# Does equivalence of Lie algebras imply equivalence of the corresponding Lie groups?

Ok so I'm confused about the relation between these two concepts. If I have a Lie Group $G$ I can associate a Lie algebra $\mathfrak{g}$ by taking his tangent space in the identity, with the appropriate bracket.

Now I have three questions (it's actually one):

1) Say I have $\mathfrak{g}$ and $\mathfrak{h}$ Lie algebras of two Lie groups $G$ and $H$. Does $\mathfrak{g} \simeq \mathfrak{h}$ imply $G \simeq H$ (as Lie groups, obviously) ? (I think no)

2) Now the opposite: if I have $G \simeq H$, can I say $\mathfrak{g} \simeq \mathfrak{h}$? (I think yes)

3) What can I say in case one, if the answer is no?

Let me give you an example: talking about Spin groups, wikipedia states that the accidental isomorphisms between low dimensional Spin groups and classic Lie groups are due to root system isomorphisms. But those are a thing of algebras, I don't understand. Can someone give me a brief overview of those concepts? Thanks

• Question 1 can be answered in the negative with the following example: let $G$ be $\mathbb{R}$ under addition and let $H$ be the unit circle in $\mathbb{C}$ under complex multiplication. We have $\mathfrak{g}\cong\mathbb{R}\cong\mathfrak{h}$, but certainly $G$ is not isomorphic to $H$. – 211792 Feb 3 '15 at 18:49
• Another example to how question 1 is negative is that $SO(n)$ and $O(n)$ have the same Lie algebra, but they are not diffeomorphic as Lie groups since one is connected (in fact a subspace of the other) and the other isn't. – Mnifldz Feb 3 '15 at 18:54
• Thanks! Now about this last thing... is there some particular kind of Lie groups that is 1:1 with Lie algebras? Like, I don't know, simply connected ones? – AnalysisStudent0414 Feb 3 '15 at 18:55

1) was answered by AustinC and Mnifldz in the comments: there are many examples showing the answer is "no."

2) This is true. The key idea of the proof is the following lemma: Any group homomorphism $f:G\rightarrow H$ induces a map $f_\ast:\mathfrak{g}\rightarrow \mathfrak{h}$ in a functorial fashion. That is, if the group homomorphism is the identity map, then so is the induced map. Further, the induced map of a composition is the composition of the induced maps.

Assuming this, suppose $f:G\rightarrow H$ is an isomorphism with inverse $g:H\rightarrow G$. Then $g\circ f$ is the identity on $G$, so the induced map is the identity on $\mathfrak{g}$. On the other hand, the induced map is the composition $g_\ast \circ f_\ast$, so $g_\ast \circ f_\ast = Id_{\mathfrak{g}}$. On the other hand, applying the same argument to $f\circ g$ shows that $f_\ast \circ g_\ast = Id_{\mathfrak{h}}$. This implies that $f_\ast$ is an isomorphism of Lie algebras with inverse $g_\ast$.

3) You can say a lot, but not everything. First, since $\mathfrak{g}$ and $\mathfrak{h}$ are computed using the tangent space as the identity, they are indifferent to components other than the identity component of $H$ and $G$. (This is how Mnifldz's example arises). But even if $G$ and $H$ are connected, you can still have $\mathfrak{g}$ and $\mathfrak{h}$ isomorphic without $G$ and $H$ being isomorphic.

You can still save something: If $G$ and $H$ have isomorphic Lie algebras, then there is another (connected) Lie group $K$ with is simultaneously covers both $G$ and $H$. So, up to covers, $G$ and $H$ are the same. (This is how AustinC's example arises: $\mathbb{R}$ covers $S^1$.)

As a simple corollary to this if $G$ and $H$ are simply connected, then an isomorphism on the algebra level does imply that $G$ and $H$ are isomorphic.

There are a few other instances where you can recover something like this. For example, if $G$ and $H$ are compact Lie groups which are centerless, then they have isomorphic Lie algebras iff they are isomorphic Lie groups.