F1 AND THE BRUHAT DECOMPOSITION MATTHEW TOWERS

These are the notes for an expository talk given at the University of Kent PGR seminar on December 18th, 2015.

Introduction Literally speaking, there is no field of order one. However thinking of finite sets as the F1 -analogue of finite-dimensional vector spaces has a long history. In this talk I will explain this idea a little more and try to explain one of the numerical coincidences it generates. Though I won’t discuss it here (not least because I don’t understand it), “algebraic geometry over F1 ” is really serious business, connected to attempts to prove the Riemann Hypothesis. People define field extensions F1n , schemes over F1 ,. . . The literature is very large: [Sou04; CC09; Coh04; CC09; CCM08; PL09; Man08] is only a tiny fragment, and I have given some links at the end. 1. Fields and vector spaces Fields are places you can add, multiply, and divide by things that aren’t zero, just like the ordinary real numbers or complex numbers or rational numbers. They are what you need to do linear algebra — vectors, matrices, linear maps. . . We need one particular example: Definition 1.1. Let q be a prime number. The field with q elements, written Fq , is the set {0, 1, . . . , q − 1} with addition and multiplication mod q.  The field axioms imply that a field F has an additive identity 0 and a multiplicative identity 1 and 1 6= 0, so in particular |F| > 1. Therefore there’s no field with one element. But you can’t go home yet. Finite fields are important here because we can make elementary counting arguments. Definition 1.2. Fnq is the set of all column vectors with height n and entries from Fq .  We will not attempt to define F1 , instead we make a guiding analogy subsets of sets ↔ subspaces of vector spaces and see where this takes us. Roughly, we think of a set as a vector space (actually, projective space) over (the non-existent) F1 . As evidence for the strength of our analogy we obtain formulas counting certain objects in terms of the size of the field they are defined over, and show that setting the size of the field equal to one — which doesn’t make any sense — gives a formula for the size of the F1 -analogue of those objects. Date: Thursday 17th December, 2015. 1

2

MATTHEW TOWERS

2. Grassmannians To sharpen our subsets-subspaces analogy we say that a size k subset of a set of size n is the F1 -analogue of a k-dimensional subspace of a n-dimensional vector space. Definition 2.1. Gr(k, Fnq ) is the set of all k-dimensional subspaces of Fnq .



Definition 2.2. Gr(k, Fn1 ) is the set of all size k subsets of {1, . . . , n}.



Gr(k, Fnq ) is called a Grassmannian, they are a very famous and classical object of study in algebraic geometry, usually with C instead of a finite field. Example 2.3. q = 2, k = 1, n= Fq = multiplication  2, {0,1}with  addition   and  0 1 0 0 0 1 modulo 2, Gr(1, F22 ) = , , , , , has size 3. 0 0 0 1 0 1  To show that our analogy is working, we prove (1) there is a formula for the size of Gr(k, Fnq ) in terms of the size q of the field involved, and (2) when you set q = 1 in that formula, which doesn’t make sense
as there is no field of size one, you get the size of Gr(k, Fn1 ), namely nk . n

−1 = 1+q +· · ·+q n−1 . This is called a q-integer. Definition 2.4. (1) [n]q = qq−1 (2) [n]q ! = [n]q [n − 1]q · · · [2]q [1]q . This is a q-factorial.
[n]q ! . This is a q-binomial coefficient.  (3) nk q = [k]q ![n−k] q

Putting q = 1 in these formulas recovers the ordinary integers, factorials, and binomial coefficients.
Theorem 2.5. | Gr(k, Fnq )| = nk q .
Returning to the example we just did, 21 q = [2]q = 1 + q which evaluates to 3 when q = 2. Proof. It’s the last day of term, let’s just do the case k = 1. Each line consists of q − 1 nonzero vectors (a line is the set of all scalar multiples of a given nonzero vector) and any two distinct lines meet in just the zero vector.
So the number of lines is the size of Fnq \{0} divided by q −1 which equals [n]q = n1 q as required.  This fact, that if you put q = 1 in the formula for the number of k-dimensional subspaces of a n-dimensional Fq -vector space you get the number of k-subsets of a set of size n, is so odd that you should demand an explanation. Why should it work? The rest of the talk is dedicated to finding out. 3. Action of the general linear group Definition 3.1. GL(n, q) is the set of all n × n invertible matrices with entries from Fq .1  We can use GL(n, q) to count the size of the Grassmannians in a different and hopefully more enlightening way. Let V be the standard k-dimensional subspace of Fnq : V = {(a1 , . . . , ak , 0, . . . , 0)T : a1 , . . . , ak ∈ Fq }. If A is an n × n invertible matrix then so is AV = {Av : v ∈ V }. 1What should GL(n, 1) be?

