I'm considering the following example/application. Let $k$ be a field, such that $char(k)\nmid\mid G\mid$.Let $G=S_3$ then we have the following representation on it:
- TRIVIAL REPRESENTATION: $V_0=k$, $G\rightarrow GL_1(k)=k^\times,\sigma\mapsto1$
- SIGN REPRESENTATION: $V_1=k$, $G\rightarrow GL_1(k)=k^\times,\sigma\mapsto sgn(\sigma)\cdot 1$
- STANDARD REPRESENTATION: $V=\{(x_1,…,x_n)\in k^n\mid\sum_ix_i=0\}\subset k^n$, $G\rightarrow GL_n(k),\sigma\mapsto P_\sigma$
CLAIM: $V_0,V_1,V$ are all the irreducible representations up to isomorphism.
To prove this we want to show that:
The regular representation of $G$ is isomorph to $V_0\oplus V_1\oplus V^2$
I have a theorem which states that every finite dimensional representation can be written as direct sum of irreducible representations (and since $\mid G\mid<\infty$ then the regular representation is finite dimensional). I also have that, up to isomorphism, there are finitely many irreducible representations and they all occur as subrepresentations of the regular representations. But this states only the existence of such a decomposition, it does not tell that my direct sum has to be irreducible. Which is the argument used to state this?
The prove states that it is easy to find copies of $V_0$ and $V_1$ in the regular representation ($k\cdot\sum_{\sigma\in S_3}{\sigma}$ resp. $k\sum_{\sigma\in S_3}{\sigma}$).
I have absolut no idea of what is going on here. why we want or need copies of $V_0,V_1$? How one comes to such copies?
Any help would be very appreciated. thaks
Actually now I'm seeing that the first two representations are irreducible since there are no non-trivial subrepresentation. And about $V$ I found a note stating that it is irreducible since $Char(k)\nmid \mid G\mid$. But why should this hold?
Best Answer
It is a remarkable and beautiful fact that the irreducible representations of the symmetric group $S_n$ are in correspondence with the partitions of $\lambda \vdash n$. For example, in the case of $S_3$, the irreducible partitions correspond to all the partitions of 3, namely $$(3) \quad (1,1,1) \quad (2,1)$$ The full story is too long for this post, but details can be found in Chapter 4 of 'Representation Theory: A first course' by Fulton and Harris.
To see the decomposition of the regular representation into its irreducible components is most easily done via character theory. Let $\chi$ be the character of the left regular representation (that is, let $S_3$ act on $K[S_3]$ on the left); $\chi(\sigma)$ is the number of fixed points of the action of $\sigma$ on $K[S_3]$ (as these contribute to the trace of this action). It is plain to see that the only element in $S_3$ which fixes anything in $K[S_3]$ is the identity element $e$, and moreover, this fixes every element in $K[S_3]$. Thus we have $$\chi(e) = |S_3|=6 \qquad \chi(\sigma)=0 \quad \forall \ \sigma \in S_3 \backslash \{e\}$$ Let $\chi_\lambda$ be the character of the irreducible representation of $S_3$ corresponding to the partition $\lambda \vdash 3$. It is a fact from character theory that the inner product of characters of a representation $A$ and an irreducible representation $B$, defined by, $$\langle \chi_A, \chi_B \rangle = \frac{1}{|G|}\left( \sum_{g \in G} \chi_A(g)\chi_B(g) \right)$$ gives the multiplicity of $B$ in $A$. For your question then, we need to compute this inner product with the character $\chi_\lambda$. For this we only need to know one more fact: that $\chi_\lambda(e)$ is the dimension of the corresponding irreducible representation of $S_3$ corresponding to $\lambda$. We can now compute $$\langle \chi, \chi_\lambda \rangle = \frac{1}{|S_3|}\left(\chi(e)\chi_\lambda(e) \right) = \frac{1}{6}(6\cdot \chi_\lambda(e)) = \chi_\lambda(e)$$ We see that the irreducible representation corresponding to $\lambda$ appears in the decomposition of the regular representation exactly the 'dimension of representation' number of times.
Here are the correspondences in your case:
$$\lambda = (3) \rightarrow V_0, \dim = 1$$ $$\lambda = (1,1,1) \rightarrow V_1, \dim = 1$$ $$\lambda = (2,1) \rightarrow V, \dim = 2$$
Therefore $$k[S_3] = V_0 \oplus V_1 \oplus V^{\oplus 2}$$ as desired.
If this is new to you, then there are a lot of details to check here, all of which can be found in Fulton Harris. You should know that this story works for any $n$ and we have in general that $$K[S_n] = \bigoplus_{\lambda \vdash n} V_\lambda^{\oplus \dim V_\lambda}$$ where $V_\lambda$ is the irreducible representation of $S_n$ corresponding to the partition $\lambda$ of $n$.