1.4. Lecture 4

1.4.1. Lie algebras

There are two ways one can introduce Lie algebras; either as an axiomatic definition or by deriving them from Lie groups (which is where they get their name). We choose the former.

Definition 1.4.1.

A Lie algebra $\mathfrak{g}$ over a field $k$ is a $k$ -vector space together with a bilinear map (Lie bracket)

[\cdot,\cdot]:\mathfrak{g}\times\mathfrak{g}\longrightarrow\mathfrak{g}

that satisfies the following properties.

(i)

It is alternating: $[Y,X]=-[X,Y]$ for all $X,Y\in\mathfrak{g}$ .
(ii)

The Jacobi identity holds:

$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$

for all $X,Y,Z\in\mathfrak{g}$ .

We will now proceed by investigating the motivating example (which is also the source of the Lie in the name Lie algebra).

Theorem 1.4.2.

Let $G$ be a (linear) Lie group. Then $\mathfrak{g}$ is a real Lie algebra with Lie bracket given by

[X,Y]=XY-YX.

Proof.

By Proposition 1.3.8(i), $\mathfrak{g}$ is a real vector space. By (iii) of the same result $[X,Y]\in\mathfrak{g}$ so it remains to show the bracket is bilinear and the two axioms are satisfied by it. This is left as an exercise (Problem 1.4.2). ∎

We call $\mathfrak{g}$ the Lie algebra of $G$ (also sometimes denoted $\operatorname{Lie}(G)$ ) and define the dimension of the Lie group $G$ to be the dimension of $\mathfrak{g}$ . We now compute the Lie algebras of many of the groups that we are interested in:

Proposition 1.4.3.

The Lie algebras of $\operatorname{GL}_{n}(k)$ , $\operatorname{SL}_{n}(k)$ , $\operatorname{O}(n)$ , $\mathrm{SO}(n)$ , $\operatorname{U}(n)$ , and $\mathrm{SU}(n)$ are given by

(i)

$\operatorname{Lie}(\operatorname{GL}_{n}(k))=\mathfrak{gl}_{n,k}$ with $\dim_{k}\mathfrak{gl}_{n,k}=n^{2}$ ;
(ii)

$\operatorname{Lie}(\operatorname{SL}_{n}(k))=\mathfrak{sl}_{n,k}=\{X\in% \mathfrak{gl}_{n,k}\,|\,\operatorname{tr}(X)=0\}$ , the traceless matrices, with $\dim_{k}\mathfrak{sl}_{n,k}=n^{2}-1$ ;
(iii)

$\operatorname{Lie}(\operatorname{O}(n))=\mathfrak{o}_{n}=\operatorname{Lie}(% \mathrm{SO}(n))=\mathfrak{so}_{n}=\{X\in\mathfrak{gl}_{n,\mathbb{R}}\,|\,X+X^{% T}=0\}$ , the skew symmetric real matrices, with $\dim_{\mathbb{R}}\mathfrak{o}_{n}=\dim_{\mathbb{R}}\mathfrak{so}_{n}=\frac{n(n% -1)}{2}$ ;
(iv)

$\operatorname{Lie}(\operatorname{U}(n))=\mathfrak{u}_{n}=\{X\in\mathfrak{gl}_{% n,\mathbb{C}}\,|\,X+X^{*}=0\}$ , the skew Hermitian matrices, with $\dim_{\mathbb{R}}\mathfrak{u}_{n}=n^{2}$ ;
(v)

$\operatorname{Lie}(\mathrm{SU}(n))=\mathfrak{su}_{n}=\{X\in\mathfrak{u}_{n}\,|% \,\operatorname{tr}(X)=0\}$ , the traceless skew Hermitian matrices, with $\dim_{\mathbb{R}}\mathfrak{su}_{n}=n^{2}-1$ .
(vi)

$\operatorname{Lie}(\operatorname{Sp}(2n))=\mathfrak{sp}_{2n}=\{X\in\mathfrak{% gl}_{n,\mathbb{R}}\,|\,X^{T}J+JX=0\}$ where $J$ is the matrix of the alternating bilinear form.

Proof.

(i)

This obvious for $k=\mathbb{C}$ and left as an exercise for $k=\mathbb{R}$ .
(ii)

First suppose that $\operatorname{tr}(X)=0$ . Then $\det\exp(tX)=\exp\operatorname{tr}(tX)=1$ so $X\in\mathfrak{sl}_{n,k}$ . Conversely, if $X\in\mathfrak{sl}_{n,K}$ then $1=\det\exp(tX)=\exp(t\operatorname{tr}(X))$ for all $t$ ; differentiating at $t=0$ gives $\operatorname{tr}(X)=0$ as required.
(iii)

We need to find all $X$ such that

$\exp(tX)\exp(tX)^{T}=\exp(tX)\exp(tX^{T})=I$ (1.4.4)

for all $t$ . Taking the derivative for both sides with respect to $t$ , we obtain

$X\exp(tX)\exp(tX^{T})+\exp(tX)\exp(X^{T})X^{T}=0.$

Evaluating at $t=0$ , we get

$X+X^{T}=0.$

Conversely, if $X+X^{T}=0$ , then clearly equation (1.4.4) holds because

$\exp(tX)^{T}=\exp(tX^{T})=\exp(-tX)=\exp(X)^{-1}.$

For the dimension, notice that $X$ satisfying $X=-X^{T}$ is determined by its upper triangular part and that the diagonal entries must be all zeros; as their are $\frac{n(n-1)}{2}$ entries strictly above the diagonal, that is the dimension of $\mathfrak{o}_{n}$ . Since these matrices already have trace zero, we see that $\mathfrak{so}_{n}=\mathfrak{o}_{n}$ .

