Lang's theorem explained
, then, writing
for the Frobenius, the
morphism of varietiesG\toG,x\mapstox-1\sigma(x)
is surjective. Note that the
kernel of this map (i.e.,
G=G(\overline{Fq})\toG(\overline{Fq})
) is precisely
.
The theorem implies that
}^1(\operatorname\mathbf_q, G) vanishes,
[1] and, consequently, any
G-bundle on
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
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
that induces a
nilpotent operator on the Lie algebra of
G.
The Lang–Steinberg theorem
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:
fa:G\toG, fa(x)=x-1a\sigma(x).
Then, by identifying the tangent space at
a with the tangent space at the identity element, we have:
(dfa)e=d(h\circ(x\mapsto(x-1,a,\sigma(x))))e=dh(e,\circ(-1,0,d\sigmae)=-1+d\sigmae
where
. It follows
is bijective since the differential of the Frobenius
vanishes. Since
, we also see that
is bijective for any
b.
[2] Let
X be the closure of the image of
. The
smooth points of
X form an open dense subset; thus, there is some
b in
G such that
is a smooth point of
X. Since the tangent space to
X at
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
then contains an open dense subset
U of
G. Now, given an arbitrary element
a in
G, by the same reasoning, the image of
contains an open dense subset
V of
G. The intersection
is then nonempty but then this implies
a is in the image of
.
References
- Book: Springer, T. A. . Linear algebraic groups . Birkhäuser . 1998 . 0-8176-4021-5 . 38179868 . 2nd.
Notes and References
- This is "unwinding definition". Here,
G)=
| 1(\operatorname{Gal}(\overline{F |
H | |
| q}/F |
q),G(\overline{Fq}))
is Galois cohomology; cf. Milne, Class field theory.
- This implies that
is étale.