Finitely generated matrix groups whose eigenvalues are all algebraic
$begingroup$
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
asked Apr 1 at 19:06
EmilyEmily
783
New contributor
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add a comment |
$begingroup$
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
asked Apr 1 at 19:06
EmilyEmily
783
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.
$endgroup$
add a comment |
$begingroup$
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
asked Apr 1 at 19:06
EmilyEmily
783
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
asked Apr 1 at 19:06
EmilyEmily
783
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.
asked Apr 1 at 19:06
EmilyEmily
783
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.
asked Apr 1 at 19:06
EmilyEmily
783
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.
asked Apr 1 at 19:06
EmilyEmily
783
asked Apr 1 at 19:06
EmilyEmily
783
783
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.
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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add a comment |
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3 Answers
3
active
oldest
votes
$begingroup$
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.
answered Apr 1 at 22:00
YCorYCor
29k485140
$endgroup$
add a comment |
$begingroup$
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).
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
$endgroup$
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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Apr 1 at 22:00
YCorYCor
29k485140
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 1 at 22:00
YCorYCor
29k485140
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 1 at 22:00
YCorYCor
29k485140
$endgroup$
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.
answered Apr 1 at 22:00
YCorYCor
29k485140
answered Apr 1 at 22:00
YCorYCor
29k485140
answered Apr 1 at 22:00
YCorYCor
29k485140
answered Apr 1 at 22:00
YCorYCor
29k485140
29k485140
add a comment |
add a comment |
$begingroup$
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).
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
$endgroup$
add a comment |
$begingroup$
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).
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
$endgroup$
add a comment |
$begingroup$
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).
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
$endgroup$
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).
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
answered Apr 1 at 19:23
Robert IsraelRobert Israel
43.5k53123
43.5k53123
add a comment |
add a comment |
$begingroup$
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.
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
$endgroup$
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
add a comment |
$begingroup$
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.
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
$endgroup$
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
add a comment |
$begingroup$
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.
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
$endgroup$
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.
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
edited Apr 2 at 14:59
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
answered Apr 2 at 13:38
Geoff RobinsonGeoff Robinson
30k282111
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
add a comment |
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
add a comment |
Emily is a new contributor. Be nice, and check out our Code of Conduct.
Emily is a new contributor. Be nice, and check out our Code of Conduct.
Emily is a new contributor. Be nice, and check out our Code of Conduct.
Emily is a new contributor. Be nice, and check out our Code of Conduct.
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