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 7 hours ago
EmilyEmily
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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 7 hours ago
EmilyEmily
361
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 7 hours ago
EmilyEmily
361
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 7 hours ago
EmilyEmily
361
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 7 hours ago
EmilyEmily
361
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 7 hours ago
EmilyEmily
361
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 7 hours ago
EmilyEmily
361
asked 7 hours ago
EmilyEmily
361
361
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.
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2 Answers
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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 7 hours ago
Robert IsraelRobert Israel
43.3k52123
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$begingroup$
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 4 hours ago
YCorYCor
28.7k484139
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2 Answers
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oldest
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Here's one easy example. Let $G$ be generated by $pmatrix{1 & xcr 0 & 1}$
for $x$ in some finite set $X$ of complex
numbers. All eigenvalues are $1$, so we can take $k = mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of
$GL(2,mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).
answered 7 hours ago
Robert IsraelRobert Israel
43.3k52123
$endgroup$
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 7 hours ago
Robert IsraelRobert Israel
43.3k52123
$endgroup$
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 7 hours ago
Robert IsraelRobert Israel
43.3k52123
$endgroup$
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 7 hours ago
Robert IsraelRobert Israel
43.3k52123
answered 7 hours ago
Robert IsraelRobert Israel
43.3k52123
answered 7 hours ago
Robert IsraelRobert Israel
43.3k52123
answered 7 hours ago
Robert IsraelRobert Israel
43.3k52123
43.3k52123
add a comment |
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 4 hours ago
YCorYCor
28.7k484139
$endgroup$
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 4 hours ago
YCorYCor
28.7k484139
$endgroup$
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 4 hours ago
YCorYCor
28.7k484139
$endgroup$
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 4 hours ago
YCorYCor
28.7k484139
answered 4 hours ago
YCorYCor
28.7k484139
answered 4 hours ago
YCorYCor
28.7k484139
answered 4 hours ago
YCorYCor
28.7k484139
28.7k484139
add a comment |
add a comment |
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