Help me find representations of $S_4$as a direct sum of irreducible representations Im taking a graduate course in Representation theory and struggling with one exercise. The task is to:

*

*Compute the characters of irreducible representations of $S_4$.

*For each $\lambda$, write a decomposition of $\varphi^{\lambda} \rightarrow Aut_{\mathbb{C}}(M^{\lambda})$ as a dirrect sum of irreducible representations (that is, for each $\mu$, determine the multiplicity of $\psi^{\mu}$ in $\varphi$).

Please, could you help me with the part 2., especially writing the decomposition?
I have computed the representations and characters (as can be seen on the pictures),  but I am struggling with thow to write the representations for $\lambda_1$, $\lambda_2$, ... $\lambda_5$ as direct sums.
Thank you for your help.



 A: Question: "Please, could you help me with the part 2., especially writing the decomposition?"
Answer: There is an operator - the Reynolds operator - that might be helpful. Let $G$ be any finite group and $V$ any finite dimensional $k$-vector space with a representation $\rho: G \rightarrow GL_k(V)$. Define for any $v\in V$
$$R_V(v):=R(v):= \frac{1}{n}\sum_{g\in G} gv$$
wher $n:=\# G$ is the number of elements of $G$. The operator $R$ is in $R\in Hom_G(V,V)$ and $Im(R) \cong V^G$ is the sub-$k$-vector space of elements fixed by $G$. You get a short exact sequence of $G$-modules
$$S1\text{  } 0 \rightarrow ker(R) \rightarrow V \rightarrow V^G \rightarrow 0.$$
This is because $R^2=R$ is an idempotent endomorphism. The sequence $S1$ splits hence $V \cong V^G \oplus ker(R)$. This may be done to any $G$-module $V$.
Example: When you do this to your $G:=S_n$-module $V$ you get $V \cong V^G \oplus ker(R)$. Then you may do a similar procedure to $V^G$ and $ker(R)$ etc. If $V$ is irreducible it follows $ker(R)=(0)$ and $Im(R)=V$, hence $R$ is an automorphism of $G$-modules. Moreover if $V \cong V_1 \oplus V_2$ is a direct sum of $G$-modules it follows for any $(u,v)\in V_1 \oplus V_2$
$$R_V(u,v):=\frac{1}{n}\sum_g g(u,v)= \frac{1}{n}\sum_g (gu,gv)=$$
$$(\frac{1}{n}\sum_g gu, \frac{1}{n}\sum_g gv)=(R_{V_1}(u), R_{V_2}(v)).$$
Hence there is an equality $R_V \cong R_{V_1}\oplus R_{V_2}$. By Schur Lemma it follows $R_V \cong \lambda Id$ if $V$ is irreducible. Hence if $V \cong V_1 \oplus \cdots \oplus V_d$ is a decomposition of $V$ into irreducibles it follows $R_V \cong \oplus R_{V_i}$ and each $R_{V_i} = \lambda_i Id$ is multiplication with $\lambda_i$ - a complex number.
Note: If $V_{\lambda_i} \subseteq V$ is an eigenspace for $R$ with eigenvalue $\lambda_i$, it follows $V_{\lambda_i}$ is a $G$-submodule and there is a direct sum decomposition $V \cong V_{\lambda_i}\oplus W$ where $W$ is a "complementary" $G$-module. If $V \cong W^d$ where $W$ is irreducible, it follows $R$ acts on $V$   with one eigenvalue, hence $R$ does not detect the direct sum decomposition of $V$. Still $R$ gives some information on the decomposition of $V$. If $k[G]$ is the group algebra of $G$, you get an element
$$R:=\frac{1}{n}\sum_{g\in G} g \in Z(k[G])$$
in the center $Z(k[G])$ of $k[G]$.
If you choose any vector $v\in V$ and consider the $G$-module $V(v):=k[G]v \subseteq V$ generated by $v$, it follows there is a direct sum decomposition $V \cong V(v)\oplus W(v)$ where $W(v)$ is complementary. This gives an "algorithm" for calculation of the direct sum decomposition.
You find some information below:
Regular representation and matrix coefficients
