Sylow の定理と代数幾何

Daniel Litt @littmath

In celebration of reaching 3k followers, here's a thread on the Sylow theorems and algebraic geometry. I'll start by recalling the Sylow theorems, and then explain how they are in some cases manifestations of the geometry of certain algebraic varieties. 1/n pic.twitter.com/qbWDxxqxQx

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Daniel Litt @littmath

Let G be a finite group (say, the set of symmetries of some object) and p a prime. If the size |G| of G is p^k*m, where m is not divisible by p, we call a subgroup of G a *p-Sylow subgroup* if it has size p^k. The first Sylow theorem says p-Sylow subgroups always exist. 2/n

Daniel Litt @littmath

That is, there is always a subgroup of G of order p^k, where k is as big as can be (since the size of a subgroup always divides the size of the group). 3/n

Daniel Litt @littmath

For example, let G be the set of symmetries of a regular tetrahedron. this has size 4!=24=2^3*3, so we want to construct a subgroup of size 3 and a subgroup of size 2^3=8. 4/n pic.twitter.com/1gJzXyPdO8

Daniel Litt @littmath

Making the four 3-Sylow subgroups is easy; they're just given by rotation 0, 120, or 240 degrees along the axis between a vertex and the center of the opposite face (that is, the dotted blue line in the picture below). 5/n pic.twitter.com/m5i0A4Ypw4

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Daniel Litt @littmath

The 2-Sylow subgroups, of order 8, are harder to see. Divide the six edges of the tetrahedron into three sets of two disjoint edges each (labeled {u,U}, {v, V}, {w,W} in the picture). Then fix one of these pairs, say (v,V). The set of symmetries preserving {v,V} is a 2-Sylow. 6/n pic.twitter.com/AtPnONHw0H

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Daniel Litt @littmath

The second Sylow theorem says that any two p-Sylow subgroups are conjugate to one another. In our example, a 3-Sylow is determined by an axis of rotation. So if a symmetry takes one axis of rotation to another, it conjugates the corresponding 3-Sylows into one another. 7/n

Daniel Litt @littmath

A 2-Sylow is determined by a pair of disjoint edges of the tetrahedron; again, a symmetry sending one such pair into another conjugates the corresponding 2-Sylows into one another. 8/n

Daniel Litt @littmath

Finally, the third Sylow theorem says that the *number* of p-Sylow subgroups is always 1 modulo p. For example, their are four 3-Sylow subgroups above (as there are four possible axes of rotation) and three 2-Sylow subgroups (as there are three pairs of disjoint edges). 9/n

Daniel Litt @littmath

OK, now I want to talk about how this is related to algebraic geometry. Let p be a prime, let F_p be the field with p elements, and let G=GL_n(F_p) be the group of n x n invertible matrices with coefficients in F_p. 10/n

Daniel Litt @littmath

This is the set of automorphisms of an n-dimensional F_p-vector space V=F_p^n. What is the size of this group? 11/n

Daniel Litt @littmath

Well, the first column of an invertible matrix can be any non-zero vector, so there are p^n-1 choices. The second can be any vector not in the span of the first, so there are p^n-p choices. The third can be anything not in the span of the first two, so p^n-p^2 choices. 12/n

Daniel Litt @littmath

Continuing in this manner, we see that the size of GL_n(F_p) is (p^n-1)(p^n-p)(p^n-p^2)...(p^n-p^{n-1}). The biggest power of p dividing this number is p^{1+2+...+(n-2)+(n-1)}=p^{n(n-1)/2}, so a p-Sylow of G should have size p^{n(n-1}/2}. 13/n

Daniel Litt @littmath

Let's make such a subgroup! Take upper-triangular matrices with 1's on the diagonal. Such matrices are always invertible, so we can fill in the upper right triangle however we want. There are n(n-1)/2 entries, and we have p choices for each, so this is indeed a p-Sylow. 14/n pic.twitter.com/0LFIghML1v

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Daniel Litt @littmath

Now the second Sylow theorems tells us that any other p-Sylow is conjugate to this one, and the third tells us that the number of such p-Sylows is equal to 1 mod p. But what does this mean geometrically? 15/n

Daniel Litt @littmath

Suppose you're given a *full flag* of your vector space V. That is, a sequence of subspace V_1 ⊂ V_2 ⊂ V_3 ⊂ ... ⊂ V_n=V, where V_i is i-dimensional. Consider the set of matrices which preserve this flag and act trivially on V_i/V_{i-1} for each i. This is a p-Sylow! 16/n

Daniel Litt @littmath

Indeed, our original p-Sylow is the case where each V_i is spanned by the first i standard basis vectors. 17/n

Daniel Litt @littmath

Given an arbitrary full flag, choosing a basis where V_i is spanned by the first i basis vectors shows that any full flag can be sent to any other by some element of G; hence the same element of G conjugates the corresponding p-Sylows into one another. 18/n

Daniel Litt @littmath

Finally, let's count full flags, and hence p-Sylows. How many choices are there for V_1? Well, if we choose a vector spanning V_1, there are p^n-1 choices. But rescaling our vector gives the same V_1, so there are really (p^n-1)/(p-1) = 1+p+...+p^{n-1} choices. 19/n

Daniel Litt @littmath

Now given any full flag V_1 ⊂ V_2 ⊂ V_3 ⊂ ..., quotienting by V_1 gives a full flag of an (n-1)-dimensional vector space. So induction tells us that the number of full flags is (p^n-1)(p^{n-1}-1)...(p^2-1)/(p-1)^{n-1}= (1+p+...+p^{n-1})(1+p+...+p^{n-2})...(1+p). 20/n

Daniel Litt @littmath

And this is indeed 1 modulo p, as claimed. 21/n

Daniel Litt @littmath

OK, but I promised you some geometry. So let's find a version of the Sylow theorems over the complex numbers! Set G=GL_n(C), the set of n x n invertible matrices with complex coefficients. This is the set of automorphisms of the vector space V=C^n. 22/n

Daniel Litt @littmath

This is an infinite group, so we can't look for p-Sylows; instead, taking a page out of our GL_n(F_p) example, we'll look for subgroups which stabilize a *full flag* V_1 ⊂ V_2 ⊂ V_3 ⊂ ... ⊂ V_n=V and act trivially on V_i/V_{i-1}. 23/n

Daniel Litt @littmath

These subgroups have a name and an intrinsic description; they're called "maximal unipotent subgroups." 24/n

Daniel Litt @littmath

The same argument as before -- that GL_n(C) acts transitively on the set of full flags -- tells us that any two such subgroups are conjugate to one another. But what's the analogue of the third Sylow theorem? 25/n