In a lecture, I saw the claim that the ring of invariant polynomials for the action of $SL_2(\mathbb{C})$ on binary quadratic forms is $\mathbb{C}[disc]$ - i.e that essentially the only invariant for this action is the discriminant. While it is clear the discriminant is an invariant, having little knowledge in representation theory, I'm having problems understanding solutions to this problem. I wonder if perhaps there's an elementary solution? Or maybe, if a rigorous solution is too involved, could anyone provide some intuition as to why should this work?
Thanks in advance!
Certainly there is a more scholarly argument but let me try by hand. Let $q_d(x,y)=x^2+dy^2$ and if $f$ is an invariant polynomial then let $\phi(d)=f(q_d)$. Then I say that $f=\phi\circ disc$.
Indeed a generic quadratic form is given by $q(x,y)=ax^2+2bxy+cy^2=q_d(\frac{ax+by}{\sqrt{a}},\frac{y}{\sqrt a})$ where $d=disc(q)$. This is clearly the composition with an element of $SL_2(\mathbb C)$ so $f(q)=f(q_d)=\phi\circ disc(q)$. Then continuity of $f$ implies that this is true for any $q$.