2
$\begingroup$

These notes on affine algebraic groups mention the following theorem.

Let $G$ be a connected nilpotent affine algebraic group (over an algebraically closed field $k$), and denote $G_s$ and $G_u$ its respective semisimple and unipotent parts. Then, $G_s$ and $G_u$ are closed normal subgroups of $G$, and the natural map $G_s \times G_u \to G$ is a isomorphism of algebraic groups.

Then it makes the remark (page 24) that in such a $G$, the semisimple part $G_s$ is then a torus (1). Which I don't get.

I get that with Lie-Kolchin's theorem, $G_s \simeq G/G_u$ is identified with a subgroup of $D_n$ (the diagonal matrices of $\mathrm{GL}_n(k)$, which is torus). But I don't see why it must be a torus. The remark seems to point out that it comes from the connectedness of $G_s$ (as a quotient of the connected group $G$), but it does not help me. Any hint ?


(1) A torus is a direct power of the multiplicative group $\mathbb G_m = (k^\times,\cdot)$.

$\endgroup$
  • 1
    $\begingroup$ I think you are asking why a connected subgroup of a torus is a torus, right? Theorem 7.7 should be enough: the only algebraic subgroups of the torus are $G_m^n \times \mu_{k_1} \times \ldots \times \mu_{k_j}$, all but the first factor are disconnected. $\endgroup$ – Jack Schmidt Feb 2 '14 at 17:10
  • $\begingroup$ @JackSchmidt That's it thanks. I thought about using character groups but didn't see it. (If you develop it as an answer, I would gladly accept it.) $\endgroup$ – Pece Feb 2 '14 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.