4 Linear Lie groups and their Lie algebras 4.5 Lie group and Lie algebra homomorphisms 4.7 The example of SU(2) and SO(3)

4.6 Topological properties

While we only defined linear Lie groups to be closed subgroups of $\operatorname{GL}_{n}(\mathbb{C})$ , in fact they have much nicer topological properties than arbitrary closed subsets (which can be pretty wild, like the Cantor set).

Theorem 4.40.

(closed subgroup theorem) Let $G\subset\operatorname{GL}_{n}(\mathbb{C})$ be a closed subgroup. Then $G$ is actually a smoothly embedded submanifold: for every $g\in G$ there is an open set $g\in U\subset\operatorname{GL}_{n}(\mathbb{C})$ , an open subset $0\in V\subset\mathfrak{gl}_{n,\mathbb{C}}$ , and a diffeomorphism $\phi:V\rightarrow U$ such that $\phi(0)=g$ and $\phi(\mathfrak{g}\cap V)=G\cap U$ . See Figure 3.

Figure 3: Chart $\phi$ in closed subgroup theorem.

Proof.

(Sketch.) Since, for any $g$ , the map ‘multiply by $g$ ’ is smooth, it suffices to prove this when $g$ is the identity element. In this case, one can show that, for a sufficiently small neighbourhood $V$ of $0$ , the exponential map satisfies $\exp(V\cap\mathfrak{g})=\exp(V)\cap G$ — the tricky point is to show that, if $g\in G$ is sufficiently close to the identity, then $\log(g)\in\mathfrak{g}$ . ∎

If you don’t want to take this theorem on faith, then feel free to include its conclusion as part of the definition of a linear Lie group (in all our examples, it would be straightforward to verify).

We say that $G$ is connected if, for every $x,y\in G$ , there is a continuous function $\gamma:[0,1]\to G$ with $\gamma(0)=x$ and $\gamma(0)=y$ . For those of you taking courses in topology, this is actually the definition of path-connected; however, it follows from the closed subgroup theorem that Lie groups are locally path-connected, and being path-connected is equivalent to being connected for such spaces.

Let $G$ be a linear Lie group and let $G^{0}$ be the set of all $g\in G$ such that there is a continuous path $\gamma:[0,1]\to G$ with $\gamma(0)=I$ and $\gamma(1)=g$ .

Proposition 4.41.

The subset $G^{0}$ is a normal subgroup of $G$ .

Proof.

Omitted from lectures.

Let $g,h\in G^{0}$ , and let $\gamma_{1},\gamma_{2}$ be paths with $\gamma_{i}(0)=I$ and $\gamma_{1}(1)=g,\gamma_{2}(1)=h$ .

Then define a path from $I$ to $g h$ by following $\gamma_{1}$ and then $g\gamma_{2}$ . Concretely, define $\gamma:[0,1]\to G$ by

\gamma(t)=\begin{cases}\gamma_{1}(2t)&0\leq t\leq 1/2\\ g\gamma_{2}(2t)&1/2\leq t\leq 1\end{cases}

and observe that this is a continuous path from $I$ to $g h$ . This shows $G^{0}$ is closed under multiplication.

The identity and inverse axioms, and the normality, are left as exercises. ∎

Proposition 4.42.

The subgroup $G^{0}$ is an open and closed subset of $G$ .

Proof.

Omitted from lectures.

Firstly, if $H\subset G$ is an open subgroup then

G\setminus H=\bigcup_{g\in G\setminus H}gH

is a union of open subsets, so $H$ is also closed.

To show $G^{0}$ is open, it suffices to show that it contains an open subset $U$ containing the identity, as then $g U$ is an open subset containing $g$ , for all $g\in G$ . If $V$ is a sufficiently small open ball around $0\in\mathfrak{g}$ then by Theorem 4.40 $\exp(V)$ is an open subset around $I\in G$ , and $\exp(V)$ is path-connected since $V$ is. Thus $V\subset G^{0}$ as required. ∎

It follows from this result that the quotient topology on $G/G^{0}$ is discrete. ⁴⁴ 4 If you don’t know the definition of the quotient topology, please ignore this.

