# (When) can the presentation in Steinberg's Yale notes fail to give an algebraic group?

I'm trying to understand a remark which appears on p. 1483 of Cohen, Murray and Taylor's "Computing in Groups of Lie Type." It says, "We have not used the presentations described in [7] or [30] because they define groups which are not necessarily algebraic when $\mathbb{F}$ is not algebraically closed." Reference [7] alluded to in the quotation is Carter's Finite Groups of Lie Type and reference [30] is Steinberg's Lectures on Chevalley groups, Tech. report, Yale University, 1968. Is there a nice example which illustrates what is meant by this remark, where the group given by the presentation in the Steinberg notes fails to be algebraic, in a suitable sense? Guidance on different relevant notions of "algebraic" and which is most likely to be operative here would also be appreciated.

• the two sheeted cover of $SL(2,\mathbb{R})$ is one such example, which is not an algebraic group (it is not even linear). – Venkataramana Aug 2 '16 at 3:35
• This question: mathoverflow.net/questions/69741/… is relevant to the above comment. – Robert Furber Aug 2 '16 at 11:31
• The question needs a more precise formulation: (1) Chevalley oriiginally looked at groups which are simple as abstract groups; Steinberg's lecture notes adopt a more general framework. (2) What are the "Steinberg presentations"? These developed out of papers he wrote before the Yale lectures and distinguish rank 1 groups from the rest. (3) What notion of "algebraic group" are you using? Steinberg for example relied on the 1950s Chevalley-Borel notions. (4) Aside from groups over the real numbers, Steinberg's work led to study of matrix groups over number fields or local fields. – Jim Humphreys Aug 2 '16 at 14:32
• Also, your tags could be broadened by adding 'gr.group-theory' and 'algebraic-k-theory'. The latter is the area especially stimulated by Steinberg's work on generators and relations. – Jim Humphreys Aug 2 '16 at 14:34
• Is this just referring to variants of the basic fact that quotient groups such as ${\rm{SL}}_n(k)/\mu_n(k)$ (the commutator subgroup of ${\rm{PGL}}_n(k)$, generated by the "root groups") are not "algebraic" in the sense that they do not arise as the group of $k$-points of a $k$-group quotient of ${\rm{SL}}_n$? Can you just ignore the remark which is unclear to you and see how the authors of the book you are reading use whatever it is that they are doing? – nfdc23 Aug 3 '16 at 3:01