# Are unipotent algebraic groups connected?

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!

• this answer on mathoverflow strongly suggests that your argument is solid. – Jesko Hüttenhain Mar 19 '15 at 8:53
• @Diego, Do you have a proof about your last statement: "...all subalgebras of $\mathfrak u_n$ are connected."? Thanks. – user328669 May 25 '19 at 23:28

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.