The unitary and symplectic Lie algebras can be computed in a similar way. This is an exercise (Problem 1.4.2). ∎

We now give an example of a Lie algebra that does not initially appear to be of the form $\operatorname{Lie}(G)$ for some Lie group $G$ .

Example 1.4.5.

Let $\mathfrak{g}=\mathbb{R}^{3}$ and let $[{\bf v},{\bf w}]={\bf v}\times{\bf w}$ . Then this is a Lie algebra (just check the axioms).

In fact, $\mathfrak{g}\cong\mathfrak{so}_{3}$ . To see this, send the vector ${\bf v}$ to the infinitesimal generator of the one parameter subgroup of $\mathrm{SO}(3)$ given by ‘rotating around the axis ${\bf v}$ at speed $|{\bf v}|$ ’.

Definition 1.4.6.

If $\mathfrak{g}$ is a Lie algebra, then a Lie subalgebra is a subspace $\mathfrak{h}\subseteq\mathfrak{g}$ that is closed under the Lie bracket.

Proposition 1.4.7.

If $G\subseteq\operatorname{GL}_{n}(k)$ is a Lie group, then $\mathfrak{g}$ is a (real) Lie subalgebra of $\mathfrak{gl}_{n,k}$ .

Proof.

By Proposition 1.3.8(i) and (iii) we have $\mathfrak{g}\in\mathfrak{gl}_{n,k}$ and $\mathfrak{g}$ is closed under the Lie bracket. ∎

Definition 1.4.8.

A Lie algebra $\mathfrak{g}$ is called abelian if $[X,Y]=0$ for all $X,Y\in\mathfrak{g}$ .

Definition 1.4.9.

The centre of a Lie algebra $\mathfrak{g}$ is

Z(\mathfrak{g})=\{Z\in\mathfrak{g}\,|\,[Z,X]=0\text{ for all }X\in\mathfrak{g}\}.

Lemma 1.4.10.

The centre of a Lie algebra is an abelian Lie subalgebra.

Proof.

This is Exercise 1.4.2. ∎

Definition 1.4.11.

A complex linear Lie group is a closed subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ whose Lie algebra is a complex subspace of $\mathfrak{gl}_{n,\mathbb{C}}$ (as opposed to just a real subspace).

Note that $\mathfrak{u}_{n}$ and $\mathfrak{su}_{n}$ are only real Lie algebras and correspondingly $\mathrm{U}(n)$ and $\mathrm{SU}(n)$ are only real Lie groups, even though they consist of complex matrices. On the other hand, $\mathfrak{gl}_{n,\mathbb{C}}$ and $\mathfrak{sl}_{n,\mathbb{C}}$ are complex Lie algebras, so $\operatorname{GL}_{n}(\mathbb{C})$ and $\operatorname{SL}_{n}(\mathbb{C})$ are complex Lie groups.

A complex Lie algebra is a $\mathbb{C}$ -vector space with a $\mathbb{C}$ -bilinear Lie bracket satisfying the same axioms as for a Lie algebra. Thus one can show the Lie algebra of a complex Lie group is a complex Lie algebra (since the Lie bracket on $\mathfrak{gl}_{n,\mathbb{C}}$ is clearly $\mathbb{C}$ -bilinear).

1.4.2. Exercises

Problem 7. Complete the proof of Theorem 1.4.2.

Problem 8. Complete the proof of Proposition 1.4.3, i.e. prove parts (i), (iv), (v) and (vi).

Problem 9. Prove Lemma 1.4.10.

Problem 10. Let $I_{p,q}=\begin{pmatrix}\operatorname{Id}_{p}&\\ &-\operatorname{Id}_{q}\end{pmatrix}$ , where $\operatorname{Id}_{k}$ denotes the identity matrix of size $k$ . Let $n=p+q$ . Let

\operatorname{O}(p,q)=\{g\in\operatorname{GL}_{n}(\mathbb{R})\,|\,gI_{p,q}g^{T% }=I_{p,q}\}

be the orthogonal group of signature $(p,q)$ . Let $\mathrm{SO}(p,q)=\operatorname{O}(p,q)\cap\operatorname{SL}_{n}(\mathbb{R})$ . We let $\mathfrak{o}_{p,q}$ and $\mathfrak{so}_{p,q}$ be their Lie algebras.

Show that the Lie algebra $\mathfrak{o}_{p,q}$ is given by

\mathfrak{o}(p,q)=\{X\in M_{n}(\mathbb{R})\,|\,XI_{p,q}+I_{p,q}X^{T}=0\}

and that $\mathfrak{so}_{p,q}=\mathfrak{o}_{p,q}$ .