Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My reference:

I have two question about Proposition 3.3.:

Proposition3.3. says that $1 \rightarrow Aut(E) \xrightarrow{j} G(E) \xrightarrow{p} T \rightarrow 1 $ is a short exact sequence of algebraic groups over an algebraically closed field $k$, where T is a torus over $k$ and $E$ is a vector bundle of a almost homogeneous variety $X$. I have no question about exactness, but the author says that since both $Aut(E) $ and $T$ are linear algebraic groups, $G(E)$ is a linear algebraic group.


Is that means that if $1 \rightarrow G \xrightarrow{f} H \xrightarrow{g} K \rightarrow 1 $ is a short exact sequence of algebraic groups, then $H$ is linear if, and only if $G$ and $K$ are linear? Is that easy to see?

In the proof of Proposition3.3., the author says that a surjective algebraic group homomorphism from a torus to a torus always has a section (see Borel [1]).


I already found this reference :A. Borel, "Linear algebraic groups", Benjamin (1969) MR0251042 Zbl 0206.49801 Zbl 0186.33201. But I can't find where prove this fact. Who can tell me where is the proof?

Thank you very much!

share|cite|improve this question
up vote 0 down vote accepted

The title of your question is not very good and doesn't give any idea on the question.

To answer your Question 1, note first that if $L$ is a linear group and $A$ is an abelian variety, then all group homomorphisms $A\to L$ and $L\to A$ are constant. The first part is easy because the image of $A$ in $L$ is closed hence affine, and is proper being image of a proper variety, so the image is just one point. For the second part, I don't have an easy proof. One can say either that $L/\mathrm{ker}(L\to A)$ is affine and closed in $A$, hence reduced to one point, or that $L$ is rational, and any morphism from a rational variety to an abelian variety is constant. This can also considered as part of Chevalley's follow theorem: over a perfect field, any smooth connected algebraic group $G$ is uniquely the extension of an abelian variety $A(G)$ by a smooth linear algebraic group $L(G)$: $$ 1\to L(G)\to G \to A(G) \to 1.$$

Now the (affirmative) answer to Question 1 is an easy consequence of the above observation and Chevalley's theorem.

The claim in Question 2 (as you stated) is not correct: consider the isogeny $x\to x^2$ in the simplest torus $\mathbb G_m$. It doesn't have a section. In fact, if $T_2\to T_1$ is a surjective group homomorphism of tori, then there exists a section up to isogeny: $T_1\to T_2$ whose composition with $T_2\to T_1$ is given by $x\to x^n$ for some positive integer $n$. This can be proved using the exact sequence $$ 1\to H\to T_2 \to T_1 \to 1$$ where the connected component of $1$ in $H$, endowed with the reduced structure, is a torus $T_3$. If we denote by $\chi(T)$ the group $\mathrm{Hom}(T, \mathbb G_{m})$ of characters of $T$ (the base field $k$ is algebraically closed here, otherwise, consider the characters over $\bar{k}$), then $$ 0\to \chi(T_3) \to \chi(T_2)\to \chi(T_1) \to 0$$ is exact up to tensoring by $\mathbb Q$. The exact sequence splits up to tensoring by $\mathbb Q$. By the equivalence of categories between the tori and the finite rank free $\mathbb Z$-modules, $T_2$ is isogeneous to $T_1\times T_3$, so the morphism $T_2\to T_1$ has an "isogeny" section.

share|cite|improve this answer
Thank you very much! I will try to realize. (Because there are some definitions, properties and details I have to know) – Peter Hu May 18 '12 at 2:19
Excuse me, I have questions: 1."The exact sequence splits up to tensoring by $\mathbb{Q}$." Dose it means splits by vector spaces over $\mathbb{Q}$? 2.Sorry, could you explain the meaning of "isogeny" section? I can't find proper reference about this concept and definition. Thank you very much! – Peter Hu May 21 '12 at 16:14
@PeterHu: By spliting up to tensoring by $\mathbb Q$, I mean the complex tensored over $\mathbb Z$ by $\mathbb Q$ is an exact sequence (of groups of vector spaces, it doesn't really matter). An isogeny here means a finite surjective homomorphism of (connected) algebraic groups. An "isogeny section" (not really standard terminology) of $f: X\to Y$ means a morphism $g : Y\to X$ such that $g\circ f$ is the multiplication map $Y$ by some positive integer. – user18119 May 22 '12 at 19:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.