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 $n\geq 3$.

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:
$$
\begin{bmatrix}-1&1\\0&-1\end{bmatrix},\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&1\end{bmatrix},\begin{bmatrix}-1&1&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}\text,{etc.}
$$
