1.6. Lecture 6

1.6.1. Topological properties: answer to question (1)

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 1.6.1.

(closed subgroup theorem) Let $G\subseteq\operatorname{GL}_{n}(\mathbb{C})$ be a closed subgroup. Then $G$ is actually a smoothly embedded submanifold: for every $g\in G$ there is an open set $g\in U\subseteq\operatorname{GL}_{n}(\mathbb{C})$ , an open subset $0\in V\subseteq\mathfrak{gl}_{n,\mathbb{C}}$ , and a diffeomorphism $\phi:V\rightarrow U$ such that $\phi(0)=g$ and $\phi(\mathfrak{g}\cap V)=G\cap U$ . See Figure 1.1.

Refer to caption — Figure 1.1. 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 $\gamma(0)=\operatorname{Id}$ and $\gamma(1)=g$ .

Proposition 1.6.2.

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 $\gamma_{i}(0)=\operatorname{Id}$ and $\gamma_{1}(1)=g,\gamma_{2}(1)=h$ .

Then define a path from $\operatorname{Id}$ to $gh$ 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 $\operatorname{Id}$ to $gh$ . This shows $G^{0}$ is closed under multiplication.

The identity and inverse axioms, and the normality, are left as exercises (Problem 1.6.2). ∎

Proposition 1.6.3.

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

Proof.

Omitted from lectures.

Firstly, if $H\subseteq G$ 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 $gU$ 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 1.6.1 $\exp(V)$ is an open subset around $\operatorname{Id}\in G$ , and $\exp(V)$ is path-connected since $V$ is. Thus $V\subseteq G^{0}$ as required. ∎

It follows from this result that the quotient topology on $G/G^{0}$ is discrete. ²² 2 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 1.6.4.

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 $\operatorname{Id}\in G$ . 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 1.6.1), 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 non-empty open and closed subsets,³³ 3 This argument uses the basic fact that a path-connected topological space is connected. and we obtain:

Theorem 1.6.5.

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 1.6.6.

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 $G^{0}=G$ . ∎

Example 1.6.7.

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 1.6.8.

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 ( $g\mapsto 1\in\mathbb{R}^{\times}$ ).

Proposition 1.6.9.

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

Proof.

Omitted, but here is a sketch.

(a)

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.
(b)

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

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

∎

Remark 1.6.10.

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.

1.6.2. Exercises

Problem 13. Complete the proof of Proposition 1.6.2, i.e. that if $G$ is a Lie group and $G^{0}$ is the connected component of the identity, then $G^{0}$ is a normal subgroup.

Problem 14.

(a)

Give a direct proof that $\mathrm{SO}(3)$ is connected, by constructing a path from an arbitrary element of $\mathrm{SO}(3)$ to the identity. Hint: every element of $\mathrm{SO}(3)$ is rotation by some angle about some axis.
(b)

Prove by induction on $n$ that $SO(n)$ is connected for all $n\geq 1$ .

Problem 15. 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 1.4.8). Hint: for the converse, consider the adjoint map $G\to GL(\mathfrak{g})$ .

What goes wrong if $G$ is not connected?

Solution: see Proposition 2.10.6.