Notation on symmetric groups and their representations

Question

I am trying to do an exercise of proving that the representation:

$$\mathcal{H} = \{ \left(\begin{smallmatrix}a \\ b \\ c \end{smallmatrix}\right) \in \mathbb{C}^3 : a + b + c = 0 \}$$

is an irreducible representation of the symmetric group $S_3$, but i must admit that i am not comfortable with the notation above. What exactly does it mean? Should it be read as 'Let the Hilbert space contain the set of vectors $\left(\begin{smallmatrix}a \\ b \\ c \end{smallmatrix}\right)$ spanning the three dimensional complex space where each vectors elements must sum to zero'? If this is not correct, then what does this notation really mean? Why does $a+b+c=0$ need to be fulfilled, and what consequences does it come with? How is this even a representation?

I understand that when we let a group act on a vector space we get a representation of the group. How is the above equal to letting the group $S_3$ act on the vector space $\left(\begin{smallmatrix}a \\ b \\ c \end{smallmatrix}\right) \in \mathbb{C}^3$ ?

I also understand that the $S_3$ group has 6 elements, and that these elements do not commute mutually, therefore the group is non-abelian. Apparently the $S_3$ has the trivial irreducible representation $\mathcal{H} = \mathbb{C} |0\rangle$ which means that each group element acts trivially with the 1x1 identity: $R_{\pi}|0\rangle = |0\rangle$ where i assume that $R_{\pi}$ is the group generator. Once again i do not see how this notation makes sense how can $\mathbb{C}$ act on a vector? Some clear examples and/or explanations of the notation in general is very welcome. Also some hints on how to solve my exercise are welcome.

EDIT: This question is quite similar, but does not answer my question regarding notation, and getting a deeper understanding through some examples of how this works overall.

Answer 1

The symmetric group $S_3$ acts on $\mathbb{C}^3$ by permuting the variables $\sigma(e_i)=e_{\sigma(i)}$ where $e_i$ are the standard coordinate vectors $$e_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\;,\;e_2=\begin{pmatrix}0\\1\\0\end{pmatrix}\;,\;e_3=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

Now, the subspace spanned by $e_1+e_2+e_3$ is obviously $S_3$ invariant, and the complement $\mathcal{H}$ at the top of your post is also clearly $S_3$-invariant. For example $$(12).\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}b\\a\\c\end{pmatrix}$$ and $a+b+c=0$ if, and only if, $b+a+c=0$.

It has a basis given by the vectors $$ \alpha_1=e_1-e_2=\begin{pmatrix}1\\-1\\0\end{pmatrix}\;,\;\mbox{and}\;\alpha_2=e_2-e_3=\begin{pmatrix}0\\1\\-1\end{pmatrix}. $$ If you are careful, you will find that the action of $(12)$ on the basis $\{\alpha_1,\alpha_2\}$ is $\begin{pmatrix}-1&1\\0&1\end{pmatrix}$ and the matrix for $(23)$ is $\begin{pmatrix}1&0\\1&-1\end{pmatrix}$.

Answer 2

Notation-wise, $\mathcal{H}$ is simply the vector space consisting of solutions to the homogeneous linear equation $a+b+c=0$. More compactly, you might see something like $$\mathcal{H}=\{v\in\mathbb{C}^3:\begin{bmatrix}1&1&1\end{bmatrix}v=0\}$$

Recall that $S_3$ is the group of permutations of three objects. Linearizing this, $S_3$ acts on $\mathbb{C}^3$ by permuting the standard basis vectors. The thing about this particular representation is that the vector $(1,1,1)\in \mathbb{C}^3$ is invariant under the $S_3$ action (and in particular it is simultaneously an eigenvector of all six matrices). This one-dimensional subspace is called the trivial representation, and it turns out one can eliminate it by taking the orthogonally complementary subspace to it, since the $S_3$ action consists of unitary matrices. The result is your $\mathcal{H}$.

To check more directly that it is a representation, just observe that if you take a vector whose coordinates sum to zero, and then permute the coordinates, the coordinates still sum to zero.

I'll leave irreducibility to you, but a possible hint is that, given any nonzero vector in $\mathcal{H}$, there is a sequence of permutations and linear combinations you can do to create a basis for all of $\mathcal{H}$.