Problem of invariants of binary forms of degree $d$

Question

This is my second pass going over Eisenbud for parts that I did not fully understand in the first round. This is chapter 1, Section 1.3 Invariant theory. It is related to problem of invariants of binary forms of degree $d$

It is suggested that consider $F=\sum_{i\leq d}x_is^{d-i}t^i$ general degree $d$ form in variables $s,t$ with $x_i\in k[x_0,\dots, x_d]$. Consider $s=as'+bt',t=cs'+dt'$ transformation on $s,t$. One obtains a new linear combination of $x_i$ with a different set of coefficients. If one restricts to matrix formed by $a,b,c,d$ in $SL_2(k)$, one obtains a $SL_2(k)$ action on $k[x_0,\dots, x_d]$.

Q1. Which ring is $F$ lying here? $k[x_0,\dots, x_d,s,t]$ or $k[x_0,\dots, x_d]$? I guess one transform $F$ and obtain a new coefficient of $x_i$ and this defines an action of $SL_2(k)$ on $k[x_0,\dots, x_d]$. I think this is what author means.

Q2. Why it is natural to consider $F$ here, rather than $x_0s^d$ or other degree $d$ terms in terms of $s,t$ as coefficient of $x_i$?

Q3. What is the significance of $k=C$(complex number) here? Diagonalizability or similarity?

Q4. What is the significance of this part of theory in relation to geometry?

Answer 1

1. You should not think of $F$ as a single fixed polynomial, but rather as a general vector in the Vector space $V_d$ of dimension $d+1$ which contains all homogeneous degree $d$-forms in the variables $s$ and $t.$ The $x_i$ are its coordinates in the standard basis given by the monomials $s^{d-i}t^i.$ Hence, the author also chooses $x_i$ as the name for the coordinate functions on the space $V_d\cong k^{d+1}.$ The group $\mathrm{SL}_2$ acts on $V_d$ from the right via \begin{align*} V_d \times \mathrm{SL}_d & \longrightarrow V_d \\ (F,A) &\longmapsto F\circ A \end{align*} Denoting by $R_A:V_d\to V_d$ the map $F\mapsto F\circ A$, we get an induced left action on $k[V_d]=k[x_0,\ldots,x_d]$ via \begin{align*} \mathrm{SL}_d \times k[V_d] & \longrightarrow k[V_d] \\ (A,\phi) &\longmapsto \phi\circ R_A \end{align*} This can be quite confusing (with the left and the right action), but we can check that $$ (A.B.\phi)(F) = (\phi\circ R_B \circ R_A)(F) = \phi(F\circ A\circ B) = (\phi\circ R_{AB})(F) = (AB.\phi)(F). $$
2. As I said, do not think of $F$ as one particular element.
3. $k=\Bbb C$ is usually chosen for two reasons in this context to keep it simple: It is algebraically closed and of characteristic zero. Neither is required to consider this group action, you can defined it over any field or even commutative ring: However, as soon as you start to ask questions, you get a lot further if your base field has these two perks. Note that not even over $\Bbb C$, every matrix is automatically diagonalizable. However, every matrix is triangularizable, and this is because it is algebraically closed.
4. Quite a broad question. However, the natural left action of $\operatorname{SL}_2$ on $W=k^2$ should strike you as quite geometric. The first action we consider here is the action on $k[W]_d=k[s,t]_d=V_d$ induced by that action. You are considering how polynomial forms on $W$ behave under precomposition with a linear operator. Studying $k[V_d]$ is then the natural algebro-geometric step in this endeavour: In order to understand how the group acts on $V_d$, you understand how it acts on the polynomial functions on that space. I sympathize if this seems contorted. However, with some patience you will get used to it.