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Consider the permutation action of $\operatorname{PGL}_2(\mathbb F_q)$ on $\mathbb P^1(\mathbb F_q)$ by fractional linear transformations. We can consider the associated (complex) representation of dimension $q+1$.

What can we say about the irreducible representations occurring in this representation? This is probably doable because our representation is an induced representation. What I am really interested in is the following question:

What can we say about the invariant polynomials for this representation?

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  • $\begingroup$ Do you mean to ask about composition factors? It's not obvious to me that the representation should be completely reducible …. I guess you also have to decide whether you really mean to ask about irreducibility, or only about indecomposability. $\endgroup$
    – LSpice
    Commented Sep 28, 2020 at 21:18
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    $\begingroup$ @LSpice The question is about a complex representation. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 28, 2020 at 21:57
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    $\begingroup$ The action is doubly transitive, so the representation is the sum of the trivial representation and an irreducible representation of degree $q$. $\endgroup$
    – Derek Holt
    Commented Sep 28, 2020 at 22:01
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    $\begingroup$ Isn't the repn just the (naively normalized) induced repn of the trivial repn on the subgroup $P$ (upper-triangular) to the whole group? Among principal series, this is a little unusual in that it is reducible, containing (as @DerekHolt notes) a copy of the trivial repn, and then another irred $q$-dimensional repn (akin to a "Steinberg repn" in fancier scenarios). $\endgroup$
    – paul garrett
    Commented Sep 28, 2020 at 22:04
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    $\begingroup$ For polynomial invariants, because it's $3$-transitive, the only degree $3$ invariant polynomials are the symmetric polynomials, of which there are $3$. But starting in degree $4$ there are way more $\endgroup$
    – Will Sawin
    Commented Sep 29, 2020 at 1:50

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