# Lang's theorem

(Redirected from Lang map)

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

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

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

The theorem implies that ${\displaystyle H^{1}(\mathbf {F} _{q},G)=H_{\mathrm {{\acute {e}}t} }^{1}(\operatorname {Spec} \mathbf {F} _{q},G)}$   vanishes,[1] and, consequently, any G-bundle on ${\displaystyle \operatorname {Spec} \mathbf {F} _{q}}$ 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 ${\displaystyle \sigma }$ 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 ${\displaystyle \sigma }$ 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:

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

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

${\displaystyle (df_{a})_{e}=d(h\circ (x\mapsto (x^{-1},a,\sigma (x))))_{e}=dh_{(e,a,e)}\circ (-1,0,d\sigma _{e})=-1+d\sigma _{e}}$

where ${\displaystyle h(x,y,z)=xyz}$. It follows ${\displaystyle (df_{a})_{e}}$ is bijective since the differential of the Frobenius ${\displaystyle \sigma }$ vanishes. Since ${\displaystyle f_{a}(bx)=f_{f_{a}(b)}(x)}$, we also see that ${\displaystyle (df_{a})_{b}}$ is bijective for any b.[4] Let X be the closure of the image of ${\displaystyle f_{1}}$. The smooth points of X form an open dense subset; thus, there is some b in G such that ${\displaystyle f_{1}(b)}$ is a smooth point of X. Since the tangent space to X at ${\displaystyle f_{1}(b)}$ 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 ${\displaystyle f_{1}}$ then contains an open dense subset U of G. Now, given an arbitrary element a in G, by the same reasoning, the image of ${\displaystyle f_{a}}$ contains an open dense subset V of G. The intersection ${\displaystyle U\cap V}$ is then nonempty but then this implies a is in the image of ${\displaystyle f_{1}}$.

## Notes

1. ^ This is "unwinding definition". Here, ${\displaystyle H^{1}(\mathbf {F} _{q},G)=H^{1}(\operatorname {Gal} ({\overline {\mathbf {F} _{q}}}/\mathbf {F} _{q}),G({\overline {\mathbf {F} _{q}}}))}$ 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 ${\displaystyle f_{a}}$ is étale.