Finitely generated matrix groups whose eigenvalues are all algebraic












7












$begingroup$


Let $G$ be a finitely generated subgroup of $GL(n,mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbb{Q}$) such that for all $g in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$.



Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?






gr.group-theory algebraic-groups algebraic-number-theory







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    $begingroup$


    Let $G$ be a finitely generated subgroup of $GL(n,mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbb{Q}$) such that for all $g in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$.



    Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?






    gr.group-theory algebraic-groups algebraic-number-theory







    share|cite|improve this question







    New contributor




    Emily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      7












      7








      7





      $begingroup$


      Let $G$ be a finitely generated subgroup of $GL(n,mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbb{Q}$) such that for all $g in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$.



      Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?






      gr.group-theory algebraic-groups algebraic-number-theory







      share|cite|improve this question







      New contributor




      Emily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $G$ be a finitely generated subgroup of $GL(n,mathbb{C})$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbb{Q}$) such that for all $g in G$, the eigenvalues of $g$ all lie in $k$. This implies that $g$ is conjugate to an element of $GL(n,k)$.



      Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$? Or at least $GL(n,k')$ for some finite extension $k'$ of $k$? If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?






      gr.group-theory algebraic-groups algebraic-number-theory




      gr.group-theory algebraic-groups algebraic-number-theory






      share|cite|improve this question







      New contributor




      Emily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











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      asked 7 hours ago









      EmilyEmily

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          2 Answers
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          $begingroup$

          Here's one easy example. Let $G$ be generated by $pmatrix{1 & xcr 0 & 1}$
          for $x$ in some finite set $X$ of complex
          numbers. All eigenvalues are $1$, so we can take $k = mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of
          $GL(2,mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).






          share|cite|improve this answer









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            4












            $begingroup$

            At the positive side, if $G$ acts irreducibly on $mathbf{C}^n$ and $k$ is an arbitrary subfield of $mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(mathbf{C})$ (keeping the irreducibility assumption).



            See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)



            Robert Israel's simple example shows that some assumption such as irreducibility has to be done.






            share|cite|improve this answer









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              2 Answers
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              2 Answers
              2






              active

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              active

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              12












              $begingroup$

              Here's one easy example. Let $G$ be generated by $pmatrix{1 & xcr 0 & 1}$
              for $x$ in some finite set $X$ of complex
              numbers. All eigenvalues are $1$, so we can take $k = mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of
              $GL(2,mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).






              share|cite|improve this answer









              $endgroup$


















                12












                $begingroup$

                Here's one easy example. Let $G$ be generated by $pmatrix{1 & xcr 0 & 1}$
                for $x$ in some finite set $X$ of complex
                numbers. All eigenvalues are $1$, so we can take $k = mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of
                $GL(2,mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).






                share|cite|improve this answer









                $endgroup$
















                  12












                  12








                  12





                  $begingroup$

                  Here's one easy example. Let $G$ be generated by $pmatrix{1 & xcr 0 & 1}$
                  for $x$ in some finite set $X$ of complex
                  numbers. All eigenvalues are $1$, so we can take $k = mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of
                  $GL(2,mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).






                  share|cite|improve this answer









                  $endgroup$



                  Here's one easy example. Let $G$ be generated by $pmatrix{1 & xcr 0 & 1}$
                  for $x$ in some finite set $X$ of complex
                  numbers. All eigenvalues are $1$, so we can take $k = mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of
                  $GL(2,mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  Robert IsraelRobert Israel

                  43.3k52123




                  43.3k52123























                      4












                      $begingroup$

                      At the positive side, if $G$ acts irreducibly on $mathbf{C}^n$ and $k$ is an arbitrary subfield of $mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(mathbf{C})$ (keeping the irreducibility assumption).



                      See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)



                      Robert Israel's simple example shows that some assumption such as irreducibility has to be done.






                      share|cite|improve this answer









                      $endgroup$


















                        4












                        $begingroup$

                        At the positive side, if $G$ acts irreducibly on $mathbf{C}^n$ and $k$ is an arbitrary subfield of $mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(mathbf{C})$ (keeping the irreducibility assumption).



                        See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)



                        Robert Israel's simple example shows that some assumption such as irreducibility has to be done.






                        share|cite|improve this answer









                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          At the positive side, if $G$ acts irreducibly on $mathbf{C}^n$ and $k$ is an arbitrary subfield of $mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(mathbf{C})$ (keeping the irreducibility assumption).



                          See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)



                          Robert Israel's simple example shows that some assumption such as irreducibility has to be done.






                          share|cite|improve this answer









                          $endgroup$



                          At the positive side, if $G$ acts irreducibly on $mathbf{C}^n$ and $k$ is an arbitrary subfield of $mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(mathbf{C})$ (keeping the irreducibility assumption).



                          See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)



                          Robert Israel's simple example shows that some assumption such as irreducibility has to be done.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 4 hours ago









                          YCorYCor

                          28.7k484139




                          28.7k484139






















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