Timeline for Points of reductive groups
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1, 2010 at 17:49 | vote | accept | user1594 | ||
| Jun 1, 2010 at 17:49 | vote | accept | user1594 | ||
| Jun 1, 2010 at 17:49 | |||||
| May 28, 2010 at 6:38 | answer | added | Xandi Tuni | timeline score: 1 | |
| May 27, 2010 at 20:40 | answer | added | George McNinch | timeline score: 1 | |
| May 27, 2010 at 17:40 | comment | added | BCnrd | To give a concrete example of the problems with non-s.c. $G$ in a "classical" setting, consider the degree-$n$ central isogeny $f:{\rm{SL}}_ n \rightarrow {\rm{PGL}}_ n$ over $k = \mathbf{R}$ with $n$ odd. Since the kernel is $\mu_ n$, the induced map on $k$-points is a real-analytic isomorphism, so the inverse map on $k$-points defines a real-analytic linear representation of ${\rm{PGL}}_ n(k)$ which is certainly not "algebraic" (since $f$ has no algebraic inverse, as its degree is $> 1$). Note here that ${\rm{PGL}}_ n$ is not simply connected. | |
| May 27, 2010 at 17:32 | comment | added | BCnrd | Assuming you use finite-dimensional $k$-rational representations and connected $G$, faithfulness holds for infinite $k$ because conn'd reductive groups are unirational over any field (so $G(k)$ is Zar.-dense in $G$ when $k$ is infinite). Must assume connectedness, as otherwise if $k$ isn't sep. closed then $G$ could correspond to nontrivial finite Galois module with no nonzero Gal-fixed points, so $G(k) = \{1\}$. Should assume $G$ semisimple, or else nontrivial torus quotients will make a mess, and even simply connected or else $G(k)$ can have nontrivial finite commutative quotients. | |
| May 27, 2010 at 17:24 | comment | added | user1594 | Thanks - I edited to hopefully make the question clearer. | |
| May 27, 2010 at 17:22 | history | edited | user1594 | CC BY-SA 2.5 |
added 359 characters in body; added 61 characters in body
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| May 27, 2010 at 17:14 | comment | added | Jim Humphreys |
It would be helpful to make the language more precise: what is meant here by a "representation" of $G$ (rational?) and does that relate to $k$. Are you identifying a group scheme $G$ with its group of rational points over an algebraic closure of $k$? And is that group connected? Also, are the representations of $G(k)$ taken over $k$ or its algebraic closure or some other field? The extreme case when $k$ is finite shows some of the problems that can come up with the restriction functor. Even going down from $\mathbb{C}$ to $\mathbb{R}$ requires some care.
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| May 27, 2010 at 16:46 | history | asked | user1594 | CC BY-SA 2.5 |