Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've heard a few times that the symmetric group is an algebraic group over a field with one element, and that the alternating group is quite specifically $SO_n(\mathbb{F}_1)$. This does make a lot of sense intuitively, and actually helps to explain to non-specialists the relations between different things I have done.

However, objects over the (non-existent) field with one element aren't just a metaphor - they are objects that can be defined properly, though that may not have happened yet. What is the status for the correspondence between $Alt(n)$ and $SO_n(\mathbb{F}_1)$? Is there really a well-defined homomorphism of some sort, and, if so, are there references where this is worked out?

share|improve this question
5  
Are you sure it is $SO_n(F_1)$ and not $SL_n(F_1)$? The latter would make sense if you consider buildings associated with algebraic groups over $F_q$: Cardinality of $F_q$ gives you thickness of the building. Setting $q=1$ gives you thin buildings, i.e., apartments. One then can argue that the "algebraic group" corresponding to the thin building is just the Weyl group. –  Misha Jan 9 '13 at 17:39
2  
Does arxiv.org/pdf/0907.3824.pdf answer your question? –  Martin Brandenburg Jan 9 '13 at 18:02
1  
By the way, I would not say that $\mathbb{F}_1$ is non-existent. In fact, there are already many precise definitions. The problem is that there is no prefered one, and up to know no theory has led to a proof of the Riemann hypothesis. –  Martin Brandenburg Jan 9 '13 at 18:04
1  
@Z254R: Such question will be surely closed as "subjective and argumentative." –  Misha Jan 9 '13 at 20:51
2  
It is not the anger of the Gods of mathematics that we fear. It is the chaos of people arguing subjectively, in what ought to be an orderly forum, over the merits of inconclusive citations. –  Lee Mosher Jan 9 '13 at 21:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.