F1 AND THE BRUHAT DECOMPOSITION

3

What’s more, for any other k-dimensional subspace U there’s an A ∈ GL(n, q) such that AV = U . (why?) So there is a surjective map π : GL(n, q) → Gr(k, Fnq ) A 7→ AV. This is no good for counting unless we know to what extent π fails to be one-toone. AV = BV iff A = BX for some X with XV = V , so if P is the subgroup of GL(n, q) consisting of matrices X such that XV = V then our map is exactly |P |-to-one everywhere. So | Gr(k, Fnq )| = |GL(n, q)|/|P | = |GL(n, q)/P |. P is all invertible matrices of the form  A 0

B C



where A is k × k; this is an example of a parabolic subgroup of GL(n, q). This gives another way of computing the size
of the Grassmannian2 that will shed more light on why it has to specialize to nk when we put q = 1. To make it work we need more information on the structure of GL(n, q). 4. The Bruhat decomposition Definition 4.1. A permutation matrix is one obtained by rearranging the columns of the identity matrix. Sn is the set of all n × n permutation matrices.  Example 4.2. When n = 2, the  1 0

only permutation matrices are    0 0 1 and . 1 1 0



These matrices form a subgroup of GL(n, q) of size n!. Let B be the subgroup of GL(n, q) consisting of all the upper triangular matrices in that group. We can compute the size of B easily: the determinant of an upper triangular matrix is the product of its diagonal entries, so such a matrix is invertible if and only if all the diagonal entries are nonzero. So there are q − 1 choices for each of the n diagonal entries and q for each of the n(n − 1)/2 entries above the diagonal, giving |B| = (q − 1)n q n(n−1)/2 . G Theorem 4.3. (Bruhat decomposition) GL(n, q) = BwB. w∈Sn

Example 4.4. For n = 2 this is just GL(2, q) = B t (G \ B).



The Bruhat decomposition can be thought of as a version of the result you probably saw in first year linear algebra, that you can reduce any invertible matrix to the identity by doing row operations and column operations. A matrix product XY can be interpreted as “do certain column ops described by Y to X,” or “do certain row ops determined by X to Y ” — what the Bruhat decomposition says is that if you’re only allowed to do “upper triangular” row and column ops of the form The case n = 2 is extremely instructive. There’s a generalization of the Bruhat decomposition which says G GL(n, q) = BwP Sn /Sn ∩P 2Another way is to write down a two-variable recurrence relation for s n nk = | Gr(k, Fq )| for
fixed q and show that it specializes to the Pascal’s triangle recurrence relation for bnk = n when k q = 1.

4

MATTHEW TOWERS

Sn ∩ P consists of the permutation matrices in P , which are those in which you permute the first k columns amongst themselves. There are
k!(n − k)! of these, so the number of cosets of Sn ∩ P in Sn is n!/(k!(n − k)!) = nk . We can say something about the size of BwP : |BwP | = |w−1 BwP | = |w−1 Bw||P |/|w−1 Bw ∩ P | = |B||P |/|w−1 Bw ∩ P | Recall |B| = (q − 1)n q n(n−1)/2 , with the (q − 1)n part coming from the diagonal subgroup T . Since w−1 T w = T we have T ≤ w−1 Bw ∩ P , so (q − 1)n divides |w−1 Bw ∩ P | and |BwP | = q dw |P | for some positive integer dw (exercise: what is it?). Finally, G X X n |BwP |/|P | = q dw BwP/P = | Gr(k, Fq )| = |GL(n, q)/P | = Sn /Sn ∩P Sn /Sn ∩P Sn /Sn ∩P
n is a sum of |Sn /Sn ∩ P | = k terms, each of which specializes to 1 when you put q = 1. This explains why putting q = 1 in the number of k-dimensional subspaces of Fnq recovers the number of k-subsets of {1, . . . , n}. 5. Links • • • • • •

http://en.wikipedia.org/wiki/Field_with_one_element http://www.its.caltech.edu/~matilde/F1msri.pdf http://ncatlab.org/nlab/show/field+with+one+element http://mathoverflow.net/q/2300 What is the field with one element? http://www.neverendingbooks.org/the-f_un-folklore http://mathoverflow.net/q/69389 on how to prove the Riemann hypothesis with F1 . References

[CC09]

Alain Connes and Caterina Consani. Schemes over F1 and zeta functions. Version 3. Mar. 11, 2009. arXiv: http://arxiv.org/abs/0903. 2024v3 [math.AG, math.NT]. [CCM08] Alain Connes, Caterina Consani, and Matilde Marcolli. Fun with F1 . Version 1. June 14, 2008. arXiv: http://arxiv.org/abs/0806.2401v1 [math.AG, math.NT, math.OA, math.QA, 14G40, 14G10, 11G25]. [Coh04] Henry Cohn. “Projective geometry over F1 and the Gaussian binomial coefficients”. In: Amer. Math. Monthly 111.6 (2004), pp. 487–495. issn: 0002-9890. doi: 10 . 2307 / 4145067. url: http : / / dx . doi . org / 10 . 2307/4145067. [Man08] Yuri I. Manin. Cyclotomy and analytic geometry over F1 . Version 2. Sept. 9, 2008. arXiv: http://arxiv.org/abs/0809.1564v2 [math.AG, 14H10]. [PL09] Javier L´ opez Pe˜ na and Oliver Lorscheid. Mapping F1 -land: An overview of geometries over the field with one element. Version 1. Sept. 1, 2009. arXiv: http://arxiv.org/abs/0909.0069v1 [math.AG]. [Sou04] Christophe Soul´e. “Les vari´et´es sur le corps `a un ´el´ement”. In: Mosc. Math. J. 4.1 (2004), pp. 217–244, 312. issn: 1609-3321. E-mail address: [email protected]