Lang's theorem

From Wikipedia, the free encyclopedia
  (Redirected from Lang map)
Jump to: navigation, search

In algebraic geometry, Lang's theorem, introduced by Serge Lang, states: if G is a connected smooth algebraic group over a finite field \mathbf{F}_q, then, writing \sigma: G \to G, \, x \mapsto x^q for the Frobenius, the morphism of varieties

G \to G, \, x \mapsto x^{-1} \sigma(x) 

is surjective. Note that the kernel of this map (i.e., G = G(\overline{\mathbf{F}_q}) \to G(\overline{\mathbf{F}_q})) is precisely G(\mathbf{F}_q).

The theorem implies that H^1(\mathbf{F}_q, G) = H_{\text{ét}}^1(\operatorname{Spec}\mathbf{F}_q, G)   vanishes,[1] and, consequently, any G-bundle on \operatorname{Spec} \mathbf{F}_q is isomorphic to the trivial one. Also, the theorem plays a basic role in the theory of finite groups of Lie type.

It is not necessary that G is affine. Thus, the theorem also applies to abelian varieties (e.g., elliptic curves.) In fact, this application was Lang's initial motivation. If G is affine, the Frobenius \sigma may be replaced by any surjective map with finitely many fixed points (see below for the precise statement.)

The proof (given below) actually goes through for any \sigma that induces a nilpotent operator on the Lie algebra of G.[2]

The Lang–Steinberg theorem[edit]

Steinberg (1968) gave a useful improvement to the theorem.

Suppose that F is an endomorphism of an algebraic group G. The Lang map is the map from G to G taking g to g−1F(g).

The Lang–Steinberg theorem states[3] that if F is surjective and has a finite number of fixed points, and G is a connected affine algebraic group over an algebraically closed field, then the Lang map is surjective.

Proof of Lang's theorem[edit]

Define:

f_a: G \to G, \quad f_a(x) = x^{-1}a\sigma(x).

Then (identifying the tangent space at a with the tangent space at the identity element) we have:

(d f_a)_e = d(h \circ (x \mapsto (x^{-1}, a, \sigma(x))))_e = dh_{(e, a, e)} \circ (-1, 0, d\sigma_e) = -1 + d \sigma_e 

where h(x, y, z) = xyz. It follows (d f_a)_e is bijective since the differential of the Frobenius \sigma vanishes. Since f_a(bx) = f_{f_a(b)}(x), we also see that (df_a)_b is bijective for any b.[4] Let X be the closure of the image of f_1. The smooth points of X form an open dense subset; thus, there is some b in G such that f_1(b) is a smooth point of X. Since the tangent space to X at f_1(b) and the tangent space to G at b have the same dimension, it follows that X and G have the same dimension, since G is smooth. Since G is connected, the image of f_1 then contains an open dense subset U of G. Now, given an arbitrary element a in G, by the same reasoning, the image of f_a contains an open dense subset V of G. The intersection U \cap V is then nonempty but then this implies a is in the image of f_1.

Notes[edit]

  1. ^ This is "unwinding definition". Here, H^1(\mathbf{F}_q, G) = H^1(\operatorname{Gal}(\overline{\mathbf{F}_q}/\mathbf{F}_q), G(\overline{\mathbf{F}_q})) is Galois cohomology; cf. Milne, Class field theory.
  2. ^ Springer 1998, Exercise 4.4.18.
  3. ^ Steinberg 1968, Theorem 10.1
  4. ^ This implies that f_a is étale.

References[edit]