Mixing
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 Jan 17 at 11:46
Martin Sleziak
3,07032130
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
$endgroup$
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 Jan 17 at 11:46
Martin Sleziak
3,07032130
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
$endgroup$
5
$begingroup$
For me, the most characteristic feature to remember is that reductive groups are quotients of a product of a torus and a semisimple group by a finite central subgroup.
$endgroup$
– მამუკა ჯიბლაძე
Jan 17 at 15:04
7
$begingroup$
See also mathoverflow.net/questions/223584/….
$endgroup$
– Guntram
Jan 17 at 20:07
4
$begingroup$
For me, a striking feature of reductive groups is that reductive groups over (any) algebraically closed field are classified by root data - a purely combinatorial structure (the same data also classifies compact Lie groups). This gives the subject a highly structured and combinatorial flavour. In particular, the Langlands dual reductive group (which plays a key role in the Langlands program) is defined in terms of the root data.
$endgroup$
– Sam Gunningham
Jan 18 at 0:05
2
$begingroup$
One of the answers to the question linked to by @Guntram above is also very good, I think: reductive complex algebraic groups are precisely the complexifications of compact Lie groups. In fact, (that answer claims that) the corresponding categories are equivalent.
$endgroup$
– მამუკა ჯიბლაძე
Jan 18 at 5:42
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 Jan 17 at 11:46
Martin Sleziak
3,07032130
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
$endgroup$
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 Jan 17 at 11:46
Martin Sleziak
3,07032130
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
edited Jan 17 at 11:46
Martin Sleziak
3,07032130
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
edited Jan 17 at 11:46
Martin Sleziak
3,07032130
edited Jan 17 at 11:46
Martin Sleziak
3,07032130
edited Jan 17 at 11:46
Martin Sleziak
3,07032130
3,07032130
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
asked Jan 17 at 9:56
John R RamsdenJohn R Ramsden
976516
976516
5
$begingroup$
For me, the most characteristic feature to remember is that reductive groups are quotients of a product of a torus and a semisimple group by a finite central subgroup.
$endgroup$
– მამუკა ჯიბლაძე
Jan 17 at 15:04
7
$begingroup$
See also mathoverflow.net/questions/223584/….
$endgroup$
– Guntram
Jan 17 at 20:07
4
$begingroup$
For me, a striking feature of reductive groups is that reductive groups over (any) algebraically closed field are classified by root data - a purely combinatorial structure (the same data also classifies compact Lie groups). This gives the subject a highly structured and combinatorial flavour. In particular, the Langlands dual reductive group (which plays a key role in the Langlands program) is defined in terms of the root data.
$endgroup$
– Sam Gunningham
Jan 18 at 0:05
2
$begingroup$
One of the answers to the question linked to by @Guntram above is also very good, I think: reductive complex algebraic groups are precisely the complexifications of compact Lie groups. In fact, (that answer claims that) the corresponding categories are equivalent.
$endgroup$
– მამუკა ჯიბლაძე
Jan 18 at 5:42
add a comment |
5
$begingroup$
For me, the most characteristic feature to remember is that reductive groups are quotients of a product of a torus and a semisimple group by a finite central subgroup.
$endgroup$
– მამუკა ჯიბლაძე
Jan 17 at 15:04
7
$begingroup$
See also mathoverflow.net/questions/223584/….
$endgroup$
– Guntram
Jan 17 at 20:07
4
$begingroup$
For me, a striking feature of reductive groups is that reductive groups over (any) algebraically closed field are classified by root data - a purely combinatorial structure (the same data also classifies compact Lie groups). This gives the subject a highly structured and combinatorial flavour. In particular, the Langlands dual reductive group (which plays a key role in the Langlands program) is defined in terms of the root data.
$endgroup$
– Sam Gunningham
Jan 18 at 0:05
2
$begingroup$
One of the answers to the question linked to by @Guntram above is also very good, I think: reductive complex algebraic groups are precisely the complexifications of compact Lie groups. In fact, (that answer claims that) the corresponding categories are equivalent.
