# Connected linear Lie groups $G$ generated by $\exp (\mathfrak{g})$.

Let $$G$$ be a connected linear Lie group with Lie algebra $$\mathfrak{g}$$. I understand that any open neighborhood of the identity of $$G$$ generates it, but, why does $$\exp(\mathfrak{g})$$ also generate it? Is $$\exp(\mathfrak{g})$$ open?

• $\exp$ is a local diffeomorphism at $0$, so it is open around $0$, hence for a neighbourhood $V$ of $0$, $\exp(V)$ is a neighbourhood of $1$, which thus generates $G$; therefore so does $\exp(\mathfrak{g})$ Jan 16, 2019 at 23:03
• @Max I wrote a comment exactly like yours at first but then I deleted it. I think the statement given by the OP is wrong. See my answer and the comments below it. What do you think? Jan 16, 2019 at 23:09
• @stressedout : you only proved that $\exp (\mathfrak{g})$ is not $G$, but that doesn't prove anything, the subgroup generated by $\exp(\mathfrak{g})$ can (and will in many cases) be strictly larger. Note that $\exp(\mathfrak{g})$ is not necessarily a subgroup ! Jan 16, 2019 at 23:12
• @stressedout : read the first answer to the post. It begins with "This is not true in general" Jan 16, 2019 at 23:14
• @Max You're right. I misread his post. Sorry. Jan 16, 2019 at 23:19

$$\exp(\cdot): \mathfrak{g}\to G$$ is a local diffeomorphism. Hence, it is a local homeomorphism and it's open. This means that $$\exp(\cdot)$$ maps an open neighborhood $$0\in V \subseteq \mathfrak{g}$$ to an open neighborhood around the identity of $$G$$. Now, $$H=\langle \exp(V)\rangle$$ is a connected component of $$G$$ containing the identity. Since $$G$$ is connected, this subgroup must be $$G$$.
• By generating I mean that $G = \bigcup_{k=1}^{\infty}\exp(\mathfrak{g})^{k}$ Jan 16, 2019 at 22:59
• @ stressed out - I agree with what you just proved, it happens that in the book of JACQUES FARAUT, Analysis on Lie Groups, page 43, Corollary 3.3.5., says that the connected component $G_{0}$ of the identity is generated by $\exp(\frak{g})$, so we can deduce that if $G$ is connected then it would be generated by $\exp(\frak{g})$. Is it probable that Fauraut is wrong with the statement $G_{0} = \bigcup_{k=1}^{\infty}{\exp(\frak{g})}^k$? Jan 17, 2019 at 0:09
• @fer6268 Unfortunately, I don't have access to that book right now. But I don't think that he's wrong. Saying that $\langle \exp(\mathfrak{g}) \rangle = G_0$ means that $G_0$ must contain all elements of $\exp(\mathfrak{g})$ and their powers because it's a group under multiplication. In fact, if you remember the previous example where I made a mistake, you need to be able to take square roots. So, you definitely need $\exp(\mathfrak{g})^2$ to be in $G_0$ and so forth. So, $G_0 = \langle \exp(\mathfrak{g}) \rangle = \cup_{k=1}^{\infty} \exp(\mathfrak{g})^k$ and I see nothing wrong with it. Jan 17, 2019 at 6:11
• @fer6268 I checked the book you mentioned. Check the proposition 1.1.1 for more information. But my last comment is also true. Proposition 1.1.1 will tell you why $\langle \exp(\mathfrak{g}) \rangle$ is the identity component $G_0$. But the algebraic reason I gave you also tells you why it is equal to the union of the powers of $\exp(\mathfrak{g})$ and there's no contradiction. Do you still have a question? Jan 19, 2019 at 5:36
• @ stressed out Thank you for reviewing the statement, and thank you for your algebraic explanations, which helped me understand why the statement is correct. What causes me confusion is that the proposition 1.1.1 says that the CONNECTED NEIGHBORS of identity generate $G_{0}$, and I am not very convinced that $\exp (\mathfrak{g})$ is a connected neighborhood of identity. Jan 19, 2019 at 5:48