Timeline for Is there an $SL_n$-invariant functional on the space of rational functions on the projective space $\mathbb P^{n-1}$?
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| Jan 10, 2021 at 11:28 | comment | added | Anton Mellit | 5. We get $0$ functional because functions on $\P^{n}$ appearing by pullback from $P^{n-1}$ have polar divisor invariant under scaling, i.e. never has trivial stabilizer, but my $P$ is chosen so that it has trivial stabilizer. | |
| Jan 10, 2021 at 11:24 | comment | added | Anton Mellit | 4. As pointed out in 3. polynomials vanishing at $x_0$ are necessarily divisible by $P$. So for any homogeneous polynomial with $\C$-coefficients there is only finitely many elements of $X$ at which $P$ vanishes. | |
| Jan 10, 2021 at 11:23 | comment | added | Anton Mellit | 3. Since $x_0$ has trivial stabilizer, you don't need to worry about invariance w.r.t. change of coordinates. Choosing local coordinates at $x_0$, for any point $x$ of the orbit of $x_0$ there is unique element $g\in G$ such that $g(x_0)=x$. So you obtain local coordinates around $x$ by translating by $g$. So this is automatically $G$-invariant. | |
| Jan 10, 2021 at 11:19 | comment | added | Anton Mellit | 2. By definition, $K$ is the field of functions of the hypersurface $P(t_0,\cdots,t_n)=0$. So it contains the functions $y_0=t_0/t_n$, $y_1=t_1/t_n$,...,$y_{n-1}=t_{n-1}/t_n$ satisfying equation $P(y_0,\cdots,y_{n-1}, 1)=0$. The point $x_0$ is the point with coordinates $(y_0,\ldots,y_{n-1},1)$. Homogeneous polynomials with $\C$-coefficients vanishing at $x_0$ are precisely polynomials divisible by $P$. So if $P$ has trivial stabilizer, $x_0$ also has a trivial stabilizer. | |
| Jan 10, 2021 at 9:51 | history | edited | Anton Mellit | CC BY-SA 4.0 |
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| Jan 10, 2021 at 9:50 | comment | added | Anton Mellit | 1. On $\P^1$ you can use a partial fraction decomposition to see if such functional $\chi$ exist it must be non-trivial on polynomials. But any polynomial $f$ can be written as $f(x)=p(x+1)-p(x)$ for some polynomial $x$. So $\chi(f)=\chi(p(x+1))-\chi(p)=0$. | |
| Jan 7, 2021 at 21:15 | comment | added | Rami | 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 $0$? | |
| Jan 7, 2021 at 21:15 | comment | added | Rami | 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? | |
| Jan 7, 2021 at 21:14 | comment | added | Rami | 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 $n=2$. | |
| Jan 7, 2021 at 21:14 | comment | added | Rami | 2. Why is there $x_0$ as above? | |
| Jan 7, 2021 at 21:14 | comment | added | Rami | Thank you very much. I have some follow up questions: 1. Why is it true that for n=2 there is no such functional? | |
| Dec 29, 2020 at 22:43 | history | edited | Anton Mellit | CC BY-SA 4.0 |
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| Dec 29, 2020 at 22:19 | history | answered | Anton Mellit | CC BY-SA 4.0 |