Finitely generated matrix groups whose eigenvalues are all algebraic












15












$begingroup$


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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    15












    $begingroup$


    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.







    $endgroup$















      15












      15








      15


      1



      $begingroup$


      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.











      share|cite|improve this question







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      asked Apr 1 at 19:06









      EmilyEmily

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      New contributor





      Emily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          3 Answers
          3






          active

          oldest

          votes


















          20












          $begingroup$

          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









          $endgroup$





















            25












            $begingroup$

            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









            $endgroup$





















              2












              $begingroup$

              Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${rm GL}(n,mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.



              Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.



              In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${rm GL}(n,mathbb{C})$ is conjugate within ${rm GL}(n,mathbb{C})$ to a subgroup of ${rm GL}(n,mathbb{Q}[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






              share|cite|improve this answer











              $endgroup$









              • 2




                $begingroup$
                $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                $endgroup$
                – YCor
                Apr 2 at 14:22






              • 2




                $begingroup$
                and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                $endgroup$
                – YCor
                Apr 2 at 14:26










              • $begingroup$
                @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                $endgroup$
                – Geoff Robinson
                Apr 2 at 14:42










              • $begingroup$
                If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                $endgroup$
                – YCor
                Apr 2 at 15:03










              • $begingroup$
                That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                $endgroup$
                – Geoff Robinson
                Apr 2 at 16:25












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              3 Answers
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              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              20












              $begingroup$

              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









              $endgroup$


















                20












                $begingroup$

                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









                $endgroup$
















                  20












                  20








                  20





                  $begingroup$

                  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 Apr 1 at 22:00









                  YCorYCor

                  29k485140




                  29k485140























                      25












                      $begingroup$

                      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









                      $endgroup$


















                        25












                        $begingroup$

                        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









                        $endgroup$
















                          25












                          25








                          25





                          $begingroup$

                          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 Apr 1 at 19:23









                          Robert IsraelRobert Israel

                          43.5k53123




                          43.5k53123























                              2












                              $begingroup$

                              Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${rm GL}(n,mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.



                              Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.



                              In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${rm GL}(n,mathbb{C})$ is conjugate within ${rm GL}(n,mathbb{C})$ to a subgroup of ${rm GL}(n,mathbb{Q}[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






                              share|cite|improve this answer











                              $endgroup$









                              • 2




                                $begingroup$
                                $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                                $endgroup$
                                – YCor
                                Apr 2 at 14:22






                              • 2




                                $begingroup$
                                and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                                $endgroup$
                                – YCor
                                Apr 2 at 14:26










                              • $begingroup$
                                @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 14:42










                              • $begingroup$
                                If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                                $endgroup$
                                – YCor
                                Apr 2 at 15:03










                              • $begingroup$
                                That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 16:25
















                              2












                              $begingroup$

                              Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${rm GL}(n,mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.



                              Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.



                              In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${rm GL}(n,mathbb{C})$ is conjugate within ${rm GL}(n,mathbb{C})$ to a subgroup of ${rm GL}(n,mathbb{Q}[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






                              share|cite|improve this answer











                              $endgroup$









                              • 2




                                $begingroup$
                                $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                                $endgroup$
                                – YCor
                                Apr 2 at 14:22






                              • 2




                                $begingroup$
                                and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                                $endgroup$
                                – YCor
                                Apr 2 at 14:26










                              • $begingroup$
                                @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 14:42










                              • $begingroup$
                                If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                                $endgroup$
                                – YCor
                                Apr 2 at 15:03










                              • $begingroup$
                                That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 16:25














                              2












                              2








                              2





                              $begingroup$

                              Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${rm GL}(n,mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.



                              Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.



                              In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${rm GL}(n,mathbb{C})$ is conjugate within ${rm GL}(n,mathbb{C})$ to a subgroup of ${rm GL}(n,mathbb{Q}[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






                              share|cite|improve this answer











                              $endgroup$



                              Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${rm GL}(n,mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.



                              Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.



                              In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${rm GL}(n,mathbb{C})$ is conjugate within ${rm GL}(n,mathbb{C})$ to a subgroup of ${rm GL}(n,mathbb{Q}[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Apr 2 at 14:59

























                              answered Apr 2 at 13:38









                              Geoff RobinsonGeoff Robinson

                              30k282111




                              30k282111








                              • 2




                                $begingroup$
                                $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                                $endgroup$
                                – YCor
                                Apr 2 at 14:22






                              • 2




                                $begingroup$
                                and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                                $endgroup$
                                – YCor
                                Apr 2 at 14:26










                              • $begingroup$
                                @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 14:42










                              • $begingroup$
                                If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                                $endgroup$
                                – YCor
                                Apr 2 at 15:03










                              • $begingroup$
                                That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 16:25














                              • 2




                                $begingroup$
                                $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                                $endgroup$
                                – YCor
                                Apr 2 at 14:22






                              • 2




                                $begingroup$
                                and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                                $endgroup$
                                – YCor
                                Apr 2 at 14:26










                              • $begingroup$
                                @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 14:42










                              • $begingroup$
                                If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                                $endgroup$
                                – YCor
                                Apr 2 at 15:03










                              • $begingroup$
                                That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                                $endgroup$
                                – Geoff Robinson
                                Apr 2 at 16:25








                              2




                              2




                              $begingroup$
                              $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                              $endgroup$
                              – YCor
                              Apr 2 at 14:22




                              $begingroup$
                              $k$ cyclotomic (including $k=mathbf{Q}$) doesn't mean that eigenvalues have finite order...
                              $endgroup$
                              – YCor
                              Apr 2 at 14:22




                              2




                              2




                              $begingroup$
                              and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                              $endgroup$
                              – YCor
                              Apr 2 at 14:26




                              $begingroup$
                              and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element.
                              $endgroup$
                              – YCor
                              Apr 2 at 14:26












                              $begingroup$
                              @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                              $endgroup$
                              – Geoff Robinson
                              Apr 2 at 14:42




                              $begingroup$
                              @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either.
                              $endgroup$
                              – Geoff Robinson
                              Apr 2 at 14:42












                              $begingroup$
                              If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                              $endgroup$
                              – YCor
                              Apr 2 at 15:03




                              $begingroup$
                              If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics).
                              $endgroup$
                              – YCor
                              Apr 2 at 15:03












                              $begingroup$
                              That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                              $endgroup$
                              – Geoff Robinson
                              Apr 2 at 16:25




                              $begingroup$
                              That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite.
                              $endgroup$
                              – Geoff Robinson
                              Apr 2 at 16:25










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