It is clear that $G$ is connected if and only if $G=G^{0}$ .

Proposition 4.43.

If $X\in\mathfrak{g}$ , then $\exp(X)\in G^{0}$ .

Proof.

Note that for any $X\in\mathfrak{g}$ , the image $\{\exp(tX):t\in\mathbb{R}\}$ defines a curve in $G$ containing the identity $I\in G$ . This curve is in the connected component of the identity. So $\exp(tX)\in G^{0}$ for all $t$ . ∎

Since $\exp(\mathfrak{g})$ contains an open neighbourhood of the identity (by Theorem 4.40), it follows that $\exp(\mathfrak{g})$ generates an open subgroup of $G^{0}$ , which is then necessarily closed. But since $G^{0}$ is connected it has no proper nonempty open and closed subsets,⁵⁵ 5 This argument uses the basic fact that a path-connected topological space is connected. and we obtain:

Theorem 4.44.

Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. Then the subgroup generated by $\exp(\mathfrak{g})$ is $G^{0}$ .

In particular, if $G$ is connected, then each element of $G$ is a (non-unique) product of a finite number of exponentials.

As a corollary we immediately obtain the answer to the first question above.

Proposition 4.45.

Let $G$ be a connected (linear) Lie group and let $\phi:G\to G^{\prime}$ be a Lie group homomorphism. Then the differential $D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}$ uniquely determines $\phi$ .

Proof.

Since $\phi(\exp(X))=\exp(D\phi(X))$ , the values $D\phi(X)$ determine $\phi$ on the subgroup generated by the $\exp(X)$ , which is exactly $G^{0}=G$ . ∎

Exercise 4.46.

Show that, if $G$ is a connected (linear) Lie group with Lie algebra $\mathfrak{g}$ , then $G$ is abelian if and only if $\mathfrak{g}$ is (see Definition 4.29).

What goes wrong when $G$ is not connected?

Example 4.47.

Any finite group $G$ can be embedded in $\operatorname{GL}_{n}(\mathbb{C})$ for some $n$ , and so regarded as a linear Lie group. Its Lie algebra is the zero vector space, so the derivative of a homomorphism $G\rightarrow H$ is always zero. In other words, the Lie algebra knows nothing in this case.

Example 4.48.

Recall that on the orthogonal group $\operatorname{O}(n)$ , the determinant (which is a continuous map!) takes the values $\{\pm 1\}$ . Hence $\operatorname{O}(n)$ is not connected; in fact, it is not too hard to show that $\mathrm{SO}(n)$ is the connected component of the identity. (This is related to $\mathfrak{so}_{n}=\mathfrak{o}_{n}$ ; that is, the condition $X=-{{}^{t}X}$ automatically implies that $X$ has trace zero and hence that $\exp(X)$ has determinant $1$ .)

Correspondingly, the determinant $\det$ on $\operatorname{O}(n)$ has zero differential, as it is constant on an open neighbourhood of the identity. This means that the differential on $\operatorname{O}(n)$ cannot distinguish the determinant from the trivial map ( $g\mapsto 1\in\mathbb{R}^{\times}$ ).

Exercise 4.49.

Prove that $\mathrm{SO}(n)$ is connected for $n\geq 2$ . Hint: show that it is path-connected, by induction.

Proposition 4.50.

The group $\operatorname{SL}_{n}(\mathbb{R})$ is connected.

Proof.

Omitted, but here is a sketch.

1.

Use Gram–Schmidt orthogonalisation to show that $SL_{n}(\mathbb{R})=SO(n)N_{+}$ where $N_{+}$ is the group of upper triangular matrices with positive diagonal entries.
2.

Show that $SO(n)$ is connected (see previous exercise) and $N_{+}$ is connected.
3.

Deduce that $\operatorname{SL}_{n}(\mathbb{R})$ is connected.

∎

Remark 4.51.

There is an alternative proof: show that $\operatorname{SL}_{n}(\mathbb{R})$ is generated by elementary matrices, and then connect every elementary matrix to the identity.

