The following theorem can be found in Lee's ``Introduction to Smooth Manifold".

Suppose $G$ is a Lie group with Lie algebra $\mathfrak g.$ If $\mathfrak h$ is a Lie subalgebra of $\mathfrak g$ then there is a unique connected Lie subgroup $H$ of $G$ with Lie algebra $\mathfrak h.$

Are the following statements true? Given the hypothesis of the above theorem, there is a subgroup $H$ of $G$ such that $H$ can be given a topology and a smooth structure such that the natural inclusion map $i:H\to G$ is a Lie group homomorphism and $Di(e)T_{e}H=\mathfrak h.$

My question is in which topology $H$ is connected? Does uniqueness mean that if there is another subgroup $H^\prime$ with a topology and smooth structure, connected in its own topology and $Di(e)T_{e}H=\mathfrak h$ then $H=H^\prime$ and the identity map from $H$ to $H^\prime$ is a Lie group isomorphism?

Can there be some natural counterexamples violating the asumptions?

Definition: Let $M$ be a smooth manifold. Then $S\subseteq M$ is said to be a submanifold if $S$ has its own topology and smooth structure with respect to the inclusion map is a smooth immersion. Definition: Let $G$ be a Lie group. A subgroup $H\subseteq G$ is called a Lie subgroup if $H$ is a submanifold and Lie group with respect to the smooth structure of being a submanifold.

  • $\begingroup$ Surely it means in the subspace topology, considering that $H \subseteq G$ as sets. This is equivalent to the inclusion $i: H \to G$ you consider, as long as you require that $i$ is smooth. $\endgroup$
    – Joppy
    Jan 13, 2021 at 13:00
  • $\begingroup$ @Joppy, please see the "topological" example in my answer. $\endgroup$
    – Malkoun
    Jan 13, 2021 at 13:53

1 Answer 1


I don't have a copy of the book accessible to me at the moment. I think it is a good question. Let me explain a standard "pathological" example first.

Let $V = \mathbb{R}^2$ and $\mathbb{Z}^2 \subset \mathbb{R}^2$ be the integer lattice inside $V$. Then $T^2 := V/\mathbb{Z}^2 \simeq U(1) \times U(1)$ is a (abelian) two-dimensional Lie group, where the Lie group "product" is induced by addition of vectors in $V$.

At $\mathbf{0} \in V$, consider a tangent line $l \subset T_\mathbf{0}(\mathbb{R}^2) = \mathbb{R}^2$. If the slope of $l$ is rational, then $l \mod \mathbb{Z}^2$ defines a Lie subgroup of $T^2$ with the subspace topology (it is a closed submanifold of $T^2$). But if the slope of $l$ is irrational, then $l \mod \mathbb{Z}^2$ defines a Lie subgroup of $T^2$ which is not a closed submanifold.

It is due to pathological examples such as this one that many authors define the notion of a Lie subgroup without requiring that a Lie subgroup have the induced topology.

I don't know the precise definitions that Lee is using in his book, so I cannot discuss further before knowing the precise definitions he uses for a "Lie subgroup" and for a "submanifold".

If the OP wishes to discuss further, please add these definitions to your post, and then I may be able to discuss in more detail.

  • $\begingroup$ > Definitions added. I am particularly interested to known about the connected part. When they say that connected subgroup... does it mean in the induced topology or the other one? $\endgroup$ Jan 14, 2021 at 3:30
  • 1
    $\begingroup$ It is not in the induced topology. It is in the other topology. The induced topology is bad to use for lines with irrational slopes in the example I provided, for instance. For instance, a line with irrational slope is dense in the $2$-torus. $\endgroup$
    – Malkoun
    Jan 14, 2021 at 3:44
  • 1
    $\begingroup$ I mean, it may happen that it is also connected in the induced topology, but the topology they mean when talking about Lie subgroups is in general not the induced topology. It coincides with the induced topology for closed Lie subgroups though. $\endgroup$
    – Malkoun
    Jan 14, 2021 at 3:50

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