SL(n;R) is connected

The special feature is that although not every element of S⁒L⁒(n,ℝ) is in the image of the exponential map of 𝔰⁒𝔩⁒(n,ℝ), S⁒L⁒(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.

(-110-1)βˆ‰exp⁑𝔰⁒𝔩⁒(2,ℝ), but it is in S⁒L⁒(2,R).


detx=:det(-110-1)=1, so x∈S⁒L⁒(2,ℝ). We see that x is not diagonalizable, it already is in Jordan normal form. Moreover, it has a double eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath, -1. Suppose that x=exp⁑X,Xβˆˆπ”°β’π”©β’(2,ℝ), then 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 contradictionMathworldPlanetmathPlanetmath. ∎

Lemma 0.2.

We have βˆ€x∈S⁒L⁒(n,R):x=exp⁑(Xa)⁒exp⁑(Xs) with Xat=-Xa,Xst=Xs∈s⁒l⁒(n,R).


The keyword here is polar decomposition. We notice that xt⁒x is symmetricPlanetmathPlanetmathPlanetmathPlanetmath and positive definitePlanetmathPlanetmath, since βˆ€Οˆβˆˆβ„n:⟨ψ,xt⁒x⁒ψ⟩>0, with the standard inner product on ℝn. Hence, we can write x=R⁒P, with P=(xt⁒x)12 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⁒Rt=idn, hence R∈O⁒(n). We had det⁑P>0 and det⁑x=1, hence det⁑(R)>0β‡’det⁑R=1β‡’R∈S⁒O⁒(nβ’π•Ÿ) and so det⁑P=1. Since the choice of positive root is unique, R and P are unique. Moreover, S⁒O⁒(n) is exactly generated by the set {X∈G⁒L⁒(n,ℝ)|Xt=-X} and Ξ©, the set of real symmetric matrices of determinantMathworldPlanetmath 1, by {X∈G⁒L⁒(n,ℝ)|Xt=X,tr⁑X=0}, we have the wanted statement: SL(n,β„βŠ‚SO(n)Γ—expΞ©. ∎

The reverse inclusion is simply shown: any such combinationPlanetmathPlanetmath is trivially in S⁒L⁒(n,ℝ).

Corollary 0.3.

S⁒L⁒(n,ℝ) is connected.


This is now clear from the fact that both S⁒O⁒(n) and Ξ© are connected and so βˆ€s,t∈[0,1]:exp⁑s⁒X⁒exp⁑t⁒Y∈S⁒L⁒(n,ℝ), a fact easily checked by taking the determinant. So S⁒L⁒(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