$endgroup$
– მამუკა ჯიბლაძე
Jan 18 at 5:42
5
5
$begingroup$
For me, the most characteristic feature to remember is that reductive groups are quotients of a product of a torus and a semisimple group by a finite central subgroup.
$endgroup$
– მამუკა ჯიბლაძე
Jan 17 at 15:04
$begingroup$
For me, the most characteristic feature to remember is that reductive groups are quotients of a product of a torus and a semisimple group by a finite central subgroup.
$endgroup$
– მამუკა ჯიბლაძე
Jan 17 at 15:04
7
7
$begingroup$
See also mathoverflow.net/questions/223584/….
$endgroup$
– Guntram
Jan 17 at 20:07
$begingroup$
See also mathoverflow.net/questions/223584/….
$endgroup$
– Guntram
Jan 17 at 20:07
4
4
$begingroup$
For me, a striking feature of reductive groups is that reductive groups over (any) algebraically closed field are classified by root data - a purely combinatorial structure (the same data also classifies compact Lie groups). This gives the subject a highly structured and combinatorial flavour. In particular, the Langlands dual reductive group (which plays a key role in the Langlands program) is defined in terms of the root data.
$endgroup$
– Sam Gunningham
Jan 18 at 0:05
$begingroup$
For me, a striking feature of reductive groups is that reductive groups over (any) algebraically closed field are classified by root data - a purely combinatorial structure (the same data also classifies compact Lie groups). This gives the subject a highly structured and combinatorial flavour. In particular, the Langlands dual reductive group (which plays a key role in the Langlands program) is defined in terms of the root data.
$endgroup$
– Sam Gunningham
Jan 18 at 0:05
2
2
$begingroup$
One of the answers to the question linked to by @Guntram above is also very good, I think: reductive complex algebraic groups are precisely the complexifications of compact Lie groups. In fact, (that answer claims that) the corresponding categories are equivalent.
$endgroup$
– მამუკა ჯიბლაძე
Jan 18 at 5:42
$begingroup$
One of the answers to the question linked to by @Guntram above is also very good, I think: reductive complex algebraic groups are precisely the complexifications of compact Lie groups. In fact, (that answer claims that) the corresponding categories are equivalent.
$endgroup$
– მამუკა ჯიბლაძე
Jan 18 at 5:42
add a comment |
3 Answers
3
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 it has no connected normal unipotent abelian subgroup except ${I}$.
See A. Borel and J. Tits, Groupes reductifs, IHES Publ Math. 1965 p. 59 or 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 normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.
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 Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
$endgroup$
1
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
2
$begingroup$
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
Jan 17 at 13:05
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
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 Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
$endgroup$
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
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– Sylvain JULIEN
Jan 17 at 12:55
1
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@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
add a comment |
$begingroup$
I understand two reasons why reductive groups play such a central role nowadays.
First:
A maybe not so satisfying answer to "Why are reductive groups important in the Langlands program" is: $operatorname{GL}_n$ is important.
It happens that many conjectures and motivating ideas in the Langlands philosophy for $operatorname{GL}_n$ can be easily formulated in the more general setting of arbitrary reductive groups. Proving things in this more general setting is another matter entirely.
I don't believe there is any definition of reductive group which motivated Langlands to formulate his conjectures that way. Rather, the structure theory of arbitrary reductive groups is so similar to that of $operatorname{GL}_n$ that the important ideas for the general linear group carry right over.
An example is parabolic induction. A reductive group $G$ has various Levi subgroups $M$, which are reductive groups in their own right, but generally less complicated than $G$. Many representations of $G$ are parabolically induced from representations of Levi subgroups. So the idea is that if we understand representations of "smaller" reductive groups $M$, we will know more about representations of the bigger group $G$. This is especially evident in the case $G = operatorname{GL}_n$, where the Levi subgroups look like products of smaller $operatorname{GL}$s.
Second:
Number theoretic objects associated to arbitrary reductive groups are expected to be special cases of those associated to the general linear group. I sketch this argument below. Breaking up these objects associated to $operatorname{GL}_n$ into subcases might be insightful.