Among the Lie groups related to this course, $\operatorname{GL}_{n}(\mathbb{C})$ , $\operatorname{SL}_{n}(\mathbb{C})$ , $\operatorname{SL}_{n}(\mathbb{R})$ , $\operatorname{U}(n)$ , $\mathrm{SU}(n)$ , $\mathrm{SO}(n)$ , and $\operatorname{Sp}(2n)$ are connected, while $\operatorname{GL}_{n}(\mathbb{R})$ and $\operatorname{O}(n)$ are not connected, with their connected components being $\operatorname{GL}^{+}_{n}(\mathbb{R})=\{g\in\operatorname{GL}_{n}(\mathbb{R}):% \det g>0\}$ and $\mathrm{SO}(n)$ respectively.

We now turn to the second question, whether every Lie algebra homomorphism exponentiates to a Lie group homomorphism. In the light of what we have seen, it is sensible to restrict to the case of connected Lie groups. However, even with this restriction, the answer is in general no, as the next example shows!

Example 4.52.

The linear Lie groups $\operatorname{GL}_{1}^{+}(\mathbb{R})=\mathbb{R}_{>0}$ and $U(1)=\{z\in\mathbb{C}:|z|=1\}$ both have Lie algebra $\mathbb{R}$ with trivial Lie bracket; in the second case we get the subspace $i\mathbb{R}\subset\mathbb{C}$ of $\mathfrak{gl}_{2,\mathbb{C}}$ and identify it with $\mathbb{R}$ by dividing by $i$ .

The Lie algebra homomorphisms $\mathbb{R}\to\mathbb{R}$ are all of the form $\phi_{a}:t\mapsto at$ for some $a\in\mathbb{R}$ . We consider which of these exponentiate to homomorphisms of Lie groups.

1.

The map $\phi_{a}$ always exponentiates to a map $\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ , specifically the map

$x\mapsto e^{ax}.$
2.

The map $\phi_{a}$ always exponentiates to a map $\mathbb{R}_{>0}\to U(1)$ , specifically the map

$x\mapsto e^{iax}.$
3.

The map $\phi_{a}$ never exponentiates to a map $U(1)\to\mathbb{R}_{>0}$ if $a\neq 0$ . If it did, the map would have to send

$e^{ix}\mapsto e^{ax},$

and setting $x=2\pi$ gives $a=0$ .
4.

The map $\phi_{a}$ exponentiates to a map $U(1)\to U(1)$ if and only if $a\in\mathbb{Z}$ , in which case the map is

$z\mapsto z^{a}.$

Indeed, the map would have to be

$e^{ix}\to e^{iax}$

and setting $x=2\pi$ shows that $a\in\mathbb{Z}$ , when the map is as claimed.

Exercise 4.53.

Check that the Lie algebra of $\mathrm{SO}(2)$ is also isomorphic to $\mathbb{R}$ . Write down an isomorphism $U(1)\to\mathrm{SO}(2)$ ; what is the identification of Lie algebras it induces?

The key difference between $\mathbb{R}_{>0}$ and $U(1)$ is that the former is simply connected while the latter is not (it has fundamental group $\mathbb{Z}$ ). We explain this a bit further.

Recall we call a topological/metric space $X$ simply connected if it is path-connected and if every loop can be continuous shrunk to a single point; rigorously, if every continuous map from the unit circle to $X$ can be extended to a continuous map from the unit disc to $X$ . In topology, the failure of a space to be simply connected is measured by the ’fundamental group’ $\pi_{1}(X)$ : $X$ is simply-connected if and only if $\pi_{1}(X)$ is trivial.

Theorem 4.54.

Let $G$ be a simply connected (linear) Lie group. Let $G^{\prime}$ be any other (linear) Lie group. Let $\mathfrak{g}$ and $\mathfrak{g}^{\prime}$ be their Lie algebras. Then every homomorphism $\mathfrak{g}\rightarrow\mathfrak{g}^{\prime}$ exponentiates to a unique homomorphism $G\rightarrow G^{\prime}$ .

