Lang's theorem

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

[math]\displaystyle{ G \to G, \, x \mapsto x^{-1} \sigma(x) }[/math] 

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

The theorem implies that [math]\displaystyle{ H^1(\mathbf{F}_q, G) = H_{\mathrm{\acute{e}t}}^1(\operatorname{Spec}\mathbf{F}_q, G) }[/math]   vanishes,[1] and, consequently, any G-bundle on [math]\displaystyle{ \operatorname{Spec} \mathbf{F}_q }[/math] 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 [math]\displaystyle{ \sigma }[/math] 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 [math]\displaystyle{ \sigma }[/math] that induces a nilpotent operator on the Lie algebra of G.[2]

The Lang–Steinberg theorem

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

Define:

[math]\displaystyle{ f_a: G \to G, \quad f_a(x) = x^{-1}a\sigma(x). }[/math]

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

[math]\displaystyle{ (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 }[/math] 

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

Notes

  1. This is "unwinding definition". Here, [math]\displaystyle{ H^1(\mathbf{F}_q, G) = H^1(\operatorname{Gal}(\overline{\mathbf{F}_q}/\mathbf{F}_q), G(\overline{\mathbf{F}_q})) }[/math] 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 [math]\displaystyle{ f_a }[/math] is étale.

References