Lang's theorem

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 $H^1({\mathbf{F}}_q,G)=H_{\acute{\mathrm{e}}\mathrm{t}}^1(\mathrm{Spec}{\mathbf{F}}_q,G)$   vanishes, and, consequently, any G-bundle on $\mathrm{Spec}{\mathbf{F}}_q$ 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.

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 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:

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

Then, by identifying the tangent space at a with the tangent space at the identity element, we have:

$(d{f}_a{)}_e=d(h\circ (x\mapsto (x^{-1},a,\sigma (x))){)}_e=d{h}_{(e,a,e)}\circ (-1,0,d{\sigma }_e)=-1+d{\sigma }_e$ 

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 $(d{f}_a{)}_b$ is bijective for any b. 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 $f_1$.

References

• Springer, T. A. (1998). Linear algebraic groups (2nd ed.). Birkhäuser. ISBN 0-8176-4021-5. OCLC 38179868.
• Lang, Serge (1956), "Algebraic groups over finite fields", American Journal of Mathematics, 78: 555–563, doi:10.2307/2372673, ISSN 0002-9327, JSTOR 2372673, MR 0086367
• Steinberg, Robert (1968), Endomorphisms of linear algebraic groups, Memoirs of the American Mathematical Society, No. 80, Providence, R.I.: American Mathematical Society, MR 0230728