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Let $K$ be a number field and $(\rho,V)$, $(\rho',V')$ be two Galois representations of $\mathrm{Gal}(\overline{\mathbb{Q}}/K)$. Suppose that for some positive integer $n$ we have $\mathrm{Sym}^n\rho\cong\mathrm{Sym}^n\rho'$. Does this imply the existence of a finite algebraic extension $L/K$ such that $\rho_{|\mathrm{Gal}(\overline{\mathbb{Q}}/L)}\cong\rho'_{|\mathrm{Gal}(\overline{\mathbb{Q}}/L)}$?

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  • $\begingroup$ Is the converse true? $\endgroup$ – Asvin Dec 25 '18 at 23:50
  • $\begingroup$ @Asvin, are you able to prove the direct sense? $\endgroup$ – User1930752648 Dec 26 '18 at 0:02
  • $\begingroup$ No! I was just curious as to why you expect it to be true. $\endgroup$ – Asvin Dec 26 '18 at 0:03
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    $\begingroup$ Is it ok to assume the reps are semisimple? $\endgroup$ – Will Sawin Dec 26 '18 at 2:43
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    $\begingroup$ @Asvin the converse is obviously false, as one can see for instance with two-dimensional irreducible representations factoring through $S_3$, all of which match after finite covers but none of which match after symmetric powers. $\endgroup$ – Will Sawin Dec 26 '18 at 17:54
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Assume $\rho$ and $\rho’$ are semi simple. Then the monodromy group of $\rho + \rho’$ is reductive. It suffices to answer the question in the category of representations of this reductive group. By passing to a finite index subgroup, we may assume that it is connected. For connected reductive groups, two representations are equivalent if and only if they are equivalent as representations of the maximal torus. So it suffices to prove, if the $\operatorname{Sym}^n$s of two reps of a torus are equivalent then they are equivalent. In other words, given two finite multisets of lattice points, if the sums of $n$ elements sampled without replacement from each are equal then they are equal. This can be proved inductively by totally ordering the lattice and checking that the largest elements are equal (division by $n$), then the next largest (subtract the second largest element of the set of sums from the largest) then the third largest (remove all sums of the first two and look at the largest remaining) and so on.

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  • $\begingroup$ Thanks a lot for your answer. $\endgroup$ – User1930752648 Dec 26 '18 at 21:18

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