Show that $\exp: \mathfrak{sl}(n,\mathbb R)\to \operatorname{SL}(n,\mathbb R)$ is not surjective It is well known that for $n=2$, this holds. The polar decomposition provides the topology of $\operatorname{SL}(n,\mathbb R)$ as the product of symmetric matrices and orthogonal matrices, which can be written as the product of exponentials of skew symmetric and symmetric traceless matrices. However I could not find  out the proof that  $\exp: \mathfrak{sl}(n,\mathbb R)\to\operatorname{SL}(n,\mathbb R)$ is not surjective for $n\geq 3$.
 A: Over $\mathbb{R}$, for a general $n$, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, you may verify that $\pmatrix{-1&1\\ 0&-1}$ (as given by rschwieb's answer) and $\operatorname{diag}(-2,-\frac12,1,\ldots,1)$ (as given by Gokler's answer) are not matrix exponentials of a real traceless matrix.
A: I remember learning somewhere that every matrix in the image of the exponential from $\mathfrak{sl}(n,\Bbb R)$ is either diagonalizable or is unipotent. So, any defective element which is not unipotent would be missed.
Defective matrices with determinant 1 can be built by installing a defective block. For example:
$$
\begin{bmatrix}-1&1\\0&-1\end{bmatrix},\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&1\end{bmatrix},\begin{bmatrix}-1&1&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}\text,{etc.}
$$
A: I came up with an example. Consider diag(-2,-1/2, 1, ..., 1) in SL(n,R). Clearly the lattice of logarithm of this does not intersect sl(n,R).
