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 $\mathbf_q$, then, writing $\sigma: G \to G, \, x \mapsto x^q$ for the Frobenius, the morphism of varieties
:$G \to G, \, x \mapsto x^ \sigma(x)$ 
is surjective. Note that the kernel of this map (i.e., $G = G(\overline) \to G(\overline)$) is precisely $G(\mathbf_q)$.

The theorem implies that $H^1(\mathbf_q, G) = H_^1(\operatorname\mathbf_q, G)$  
vanishes, and, consequently, any ''G''-bundle on $\operatorname \mathbf_q$ 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

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''^−1''F''(''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:
:$f_a: G \to G, \quad f_a(x) = x^a\sigma(x).$
Then (identifying the tangent space at ''a'' with the tangent space at the identity element) we have:
:$(d f_a)_e = d(h \circ (x \mapsto (x^, a, \sigma(x))))_e = dh_ \circ (-1, 0, d\sigma_e) = -1 + d \sigma_e$ 
where $h(x, y, z) = xyz$. It follows $(d f_a)_e$ is bijective since the differential of the Frobenius $\sigma$ vanishes. Since $f_a(bx) = f_(x)$, we also see that $(df_a)_b$ is bijective for any ''b''.This implies that $f_a$ is étale. 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

References

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*{{Citation , last1=Steinberg , first1=Robert , title=Endomorphisms of linear algebraic groups , url=https://books.google.com/books?id=54HO1wDNM_YC , publisher= American Mathematical Society , location=Providence, R.I. , series=Memoirs of the American Mathematical Society, No. 80 , mr=0230728 , year=1968

Algebraic groups
Theorems in algebraic geometry