but I think you probably want the ${\mathbb{Q}}$-representations. The very slightest modification of the ${\mathbb{F}}_2$-representation that you found will give you something over the rationals. –  Lubin Apr 24 '12 at 16:57
    
I thinkg I need the $\mathbb C$ one actually, but that just changest the coefficients of the formal linear combinations of matrices (or permutations if you don't use the isomorphism with $SL_2(\mathbb F_2)$) –  Artem Apr 24 '12 at 17:40
    
But once you have the rational representations, you look at them and see that they are still irreducible over the complex numbers. This is a special property of the symmetric groups, not even true for $A_3$, for example. –  Lubin Apr 24 '12 at 18:58
    
I do know the irreducible representations. What I need is to find them all as the subrepresentations of the regular one (apperently the two-dimensional twice). As I understand it, I need to describe the invariant subspases of $\mathbb CS_3$ (by providing a basis, say). –  Artem Apr 24 '12 at 19:16

2 Answers 2

up vote 3 down vote accepted

This is how it can be done "by hand", that is without charachter theory. All of this can be found (in more compressed form) at the end of first lecture in Fulton, Harris "Representation theory: the first course". I'll write an outline.

First we classify the irreducible representatins of $S_3$: if $\tau = (123)$, $\sigma = (12)$, then $\tau$ has an eigenvector $v$ with an eigenvalue $\omega$ which is one of the three cubic roots of 1. Then $\sigma(v)$ is also an eigenvector with an eigenvalue $\omega^2$. This follows from $\sigma\tau\sigma = \tau^2$. Now, if $\omega\ne 1$, the representation $<v,\sigma(v)>$ is irreducible. This shows that an irreducible representation of $S_3$ has degree at most 2. Also, every two-dimensional irreducible representation of $S_3$ is isomorphic to this. Now, if the representation is one-dimensional, then it's generator is an eigenvalue of $\sigma$ and since the order of $\sigma$ is 2, it's eigenvalues are $\pm 1$. Therefore, there are two one-dimensional representations - the trivial one and the one for which $\sigma(v) = -v$. The last one is easily seen to be the sign representation.

To summarise: two one-deminsional representations and one two-dimensional one (up to isomorphism). Also, any two-dimensional representation is generated by eigenvectors $\{x,y\}$ of $\tau$. If the corresponding eigenvalues are 1, then it is decomposable into one-dimensional subrepresentations generated either by $x$ and $y$ (if $\sigma(x)\in<x>$, $\sigma(y) = y$) or by $x+\sigma(x)$ and $x-\sigma(x)$ (if $\sigma$ permutes $<x>$,$<y>$). In both cases one of these is the trivial and the other one is the sign representation.

It is left to notice that when acting on the regular representation (dimension 6) $\tau$ cyclicly permutes two sets of basis vectors as follows: $$ (12)\mapsto(13)\mapsto(23)\mapsto(12) $$ and $$ Id\mapsto\tau\mapsto\tau^2\mapsto Id. $$ This means that the matrix of $\tau$ in the basis $\{(12),(13),(23),Id,(123) = \tau, (132) = \tau^2\}$ has a block-diagonal form (easy to write out, but tideous to type) and the charachteristic polynomial of $\tau$ is $(t^3 - 1)^2$. This shows that $\tau$ has every cubic root of 1 as an eigenvalue and the multiplicity of each is 2. Combine this with the fact that $\tau$ is diagonalisable and the reasoning above, and we get that the regular representation decomposes into three 2-dimensional ones, 2 of which are siomorphic irreducible and the third itself decomposes into the trivial and the sign.

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The number of 1-dimensional representations is given by the number of elements in the abelianisation of $S_3$. The commutator being $A_3$, we have two 1-dimensional representations. But you can even get more: since $S_3/A_3$ is cyclic of order 2, the non-trivial representation is given by sending all elements not in $A_3$ to $-1$.

Now, you have 3 irreducible representations, and $S_3$ has 3 conjugacy classes, so you have found them all. To conclude, recall that for a finite group $G$, the regular representation decomposes as $$k[G]=\oplus_{n=1}^r V_n^{m_n},$$ where $V_1,\ldots,V_r$ are the distinct irreducible representations of $G$ and $m_n=\dim V_n$.

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I know (from reading a couple commenta on this site to a similar question) that the regular representation of $S_3$ has all it's irreducible representations in it's decomposition. But I can not prove this so far. I also know all these representations. What I need, however, is to start with the regular representation and actually find all the irreducible ones within it. –  Artem Apr 24 '12 at 16:56
    
What tools do you have? Character theory? –  M Turgeon Apr 24 '12 at 16:59
    
No. That's the next lecture. For this - basic definitions, linear algebra, Maschke's theorem, Schur's lemma. I think that's about it. –  Artem Apr 24 '12 at 17:04

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