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Francisco García Cortés
My email is: (underlined parts of my name)3 at alum.us.es
Poster presentation at Ramification in geometric Langlands and non-abelian Hodge theory
Here is the list of families of pairs $(d,e)$ potentially giving finite monodromy for $\mathcal{F}_{x^d(x-1)^e}$, that is, such that $d$, $e$ and $d+e$ are FM-exponents.
Theorem. Let $d,e$ be positive integers such that $d$ is prime to $p$ and all of $d$, $e$ and $d+e$ are FM-exponents. Then $(d,e)$ is one of the following pairs (or their reversed pairs):
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$(\frac{p^a+1}{p^b+1},p^b\frac{p^a+1}{p^b+1})$ for $a\geq1$ an odd multiple of $b$.
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$(\frac{p^{a+2b}+1}{p^b+1},p^b\frac{p^a+1}{p^b+1})$ for $a\geq1$ an odd multiple of $b$.
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$p\geq3$, $(\frac{p^a+1}{2},\frac{p^a+1}{2})$ for $a\geq0$.
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$p=2$, $(3,10)$, $(5,6)$, $(5,8)$, $(5,17)$, $(5,52)$, $(9,2)$, $(9,4)$, $(9,11)$, $(9,13)$, $(9,17)$, $(9,34)$, $(9,43)$, $(9,48)$, $(11,2)$, $(11,13)$, $(11,57)$, $(13,44)$, $(13,228)$, $(17,26)$, $(17,40)$, $(33,10)$, $(33,24)$, $(33,172)$, $(33,208)$, $(65,176)$, $(171,34)$ or $(205,36)$.
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$p=2$, $(2^a+1,2^a+1)$ for $a\geq1$.
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$p=2$, $(\frac{2^a+1}{2^b+1},\frac{2^a+1}{2^b+1})$ for $b\geq1$, $a$ an odd multiple of $b$.
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$p=2$, $(1,2^a+1)$ for $a\geq1$.
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$p=2$, $(2^a+1,2^a)$ for $a\geq1$.
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$p=2$, $(3,2^a+1)$ for $a\geq1$.
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$p=2$, $(2^a+1,3\cdot2^a)$ for $a\geq1$.
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$p=2$, $(2^a+1,\frac{2^{3a}+1}{2^a+1})$ for $a\geq1$.
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$p=2$, $(\frac{2^{3a}+1}{2^a+1},2^a(2^a+1))$ for $a\geq1$.
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$p=2$, $(2^a+1,\frac{2^{a}+1}{3})$ for odd $a\geq1$.
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$p=2$, $(1,\frac{2^{a}+1}{3})$ for odd $a\geq1$.
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$p=2$, $(\frac{2^{a}+1}{3},2^a)$ for odd $a\geq1$.
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$p=2$, $(3,\frac{2^{2a}+1}{5})$ for odd $a\geq1$.
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$p=2$, $(\frac{2^{2a}+1}{5},2^{2a}\cdot3)$ for odd $a\geq1$.
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$p=2$, $(5,\frac{2^{a}+1}{3})$ for odd $a\geq1$.
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$p=2$, $(\frac{2^{a}+1}{3},5\cdot2^a)$ for odd $a\geq1$.
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$p=2$, $(\frac{2^{3a}+1}{9},\frac{2^{3a}+1}{3})$ for odd $a\geq1$.
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$p=2$, $(\frac{2^{3a}+1}{9},2\frac{2^{3a}+1}{9})$ for odd $a\geq1$.
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$p=3$, $(2,5)$, $(4,3)$, $(4,10)$, $(5,7)$, $(10,63)$, $(28,45)$ or $(61,12)$.
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$p=3$, $(2,3^a+1)$ for $a\geq0$.
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$p=3$, $(3^a+1,2\cdot3^a)$ for $a\geq0$.
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$p=3$, $(3^a+1,\frac{3^a+1}{2})$ for $a\geq0$.
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$p=3$, $(1,\frac{3^a+1}{2})$ for $a\geq0$.
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$p=3$, $(\frac{3^a+1}{2},3^a)$ for $a\geq0$.
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$p=3$, $(4,\frac{3^a+1}{2})$ for $a\geq0$.
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$p=3$, $(\frac{3^a+1}{2},4\cdot3^a)$ for $a\geq0$.
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$p=3$, $(\frac{3^a+1}{2},\frac{3^a+1}{4})$ for odd $a\geq1$.
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$p=3$, $(\frac{3^a+1}{4},\frac{3^a+1}{4})$ for odd $a\geq1$.
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$p=3$, $(2,\frac{3^a+1}{4})$ for odd $a\geq1$.
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$p=3$, $(\frac{3^a+1}{4},2\cdot3^a)$ for odd $a\geq1$.
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$p=5$, $(1,6)$, $(2,5)$, $(3,7)$, $(6,7)$ or $(6,15)$.
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$p=5$, $(2,\frac{5^a+1}{2})$ for $a\geq0$.
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$p=5$, $(\frac{5^a+1}{2},2\cdot5^a)$ for $a\geq0$.
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$p=7$, $(2,2)$.
Proof.
This is Theorem 5 in the paper.