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Is there anywhere where I can read a complete proof in English of this theorem by Borel and Tits:

Suppose that $G$ is a simple algebraic group over an infinite field $k$, and that $H$ is a subgroup of $G(k)$ containing the subgroup of $G(k)$ generated by the rational points of the unipotent radicals of the $k$-parabolic subgroups, and that $\alpha \colon H \to G'(k')$ is a homomorphism, where $G'$ is a simple algebraic group over an infinite field $k'$, such that $\alpha(G'')$ is Zariski dense in $G'$. Then there exists a homomorphism $\phi\colon k \to k'$, a $k'$-isogeny $\beta\colon G^\phi\to G'$ with $d \beta \ne 0$, and a homomorphism $\gamma\colon H \to Z_{G'(k')}$ to the centre, all three unique, such that $\alpha(h)=\gamma(h)\beta(\phi^0(h))$ for all $h \in H$.

There is a proof in French in Borel and Tits, Homomorphismes 'abstraits' de groupes algebriques simples, Annals of Mathematics, Second Series, Vol. 97, No. 3 (May, 1973), pp. 499-571.

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    $\begingroup$ I am doubtful... On the plus side, French isn't that hard. $\endgroup$ – Victor Protsak Jul 26 '10 at 3:28
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    $\begingroup$ I too am doubtful. Learn (how to read mathematical) French. Anyone who is doing enough math to ask questions at that level needs to learn French. Pronto. $\endgroup$ – Marty Jul 26 '10 at 5:31
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    $\begingroup$ As I've said many times, math French is not real French; it is far far easier to learn (especially with the motivation of trying to understand something interesting!). Considering the large number of important articles in this area of math written in French, it seems apt to quote Darth Vader (in English!): resistance is futile. $\endgroup$ – BCnrd Jul 26 '10 at 15:09
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    $\begingroup$ @Georges: Perhaps I should have said "rapidement" or "prêt"? Or I could have been really obnoxious and just commented entirely in French (probably with many mistakes) :) $\endgroup$ – Marty Jul 26 '10 at 16:54
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    $\begingroup$ BCnrd may have said this many times, but he is right every time. French mathematics is not French symbolist poetry, notwithstanding the tendency of outsiders to confuse the two. – Jim $\endgroup$ – Jim Humphreys Jul 26 '10 at 19:16
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This is too long for a comment, but like others who have commented I don't expect to find anything like a "complete proof" of the Borel-Tits theorem written in English. Borel and Tits have each written at times in French, English, German, but their serious joint work has been in French and is not especially hard to follow. The actual mathematics is difficult, however, requiring a lengthy technical treatment in their 1973 Annals paper on abstract homomorphisms. (Even so, they stopped short of allowing anisotropic groups even while suspecting there were would be some good results in that case; perhaps later work by people like Gopal Prasad sheds light there.)

The underlying question originates with work of Dieudonne, O'Meara, and others on the automorphisms of various classical groups over fields (then more general rings), including compact groups. On the other hand, Steinberg gave in 1960 a unified treatment of the automorphisms of finite Chevalley groups, which I imitated for infinite Chevalley groups in 1967 using more from algebraic groups. What Borel and Tits did was far more comprehensive and sophisticated than any of these concrete investigations. There was a short survey given in French, but also an earlier 1968 preview in English by both authors (not indexed in MathSciNet) which gives a short overview of the method of proof:

On "abstract" homomorphisms of simple algebraic groups, pp. 75-82, Proc. of the Bombay Colloquium in Algebraic Geometry, 1968. (This was published in book form for Tata Institute).

As the time lag between 1968 and 1973 suggests, the full proofs took a lot of work but achieved near-definitive results in a reasonably unified framework.

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There is also a model theoretic proof of a simple version of Borel-Tits on page 90 of "Stable Groups", by Bruno Poizat (Mathematical Surveys and Monographs, 87. American Mathematical Society, Providence, RI, 2001. xiv+129 pp. ISBN: 0-8218-2685-9):

THEOREM 4.17 Every pure group isomorphism $s$ between $G$, a simple algebraic group over the algebraically closed field $K$, and $H$, a simple algebraic group over the algebraically closed field $L$, can be decomposed into a transfer of structures induced by an isomorphism between the fields $K$ and $L$, followed by a quasi-rational function relative to $L$.

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    $\begingroup$ This is interesting to know, but is far weaker than the Borel-Tits theorem and has its own probably substantial prerequisites. $\endgroup$ – Jim Humphreys Aug 2 '10 at 22:04
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The paper Abstract homomorphisms of simple algebraic groups by Robert Steinberg (Semináire N. Bourbaki, 1972-1973), is written in English and includes a proof of the Borel-Tits theorem.

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