Finitely generated matrix groups whose eigenvalues are all algebraic Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is a normal subgroup of a finitely presented group finitely generated or normal finitely genrated?Finitely generated, infinite, residually finite groups whose finite quotients are $p$-groups.Are affine groups over rings of integers finitely generated?Finitely generated Galois groupsAre all free factors of finitely generated subgroups of free groups geometric?Is $SL_1(D)$ toplogically finitely generated, for $D$ a division algebra over a local field?Is a finitely generated residually free group “almost LERF”?Are all finitely generated subgroups of SL2 LERF?HNN extension group with finitely generated baseNon-residually-finite finitely-presented sofic group with all finitely generated subgroups Hopfian

Finitely generated matrix groups whose eigenvalues are all algebraic



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is a normal subgroup of a finitely presented group finitely generated or normal finitely genrated?Finitely generated, infinite, residually finite groups whose finite quotients are $p$-groups.Are affine groups over rings of integers finitely generated?Finitely generated Galois groupsAre all free factors of finitely generated subgroups of free groups geometric?Is $SL_1(D)$ toplogically finitely generated, for $D$ a division algebra over a local field?Is a finitely generated residually free group “almost LERF”?Are all finitely generated subgroups of SL2 LERF?HNN extension group with finitely generated baseNon-residually-finite finitely-presented sofic group with all finitely generated subgroups Hopfian










15












$begingroup$


Let $G$ be a finitely generated subgroup of $GL(n,mathbbC)$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbbQ$) 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?










share|cite|improve this question









$endgroup$
















    15












    $begingroup$


    Let $G$ be a finitely generated subgroup of $GL(n,mathbbC)$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbbQ$) 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?










    share|cite|improve this question









    $endgroup$














      15












      15








      15


      1



      $begingroup$


      Let $G$ be a finitely generated subgroup of $GL(n,mathbbC)$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbbQ$) 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?










      share|cite|improve this question









      $endgroup$




      Let $G$ be a finitely generated subgroup of $GL(n,mathbbC)$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbbQ$) 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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 1 at 19:06









      EmilyEmily

      783




      783




















          3 Answers
          3






          active

          oldest

          votes


















          20












          $begingroup$

          At the positive side, if $G$ acts irreducibly on $mathbfC^n$ and $k$ is an arbitrary subfield of $mathbfC$, 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 $M_n(mathbfC)$ (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 $pmatrix1 & xcr 0 & 1$
            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 $rm GL(n,mathbbC)$ 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 $rm GL(n,mathbbC)$ is conjugate within $rm GL(n,mathbbC)$ to a subgroup of $rm GL(n,mathbbQ[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






              share|cite|improve this answer











              $endgroup$








              • 2




                $begingroup$
                $k$ cyclotomic (including $k=mathbfQ$) 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 $mathbfQ$ 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











              Your Answer








              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "504"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326907%2ffinitely-generated-matrix-groups-whose-eigenvalues-are-all-algebraic%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              20












              $begingroup$

              At the positive side, if $G$ acts irreducibly on $mathbfC^n$ and $k$ is an arbitrary subfield of $mathbfC$, 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 $M_n(mathbfC)$ (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 $mathbfC^n$ and $k$ is an arbitrary subfield of $mathbfC$, 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 $M_n(mathbfC)$ (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 $mathbfC^n$ and $k$ is an arbitrary subfield of $mathbfC$, 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 $M_n(mathbfC)$ (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 $mathbfC^n$ and $k$ is an arbitrary subfield of $mathbfC$, 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 $M_n(mathbfC)$ (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

                  29.1k486141




                  29.1k486141





















                      25












                      $begingroup$

                      Here's one easy example. Let $G$ be generated by $pmatrix1 & xcr 0 & 1$
                      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 $pmatrix1 & xcr 0 & 1$
                        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 $pmatrix1 & xcr 0 & 1$
                          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 $pmatrix1 & xcr 0 & 1$
                          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 $rm GL(n,mathbbC)$ 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 $rm GL(n,mathbbC)$ is conjugate within $rm GL(n,mathbbC)$ to a subgroup of $rm GL(n,mathbbQ[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






                              share|cite|improve this answer











                              $endgroup$








                              • 2




                                $begingroup$
                                $k$ cyclotomic (including $k=mathbfQ$) 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 $mathbfQ$ 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 $rm GL(n,mathbbC)$ 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 $rm GL(n,mathbbC)$ is conjugate within $rm GL(n,mathbbC)$ to a subgroup of $rm GL(n,mathbbQ[omega]),$ where $omega$ is a primitive complex $|G|$-th root of unity.






                              share|cite|improve this answer











                              $endgroup$








                              • 2




                                $begingroup$
                                $k$ cyclotomic (including $k=mathbfQ$) 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 $mathbfQ$ 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 $rm GL(n,mathbbC)$ 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 $rm GL(n,mathbbC)$ is conjugate within $rm GL(n,mathbbC)$ to a subgroup of $rm GL(n,mathbbQ[omega]),$ 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 $rm GL(n,mathbbC)$ 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 $rm GL(n,mathbbC)$ is conjugate within $rm GL(n,mathbbC)$ to a subgroup of $rm GL(n,mathbbQ[omega]),$ 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

                              30k284112




                              30k284112







                              • 2




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




                              $begingroup$
                              $k$ cyclotomic (including $k=mathbfQ$) 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 $mathbfQ$ 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 $mathbfQ$ 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

















                              draft saved

                              draft discarded
















































                              Thanks for contributing an answer to MathOverflow!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid


                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326907%2ffinitely-generated-matrix-groups-whose-eigenvalues-are-all-algebraic%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Adding axes to figuresAdding axes labels to LaTeX figuresLaTeX equivalent of ConTeXt buffersRotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizeNumerical conditional within tikz keys?adding axes to shapesAlign axes across subfiguresAdding figures with a certain orderLine up nested tikz enviroments or how to get rid of themAdding axes labels to LaTeX figures

                              Luettelo Yhdysvaltain laivaston lentotukialuksista Lähteet | Navigointivalikko

                              Gary (muusikko) Sisällysluettelo Historia | Rockin' High | Lähteet | Aiheesta muualla | NavigointivalikkoInfobox OKTuomas "Gary" Keskinen Ancaran kitaristiksiProjekti Rockin' High