SL(n;R) is connected

The special feature is that although not every element of $SL(n,ℝ)$ is in the image of the exponential map of $𝔰𝔩(n,ℝ)$, $SL(n,ℝ)$ 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.

, but it is in $S\mathit{}L\mathit{}\mathrm{(}\mathrm{2}\mathrm{,}\mathrm{R}\mathrm{)}$.

Proof.

, so $x\in SL(2,ℝ)$. We see that $x$ is not diagonalizable, it already is in Jordan normal form. Moreover, it has a double eigenvalue, $-1$. Suppose that $x=\exp X,X\in 𝔰𝔩(2,ℝ)$, then $\mathrm{tr}X=0$. 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 $\mathrm{\forall }x\mathrm{\in }S\mathit{}L\mathit{}\mathrm{(}n\mathrm{,}\mathrm{R}\mathrm{)}\mathrm{:}x\mathrm{=}\exp \mathit{}\mathrm{(}{X}_a\mathrm{)}\mathit{}\exp \mathit{}\mathrm{(}{X}_s\mathrm{)}$ with $X_a^t\mathrm{=}\mathrm{-}{X}_a\mathrm{,}{X}_s^t\mathrm{=}{X}_s\mathrm{\in }\mathrm{s}\mathit{}\mathrm{l}\mathit{}\mathrm{(}n\mathrm{,}\mathrm{R}\mathrm{)}$.

Proof.

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

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

Corollary 0.3.

$SL(n,ℝ)$ is connected.

Proof.

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

Title	SL(n;R) is connected
Canonical name	SLnRIsConnected
Date of creation	2013-03-22 18:52:05
Last modified on	2013-03-22 18:52:05
Owner	Stephaninos (23208)
Last modified by	Stephaninos (23208)
Numerical id	5
Author	Stephaninos (23208)
Entry type	Result
Classification	msc 20G15