If $^LG$ is the Langlands dual group of a reductive group $G$ over a field $k$, $r: , ^LG rightarrow operatorname{GL}_n(mathbb C)$ is a representation, and $pi$ is a representation of $G$, there should be an associated L-function $L(s,pi,r)$. The Langlands correspondence should (roughly) associate to this a homomorphism $rho: operatorname{Gal}(k_s/k) rightarrow space ^LG$. This correspondence should preserve the L-functions
$$L(s,pi,r) = L(s, r circ rho)$$
The composition $r circ rho$ is an $n$ dimensional Galois representation which by the Langlands correspondence for $operatorname{GL}_n$ ought to associate this to a representation $Pi$ of $operatorname{GL}_n(k)$. That Langlands correspondence should also preserve L-functions:
$$L(s, r circ rho) = L(s,Pi)$$
and therefore
$$L(s,pi,r) = L(s,Pi)$$
So the philosophy is that everything should eventually come back to $operatorname{GL}_n$.
answered Jan 31 at 4:15
D_SD_S
1,791619
$endgroup$
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
1
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
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Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
add a comment |
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3 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 it has no connected normal unipotent abelian subgroup except ${I}$.
See A. Borel and J. Tits, Groupes reductifs, IHES Publ Math. 1965 p. 59 or 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 normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.
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 Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
$endgroup$
1
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
2
$begingroup$
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
Jan 17 at 13:05
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
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 it has no connected normal unipotent abelian subgroup except ${I}$.
See A. Borel and J. Tits, Groupes reductifs, IHES Publ Math. 1965 p. 59 or 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 normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.
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 Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
$endgroup$
1
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
2
$begingroup$
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
Jan 17 at 13:05
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
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 it has no connected normal unipotent abelian subgroup except ${I}$.
See A. Borel and J. Tits, Groupes reductifs, IHES Publ Math. 1965 p. 59 or 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 normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.
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 Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
$endgroup$
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 it has no connected normal unipotent abelian subgroup except ${I}$.
See A. Borel and J. Tits, Groupes reductifs, IHES Publ Math. 1965 p. 59 or 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 normal subgroups like this. Clearly such a subgroup gets in the way of having all representations completely reducible.
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 Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
edited Jan 19 at 13:20
answered Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
answered Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
answered Jan 17 at 10:24
Ben McKayBen McKay
14.6k22861
14.6k22861
1
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
2
$begingroup$
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
Jan 17 at 13:05
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
add a comment |
1
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
2
$begingroup$
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
Jan 17 at 13:05
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
1
1
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
The definition should be "has no non-trivial connected normal unipotent subgroup".
$endgroup$
– spin
Jan 17 at 11:17
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
$begingroup$
@spin: thanks; corrected.
$endgroup$
– Ben McKay
Jan 17 at 11:44
2
2
$begingroup$
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
Jan 17 at 13:05
$begingroup$
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
Jan 17 at 13:05
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
@Ben: The original definition of (connected) "reductive" over any field was given by Borel and Tits in their foundational paper Groupes reductifs (IHES Publ Math. 1965), which has the advantage of being avaiiable online at numdam.org . (It's also available in Borel's 1967 Columbia lectures from W.A.Benjamin, written up by Bass and filled in by my 1975 Springer text or Borel's later second edution or T.A. Springer's book---all with same title,) The basic problem in general is to relate the classification to norrnal unipotent subgroups.
$endgroup$
– Jim Humphreys
Jan 18 at 20:37
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
$begingroup$
One also has to be careful for groups defined over non-perfect fields, where the non-trivial, connected, normal, unipotent subgroup might become 'visible' only after base change.
$endgroup$
– LSpice
Feb 12 at 0:25
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 Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
$endgroup$
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
Jan 17 at 12:55
1
$begingroup$
@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
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 Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
$endgroup$
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
Jan 17 at 12:55
1
$begingroup$
@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
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 Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
$endgroup$
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 Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
answered Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
answered Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
answered Jan 17 at 12:50
François BrunaultFrançois Brunault
13.3k23572
13.3k23572
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
Jan 17 at 12:55
1
$begingroup$
@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
add a comment |
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
Jan 17 at 12:55
1
$begingroup$
@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
Jan 17 at 12:55
$begingroup$
Is there a relationship between this self-adjointness criterion and Hilbert-Polya's conjecture ?
