4
$\begingroup$

Is a unipotent algebraic group over a field of characteristic zero always connected?. As far as I know, every unipotent algebraic group over field of characteristic zero is isomorphic to a closed subgroup of $U_n$, the group of upper triangular matrices with 1's in the diagonal. I think that every closed subgroup of $U_n$ is the image by $\exp:\mathfrak{u}_n\rightarrow U_n$ of a subalgebra of $\mathfrak{u}_n$, where $\mathfrak{u}_n$ is the Lie algebra of upper triangular matrices with 0's in the diagonal. My assertion follows because all subalgebras of $\mathfrak{u}_n$ are connected. Am I right?

Thanks!

$\endgroup$
  • 1
    $\begingroup$ this answer on mathoverflow strongly suggests that your argument is solid. $\endgroup$ – Jesko Hüttenhain Mar 19 '15 at 8:53
  • $\begingroup$ @Diego, Do you have a proof about your last statement: "...all subalgebras of $\mathfrak u_n$ are connected."? Thanks. $\endgroup$ – Amrat A May 25 at 23:28
4
$\begingroup$

A unipotent group $U$ over a field $K$ of characteristic zero is always connected; moreover, the exponential mapping $\exp\colon \mathfrak{u}\rightarrow U$, where $\mathfrak{u}$ is the Lie algebra of $U$ is an isomorphism of algebraic varieties.
This is no longer true in in prime characteristic. Over a field $K$ of characteristic $p>0$ there exist non-connected unipotent algebraic groups, e.g. the additive group $G_a$ of the ground field (which may be identified with $U_2(K)$) is a $p$-group and so contains a finite unipotent group.

$\endgroup$
  • $\begingroup$ Howd does the exponential being isomorphism imply that any subgroup is connected? Thanks. $\endgroup$ – Amrat A May 25 at 23:23

Your Answer

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

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