# Example: SU(2) to SO(3)

## The group SU(2)

I want to talk about a particular smooth homomorphism $SU(2)\to SO(3)$ .

Definition:

$SU(2)=\{M\in GL(2,\mathbf{C})\ :\ \det(M)=1,\ M^{\dagger}M=I\}$ is the group of 2-by-2 complex matrices which are unitary and have determinant 1 (special unitary group). Here $M^{\dagger}=\bar{M}^{T}$ .

We can alternatively characterise $SU(2)$ as the group of matrices $\begin{pmatrix}a&b\\ -\bar{b}&\bar{a}\end{pmatrix}\ :\ a,b\in\mathbf{C},\ |a|^{2}+|b|^{2}=1\}$ . To see this, note that unitarity implies $M^{\dagger}=M^{-1}$ , so $\begin{pmatrix}\bar{a}&\bar{c}\\ \bar{b}&\bar{d}\end{pmatrix}=\frac{1}{\det M}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix},$ and because $\det(M)=1$ we get $c=-\bar{b}$ and $d=\bar{a}$ . The condition $\det(M)=1$ then becomes $|a|^{2}+|b|^{2}=1$ .

## The Lie algebra su(2)

The Lie algebra is $\mathfrak{su}(2)=\{X\in\mathfrak{gl}(2,\mathbf{C})\ :\ \mathrm{Tr}(X)=0,\ X^{% \dagger}=-X\}$ (these conditions come from differentiating $\det(\exp(tX))=1$ and $\exp(tX)^{\dagger}\exp(tX)=I$ with respect to $t$ at $t=0$ as usual). We can write down a general element of this Lie algebra as follows:$\mathfrak{su}(2)=\left\{\begin{pmatrix}ix&y+iz\\ -y+iz&-ix\end{pmatrix}\ :\ x,y,z\in\mathbf{R}\right\}$ The fact that the diagonal elements are imaginary follows from $X^{\dagger}=-X$ . They sum to zero because the trace is zero. The bottom-left entry is determined by the top-right by $X^{\dagger}=-X$ .

Remark:

Even though $SU(2)$ is a group of complex matrices, $\mathfrak{su}(2)$ is a real vector space, not a complex vector space. That's because $SU(2)$ is cut out by a single real equation (rather than something holomorphic).

If $(v=(x,y,z)\in\mathbf{R}^{3}$ , let's write $M_{v}=\begin{pmatrix}ix&y+iz\\ -y+iz&-ix\end{pmatrix}$ . Here are some useful equations we will use.

Exercise:

$[M_{u},M_{v}]=M_{2u\times v}$ and $\mathrm{Tr}(M_{u}M_{v})=-2u\cdot v$ .

## A homomorphism

Lemma:

If $B\in SU(2)$ then $BM_{v}B^{-1}\in\mathfrak{su}(2)$ for any $M_{v}\in\mathfrak{su}(2)$ . This means that $BM_{v}B^{-1}=M_{w}$ for some $w\in\mathbf{R}^{3}$ . Moreover, $w=R(B)v$ for some rotation $R(B)$ . In other words, we have a map $R\colon SU(2)\to SO(3)$ such that $M_{R(B)v}=BM_{v}B^{-1}$ (we will only check that $R(B)$ is an orthogonal matrix, not that it's a rotation).

Proof:

To see that $BM_{v}B^{-1}\in\mathfrak{su}(2)$ , we need to show that it is tracefree and anti-Hermitian:

• Because trace is conjugation invariant, we have $\mathrm{Tr}(BM_{v}B^{-1})=\mathrm{Tr}(M_{v})=0$ .

• We have $(BM_{v}B^{-1})^{\dagger}=(B^{-1})^{\dagger}M_{v}^{\dagger}B^{\dagger}=B(-M_{v}% )B^{-1}=-BM_{v}B^{-1}$ because $B\in SU(2)$ and $M_{v}\in\mathfrak{su}(2)$ .

Now define $R(B)\colon\mathbf{R}^{3}\to\mathbf{R}^{3}$ implicitly by $M_{R(B)v}=BM_{v}B^{-1}$ . We want to show that $R(B)$ is an orthogonal matrix, i.e. that it preserves dot products.