$endgroup$
– Sylvain JULIEN
Jan 17 at 12:55
1
1
$begingroup$
@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
$begingroup$
@SylvainJULIEN I don't think so: the Riemann zeta function is associated to the trivial automorphic representation of $mathrm{GL}_1$, and self-adjointness is rather trivial in this case.
$endgroup$
– François Brunault
Jan 17 at 13:54
add a comment |
$begingroup$
I understand two reasons why reductive groups play such a central role nowadays.
First:
A maybe not so satisfying answer to "Why are reductive groups important in the Langlands program" is: $operatorname{GL}_n$ is important.
It happens that many conjectures and motivating ideas in the Langlands philosophy for $operatorname{GL}_n$ can be easily formulated in the more general setting of arbitrary reductive groups. Proving things in this more general setting is another matter entirely.
I don't believe there is any definition of reductive group which motivated Langlands to formulate his conjectures that way. Rather, the structure theory of arbitrary reductive groups is so similar to that of $operatorname{GL}_n$ that the important ideas for the general linear group carry right over.
An example is parabolic induction. A reductive group $G$ has various Levi subgroups $M$, which are reductive groups in their own right, but generally less complicated than $G$. Many representations of $G$ are parabolically induced from representations of Levi subgroups. So the idea is that if we understand representations of "smaller" reductive groups $M$, we will know more about representations of the bigger group $G$. This is especially evident in the case $G = operatorname{GL}_n$, where the Levi subgroups look like products of smaller $operatorname{GL}$s.
Second:
Number theoretic objects associated to arbitrary reductive groups are expected to be special cases of those associated to the general linear group. I sketch this argument below. Breaking up these objects associated to $operatorname{GL}_n$ into subcases might be insightful.
If $^LG$ is the Langlands dual group of a reductive group $G$ over a field $k$, $r: , ^LG rightarrow operatorname{GL}_n(mathbb C)$ is a representation, and $pi$ is a representation of $G$, there should be an associated L-function $L(s,pi,r)$. The Langlands correspondence should (roughly) associate to this a homomorphism $rho: operatorname{Gal}(k_s/k) rightarrow space ^LG$. This correspondence should preserve the L-functions
$$L(s,pi,r) = L(s, r circ rho)$$
The composition $r circ rho$ is an $n$ dimensional Galois representation which by the Langlands correspondence for $operatorname{GL}_n$ ought to associate this to a representation $Pi$ of $operatorname{GL}_n(k)$. That Langlands correspondence should also preserve L-functions:
$$L(s, r circ rho) = L(s,Pi)$$
and therefore
$$L(s,pi,r) = L(s,Pi)$$
So the philosophy is that everything should eventually come back to $operatorname{GL}_n$.
answered Jan 31 at 4:15
D_SD_S
1,791619
$endgroup$
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
1
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
$begingroup$
Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
add a comment |
$begingroup$
I understand two reasons why reductive groups play such a central role nowadays.
First:
A maybe not so satisfying answer to "Why are reductive groups important in the Langlands program" is: $operatorname{GL}_n$ is important.
It happens that many conjectures and motivating ideas in the Langlands philosophy for $operatorname{GL}_n$ can be easily formulated in the more general setting of arbitrary reductive groups. Proving things in this more general setting is another matter entirely.
I don't believe there is any definition of reductive group which motivated Langlands to formulate his conjectures that way. Rather, the structure theory of arbitrary reductive groups is so similar to that of $operatorname{GL}_n$ that the important ideas for the general linear group carry right over.
An example is parabolic induction. A reductive group $G$ has various Levi subgroups $M$, which are reductive groups in their own right, but generally less complicated than $G$. Many representations of $G$ are parabolically induced from representations of Levi subgroups. So the idea is that if we understand representations of "smaller" reductive groups $M$, we will know more about representations of the bigger group $G$. This is especially evident in the case $G = operatorname{GL}_n$, where the Levi subgroups look like products of smaller $operatorname{GL}$s.
