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So we have the group $S_4$ which has the standard representation. We also have the subgroup generated by permutation (1234). This is isomorphic to $C_4$ which has four irreducible representation. How does Res$^{S_4}_{C_4}(V)$ decompose into a direct sum of the four irreducibles of $C_4$. I know that for the trivial (and alt.) representation of $S_4$, it restricts to the trivial (and alt.) rep'n of $C_4$, but I can't seem to figure out how to do this for the other three irreducibles of $S_4$, and am hoping if you teach me how to do it for the standard I can do the rest.

I feel like the problem is that the standard rep'n is confusing in that I cannot even tell if there is a subspace of $V$ that is fixed when I only act on it with the elements of $C_4$ i.e. those generated by $(1234)$.

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Have you written down the matrix corresponding to $(1,2,3,4)$ under the standard representation with respect to some basis? –  Alex B. Apr 23 '12 at 4:12
    
I just did, I got {{0,0,-1},{1,0,-1},{0,1,-1}} in Mathematica notation. –  Steven-Owen Apr 23 '12 at 4:28
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which has eigenvalues $-1,-i,$ and $i$. So does the standard representation decompose into one of each irreducible of $C_4$ except the trivial? –  Steven-Owen Apr 23 '12 at 4:30

2 Answers 2

$\def\triv{\boldsymbol{1}}$ One way of getting the answer is to just write down the matrix of $(1,2,3,4)$ under the standard representation with respect to your favourite basis and compute eigenvalues, as you have done. Here is another way, using some more machinery, but with almost no calculation:

The standard representation can be written as $\rho=\mathrm{Ind}_{S_4/S_3}\triv_{S_3} - \triv_{S_4}$, i.e. as the 4-dimensional permutation representation $\mathbb{C}[S_4/S_3]$, but without the trivial summand. So, using Mackey decomposition and the fact that $S_4 = S_3C_4$, $S_3\cap C_4 = \{1\}$, we get $$ \mathrm{Res}_{S_4/C_4}\rho = \mathrm{Res}_{S_4/C_4}(\mathrm{Ind}_{S_4/S_3}\triv_{S_3} - \triv_{S_4}) = \mathrm{Ind}_{C_4/1}\triv - \triv_{C_4}, $$ i.e. the sum of all non-trivial representations of $C_4$.

Similar tricks work for all the other representations of $S_4$, since it just so happens, that any irreducible character of $S_n$, $n\in \mathbb{N}$, is a linear combination of permutation characters.

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$\newcommand{\Res}{{\text{Res}}} \newcommand{\Ind}{{\text{Ind}}} \newcommand{\ds}{{\displaystyle}} \newcommand{\inv}{{^{-1}}}$By Frobenius reciprocity, if $W$ and $U$ are irreducible representations of $H < G$, respectively, thenthe number of times $U$ appears in $\Ind W$ is the same as the number of times $W$ appears in $\Res U$. Let $W$ be the one dimensional representation (thus obviously irreducible) of the subgroup $H$ of $G$ generated by $(1234)$ in which $(1234)\cdot w=iw$. If we want to decompose $\Ind W$ into irreducible representations of $G=S_4$ i.e how many times an irreducible representation $U$ appears in $\Ind W$, we just need to find how many times $W$ appears in $\Res U$. The trivial and alternating representations restrict to trivial representation of $C_4$ and the alternating representation of $C_4$. We know that the induced representation is $\bigoplus_{\sigma \in S_4/C_4} \sigma W $, and since $W$ is one dimensional, the induced representation has dimension $[S_4:C_4]$ which is 6. Let us begin by writing the matrix corresponding to $(1234)$ in the standard representation of $S_4$. We get this by knowing that the standard representation is just the regular representation "minus" the trivial representation. (1234) in the regular representation is just the $$ \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix} $$ And we can write this in the basis $\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ -1 \\ 0 \\0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1 \\ -1 \end{pmatrix} $ and we would get $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1\\ \end{pmatrix} $$ And thus we have that the element (1234) acts in the standard representation by the matrix $$ \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & -1\\ \end{pmatrix} $$ which has eigenvalues $i, -i,$ and $-1$, so that the standard representation restricted to the group generated by (1234) decomposes into one of each of the irreducibles of the $C_4$ representations except for the trivial. We would find the same to be true for the standard representation tensor the alternating representation. Now since $\Res V$ and $\Res V'$ each contain a copy of $W$, we know that $\Ind W$ contains a copy of $V$ and $V'$, and since $\dim(V)=\dim(V')=3$ we need not even check the other irreducible representation of $S_4$, as $\Ind W = V \oplus V'$.

We do precisely the same for $(123)$. Let $W$ be the one dimensional representation (thus obviously irreducible) of the subgroup $H$ of $G$ generated by $(123)$ in which $(123)\cdot w=e^{2 \pi i/3} w$. We know the dimension of $\Ind W$ will be $[S_4:C_3]= 8$. In this case, We have that both the trivial and alternating representation restrict to the trivial of $C_3$.

Let us begin by writing the matrix corresponding to $(123)$ in the standard representation of $S_4$. We get this by knowing that the standard representation is just the regular representation "minus" the trivial representation. (1234) in the regular representation is just the $$ \begin{pmatrix} 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

And we can write this in the basis $\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ -1 \\ 0 \\0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1 \\ -1 \end{pmatrix} $ and we would get $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 1\\ \end{pmatrix} $$

And thus we have that the element (1234) acts in the standard representation by the matrix $$ \begin{pmatrix} 0 & -1 & 1 \\ 1 & -1 & 1 \\ 0 & 0 & 1\\ \end{pmatrix} $$ which has eigenvalues $1, e^{2 \pi i/3}, e^{4 \pi i/3}$, so that the standard representation restricted to the group generated by (123) decomposes into each of the irreducibles of the $C_4$ representations We would find the same to be true for the standard representation tensor the alternating representation. Now since $\Res V$ and $\Res V'$ each contain a copy of $W$, we know that $\Ind W$ contains a copy of $V$ and $V'$, and since $\dim(V)=\dim(V')=3$ we need not even check the other irreducible representation of $S_4$, as $\Ind W = V \oplus V' \oplus Q$ just by dimensionality where $Q$ is the representation of the quotient group, it is labeled as "Another $W$" in the text by Fulton and Harris.

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A minor terminology nitpick: the regular representation of a group $G$ is $\mathbb{C}[G]$, which is $|G|$-dimensional, i.e. 24-dimensional in the case of $S_4$. That's not what you meant. See also my answer. The beginning of the second paragraph is what you meant to write. –  Alex B. Apr 23 '12 at 5:21
    
Good catch; I still haven't mastered the terminology as of yet. –  Steven-Owen Apr 23 '12 at 5:23

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