Proposition 0.1   $\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \notin \exp{\mathfrak{sl}(2,\mathbb{R})}$ , but it is in $SL(2,\mathbb{R})$ .

Proof. $\det x =: \det \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} = 1$ , so $x\in SL(2,\mathbb{R})$ . We see that $x$ is not diagonalizable, it already is in Jordan normal form. Moreover, it has a double eigenvalue, $-1$ . Suppose that $x=\exp X, X \in \mathfrak{sl}(2,\mathbb{R})$ , then . Since $x$ had a double eigenvalue, so does $X$ , hence the eigenvalues of $X$ both are $0$ . But this implies the eigenvalues of $x$ are $1$ . This is a contradiction.

Lemma 0.2   We have $\forall x\in SL(n,\mathbb{R}): x = \exp(X_a)\exp(X_s)$ with $X_a^t=-X_a, X_s^t=X_s\in \mathfrak{sl}(n,\mathbb{R})$ .

Proof. The keyword here is polar decomposition. We notice that $x^tx$ is symmetric and positive definite, since $\forall \psi\in\mathbb{R}^n: \langle \psi,x^tx\psi\rangle >0$ , with the standard inner product on $\mathbb{R}^n$ . Hence, we can write $x=RP$ , with $P=(x^tx)^\frac{1}{2}$ and $R=xP^{-1}$ . $P$ is well defined, since any real symmetric, positive definite matrix is diagonalizable. It's easy to check that , hence $R\in O(n)$ . We had $\det{P}>0$ and $\det{x}=1$ , hence $\det(R)>0 \Rightarrow \det{R}=1 \Rightarrow R\in SO(n\mathbb{n})$ and so $\det{P}=1$ . Since the choice of positive root is unique, $R$ and $P$ are unique. Moreover, $SO(n)$ is exactly generated by the set $\{ X\in GL(n,\mathbb{R}) | X^t=-X \}$ and $\Omega$ , the set of real symmetric matrices of determinant $1$ , by , we have the wanted statement: $SL(n,\mathbb{R} \subset SO(n)\times \exp{\Omega}$ .

Corollary 0.3   $SL(n,\mathbb{R})$ is connected.

Proof. This is now clear from the fact that both $SO(n)$ and $\Omega$ are connected and so $\forall s,t\in [0,1]: \exp{sX}\exp{tY}\in SL(n,\mathbb{R})$ , a fact easily checked by taking the determinant. So $SL(n,\mathbb{R})$ is path-connected, hence connected.