4 Linear Lie groups and their Lie algebras 4.3 One-parameter subgroups 4.5 Lie group and Lie algebra homomorphisms

4.4 Lie algebras

Definition 4.20.

Let $G\subset\operatorname{GL}_{n}(\mathbb{C})$ be a linear Lie group. We define its Lie algebra by

\mathfrak{g}=\{X\in\mathfrak{gl}_{n,\mathbb{C}}:\exp(tX)\in G\text{ for all $t% \in\mathbb{R}$}\}.

In other words, it is the set of $X$ such that the one-parameter subgroup infinitesimally generated by $X$ is contained in the group $G$ .

The Lie algebra can also be defined more geometrically as the tangent space to $G$ at the identity; the above definition then becomes the “exponential characterization” of the Lie algebra. The equivalence is given by the following theorem:

Theorem 4.21.

With $G$ and $\mathfrak{g}$ as above, we have

\mathfrak{g}=\{X\in\mathfrak{gl}_{n,\mathbb{C}}:X=\gamma^{\prime}(0)\text{ for% some continuous map $\gamma:[-a,a]\rightarrow G$, $a>0$}\}.

In other words, $\mathfrak{g}$ is the set of all possible tangent vectors to curves in $G$ passing through $I$ .

Proof.

We show that $X\in\mathfrak{g}$ if and only if $\exp(tX)\in G$ for all $t\in\mathbb{R}$ .

If $\exp(tX)\in G$ for all $t\in\mathbb{R}$ then we may take the derivative of $\gamma(t)=\exp(tX)$ and conclude that $X\in\mathfrak{g}$ .

Now assume that $X\in\mathfrak{g}$ . By definition, we know that there is a differentiable map $\gamma:[-a,a]\to G$ , for some $a>0$ , such that $X=\gamma^{\prime}(0)$ . Fix any $t\in\mathbb{R}$ . As $k\rightarrow\infty$ we have an expansion

\gamma\left(\frac{t}{k}\right)=I+\frac{t}{k}X+o\left(\frac{1}{k}\right)=\exp% \left(\frac{t}{k}X+o\left(\frac{1}{k}\right)\right)

of $\gamma$ around $0$ . Then

\left(\gamma\left(\frac{t}{k}\right)\right)^{k}=\exp\left(tX+o(1)\right)\in G.

Since $G$ is closed and

\lim_{k\to\infty}\left(\gamma\left(\frac{t}{k}\right)\right)^{k}=\exp(tX),

we conclude that $\exp(tX)\in G$ . ∎

Remark 4.22.

It is not true that $\mathfrak{g}=\{X\in\mathfrak{gl}_{n,\mathbb{C}}:\exp(X)\in G\}$ . This is not even true for $G=\{1\}\subset\mathbb{C}$ , why?

We define the dimension of the Lie group $G$ to be the dimension of the associated Lie algebra $\mathfrak{g}$ . We now compute the Lie algebras of many of the groups that we are interested in:

Proposition 4.23.

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

•

$\operatorname{Lie}(\operatorname{GL}_{n}(\mathbb{K}))=\mathfrak{gl}_{n,\mathbb% {K}}$ with $\dim_{\mathbb{K}}\mathfrak{gl}_{n,\mathbb{K}}=n^{2}$ ;
•

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

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

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

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

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

Proof.

The first one is obvious for $\mathbb{K}=\mathbb{C}$ and left as an exercise for $\mathbb{K}=\mathbb{R}$ . For the second one, first suppose that $\operatorname{tr}(X)=0$ . Then $\det\exp(tX)=\exp\operatorname{tr}(tX)=1$ so $X\in\mathfrak{sl}_{n,\mathbb{K}}$ . 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.

For the third one, we need to find all $X$ such that

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

(4.1)

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 (4.1}) 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 — homework! ∎

Proposition 4.24.

Let $\mathfrak{g}$ be the Lie algebra of a (linear) Lie group $G$ . Then

1.

