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

Question

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.

Answer 1

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.

Answer 2

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 $2\pi i$.

Answer 3

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 $SL(2,\mathbb{C})$.