Finitely generated matrix groups whose eigenvalues are all algebraic The Next CEO of Stack OverflowIs 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
The Next CEO of Stack OverflowIs 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
$begingroup$
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?
gr.group-theory algebraic-groups algebraic-number-theory
asked 2 hours ago
EmilyEmily
211
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$begingroup$
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?
gr.group-theory algebraic-groups algebraic-number-theory
asked 2 hours ago
EmilyEmily
211
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 $GL(n,mathbbC)$. 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
asked 2 hours ago
EmilyEmily
211
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 $GL(n,mathbbC)$. 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
gr.group-theory algebraic-groups algebraic-number-theory
asked 2 hours ago
EmilyEmily
211
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 2 hours ago
EmilyEmily
211
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 2 hours ago
EmilyEmily
211
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Emily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago
EmilyEmily
211
asked 2 hours ago
EmilyEmily
211
211
New contributor
Emily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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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1 Answer
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$begingroup$
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).
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
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1 Answer
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1 Answer
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$begingroup$
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).
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
$endgroup$
add a comment |
$begingroup$
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).
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
$endgroup$
add a comment |
$begingroup$
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).
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
$endgroup$
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).
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
answered 2 hours ago
Robert IsraelRobert Israel
43.2k52122
43.2k52122
add a comment |
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.
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