It seems to be known that the only invariant homogeneous polynomials in $k[X_0,...,X_n]_d$ by the action of $SL(n+1)$ on $\mathbb{P}^n$, are the constants, ie, $F(gx)=F(x)$ for all $x\in\mathbb{P}^n$, $g\in SL(n+1)$ iff $F(X_0,...,X_n)=ctt$.
I'm not able to find any proof. Any simple proof, or a reference where I can find it?
Thank you!
The action of $SL_{n+1}$ on $k[X_0,X_1,\dots,X_n]$ is also induced by the action of $SL_{n+1}$ on $\mathbb{C}^{n+1}$. Let $F$ be a homogeneous invariant of degree $d$ and let $v\in\mathbb{C}^{n+1}$. I'll show that if $d\neq 0$, then $F(v)=0$.
Let $g\in SL_{n+1}$ such that the last entry of $g\cdot v$ is $0$ (try to show that such a $g$ always exists). Then, for $t\in\mathbb{C}^{\times}$, consider the diagonal matrix $$ h(t)=\begin{bmatrix} t & & & \\ & \ddots & & \\ & & t &\\ & & & t^{-n} \end{bmatrix}\in SL_{n+1} $$ Note that $h(t)\cdot v=t v$ so we have $$ F(v)=F(h(t)v)=F(tv)=t^{d} F(v). $$ Then, either $d=0$, or (as there exists $t$ such that $t^d\neq 1$) $F(v)=0$. As $v$ is chosen arbitrary, $F(v)=0$ implies that $F=0$. Thus, only invariants are constants.