Is every 3-dim self-dual Galois representation a symmetic square of 2-dim representation?

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Basically, my question as in the title. Here the Galois representation I consider is an $ell$-adic Galois representation (comes from geometry). And by the word "self-dual" I mean that representation is isomorphic to its dual representation (transpose inverse in language of matrix) up to a Tate twist.



If the answer is positive, please let me know the idea or reference. If the answer is negative, I am wondering if there is any extra condition(s) to make my statement to be true?



Thanks






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    Basically, my question as in the title. Here the Galois representation I consider is an $ell$-adic Galois representation (comes from geometry). And by the word "self-dual" I mean that representation is isomorphic to its dual representation (transpose inverse in language of matrix) up to a Tate twist.



    If the answer is positive, please let me know the idea or reference. If the answer is negative, I am wondering if there is any extra condition(s) to make my statement to be true?



    Thanks






    nt.number-theory galois-representations







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      up vote
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      down vote

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      up vote
      3
      down vote

      favorite











      Basically, my question as in the title. Here the Galois representation I consider is an $ell$-adic Galois representation (comes from geometry). And by the word "self-dual" I mean that representation is isomorphic to its dual representation (transpose inverse in language of matrix) up to a Tate twist.



      If the answer is positive, please let me know the idea or reference. If the answer is negative, I am wondering if there is any extra condition(s) to make my statement to be true?



      Thanks






      nt.number-theory galois-representations







      share|cite|improve this question







      New contributor




      Leo D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Basically, my question as in the title. Here the Galois representation I consider is an $ell$-adic Galois representation (comes from geometry). And by the word "self-dual" I mean that representation is isomorphic to its dual representation (transpose inverse in language of matrix) up to a Tate twist.



      If the answer is positive, please let me know the idea or reference. If the answer is negative, I am wondering if there is any extra condition(s) to make my statement to be true?



      Thanks






      nt.number-theory galois-representations




      nt.number-theory galois-representations






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          I'm not sure how to define "comes from geometry" precisely, but how about the following counterexample: Let $K$ be a quadratic imaginary extension of $mathbbQ$. Let $V$ be the three dimensional representation where the nontrivial element of $mathrmGal(K/mathbbQ)$ acts by $mathrmdiag(-1,1,-1)$. We claim that $V$ cannot be $mathrmSym(W)$ for a two dimensional Galois representation $W$.



          Let $sigma$ denote complex conjugation in $mathrmGal(overlinemathbbQ/mathbbQ)$. Then $sigma$ acts on $V$ with eigenvalues $(-1,1,-1)$, so it would have to act on $W$ with eigenvalues $pm i$. But $sigma$ has order $2$ in $mathrmGal(overlinemathbbQ/mathbbQ)$, so it cannot act with eigenvalues other than $pm 1$.






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            I'm not sure how to define "comes from geometry" precisely, but how about the following counterexample: Let $K$ be a quadratic imaginary extension of $mathbbQ$. Let $V$ be the three dimensional representation where the nontrivial element of $mathrmGal(K/mathbbQ)$ acts by $mathrmdiag(-1,1,-1)$. We claim that $V$ cannot be $mathrmSym(W)$ for a two dimensional Galois representation $W$.



            Let $sigma$ denote complex conjugation in $mathrmGal(overlinemathbbQ/mathbbQ)$. Then $sigma$ acts on $V$ with eigenvalues $(-1,1,-1)$, so it would have to act on $W$ with eigenvalues $pm i$. But $sigma$ has order $2$ in $mathrmGal(overlinemathbbQ/mathbbQ)$, so it cannot act with eigenvalues other than $pm 1$.






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              I'm not sure how to define "comes from geometry" precisely, but how about the following counterexample: Let $K$ be a quadratic imaginary extension of $mathbbQ$. Let $V$ be the three dimensional representation where the nontrivial element of $mathrmGal(K/mathbbQ)$ acts by $mathrmdiag(-1,1,-1)$. We claim that $V$ cannot be $mathrmSym(W)$ for a two dimensional Galois representation $W$.



              Let $sigma$ denote complex conjugation in $mathrmGal(overlinemathbbQ/mathbbQ)$. Then $sigma$ acts on $V$ with eigenvalues $(-1,1,-1)$, so it would have to act on $W$ with eigenvalues $pm i$. But $sigma$ has order $2$ in $mathrmGal(overlinemathbbQ/mathbbQ)$, so it cannot act with eigenvalues other than $pm 1$.






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                up vote
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                I'm not sure how to define "comes from geometry" precisely, but how about the following counterexample: Let $K$ be a quadratic imaginary extension of $mathbbQ$. Let $V$ be the three dimensional representation where the nontrivial element of $mathrmGal(K/mathbbQ)$ acts by $mathrmdiag(-1,1,-1)$. We claim that $V$ cannot be $mathrmSym(W)$ for a two dimensional Galois representation $W$.



                Let $sigma$ denote complex conjugation in $mathrmGal(overlinemathbbQ/mathbbQ)$. Then $sigma$ acts on $V$ with eigenvalues $(-1,1,-1)$, so it would have to act on $W$ with eigenvalues $pm i$. But $sigma$ has order $2$ in $mathrmGal(overlinemathbbQ/mathbbQ)$, so it cannot act with eigenvalues other than $pm 1$.






                share|cite|improve this answer












                I'm not sure how to define "comes from geometry" precisely, but how about the following counterexample: Let $K$ be a quadratic imaginary extension of $mathbbQ$. Let $V$ be the three dimensional representation where the nontrivial element of $mathrmGal(K/mathbbQ)$ acts by $mathrmdiag(-1,1,-1)$. We claim that $V$ cannot be $mathrmSym(W)$ for a two dimensional Galois representation $W$.



                Let $sigma$ denote complex conjugation in $mathrmGal(overlinemathbbQ/mathbbQ)$. Then $sigma$ acts on $V$ with eigenvalues $(-1,1,-1)$, so it would have to act on $W$ with eigenvalues $pm i$. But $sigma$ has order $2$ in $mathrmGal(overlinemathbbQ/mathbbQ)$, so it cannot act with eigenvalues other than $pm 1$.







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                answered 3 hours ago









                David E Speyer

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