Lie algebra homomorphism: is my understanding correct?

Using answer to my previous question I made some progress towards understanding Lie algebra homomorphisms. But of course I am unsure whether my thoughts are really correct so again I'd like to request the community to check my thoughts.

Is the following correct?

It is my goal to find a Lie algebra isomorphism from the Lie algebra of $SL_2(\mathbb C)$ to the Lie algebra of $O(3, \mathbb C)$.

The group $G = SL_2(\mathbb C)$ is both connected and simply connected. The following is a theorem (it can be found e.g. here):

For Lie groups $G, H$ with $G$ connected and simply connected, a linear map $\varphi : \mathfrak g \to \mathfrak h$ is the derivative of a homomorphism $\phi : G \to H$ if and only if φ is a Lie algebra homomorphism.

Hence if $\phi: SL_2 \to O(3,\mathbb C)$ is a Lie group homomorphism, its derivative will yield a Lie group homomorphism.

(If $\phi: SL_2 \to O(3,\mathbb C)$ is a Lie group isomorphism is its derivative a Lie algebra isomorphism?)

Therefore, to find the desired Lie algerba isomorphism I pick a basis generators for $SL_2$ and a basis generators for $O(3, \mathbb C)$, define a map on the generators and then take its derivative.

But the derivative of a linear map is the linear map itself here. This leads me to believe that I made a mistake somewhere. Or did I not?

Edit

In response to Ben's comment: Let's replace "basis" with "generators" instead.

• Lie groups are not linear spaces, so the group homomorphism will not be a linear map. The phrase "pick a basis for $SL_2$" doesn't make sense because $SL_2$ isn't a linear space. – Ben Blum-Smith Feb 3 '15 at 5:27
• @BenBlum-Smith Oops, of course. You are right. What if I replace "basis" with "generators"? It's not yet obvious to me at this moment how to take a derivative of a group homomorphism though... – learner Feb 3 '15 at 6:07
• $\mathrm{SL}_2(\mathbb C)$ is not finitely generated as a group, and even if it were generators aren't as nice as basis elements of a vector space. Generators can have relations which need to be respected by their images in the group you're mapping to. – Jim Feb 3 '15 at 6:27
• @Jim Thank you, I did not know this. So... is it not possible to specify a group homomorphism? – learner Feb 3 '15 at 6:28
• Sure it is, you can just use the matrix coordinates to do it. – Jim Feb 3 '15 at 6:29

You are confusing Lie groups with Lie algebras. The Lie group $\mathrm{SL}_2(\mathbb C)$ is $$\mathrm{SL}_2(\mathbb C) = \left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix} \ \middle| \ a, b, c, d \in \mathbb C \ \text{and} \ ad - bc = 1\right\}$$ and it's Lie algebra is $$\mathfrak{sl}_2(\mathbb C) = \left\{\begin{bmatrix} a & b \\ c & -a \end{bmatrix} \ \middle| \ a, b, c \in \mathbb C\right\}$$ Lie algebras are vector spaces and therefore have bases. Lie groups are in general not vector spaces (in particular $\mathrm{SL}_2(\mathbb C)$ is not naturally a vector space) so it does not make sense to talk about a basis of $\mathrm{SL}_2(\mathbb C)$.
As to whether you can specify an isomorphism $\mathrm{SL}_2(\mathbb C) \to \mathrm O_2(\mathbb C)$, you can't. Those groups are not isomorphic Lie groups. For example, $\mathrm{SL}_2(\mathbb C)$ is connected and $\mathrm O_2(\mathbb C)$ is not.
• Thank you for your help! Does this mean my idea doesn't work or can I instead use just a group homo $SL_2(\mathbb C)\to O(3,\mathbb C)$ and take the derivative of that to get a Lie algebra iso? – learner Feb 3 '15 at 7:00