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Let the group $SL_n$ act on the projective space $\mathbb P^{n-1}$ in the standard way (both defined over $\mathbb C$).

Is there an $SL_n$-invariant (linear) functional on the space of rational functions on $\mathbb P^{n-1}$?

How to describe the space of all such functionals?

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$\newcommand{\C}{\mathbb{C}}\newcommand{\P}{\mathbb{P}}$ For $n=2$ it doesn't exist. For $n\geq 3$ it does. Let $P$ be any irreducible polynomial with trivial stabilizer in $G:=SL_n(\C)$. Let $K$ be the function field of the variety given by the equation $P=0$. Then there is a $K$- valued point $x_0$ in $\mathbb{P}^{n-1}$ with a trivial stabilizer. Let $X$ be the $G$- orbit of $x_0$ (just as a set). We can choose a local parameter at every point $x\in X$ in a $G$-equivariant way. Now every rational function $f$ can be expanded in every point $x \in X$ as a power series with coefficients say $a_k(f,x)\in K$. Take $\varphi(f)=\sum_{x\in X} a_{-1}(f,x)$. This is a non-trivial $K$- valued $G$-invariant functional on rational functions. Composing $\varphi$ with some $\C$-linear map $K\to \C$ will produce the desired functional.

EDIT: It shouldn't be a local parameter at every $x$, it should be $k-1$ of them. Anyway, one can expand rational functions in several variables into power series and extract coefficients, so one can construct lots of functionals like this.

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  • $\begingroup$ Thank you very much. I have some follow up questions: 1. Why is it true that for n=2 there is no such functional? $\endgroup$
    – Rami
    Commented Jan 7, 2021 at 21:14
  • $\begingroup$ 2. Why is there $x_0$ as above? $\endgroup$
    – Rami
    Commented Jan 7, 2021 at 21:14
  • $\begingroup$ 3. As you mentioned one can not just take a_{-1} if n>2. one has to take some combination of the coefficients of the taylor expansion of $f$ at x. not any combination will do. one has to choose a combination which is invariant w.r.t. change of coordinates. I do not see why there is such a combination (except for the trivial one $a_0$, which is not good because then the sum will be infinite), it looks as a problem similar to the previous one but in one dimension less. This could have given a proof by induction, but for this it has to work also for $n=2$. $\endgroup$
    – Rami
    Commented Jan 7, 2021 at 21:14
  • $\begingroup$ 4. it seems that you are using the fact that f is regular in all points of X except for finitely many. why is it true? $\endgroup$
    – Rami
    Commented Jan 7, 2021 at 21:15
  • $\begingroup$ 5. rational functions on \P^{n-1} are the same as rational functions on \A^{n} of homogeneous degree 0. any such function can be pulled back to $\A^{n+1}$ under the projection. Thus we have an embedding of the space of rational functions on $\P^{n-1}$ to the space of rational functions on $\P^{n}$ . This gives, for any $SL_{n+1}$ -invariant (linear) functional on the space of rational functions on $P^n$ a $SL_{n}$ -invariant (linear) functional on the space of rational functions on $P^{n-1}$.Thus we can use the construction for n=3 and get a construction for n-2. will we get nesserly $0$? $\endgroup$
    – Rami
    Commented Jan 7, 2021 at 21:15

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