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Algebraic varieties with group operations given by morphisms, or group objects in the category of algebraic varieties, the category of algebraic schemes, or closely related categories.

2 votes

Absolutely irreducible representations of affine group schemes of finite type over a field

The classical viewpoint is captured well in the 1962 book Representation Theory of Finite Groups and Associative Algebras by Curtis and Reiner (Wiley), Corollary 29.15. This of course doesn't direc …
Jim Humphreys's user avatar
3 votes

Group cohomology of modular representations for finite groups of Lie type

As Derek Holt comments, cohomology has complications even for fimite general linear groups. Probably you are using the term "reductive" too casually and should replace it by "simple" or perhaps "semis …
Jim Humphreys's user avatar
4 votes

Diagonal automorphisms for twisted Chevalley groups

First of all, I'd inquire what role the characteristic of the field plays here. It's true that the finite twisted groups rely on characteristics 2, 3 especiallu, but infinite twisted groups include …
Jim Humphreys's user avatar
3 votes

Nilpotent elements of Lie algebra and unipotent groups

YCor's question about your definition of "nilpotent" is definitely in order, because Borel and Springer already defined this term in general for affine (=linear) algebraic groups over an arbitrary inf …
Jim Humphreys's user avatar
7 votes
Accepted

For $G$ an adjoint Chevalley group, are all of $G(\mathbb Z)$'s finite-index subgroups congr...

Chapter VI of my old Springer Lecture Notes in Mathematics 789 Arithmetic Groups (1980) is in English and gives a version of H. Matsumoto's and C. Moore's arguments for the subgroups of $\operatorname …
Jim Humphreys's user avatar
3 votes

How to prove that Chevalley groups over $\mathbb R$ have no compact factors

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevalley or the broader sense of Steinberg's lectures). One …
Jim Humphreys's user avatar
2 votes

Character of a semisimple connected Lie groups

In the original sense, Chevalley groups are generated by copies of the additive group of the field and are in fact simple as abstract groups if the field is not too small. (This was the motive for …
Jim Humphreys's user avatar
1 vote

Regular semisimple elements in $SL(n,q)$

This is just a comment but in community wiki format. Most studies of semisimple (or reductive) algebraic groups and finite groups of Lie type emphasize counting the number of classes of various el …
2 votes
Accepted

Decomposition of parabolic subgroup in reductive group

Over an algebraically closed field (of any characteristic), it is fairly obvious using the omitted definition of "standard parabolic" that the assertion here for a pair of included standard parabolics …
Jim Humphreys's user avatar
2 votes
Accepted

Element in finite number of Borel subgroups

First of all, it's probably intended that "linear algebraic groups" are semisimple or at least reductive (and connected). For example, a solvable group might have no semisimple elements except 1 (et …
Jim Humphreys's user avatar
5 votes

Reference Request: Structure constants for G2

Probably the earliest reference is the 1956-58 Chevalley seminar, available online in typed format, which has been republished in 2005 as a typeset book edited by P. Cartier: see Chapter 21. (No speci …
Jim Humphreys's user avatar
6 votes
Accepted

a question on Deligne-Lusztig characters

No, your parenthetic comment at the end indicates some confusion about the nature of Deligne-Lusztig virtual (= generallized) characters: these are defined to be $\mathbb{Z}$-linear combinations of ac …
Jim Humphreys's user avatar
1 vote

Fixed Points of the Weyl Group action on a Maximal Torus and the Center of a Reductive Group

To me the question itself (and the answers) are out of focus, starting with the claim that the ring of Weyl group invariants is somehow central. Chevalley's 1955 argument does show that this ring is …
Jim Humphreys's user avatar
1 vote

Do the absolute roots restricting to a given root form a Galois orbit?

To answer your question, it's enough to point out that every root (in a reduced root system such as $\Phi$) plays the role of simple root in some basis $\Delta$. Note however that some absolute root …
Jim Humphreys's user avatar
7 votes

Number of points of the nilpotent cone over a finite field and its cohomology

Concerning your Q1, it may be of interest to fill in some of the background due to Steinberg (which in turn had a lot of influence on Springer's work). In his 1966 ICM talk, Steinberg formulated quit …
Jim Humphreys's user avatar

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