SL(n;R) is connected

The special feature is that although not every element of $SL(n,\mathbb{R})$ is in the image of the exponential map of $\mathfrak{sl}(n,\mathbb{R})$, $SL(n,\mathbb{R})$ is still a connected Lie group. The proof below is a guideline and should be clarified a bit more at some points, but this was done intentionally.

To illustrate the point, first we show

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 $\operatorname{tr}{X}=0$. 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 $x^{t}x$ is symmetric and positive definite, since $\forall\psi\in\mathbb{R}^{n}:\langle\psi,x^{t}x\psi\rangle>0$, with the standard inner product on $\mathbb{R}^{n}$. Hence, we can write $x=RP$, with $P=(x^{t}x)^{\frac{1}{2}}$ and $R=xP^{-1}$. $P$ is well defined, since any real symmetric, positive definite matrix is diagonalizable. It’s easy to check that $RR^{t}=\operatorname{id}_{n}$, 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 $\{X\in GL(n,\mathbb{R})|X^{t}=X,\operatorname{tr}{X}=0\}$, we have the wanted statement: $SL(n,\mathbb{R}\subset SO(n)\times\exp{\Omega}$. ∎

The reverse inclusion is simply shown: any such combination is trivially in $SL(n,\mathbb{R})$.

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. ∎

Title SL(n;R) is connected SLnRIsConnected 2013-03-22 18:52:05 2013-03-22 18:52:05 Stephaninos (23208) Stephaninos (23208) 5 Stephaninos (23208) Result msc 20G15