# Why ins't this SU(4) Matrix produced by the exponential map?

I was working with the SU(4) Lie group, which is compact and simply connected. This should imply that the exponential map is sujective on the group.

However i came across the matrix $$G=\begin{pmatrix}i & 0 & 0 & 0\\0 & i & 0 & 0\\0 & 0 & i & 0\\0 & 0 & 0 & i\end{pmatrix},$$ which is special unitary, but does not seem to be the exponential of any algebra element, since $$log(G)=\begin{pmatrix}\frac{i\pi}{2} & 0 & 0 & 0\\0 & \frac{i\pi}{2} & 0 & 0\\0 & 0 & \frac{i\pi}{2} & 0\\0 & 0 & 0 & \frac{i\pi}{2}\end{pmatrix}$$ is not traceless.

How could this be? Could someone help me see what I am missing here?

The exponential map from $$\mathfrak{su}(4)$$ $$\textit{is}$$ onto $$SU(4)$$. It should be emphasized that this means for any given $$SU(4)$$ matrix, $$\textit{there exists}$$ an $$\mathfrak{su}(4)$$ matrix that maps onto it by the exponential map. I'll show you such a matrix: $$$$\mathfrak{g}=\begin{pmatrix}-\frac{3\pi i}{2} & 0 & 0 & 0\\0 & \frac{\pi i}{2} & 0 & 0\\0 & 0 & \frac{\pi i}{2} & 0\\0 & 0 & 0 & \frac{\pi i}{2}\end{pmatrix}$$$$ This $$\textit{is}$$ traceless. It's important to keep in mind that the logarithmic map from a Lie Group to the associated Lie Algebra is branching, and not every branch of its action on an element of the Lie Group must lie in the corresponding Lie Algebra.