Hence we have a 1-1 correspondence

\{\operatorname{Hom}(G,G^{\prime})\}\quad\longleftrightarrow\quad\{% \operatorname{Hom}(\mathfrak{g},\mathfrak{g}^{\prime})\}.

Proof.

This is beyond the scope of this course. Note in the above example $\operatorname{GL}_{1}^{+}(\mathbb{R})$ is simply connected while the circle group $U(1)$ is not. ∎

One can show that $\operatorname{SL}_{n}(\mathbb{C})$ and $\mathrm{SU}(n)$ are simply connected. Here is a small table showing our connected groups and their fundamental groups.

$G$	$\pi_{1}(G)$
$\operatorname{GL}_{n}(\mathbb{C})$	$\mathbb{Z}$
$\operatorname{SL}_{n}(\mathbb{C})$	$1$
$\operatorname{SL}_{2}(\mathbb{R})$	$\mathbb{Z}$
$\operatorname{SL}_{n}(\mathbb{R})$ , $n\geq 3$	$C_{2}$
$\mathrm{SO}(2)$	$\mathbb{Z}$
$\mathrm{SO}(n)$ , $n\geq 3$	$C_{2}$
$\operatorname{U}(n)$	$\mathbb{Z}$
$\mathrm{SU}(n)$	$1$
$\mathrm{Sp}(2n)$	$Z$

Remark 4.55.

It is not an accident that the fundamental groups of $\operatorname{SL}_{n}(\mathbb{R})$ and $\mathrm{SO}(n)$ are isomorphic — Gram–Schmidt orthogonalization, as used in the proof of Proposition 4.50, shows that $\operatorname{SL}_{n}(\mathbb{R})$ and $\mathrm{SO}(n)$ are homotopy equivalent. A similar remark replies to $\operatorname{SL}_{n}(\mathbb{C})$ and $\mathrm{SU}(n)$ .

If $G$ is not connected, or its identity component is not simply connected, we can work in the following way.

•

There exists a ’universal cover’ $\widetilde{G}$ of $G^{0}$ which is simply connected, and also has the structure of a Lie group (not necessarily linear, unfortunately). There is a surjective group homomorphism $\pi:\widetilde{G}\to G^{0}$ with discrete kernel $Z\cong\pi_{1}(G^{0})$ , so that $G^{0}\cong\widetilde{G}/Z$ .
•

The kernel $Z$ of $\pi$ is isomorphic to the fundamental group $\pi_{1}(G^{0})$ .
•

Homomorphisms out of $G^{0}$ are in 1-1 correspondence with homomorphisms out of $\widetilde{G}$ which are trivial on $Z$ .
•

The Lie algebras of $G$ , $G^{0}$ and $\widetilde{G}$ coincide (more precisely, the maps $\pi:\widetilde{G}\to G^{0}$ and $\iota:G^{0}\to G$ induce isomorphisms of Lie algebras).
•

In general $G/G^{0}$ can be an arbitrary finite group! For this reason, it is common to restrict attention to connected Lie groups.

The diagram looks as follows:

\begin{CD}\mathfrak{g}@>{\exp}>{}>\widetilde{G}\\ \Big{\|}@V{}V{\pi}V\\ \mathfrak{g}@>{}>{\exp}>G^{0}@>{\iota}>{}>G.\end{CD}

Example 4.56.

The group $U(1)$ is not simply connected. Here the universal cover is $(\mathbb{R},+)$ (this is a linear Lie group because it is isomorphic to the upper triangular $2x2$ matrices with 1s on the diagonal — a similar argument shows that any vector space (with addition) is a Lie group). The map $\widetilde{G}\to G^{0}$ is then

	$\displaystyle\pi:\mathbb{R}$	$\displaystyle\to U(1)$
	$\displaystyle x$	$\displaystyle\mapsto e^{2\pi ix}$

and we see that the kernel of $\pi$ is $\mathbb{Z}$ , which is indeed the fundamental group $\pi_{1}(U(1))$ .