I am trying to understand invariant algebras for nontrivial characters. The gist of my question is confusion over the definition and an equivalent condition given in Stanley's survey paper Invariants Of Finite Groups And Their Applications To Combinatorics. I worked through an example, and my conclusion seems wrong, but I don't understand why and I was hoping someone could see where I went wrong.
My work
I worked through an example of the Orlik-Solomon algebra $A$ of the rank-2 braid arrangement $$\{x_i = x_j: 1 \leq i < j\leq 3\},$$ which is a skew-commutative algebra (I take coefficients in $\mathbb{C}$) with basis $$\{1, e_{(1,2)}, e_{(1,3)}, e_{(2,3)}, e_{(1,2)}e_{(1,3)}, e_{(1,2)}e_{(2,3)} \}$$ and the additional relation $e_{(1,3)}e_{(2,3)} = e_{(1,2)}e_{(1,3)} - e_{(1,2)}e_{(2,3)}$. It is also graded by the number of generators $e_{(i,j)}$.
I want to think of the symmetric group $\mathfrak{S}_3$ acting on the coordinates of the braid arangment, so for example the involution $(1,2) \in \mathfrak{S}_3$ acts like $(1,2) \cdot e_{(1,3)} = e_{(2,3)}$. I am also treating the subscripts as a set so that $e_{(1,2)} = e_{(2,1)}$.
After going through the character computation by explicitly writing the group action as a matrix, I came to the following conclusion: $A \cong S^{(3)} \oplus S^{(3)} \oplus S^{(2,1)} \oplus S^{(2,1)}$, where $S^{\lambda}$ is the Specht module indexed by $\lambda \vdash 3$. My thinking was that $\sigma \cdot 1 = 1$, so $A_0$ (the degree-0 homogeneous piece of $A$) is isomorphic to the trivial representation. Then $A_1$ is the permutation representation, which decomposes into $S^{(3)} \oplus S^{(2,1)}$. Finally, $A_2$ is the regular representation $S^{(2,1)}$. (I feel ok about this, but would love to hear if this is right or wrong.)
Edit: it is not the regular representation, it is the standard representation, as pointed out in the comments.
Now, I want to understand the two invariant algebras $A^{\mathfrak{S}_3}_{(3)}$ and $A^{\mathfrak{S}_3}_{(2,1)}$. We can change basis for $A_1$ to write it as the span of $\{e_{(1,2)}+e_{(1,3)} + e_{(2,3)}, e_{(1,2)}- e_{(1,3)}, e_{(1,3)} - e_{(2,3)}\}$ which decomposes it into the trivial representation (on $e_{(1,2)}+e_{(1,3)} + e_{(2,3)}$) and the regular representation (on $e_{(1,2)}- e_{(1,3)}, e_{(1,3)} - e_{(2,3)}$). The basis for the $(2,1)$ isotypic component on $A_2$ is the same as the basis for $A_2$.
So then bases for the two isotypic components $A_{(3)}$ and $A_{(2,1)}$ are: $$A_{(1,1,1)} : \{ 1,\, e_{(1,2)}+e_{(1,3)} + e_{(2,3)}\}$$ and $$A_{(2,1)} : \{e_{(1,2)}- e_{(1,3)},\, e_{(1,3)} - e_{(2,3)},e_{(1,2)}e_{(1,3)}, e_{(1,2)}e_{(2,3)} \}.$$
Stanley defines the invariants to be $R_\chi^G$ (he does define them on a polynomial ring, but it is unclear to me if this distinction matters for this example) where $R_\chi^G$ is the isotypic component associated to the character $\chi$. So, $A_{(3)}^{\mathfrak{S}_3}$ should be the span of $\{ 1,\, e_{(1,2)}+e_{(1,3)} + e_{(2,3)}\}$ and $A_{(2,1)}^{\mathfrak{S}_3}$ should be the span of $\{e_{(1,2)}- e_{(1,3)},\, e_{(1,3)} - e_{(2,3)}\}$. Up to this point, I thought everything had made sense to me.
