Lie group $SU(2)$ is the universal covering group of $SO(3)$.

1.1k Views Asked by At

I need to show that Lie group $SU(2)$ is the universal covering group of $SO(3)$ using the Adjoint representation of $SU(2)$.

But I am stuck at the first step of finding the adjoint representation of $SU(2)$. Using the adjoint representation of $SU(2)$ I need to produce a continuous projection $p: SU(2)\to SO(3)$ so that $SU(2)$ is the universal covering of $SO(3)$. Then I know a result which will imply that $SU(2)$ is simply connected since $SO(3)$ is connected.

2

There are 2 best solutions below

0
On

Since $SU(2)$ is simple, the Killing form of $su(2)$is negative definite, the adjoint representation preserves the Killing form, thus its image is contained in $SO(3)$, it is a local diffeomorphism and since $dim SU(2)=dim SO(3)$ the adjoint representation is a covering (Ehresmann Lemma).

0
On

Most of this can be found in Naive Lie Theory by John Stillwell.

Proof of simple connectivity of $SU(2)$: $SU(2)$ is homeomorphic to the space $S^3$. A loop $p:[0,1]\to S^3$ can be contracted by:

  • Ideally, picking a point not in the image of $p$, stereographically projecting from that point onto $\mathbb R^3$, and then contracting in $\mathbb R^3$ (which is clearly simply connected).

  • To deal with paths that fill space, which blocks the above argument, exploit the uniform continuity of paths. Uniform continuity implies that the domain of $p$ can be broken up into intervals of size $\delta$ such that the image of such an interval doesn't fill every point on the sphere. Within each such interval, straighten out the path, then apply the above argument.

Proof that $SU(2)$ covers $SO(3)$: Proof 1: $\mathfrak{su}(2)$ is isomorphic to $\mathfrak{so}(3)$, while the former is simply connected. (Is this correct?) Proof 2: Interpret $\mathfrak{su}(2)$ as a set of matrices (which is justified by the fact that $SU(2)$ is a matrix group) and a vector space too. Let $u$ and $v$ be in $\mathfrak{su}(2)$; observe that $e^{v\theta/2}\,u\,e^{-v\theta/2}$ is a rotation of $u$. The mapping $u \in \mathfrak{su}(2) \mapsto e^{v\theta/2}\,u\,e^{-v\theta/2}$ is the adjoint representation of $e^{v\theta/2}$, and every element of $SU(2)$ can be expressed as $e^{v\theta/2}$.