$\mathfrak{g}$ is a real vector space (inside $\mathfrak{gl}_{n,\mathbb{C}}$ ).
2.

If $X\in\mathfrak{g}$ and if $g\in G$ , then $gXg^{-1}\in\mathfrak{g}$ .
3.

For $X,Y\in\mathfrak{g}$ ,

[X,Y]=XY-YX\in\mathfrak{g}.

Proof.

For the first part we must show closure under scalar multiplication and addition. For scalar multiplication, note that if $X\in\mathfrak{g}$ and $\lambda\in\mathbb{R}$ , then $\lambda X\in\mathfrak{g}$ since $\exp(\lambda tX)\in G$ for all $t$ . (You could also prove this directly from the definition of $\mathfrak{g}$ .) For addition, assume that $X,Y\in\mathfrak{g}$ , so that $\exp(tX),\exp(sY)\in G$ for all $s,t\in\mathbb{R}$ . We need $\exp(t(X+Y)\in G$ for all $t$ . By Corollary 4.14 we have

\exp(t(X+Y))=\lim\limits_{k\to\infty}\left(\exp\left(\frac{tX}{k}\right)\exp% \left(\frac{tY}{k}\right)\right)^{k}.

But now on the right hand side,

\left(\exp\left(\frac{tX}{k}\right)\exp\left(\frac{tY}{k}\right)\right)^{k}\in G

and hence so is the limit since $G$ is closed.

Part (2) follows from $\exp(t(gXg^{-1}))=g\exp(tX)g^{-1}\in G$ . Again, you could also prove this directly from the definition of $\mathfrak{g}$ .

For part (3), we know by part (2) that, for $X,Y\in\mathfrak{g}$ ,

\exp(tX)Y\exp(-tX)\in\mathfrak{g}.

Then

	$\displaystyle\left.\frac{d}{dt}\exp(tX)Y\exp(-tX)\right\|_{t=0}$	$\displaystyle=\left.\left(X\exp(tX)Y\exp(-tX)-\exp(tX)Y\exp(-tX)X\right)\right% \|_{t=0}$
		$\displaystyle=XY-YX.$

But also by definition

\left.\frac{d}{dt}\exp(tX)Y\exp(-tX)\right|_{t=0}=\lim_{t\to 0}\frac{\exp(tX)Y% \exp(-tX)-Y}{t}.

This is a limit of elements of the vector space $\mathfrak{g}$ , which is a closed subset of $\mathfrak{gl}_{n,\mathbb{K}}$ , and so must itself be an element of $\mathfrak{g}$ . ∎

Definition 4.25.

A Lie algebra $\mathfrak{g}$ is an $\mathbb{R}$ -vector space together with a bilinear map (Lie bracket)

[\,,\,]=\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}

that satisfies the following properties.

1.

It is alternating: $[Y,X]=-[X,Y]$ for all $X,Y\in\mathfrak{g}$ .
2.

The Jacobi identity holds:

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

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

Definition 4.26.

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

Proposition 4.27.

The Lie bracket

[X,Y]=XY-YX

makes $\mathfrak{gl}_{n,\mathbb{K}}$ into a Lie algebra.

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

Proof.

For the first part, simply check the axioms directly (the Jacobi identity is a bit of a pain…).

We have already shown the second part (we only have to show that $\mathfrak{g}$ is closed under the bracket). ∎

Before computing the Lie algebras of various matrix groups, we give an example that doesn’t seem to be of this form.

Example 4.28.

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

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

Definition 4.29.

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

Definition 4.30.

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

\{Z\in\mathfrak{g}:[Z,X]=0\text{ for all $X\in\mathfrak{g}$}.

It is an abelian Lie subalgebra of $\mathfrak{g}$ .

Definition 4.31.

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 the Lie algebra of a complex Lie group may be viewed as a complex Lie algebra (since the Lie bracket on $\mathfrak{gl}_{n,\mathbb{C}}$ is clearly $\mathbb{C}$ -bilinear).