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I am preparing for an exam and there is an excercise which I have to solve but I got stuck. The excercise states:

Let $V=\mathbb{C}^3$ be the permutation representation of the symmetric group $S_3$. Show that any map of representations (intertwiner) $I: V \rightarrow V$ has the form: $ \begin{pmatrix} \lambda & \mu & \mu \\ \mu & \lambda & \mu \\ \mu & \mu & \lambda \end{pmatrix} $ w.r.t the standard basis of $\mathbb{C}^3$ for some $\lambda, \mu \in \mathbb{C}$

What I did: We got a representation $\rho: \rightarrow GL_n(V)$ and another representation $\sigma:G \rightarrow GL_n(V)$ and an intertwiner $I: \rho \rightarrow \sigma$ which is defined by the linear map $I: V \rightarrow V$ which has to satisfy: $[\sigma(g)](I(v))=I([\rho(g)] (v))$ for all $v \in V$.

So our linear map is defined by the condition: $[\sigma(\pi)] \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \ \vec{v} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} ([\rho(\pi)] (\vec{v})) \\ $

My first intention now was to simply plug in the standard basis for v and see what conditions it generates for the matrix elements.This works just fine for two concrete representations but I'm not sure how to show this for every representation i.e for every map. I'm not an expert in representation theory since I'm a physics undergraduate and only now the basics. Maybe the whole idea is wrong but I definitely need a hint here. Thanks in advance.

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Fix the standard ordered basis $\{e_1,e_2,e_3\}$. A matrix $A$ intertwines with the standard permutation representation iff it is invariant under conjugation by permutation matrices. Conjugating $A$ by a permutation matrix is equivalent to rewriting it according to a different ordered basis with the same basis vectors. For example, the permutation $1\leftrightarrow 2$ yields

$$\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\longleftrightarrow \begin{pmatrix}e & d & f \\ b & a & c \\ h & g & i\end{pmatrix}$$

(the $1$st/$2$nd rows and columns were switched). Stare at a $3\times3$ array long enough to convince yourself that any two off-diagonal entries can be swapped, and any two diagonal entries can be swapped, but no diagonal entry may be swapped with a off-diagonal entry. (Prove this rigorously with transpositions if need be.) Thus all off-diagonals and diagonal entries respectively are equal.

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  • $\begingroup$ Maybe I don't exactly understand the exercise but I don't see why it should be sufficient if A is invariant under conjugation by permutation matrices? This would be sufficient if both representations are the standard permutation rep. (but then it's clear since the map is simply the identity). (Continued in next comment) $\endgroup$ – Prook Jun 26 '13 at 11:37
  • $\begingroup$ I understand the exercise in the way that we got one representation $(\rho,V)$ which is the standard permutation representation and another arbitrary one (for example an irreducible one: $W=\{v \in \mathbb{C}^3|\lambda_1=\lambda_2=\lambda_3\}=(\sigma,V)$ So our condition for A to intertwine would be: $\sigma(\pi) A = A \rho(\pi)$ So either am I really wrong about the task or I just don't get the point. $\endgroup$ – Prook Jun 26 '13 at 11:38
  • $\begingroup$ @Prook The letter $V$ is set to mean the standard permutation representation, so an intertwiner $V\to V$ is one such that $\sigma(\pi)A=A\sigma(\pi)$. There is no second representation in play. Secondly "then it's clear since the map is simply the identity" is false; if the permutation representation was irreducible, then Schur's would tell us the intertwiners are scalar multiples of the identity, but this representation is not irreducible so Schur's does not apply. $\endgroup$ – anon Jun 26 '13 at 18:30
  • $\begingroup$ Aaah! Thank you very much for pointing this out! Now it's clear. $\endgroup$ – Prook Jun 27 '13 at 11:10
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A more general approach is to look at the intertwiner on each irreducible subrepresentation.

The permutation representation of $S_3$ is a direct sum of the trivial representation and the standard representation, $T\oplus R$. By Schur's lemma, an intertwiner must leave these two representations invariant (as in the image of $I|_T$ lies in $T$ and the image of $I|_R$ lies in $R$). Over an algebraically closed field like $\mathbb{C}$, intertwiners between irreducible representations are simply scalars, so $I|_T=a$ and $I|_R=b$.

The representation $T$ is spanned by $(1,1,1)$, and $R$ is the orthogonal subspace. When scaling $T$ by $a$ and $R$ by $b$, using some linear algebra gives \begin{equation} I=a\begin{pmatrix} 1&1&1\\1&1&1\\1&1&1 \end{pmatrix} +b\begin{pmatrix} 2&-1&-1\\ -1&2&-1\\ -1&-1&2 \end{pmatrix} \end{equation} where the first matrix is $(1,1,1)(1,1,1)^T$ and the second is $3I_3$ minus the first matrix (that is, a matrix which is a multiple of the projection onto the orthogonal complement of $(1,1,1)$). Substituting $\lambda=a+2b$ and $\mu=a-b$ gives your result.

Said differently: we may use orthogonal diagonalization of $I$ knowing eigenvalues $a$ and $b$ and the basis for the first eigenspace, and that $R$ is the orthogonal complement of $T$.

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