Confused about coverings for Lie groups I'm trying to establish a cover for one lie group (SL(2,C)) to another (SO(3,C)). I can find an isomorphism from the lie algebras sl(2,c) to so(3,c) by taking $
  \left[ {\begin{array}{cc}
   a & b \\
   c & -a \\
  \end{array} } \right]
$
to 
$
  \left[ {\begin{array}{cc}
   0 & a & b\\
   -a^* & 0 & c \\
   -b^* & -c^* & 0 \\
  \end{array} } \right]
$
However, I am utterly confused as to how to begin thinking about finding the surjection from the actual lie groups. I think that the correct approach is to use Ad, but I don't know how to prove that it maps from SL(2,C) to SO(3,C). I'm also confused as to how to think about sl(2,C) and so(3,C) as vector spaces and how to compare how SL(2,C) sends these elements to how SO(3,C) does. I was wondering if there were any tips for how to think about this problem and approach it more rigorously. 
 A: Reading your question, I get the impression that you did not interpret the hint to use the adjoint action correctly. What is meant by the hint is that the adjoint representation of $SL(2,\mathbb C)$ can be interpreted as a homomorphism $SL(2,\mathbb C)\to GL(\mathfrak{sl}(2,\mathbb C))\cong GL(3,\mathbb C)$. Now it is easy to see that there is a non-degenerate complex bilinear form $b$ on $\mathfrak{sl}(2,\mathbb C)$ which is invariant under $Ad(A)$ for each $A\in SL(2,\mathbb C)$. The orthogonal group of this bilinear form is isomorphic to $SO(3,\mathbb C)$, so you can actually view $Ad$ as a homomorphism $SL(2,\mathbb C)\to SO(3,\mathbb C)$. It is easy to show that the derivative of this homomorphism is a linear isomorphism (twice the isomorphism you have written out in the question). A bit of Lie theory then shows that the homomorphism has to be a covering and it is easy to see that the kernel has two elements. 
I wouldn't say that this is "the general way to think about the question", but I am not sure that there is such a "general way". It is a general theorem that for Lie groups $G$ and $H$ with Lie algebras $\mathfrak g$ and $\mathfrak h$ with $G$ simply connected and $H$ connected and isomoprhism $\mathfrak g\to\mathfrak h$ is the derivative of a unique homomorphism $G\to H$ which automatically is a covering. If you want to write out such an isomorphism explicitly in general the starting point will be to look at exponential images and use that $\phi(exp(X))=exp(\phi'(X))$. 