My confusion
Stanley says,
In the special case $\chi$ is linear (i.e.,$\chi$ is a homomorphism $G \rightarrow \mathbb{C}\backslash \{0\}$), then the condition $f \in R^G_\chi$ is equivalent to $M(f) = \chi(M)f$ for all $M \in G$.
From my understanding this means that the invariant $A_\chi^{\mathfrak{S}_3}$ should be the following intersection of kernels: $$ \bigcap_{\sigma \in \mathfrak{S}_3} \ker ( x \mapsto \sigma \cdot d - \chi(\sigma) x). $$ However, I went to explicitly compute that intersection, and got some confusing results.
Taking $\chi$ to be the trivial character, things made sense. However, taking $\chi$ to be the character of the regular representation, and computing for conjugacy class representatives $()$, $(12)$ and $(123)$ (in cycle notation), I found that
| $\sigma$ | $\chi(\sigma)$ | $f$ | $\sigma \cdot f - \chi(\sigma)f $ |
|---|---|---|---|
| $() $ | $2$ | $e_{(1,2)}-e_{(1,3)}$ | $-e_{(1,2)}+e_{(1,3)}\neq 0$ |
| " | " | $e_{(1,3)}-e_{(2,3)}$ | $-e_{(1,2)}+e_{(1,3)}\neq 0$ |
| " | " | $e_{(1,2)}e_{(1,3)}$ | $-e_{(1,2)}e_{(1,3)} \neq 0$ |
| " | " | $e_{(1,2)}e_{(2,3)}$ | $-e_{(1,2)}e_{(2,3)} \neq 0$ |
| $(12)$ | $0$ | $e_{(1,2)}-e_{(1,3)}$ | $e_{(1,2)}-e_{(2,3)}\neq 0$ |
| " | " | $e_{(1,3)}-e_{(2,3)}$ | $e_{(2,3)}-e_{(1,3)}\neq 0$ |
| " | " | $e_{(1,2)}e_{(1,3)}$ | $e_{(1,2)}e_{(2,3)} \neq 0$ |
| " | " | $e_{(1,2)}e_{(2,3)}$ | $e_{(1,2)}e_{(1,3)} \neq 0$ |
| $(123)$ | $-1$ | $e_{(1,2)}-e_{(1,3)}$ | $e_{(2,3)}-e_{(1,2)} + e_{(1,2)} -e_{(1,3)} = e_{(2,3)} - e_{(1,3)} \neq 0$ |
| " | " | $e_{(1,3)}-e_{(2,3)}$ | $e_{(1,2)}-e_{()} + e_{(1,2)} -e_{(1,3)} = e_{(2,3)} - e_{(1,3)} \neq 0$ |
| " | " | $e_{(1,2)}e_{(1,3)}$ | $e_{(2,3)}e_{(1,2)} + 2e_{(1,2)}e_{(1,3)} \neq 0 $ |
| " | " | $e_{(1,2)}e_{(2,3)}$ | $e_{(2,3)}e_{(1,3)} + 2e_{(1,2)}e_{(2,3)} = e_{(1,2)}e_{(2,3)} + e_{(1,2)}e_{(1,3)} \neq 0$ |
This leads me to believe that $ \bigcap_{\sigma \in \mathfrak{S}_3} \ker ( x \mapsto \sigma \cdot d - \chi(\sigma) x)$ is not the same as the previously computed isotypic component.
I wish to compute the invariant algebra as a kernel. Is there a way to do this for noncommutative (or just skew-commutative) underlying algebra? Or am I somehow misunderstanding the relationship between the isotypic component and the kernel I described? Or maybe I am wrongly assuming that the character I used is linear?
Any help you could provide would be much appreciated! Thanks!!
Stanley's remark only applies to linear characters, i.e. $\chi(1) = 1$, i.e. 1-dimensional representations. The standard representation is not 1-dimensional.