Using our formula from before, we have: $(R(B)v_{1})\cdot(R(B)v_{2})=-\frac{1}{2}\mathrm{Tr}(M_{R(B)v_{1}}M_{R(B)v_{2}}),$ which is equal to $-\frac{1}{2}\mathrm{Tr}(BM_{v_{1}}B^{-1}BM_{v_{2}}B^{-1})=\mathrm{Tr}(M_{v_{1}% }M_{v_{2}})$ because the central $B^{-1}B$ cancels and the trace is unchanged by conjugation. This is then equal to $v_{1}\cdot v_{2}$ by the formula from above.

We have therefore defined a map $R\colon SU(2)\to O(3)$ . This turns out to be a homomorphism. Let's compute $R_{*}$ . We know that $R(\exp(tM_{u}))=\exp(tR_{*}(M_{u}))$ , so $R_{*}(M_{u})=\left.\frac{d}{dt}\right|_{t=0}R(\exp(tM_{u})).$

We have $M_{R(B)v}=BM_{v}B^{-1}$ , so if $B=\exp(tM_{u})$ then we get $M_{R(\exp(tM_{u})v)}=\exp(tM_{u})M_{v}\exp(-tM_{u}).$ Differentiating with respect to $t$ gives $M_{R_{*}(M_{u})v}=\left.\frac{d}{dt}\right|_{t=0}=M_{u}M_{v}-M_{v}M_{u}$ by the product rule. Therefore $M_{R_{*}(M_{u})v}=[M_{u},M_{v}]=M_{2u\times v}$ , so $R_{*}(M_{u})v=2u\times v.$

Therefore $R_{*}(M_{u})$ is the map which sends $v$ to $2u\times v$ . If we write $u=(u_{1},u_{2},u_{3})$ and $v=(v_{1},v_{2},v_{3})$ , note that $\begin{pmatrix}0&-u_{3}&u_{2}\\ u_{3}&0&-u_{1}\\ -u_{2}&u_{1}&0\end{pmatrix}\begin{pmatrix}v_{1}\\ v_{2}\\ v_{3}\end{pmatrix}=\begin{pmatrix}u_{2}v_{3}-u_{3}v_{2}\\ u_{3}v_{1}-u_{1}v_{3}\\ u_{1}v_{2}-u_{2}v_{1}\end{pmatrix}=u\times v,$ so we have $R_{*}\begin{pmatrix}ix&y+iz\\ -y+iz&-ix\end{pmatrix}=\begin{pmatrix}0&-2z&2y\\ 2z&0&-2x\\ -2y&2x&0\end{pmatrix}.$

This is a nontrivial example of a Lie algebra homomorphism $\mathfrak{su}(2)\to\mathfrak{o}(3)$ . Even though it's nontrivial, it is easy to write down: it's linear in $x$ , $y$ , and $z$ .

## Concluding remarks

Remark:

We have this homomorphism $R\colon SU(2)\to O(3)$ . This is not an isomorphism. For a start, it only hits the rotations, i.e. the orthogonal transformations with determinant $1$ ($SO(3)\subset O(3)$ ). Even if you think of it as a map $SU(2)\to SO(3)$ , it's still not an isomorphism: it is 2-to-1. For example, $B$ and $-B$ both map to the same rotation: $BM_{v}B^{-1}=(-B)M_{v}(-B)^{-1}.$

At the level of Lie algebras, $R_{*}\colon\mathfrak{su}(2)\to\mathfrak{so}(3)$ is an isomorphism: from $\begin{pmatrix}0&-2z&2y\\ 2z&0&-2x\\ -2y&2x&0\end{pmatrix}$ we can figure out $x$ , $y$ and $z$ by dividing by $2$ , so we know which matrix $\begin{pmatrix}ix&y+iz\\ -y+iz&-ix\end{pmatrix}$ to write down. That is, we have an inverse $(R_{*})^{-1}\colon\mathfrak{so}(3)\to\mathfrak{su}(2)$ . However, it doesn't correspond to a Lie group homomorphism $SO(3)\to SU(2)$ . This is OK because Lie's theorem about exponentiating homomorphisms only applies when the domain is simply-connected, and the group $SO(3)$ is not simply-connected.

## Pre-class exercises

Exercise:

We constructed a homomorphism $R\colon SU(2)\to O(3)$ . Can you see why $R$ lands in the subset $SO(3)$ of matrices with determinant 1?

Exercise:

Can you prove that $\mathfrak{su}(2)$ is the set of 2-by-2 matrices $X$ such that $X^{T}=-X$ and $\mathrm{Tr}(X)=0$ ?