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Let the permutation group $S_4$ act on $\mathbb C^4$ by permuting the coordinates. Consider the categorical quotient $\mathbb P(\mathbb C^4)/S_4$. It is a projective variety by a theorem of Mumford. Does the line bundle $\mathcal O(1)^{\otimes 12}$ on $\mathbb P(\mathbb C^4)$ descend to the quotient $\mathbb P(\mathbb C^4)/S_4$ ?

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Yes, and the good news are that there isn't anything to compute. By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$. Now $S_4$ acts on $\mathcal{O}(1)$, and $S_x$ act on $\mathcal{O}(1)_x$ through a character $\chi _x:S_x\rightarrow \mathbb{C}^*$, and on $\mathcal{O}(1)^{\otimes k}$ through the character $\chi _x^{k}$. The abelian subquotients of $S_4$ have order $2,3$ or $4$, hence taking $k=12$ kills all the characters, and therefore $\mathcal{O}(1)^{\otimes 12}$ descends.

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  • $\begingroup$ The statement "By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$." is not very precise. You probably mean : an equivariant line bundle, else the action on the fiber is not even defined. Could you give a reference ? $\endgroup$
    – Niels
    Commented Jan 27, 2015 at 13:37
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    $\begingroup$ Yes, of course, I meant an equivariant line bundle. A good reference is Theorem 2.3 in the paper of Drezet and Narasimhan, Groupe de Picard des variétés de modules de fibrés semi-stables sur les courbes algébriques, Invent. math. 97 (1989), 53-94. $\endgroup$
    – abx
    Commented Jan 27, 2015 at 13:49
  • $\begingroup$ I think the same proof will work for any $k$ a multiple of $2,3,...,n$ with the action of $S_n$ on $\mathbb P^{n-1}$. $\endgroup$
    – Ram
    Commented Jan 28, 2015 at 9:47
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    $\begingroup$ Yes, that's correct. $\endgroup$
    – abx
    Commented Jan 28, 2015 at 13:06
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    $\begingroup$ I gave the precise condition in my answer, please read it. $\endgroup$
    – abx
    Commented Jan 21, 2016 at 9:53

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