What is the defining property of reductive groups and why are they important?
$begingroup$
Having read (skimmed more like) many surveys of the Langlands Program and similar, it seems the related ideas apply exclusively to groups that are "reductive".
But nowhere, either in these surveys or elsewhere, have I been able to find a simple and compelling definition of what it means for a group to be reductive and why this property is important and why Langlands was naturally led to frame his conjectures for them and not, say, for any group.
Any suggestions or links?
gr.group-theory algebraic-groups reductive-groups
edited 1 hour ago
Martin Sleziak
2,96532028
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
$endgroup$
add a comment |
$begingroup$
Having read (skimmed more like) many surveys of the Langlands Program and similar, it seems the related ideas apply exclusively to groups that are "reductive".
But nowhere, either in these surveys or elsewhere, have I been able to find a simple and compelling definition of what it means for a group to be reductive and why this property is important and why Langlands was naturally led to frame his conjectures for them and not, say, for any group.
Any suggestions or links?
gr.group-theory algebraic-groups reductive-groups
edited 1 hour ago
Martin Sleziak
2,96532028
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
$endgroup$
add a comment |
$begingroup$
Having read (skimmed more like) many surveys of the Langlands Program and similar, it seems the related ideas apply exclusively to groups that are "reductive".
But nowhere, either in these surveys or elsewhere, have I been able to find a simple and compelling definition of what it means for a group to be reductive and why this property is important and why Langlands was naturally led to frame his conjectures for them and not, say, for any group.
Any suggestions or links?
gr.group-theory algebraic-groups reductive-groups
edited 1 hour ago
Martin Sleziak
2,96532028
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
$endgroup$
Having read (skimmed more like) many surveys of the Langlands Program and similar, it seems the related ideas apply exclusively to groups that are "reductive".
But nowhere, either in these surveys or elsewhere, have I been able to find a simple and compelling definition of what it means for a group to be reductive and why this property is important and why Langlands was naturally led to frame his conjectures for them and not, say, for any group.
Any suggestions or links?
gr.group-theory algebraic-groups reductive-groups
gr.group-theory algebraic-groups reductive-groups
edited 1 hour ago
Martin Sleziak
2,96532028
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
edited 1 hour ago
Martin Sleziak
2,96532028
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
edited 1 hour ago
Martin Sleziak
2,96532028
edited 1 hour ago
Martin Sleziak
2,96532028
edited 1 hour ago
Martin Sleziak
2,96532028
2,96532028
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
asked 3 hours ago
John R RamsdenJohn R Ramsden
881515
881515
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if its unipotent radical is trivial, or equivalently if it has no connected unipotent abelian subgroup except ${I}$.
See Armand Borel, Linear Algebraic Groups, p. 158.
But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group.
A motivating example: the matrices $begin{pmatrix}1&x\0&1end{pmatrix}$ form a group of matrices acting on the plane, preserving the vertical axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no subgroups like this.
Another: the group $GL(2)$ of all $2 times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W subset GL(2)$ with $W cap U(2) ne emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
$endgroup$
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
add a comment |
$begingroup$
For algebraic groups over $mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:
For an algebraic subgroup $G$ of $mathrm{GL}_n(mathbb{C})$, the following conditions are equivalent:
$G$ is reductive.
$G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.
$G$ is conjugate to a subgroup of $mathrm{GL}_n(mathbb{C})$ which is self-adjoint, that is stable under $g mapsto g^* := overline{g}^T$.
The basic examples to have in mind: $mathrm{GL}_n(mathbb{C})$ and $mathrm{SL}_n(mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n geq 2$).
As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
$endgroup$
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if its unipotent radical is trivial, or equivalently if it has no connected unipotent abelian subgroup except ${I}$.
See Armand Borel, Linear Algebraic Groups, p. 158.
But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group.
A motivating example: the matrices $begin{pmatrix}1&x\0&1end{pmatrix}$ form a group of matrices acting on the plane, preserving the vertical axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no subgroups like this.
Another: the group $GL(2)$ of all $2 times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W subset GL(2)$ with $W cap U(2) ne emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
$endgroup$
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
add a comment |
$begingroup$
A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if its unipotent radical is trivial, or equivalently if it has no connected unipotent abelian subgroup except ${I}$.
See Armand Borel, Linear Algebraic Groups, p. 158.
But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group.
A motivating example: the matrices $begin{pmatrix}1&x\0&1end{pmatrix}$ form a group of matrices acting on the plane, preserving the vertical axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no subgroups like this.
Another: the group $GL(2)$ of all $2 times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W subset GL(2)$ with $W cap U(2) ne emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
$endgroup$
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
add a comment |
$begingroup$
A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if its unipotent radical is trivial, or equivalently if it has no connected unipotent abelian subgroup except ${I}$.
