Why are the linear factors of $g.p$ are given by $(a_i, b_i)g^{-1}$?

Question

I am reading a part of the paper below that computes the semistable locus in case of $\operatorname{Sym^3}(\mathbb{C}^2).$ Here is the part of the paper I do not understand:

Specifically, I do not understand the following:

Why are the linear factors of $g.p$ are given by $(a_i, b_i)g^{-1}$?

Answer 1

I can only answer a. If $p, q$ are homogeneous polynomials, then we have $$ g\cdot (pq) = (g\cdot p)(g\cdot q). $$ That basically follows from your definition of the action, and the fact that evaluation is a homomorphism of the multiplicative groups.

Now, let $$ g^{-1} = \begin{bmatrix} t_1 & t_2 \\ t_3 & t_4 \end{bmatrix}. $$ Then $$ g(ax + by)= a(t_1x + t_2 y) + b(t_3x + t_4y))=(at_1 + bt_3)x + (at_2 + bt_4)y = (a,b)\cdot \begin{bmatrix} t_1 & t_2 \\ t_3 & t_4 \end{bmatrix}. $$

Answer 2

As Maciej points out, the $G'$-action preserves factorizations, for the reasons discussed in my answer to your other question. Thus it suffices to understand how $G'$ acts on linear factors.

A linear factor is an element of $\mathbb{C}[x,y]_1\cong\mathrm{Sym}^1(\mathbb{C}^2)\cong\mathbb{C}^2$. It's worth taking a second to parse out these identifications; make the implicit isomorphism between the left-hand side and right-hand side explicit as $T$. Then

• $T(x)=(1,0)$,
• $T(y)=(0,1)$, and, more generally,
• $T(ax+by)=(a,b)$ (by linearity).

That is, $T$ identifies $(a,b)=\begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$.

Now, the authors seem to identify ordered pairs with both row and column matrices. That is, if $A=\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$ is a matrix, then $$A(u,v)=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}$$ but $$(u,v)A=\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}\text{.}$$ I…don't think this is good practice (how do you distinguish between inner and outer products?), but I didn't write the paper.

With all that setup, the result is now a calculation: \begin{align*} g\cdot\begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=\begin{bmatrix}a&b\end{bmatrix}\left(g^{-1}\begin{bmatrix}x\\y\end{bmatrix}\right) \\ &=\left(\begin{bmatrix}a&b\end{bmatrix}g^{-1}\right)\begin{bmatrix}x\\y\end{bmatrix} \\ &=((a,b)g^{-1})\begin{bmatrix}x\\y\end{bmatrix} \\ &=(a,b)g^{-1} \end{align*} where the $T$-identification appears in the last step.