Second:
Number theoretic objects associated to arbitrary reductive groups are expected to be special cases of those associated to the general linear group. I sketch this argument below. Breaking up these objects associated to $operatorname{GL}_n$ into subcases might be insightful.
If $^LG$ is the Langlands dual group of a reductive group $G$ over a field $k$, $r: , ^LG rightarrow operatorname{GL}_n(mathbb C)$ is a representation, and $pi$ is a representation of $G$, there should be an associated L-function $L(s,pi,r)$. The Langlands correspondence should (roughly) associate to this a homomorphism $rho: operatorname{Gal}(k_s/k) rightarrow space ^LG$. This correspondence should preserve the L-functions
$$L(s,pi,r) = L(s, r circ rho)$$
The composition $r circ rho$ is an $n$ dimensional Galois representation which by the Langlands correspondence for $operatorname{GL}_n$ ought to associate this to a representation $Pi$ of $operatorname{GL}_n(k)$. That Langlands correspondence should also preserve L-functions:
$$L(s, r circ rho) = L(s,Pi)$$
and therefore
$$L(s,pi,r) = L(s,Pi)$$
So the philosophy is that everything should eventually come back to $operatorname{GL}_n$.
answered Jan 31 at 4:15
D_SD_S
1,791619
$endgroup$
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
1
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
$begingroup$
Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
add a comment |
$begingroup$
I understand two reasons why reductive groups play such a central role nowadays.
First:
A maybe not so satisfying answer to "Why are reductive groups important in the Langlands program" is: $operatorname{GL}_n$ is important.
It happens that many conjectures and motivating ideas in the Langlands philosophy for $operatorname{GL}_n$ can be easily formulated in the more general setting of arbitrary reductive groups. Proving things in this more general setting is another matter entirely.
I don't believe there is any definition of reductive group which motivated Langlands to formulate his conjectures that way. Rather, the structure theory of arbitrary reductive groups is so similar to that of $operatorname{GL}_n$ that the important ideas for the general linear group carry right over.
An example is parabolic induction. A reductive group $G$ has various Levi subgroups $M$, which are reductive groups in their own right, but generally less complicated than $G$. Many representations of $G$ are parabolically induced from representations of Levi subgroups. So the idea is that if we understand representations of "smaller" reductive groups $M$, we will know more about representations of the bigger group $G$. This is especially evident in the case $G = operatorname{GL}_n$, where the Levi subgroups look like products of smaller $operatorname{GL}$s.
Second:
Number theoretic objects associated to arbitrary reductive groups are expected to be special cases of those associated to the general linear group. I sketch this argument below. Breaking up these objects associated to $operatorname{GL}_n$ into subcases might be insightful.
If $^LG$ is the Langlands dual group of a reductive group $G$ over a field $k$, $r: , ^LG rightarrow operatorname{GL}_n(mathbb C)$ is a representation, and $pi$ is a representation of $G$, there should be an associated L-function $L(s,pi,r)$. The Langlands correspondence should (roughly) associate to this a homomorphism $rho: operatorname{Gal}(k_s/k) rightarrow space ^LG$. This correspondence should preserve the L-functions
$$L(s,pi,r) = L(s, r circ rho)$$
The composition $r circ rho$ is an $n$ dimensional Galois representation which by the Langlands correspondence for $operatorname{GL}_n$ ought to associate this to a representation $Pi$ of $operatorname{GL}_n(k)$. That Langlands correspondence should also preserve L-functions:
$$L(s, r circ rho) = L(s,Pi)$$
and therefore
$$L(s,pi,r) = L(s,Pi)$$
So the philosophy is that everything should eventually come back to $operatorname{GL}_n$.
answered Jan 31 at 4:15
D_SD_S
1,791619
$endgroup$
I understand two reasons why reductive groups play such a central role nowadays.
First:
A maybe not so satisfying answer to "Why are reductive groups important in the Langlands program" is: $operatorname{GL}_n$ is important.
It happens that many conjectures and motivating ideas in the Langlands philosophy for $operatorname{GL}_n$ can be easily formulated in the more general setting of arbitrary reductive groups. Proving things in this more general setting is another matter entirely.
