0
$\begingroup$

Let $X=\{1,2,3\}$ on which the subgroup of $H$ of $S_3$ acts, where $$H=\{(1)(2)(3),(12)(3)\}.$$ Let $\rho$ be the permutation representation of $G$ with respect to $X$. We then have that $$\rho:G\to \text{GL}(V),\quad V\cong \Bbb C^3.$$ Now I have here that the number of distinct orbits is the number of times the unit representation is in our representation.

I can see that there are two orbits $\{1,2\},\{3\}$, and denoting our non-identity element of our group by $s$, I can see that $$\rho_s(e_1)=e_2,\quad \rho_s(e_2)=e_1,\quad \rho_s(e_3)=e_3,$$ and hence we have the representation matrices: $$\rho_s=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix},\qquad \rho_i=I_{3\times 3},$$ There clearly having invariant subspace generated by $e_3$ means that we can decompose the space into $V=W\oplus \Bbb C$ and we thus have a unit representation contained in our representation $V$. There are no other unit representations contained here, but there are two orbits.

Why are there two orbits, yet only one unit representation contained?

  • 0
    What do you mean by "unit representation"? Do you mean one dimensional invariant subspace? If so, consider $\mathbb{C}(e_1 + e_2)$. – 2017-01-06
  • 0
    @ChocolateAndCheese Oh my. That does it, thanks! (Sort of - Serre defines it as 'trivial rep' = 'unit rep' = 'one dim space where the matrices all act by the identity' - but this answers my concern) – 2017-01-06

1 Answers 1

1

Because $$ \begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix} $$ is similar to $$ \begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}. $$ This is because $$ \begin{bmatrix}1&-1\\1&1\end{bmatrix}^{-1} \begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}1&-1\\1&1\end{bmatrix} = \begin{bmatrix}1&0\\0&-1\end{bmatrix}. $$

You see you failed to see the other, non-obvious trivial representation: possibly the best way to see these kinds of things is via characters. $\rho_{s}$ has character $\chi_{s}$ with value $3$ on $1$ and $1$ on $g = (12)$. Take the scalar product with the trivial character $\tau$ to get that $\tau$ has multiplicity $$ \frac{1}{2} ( 1 \cdot 3 + 1 \cdot 1) = 2 $$ in $\rho_{s}$.