This is a mostly philosophical question. Is it fair to think of usual binomial coefficients and their identities as an $F_1$ case of $q$-binomial coefficients and identities? Here $F_1$ is the field of one element.
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asked Jan 13, 2015 at 20:30
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2$\begingroup$ Yes, I would say you can. $\endgroup$– Ben Webster ♦Commented Jan 13, 2015 at 21:38
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$\begingroup$ This is indeed a very philosophical question. They do count something that may be interpreted as something over $\Bbb F_1$: Subsets are subspaces $\endgroup$– Alex DegtyarevCommented Jan 13, 2015 at 21:55
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1$\begingroup$ See, for example, arxiv.org/abs/math/0407093. $\endgroup$– Qiaochu YuanCommented Jan 13, 2015 at 22:34
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1 Answer
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Yes, it is not only fair, but also often very useful. The analogy is that sets are like vector spaces over $\mathbb F_1$, so for example the analog of $\binom{n}{k}$ counting $k$-element subsets of an $n$-element set, is that $\binom{n}{k}_q$ counts the number of $k$-dimensional subspaces of $\mathbb F_q^n$.
In lack of a rigorous theory of $\mathbb F_1$, this usefulness can be illustrated through numerous examples. Take for example the Erdos-Ko-Rado theorem, which is an important starting point of extremal combinatorics. Even though writing down the analogous statement over $\mathbb F_q$ is very easy, by just interchanging subsets with subspaces, the proof turned out to be much harder (the first proof was due to Frankl and Wilson).
The only uniform proofs I know Erdos-Ko-Rado over all characteristics (including characteristic 1) are "geometric" in nature and were discovered much later. However I'm afraid this isn't particularly emphasized in the literature. For example I haven't seen anywhere explicitly mention the analogy between the method of "shadows" and the study of the relevant partial flag variety over $\mathbb F_q$. Still, I think this is a very fun example of interplay between combinatorics of sets and geometry over $\mathbb F_q$ that has produced non-trivial theorems.
answered Jan 13, 2015 at 22:21
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$\begingroup$ Where would one look to see a uniform-in-characteristic-including-1 proof (of anything)? I thought that the theory was still so much in flux that such uniformity would be very hard to come by. $\endgroup$– LSpiceCommented Jan 13, 2015 at 22:56
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1$\begingroup$ @LSpice, I just meant a uniform proof for finite fields, that works when you "translate" it to $\mathbb F_1$. $\endgroup$ Commented Jan 14, 2015 at 1:14
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$\begingroup$ Henry Cohn gives such a proof in the paper I linked to above (arxiv.org/abs/math/0407093). $\endgroup$ Commented Jan 14, 2015 at 5:39