Why $\mu_3 = \operatorname{ diag} (\zeta_3, \zeta_3)$ is the kernel of the G'-action?

Question

Here is the paper I am trying to understand:

But I do not understand why $\mu_3 = \operatorname{ diag} (\zeta_3, \zeta_3)$ is the kernel of the G'-action? could anyone show me the details of the calculations please?

Answer 1

Suppose $g = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in G'$ is in the kernel of the action. Then $$p((ax+by, cx+dy)) = p(g(x,y)) = (g^{-1} \cdot p)((x,y)) = p((x,y))$$ for all $p \in \operatorname{Sym}^3(\mathbb{C}^2)$ and all $(x,y) \in \mathbb{C}^2$.

In particular, this holds for $p((x,y)) = x^3$ at the point $(x,y) = (1,0)$, yielding $a^3 = 1$.

Next, this holds for $p((x,y)) = x^3$ at the point $(x,y) = (0,1)$, yielding $b^3 = 0$ (so $b = 0$).

Likewise, using $p((x,y)) = y^3$ and $(x,y) = (1,0)$ we get $c^3 = 0$ (so $c = 0$).

Finally, using $p((x,y)) = x^2y$ and $(x,y) = (1,1)$ we get $a^2 d = 1$, so multiplying by $a$ on both sides yields $d = a$.

We now know that $g$ is a scalar matrix such that $g^3$ is the identity. The set of all such matrices is generated by $\operatorname{diag}(\zeta_3, \zeta_3)$. Conversely, the above analysis shows that all such matrices are in the kernel of the action.

I have one small nit to pick with your text: the kernel $\mu_3$ is generated by the diagonal matrix $\operatorname{diag}(\zeta_3, \zeta_3)$, but of course it also contains the identity matrix and $\operatorname{diag}(\zeta_3^2, \zeta_3^2)$.