# Element of $\mathfrak{sl}(2,\Bbb C)$ corresponding to element in $SL(2,\Bbb C)$

If I have $\begin{bmatrix}a&0\\0&a^{-1}\end{bmatrix}$ in $SL(2,\Bbb C)$, how do I find what element I would have corresponding to this in $\mathfrak{sl}(2,\Bbb C)$? I imagine it might be something like $\begin{bmatrix}a&0\\0&-a\end{bmatrix}$, but I am not sure how to find this.

I know I want to go from determinant $1$ matrices to traceless matrices. But I can't get the correspondence down yet.

The exponential map $\exp: \mathfrak{g}\rightarrow G$ gives you the matrices, however it need not be surjective (or injective) in general. Indeed, for $\mathfrak{g}=\mathfrak{sl}_2(\mathbb{C})$ and $G=SL_2(\mathbb{C})$ it is not - see here, or here. However, the diagonal matrices have preimages in the Lie algebra, as was shown already. So you can find such matrices of trace zero.
In general, there is no one-to-one correspondence between a group $G$ and its Lie algebra $\mathfrak{g}$, but we do have a map $$\exp:\mathfrak{g}\to G.$$
In case of matrix groups, we have that for diagonal matrices $$\exp\begin{pmatrix}s & 0 \\ 0 & t\end{pmatrix}=\begin{pmatrix}e^s & 0 \\ 0 & e^t\end{pmatrix}.$$ Thus, $$\begin{pmatrix}\log a & 0 \\ 0 & -\log a\end{pmatrix}\in\mathfrak{sl}(2,\mathbb{C})$$ is an element you are looking for (for any branch of $\log$ not passing through $a$). But note that it is not unique, as we may add multiples of $2\pi i$.
The correspondence between elements is given by the exponential map. This is actually a pretty deep statement in Lie theory, but for this particular case suffice it to say that the identity $$e^{\text{tr}(A)} = \det(e^A)$$ holds (not hard to show with Jordan normal form), and this shows that $$\text{tr}(A) = 0 \iff \det(e^A) = 1$$ For diagonal matrices, the exponential map (and it's inverse) are easy to calculate: you'll just apply a complex logarithm to the diagonal entries of your element of $SL(2,\mathbb{C})$.