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$G$ is a connected Lie group with Lie algebra $g$ and $l$ is an abelian ideal of $g$. If $K$ is the connected Lie subgroup of $G$ with the Lie algebra $l$, then is $K$ necessarily closed in $G$?

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up vote 2 down vote accepted

There are abelian ideals in the Lie algebra of the $2$-dimensional torus $T^2=S^1\times S^2$ which are tangent to connected Lie subgroups which are not cloed.

To build an example, let $p:(x,y)\in\mathbb R^2\to (\exp 2\pi ix,\exp 2\pi iy)\in T^2$. This is a Lie group homomorphism. Let $dp:T_e\mathbb R^2\to T_eT^2$ be the differential at the identity elements; it s in fact an isomorphism. Let $S=\{(x,x\sqrt{2}):x\in\mathbb R\}$, which is a subspace of $\mathbb R^2$, so it is a subspace of $T_e\mathbb R^2$, and, in fact, a Lie subalgebra. Consider $\mathfrak l=dp(S)$, the image under $dp$ of $S$. This is a $1$-dimensional Lie subalgebra of $T_eT^2$.

Now check that the connected Lie subgroup of $T^2$ which has $\mathfrak l$ as Lie algebra is not closed.

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This is the canonical example jumped to my mind... – Bombyx mori Apr 17 '13 at 6:23

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