I don't believe there is any definition of reductive group which motivated Langlands to formulate his conjectures that way. Rather, the structure theory of arbitrary reductive groups is so similar to that of $operatorname{GL}_n$ that the important ideas for the general linear group carry right over.
An example is parabolic induction. A reductive group $G$ has various Levi subgroups $M$, which are reductive groups in their own right, but generally less complicated than $G$. Many representations of $G$ are parabolically induced from representations of Levi subgroups. So the idea is that if we understand representations of "smaller" reductive groups $M$, we will know more about representations of the bigger group $G$. This is especially evident in the case $G = operatorname{GL}_n$, where the Levi subgroups look like products of smaller $operatorname{GL}$s.
Second:
Number theoretic objects associated to arbitrary reductive groups are expected to be special cases of those associated to the general linear group. I sketch this argument below. Breaking up these objects associated to $operatorname{GL}_n$ into subcases might be insightful.
If $^LG$ is the Langlands dual group of a reductive group $G$ over a field $k$, $r: , ^LG rightarrow operatorname{GL}_n(mathbb C)$ is a representation, and $pi$ is a representation of $G$, there should be an associated L-function $L(s,pi,r)$. The Langlands correspondence should (roughly) associate to this a homomorphism $rho: operatorname{Gal}(k_s/k) rightarrow space ^LG$. This correspondence should preserve the L-functions
$$L(s,pi,r) = L(s, r circ rho)$$
The composition $r circ rho$ is an $n$ dimensional Galois representation which by the Langlands correspondence for $operatorname{GL}_n$ ought to associate this to a representation $Pi$ of $operatorname{GL}_n(k)$. That Langlands correspondence should also preserve L-functions:
$$L(s, r circ rho) = L(s,Pi)$$
and therefore
$$L(s,pi,r) = L(s,Pi)$$
So the philosophy is that everything should eventually come back to $operatorname{GL}_n$.
answered Jan 31 at 4:15
D_SD_S
1,791619
edited Jan 31 at 14:11
answered Jan 31 at 4:15
D_SD_S
1,791619
answered Jan 31 at 4:15
D_SD_S
1,791619
answered Jan 31 at 4:15
D_SD_S
1,791619
1,791619
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
1
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
$begingroup$
Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
add a comment |
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
1
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
$begingroup$
Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
$begingroup$
This reminds that the Mumford-Tate group associated to a pure polarizable Hodge structure is always a reductive group. So the (algebraic) image of Galois representations associated to pure motives $H^i(X)$ is a reductive group.
$endgroup$
– François Brunault
Jan 31 at 7:44
1
1
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
$begingroup$
I don't think the first sentence "Langlands' conjectures originally apply to the general linear group, but later included reductive groups in general" is accurate. Already in his famous letter to Weil, Langlands formulates his programme in the generality of reductive groups.
$endgroup$
– Alex B.
Jan 31 at 9:38
$begingroup$
Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
$begingroup$
Going back only to Langlands puts the history too late. I believe it was Harish-Chandra who recognised the reductive groups as an appropriate object of study, largely because (as you mention) they are a rich class of groups closed under passage to Levi components of parabolic subgroups.
$endgroup$
– LSpice
Feb 12 at 0:27
add a comment |
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5
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For me, the most characteristic feature to remember is that reductive groups are quotients of a product of a torus and a semisimple group by a finite central subgroup.
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– მამუკა ჯიბლაძე
Jan 17 at 15:04
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See also mathoverflow.net/questions/223584/….
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– Guntram
Jan 17 at 20:07
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For me, a striking feature of reductive groups is that reductive groups over (any) algebraically closed field are classified by root data - a purely combinatorial structure (the same data also classifies compact Lie groups). This gives the subject a highly structured and combinatorial flavour. In particular, the Langlands dual reductive group (which plays a key role in the Langlands program) is defined in terms of the root data.
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– Sam Gunningham
Jan 18 at 0:05
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One of the answers to the question linked to by @Guntram above is also very good, I think: reductive complex algebraic groups are precisely the complexifications of compact Lie groups. In fact, (that answer claims that) the corresponding categories are equivalent.
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– მამუკა ჯიბლაძე
Jan 18 at 5:42