See Armand Borel, Linear Algebraic Groups, p. 158.
But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group.
A motivating example: the matrices $begin{pmatrix}1&x\0&1end{pmatrix}$ form a group of matrices acting on the plane, preserving the vertical axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no subgroups like this.
Another: the group $GL(2)$ of all $2 times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W subset GL(2)$ with $W cap U(2) ne emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
$endgroup$
A linear algebraic group is unipotent if it consists entirely of unipotent linear transformations, i.e. $I+N$ with $N$ nilpotent.
A linear algebraic group is reductive if its unipotent radical is trivial, or equivalently if it has no connected unipotent abelian subgroup except ${I}$.
See Armand Borel, Linear Algebraic Groups, p. 158.
But the essential idea is that a reductive group (over the complex numbers) has all finite dimensional representations completely reducible, so behaves like a finite group.
A motivating example: the matrices $begin{pmatrix}1&x\0&1end{pmatrix}$ form a group of matrices acting on the plane, preserving the vertical axis, but not any complementary line. So this is not reductive. A linear algebraic group is reductive if it has no subgroups like this.
Another: the group $GL(2)$ of all $2 times 2$ complex matrices is reductive, because it contains the unitary group $U(2)$, which is compact (so behaves like a finite group), as a Zariski dense subgroup, i.e. any holomorphic function $f$ defined on an open set $W subset GL(2)$ with $W cap U(2) ne emptyset$, and vanishing on that intersect, vanishes. So holomorphic representations of $GL(2)$ determine and are determined by continuous representations of $U(2)$.
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
edited 1 hour ago
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
answered 3 hours ago
Ben McKayBen McKay
14.1k22759
14.1k22759
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
add a comment |
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
2 hours ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
1 hour ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
$begingroup$
You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.
$endgroup$
– Jay Taylor
22 mins ago
add a comment |
$begingroup$
For algebraic groups over $mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:
For an algebraic subgroup $G$ of $mathrm{GL}_n(mathbb{C})$, the following conditions are equivalent:
$G$ is reductive.
$G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.
$G$ is conjugate to a subgroup of $mathrm{GL}_n(mathbb{C})$ which is self-adjoint, that is stable under $g mapsto g^* := overline{g}^T$.
The basic examples to have in mind: $mathrm{GL}_n(mathbb{C})$ and $mathrm{SL}_n(mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n geq 2$).
As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
$endgroup$
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
add a comment |
$begingroup$
For algebraic groups over $mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:
For an algebraic subgroup $G$ of $mathrm{GL}_n(mathbb{C})$, the following conditions are equivalent:
$G$ is reductive.
$G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.
$G$ is conjugate to a subgroup of $mathrm{GL}_n(mathbb{C})$ which is self-adjoint, that is stable under $g mapsto g^* := overline{g}^T$.
The basic examples to have in mind: $mathrm{GL}_n(mathbb{C})$ and $mathrm{SL}_n(mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n geq 2$).
As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
$endgroup$
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
add a comment |
$begingroup$
For algebraic groups over $mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:
For an algebraic subgroup $G$ of $mathrm{GL}_n(mathbb{C})$, the following conditions are equivalent:
$G$ is reductive.
$G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.
$G$ is conjugate to a subgroup of $mathrm{GL}_n(mathbb{C})$ which is self-adjoint, that is stable under $g mapsto g^* := overline{g}^T$.
The basic examples to have in mind: $mathrm{GL}_n(mathbb{C})$ and $mathrm{SL}_n(mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n geq 2$).
As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
$endgroup$
For algebraic groups over $mathbb{C}$, we have the following deep theorem of Cartan, Chevalley and Mostow describing concretely the reductive groups:
For an algebraic subgroup $G$ of $mathrm{GL}_n(mathbb{C})$, the following conditions are equivalent:
$G$ is reductive.
$G$ has a subgroup $K$ which is compact for the usual topology, and Zariski-dense in $G$.
$G$ is conjugate to a subgroup of $mathrm{GL}_n(mathbb{C})$ which is self-adjoint, that is stable under $g mapsto g^* := overline{g}^T$.
The basic examples to have in mind: $mathrm{GL}_n(mathbb{C})$ and $mathrm{SL}_n(mathbb{C})$ are reductive, while the subgroup of upper-triangular matrices is not reductive (when $n geq 2$).
As Ben McKay explained, a complex algebraic group is reductive if and only if all its algebraic representations are semi-simple.
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
answered 37 mins ago
François BrunaultFrançois Brunault
12.8k23569
12.8k23569
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
add a comment |
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
32 mins ago
add a comment |
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