Decomposing the tensor product representation of $S_3$ in terms of irreducibles I have a theorem which says that:
If $\rho_1,...\rho_n$ are a complete set of irreducible $K$-representations of $S_n$ then we have that:
$V^{\otimes n}=\bigoplus_1^k(V^{\otimes n}_{\rho_i})$ as $G$-reps.
We have then worked through an example with $G=S_2$ as follows:
Let $v_1,..v_n$ be a basis of $V$. Then $V^{\otimes 2}$ has a basis given by $v_i\otimes v_j$ for $1\leq i,j\leq n$
We then have that a basis for $V_{1}^{\otimes 2}$ is given by $v_i\otimes v_j+v_j\otimes v_i$ for $1\leq i\leq j\leq n$

As $1$ is the trivial representation and so $G$ is acting trivially and we just need stuff that is invariant under the permutation action

We then have that a basis for $V_{sign}^{\otimes 2}$ is given by $v_i\otimes v_j-v_j\otimes v_i$ for $1\leq i< j\leq n$

Just as the group is simply sending all the two cycles to minus themselves

Then via counting arguments we have that:
$dim V_{1}^{\otimes 2}=\frac{n(n+1)}{2}$ and
$dim V_{sign}^{\otimes 2}=\frac{n(n-1)}{2}$
Now I am trying to work the example through myself with $S_3$.
So we know that there are 3 irreducible representations on $S_3$ namely $1$, sign and the standard representation.
We also have that:
$V^{\otimes 3}$ has a basis given by $v_i\otimes v_j\otimes v_k$ for $1\leq i,j,k\leq n$
So now I am looking for a basis of $V^{\otimes 3}_{1}$, which is where the $S_3$ action is just trivial so I am looking for symmetrized $v_i\otimes v_j \otimes v_k$ which will be:
$ v_i\otimes v_j \otimes v_k + v_i\otimes v_k \otimes v_j + v_j\otimes v_i \otimes v_k+ v_j\otimes v_k \otimes v_i + v_k\otimes v_j \otimes v_i+v_k\otimes v_i \otimes v_j$

Is this right because it seems a bit off?

Now for a basis of $V^{\otimes 3} _{sign}$ as the three cycles are even am I looking for just the same thing as above?
I am now unsure how to continue?
Thanks for any explanation or help (sorry my question turns into a mess at the end)
 A: In general, if $G$ acts on some complex vector space $W$, and you want the $\chi$-isotypical component of $W$, where $\chi$ is some irreducible character, you use the idempotent corresponding to $\chi$: define
$$
e_{\chi} = \frac{\chi(1)}{|G|}\sum_{g\in G}\chi(g^{-1})\cdot g.
$$
This is an element of the group algebra $\mathbb{C}[G]$; in fact it is a primitive central idempotent in $\mathbb{C}[G]$.
Now, $W$ is a module under this group algebra, so you can consider the subspace of $W$ given by $e_{\chi}W = \{e_{\chi}(w): w \in W\}$. This is precisely the $\chi$-isotypical component of $W$. Of course, this is the same as the space spanned by $e_{\chi}(w_i)$ as $w_i$ runs over a basis of $W$.
In your example, $G$ is $S_3$ and $W$ is $V^{\otimes 3}$, and $G$ acts on elementary tensors by permuting the entries. If you write out the above in the case that $\chi$ is the trivial character, you get exactly what you wrote down (strictly speaking, my version differs from yours by the normalisation $1/|S_3|$, which doesn't matter when it comes to linear spans). For the sign character, this procedure gives you
$$
e_{\chi}(v_1\otimes v_2\otimes v_3) = \frac{1}{|G|}\sum_{\sigma\in S_3}{\rm sign}\sigma\cdot v_{\sigma^{-1}(1)}\otimes v_{\sigma^{-1}(2)}\otimes  v_{\sigma^{-1}(3)}.
$$
Just like in the case of the trivial character, if you do this for each basis element of $V^{\otimes 3}$, you get redundancies, so to get a basis, you put suitable restrictions on your indices.
See if you can work out the rest yourself; if not, then feel free to ask for more details. It is also a nice exercise to then go on to compute the characters of these components in terms of the character of $V$, like you do in the $V^{\otimes 2}$ case.
