1.3. Lecture 3

1.3.1. Linear Lie groups

Definition 1.3.1.

A (linear) Lie group is a closed subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ , for some $n$ .

Here the group is called closed if it contains the limit of any Cauchy sequence, provided that limit is invertible.

Remark 1.3.2.

The usual definition of a Lie group is a smooth manifold together with a group structure such that the group operations are smooth functions. It is a theorem (Cartan’s theorem, or the closed subgroup theorem) that every linear Lie group in the sense of definition 1.3.1 is a Lie group in this sense. Not every Lie group is a linear Lie group, but we will only be studying linear Lie groups so we will often drop the word ‘linear’.

We give various examples (note that any subgroup defined by equalities of continuous functions will be closed):

Example 1.3.3.

(i)

the real general linear group $\operatorname{GL}_{n}(\mathbb{R})$ : we simply impose the closed condition that all the entries of the matrix are real;
(ii)

the (real or complex) special linear groups $\operatorname{SL}_{n}(k)$ ;
(iii)

if $\left\langle\cdot,\cdot\right\rangle$ is a bilinear form on $\mathbb{R}^{n}$ then we obtain a linear Lie group

$\{g\in\operatorname{GL}_{n}(\mathbb{R})\,|\,\left\langle g{\bf v},g{\bf w}% \right\rangle=\left\langle{\bf v},{\bf w}\right\rangle\}.$

There is a matrix $A$ such that $\left\langle{\bf v},{\bf w}\right\rangle={\bf v}^{T}A{\bf w}$ for all ${\bf v},w$ ; the bilinear form is symmetric if and only $A$ is symmetric, alternating if and only if $A$ is skew-symmetric ( $A^{T}=-A$ ), and non-degenerate if and only if $\det A$ is non-zero. Then the group is:

$\{g\in GL_{n}(\mathbb{R})\,|\,g^{T}Ag=A\}.$

Some special cases follow.
(iv)

The orthogonal and special orthogonal groups

$\mathrm{O}(n)=\{g\in\operatorname{GL}_{n}(\mathbb{R})\,|\,gg^{T}=\operatorname% {Id}\}$

and $\mathrm{SO}(n)=\operatorname{O}(n)\cap\operatorname{SL}_{n}(\mathbb{R})$ ;
(v)

the unitary and special unitary groups

$\mathrm{U}(n)=\{g\in\operatorname{GL}_{n}(\mathbb{C})\,|\,gg^{*}=\operatorname% {Id}\}$

and $\mathrm{SU}(n)=\mathrm{U}(n)\cap\operatorname{SL}_{n}(\mathbb{C})$ (not strictly a special case of the above, but closely related);
(vi)

the symplectic groups

$\operatorname{Sp}(2n)=\{g\in GL_{n}(\mathbb{R})\,|\,gJg^{T}=J\}$

where $J=\begin{pmatrix}0&I^{\prime}\\ -I^{\prime}&0\end{pmatrix}$ and $\operatorname{Id}^{\prime}$ is the $n\times n$ matrix with $1$ s on the antidiagonal and $0$ s elsewhere. This corresponds to a non-degenerate alternating bilinear form.
(vii)

the Heisenberg group

$\left\{\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}\,\Bigg{|}\,x,y,z\in\mathbb{R}\right\};$

Example 1.3.4.

Non-examples are $\operatorname{GL}_{n}(\mathbb{Q})$ (this is a subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ , but not closed), or (if $\alpha$ is an irrational real number) the subgroup

\left\{\begin{pmatrix}e^{ix}&0\\ 0&e^{i\alpha x}\end{pmatrix}\,\bigg{|}\,x\in\mathbb{R}\right\}\subseteq% \operatorname{GL}_{2}(\mathbb{C}).

This is a subgroup, isomorphic — as a group — to $\mathbb{R}$ , but not closed. You should picture it as a string wound infinitely densely around a torus.

The idea of Lie theory is to simplify the study of these groups by just studying their structure ‘very close to the identity’. This crucially uses that they are groups with a topology. By looking at the tangent spaces of these groups at the origin, you obtain Lie algebras; the group operation then turns into a structure called the Lie bracket.

1.3.2. The set $\mathfrak{g}$

Given some Lie group $G$ we can define a subset of $\mathfrak{gl}_{n,\mathbb{C}}$ related to this $G$ .

Definition 1.3.5.

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

\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 set $\mathfrak{g}$ can also be defined more geometrically as the tangent space to $G$ at the identity; the above definition then becomes the ‘exponential characterisation’. The equivalence is given by the following theorem:

Theorem 1.3.6.

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 $\operatorname{Id}$ .

Proof.

We show that $X=\gamma^{\prime}(0)$ for some continuous map $\gamma:[-a,a]\rightarrow G$ with $a>0$ 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, by Proposition 1.1.3, $\gamma(t)=\exp(tX)$ is differentiable for small $t$ with $\gamma^{\prime}(t)=X\exp(tX)$ and thus $\gamma^{\prime}(0)=X$ .

Now assume that there exists 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)=\operatorname{Id}+\frac{t}{k}X+O\left(\frac{1}{% k^{2}}\right)=\exp\left(\frac{t}{k}X+O\left(\frac{1}{k^{2}}\right)\right)

of $\gamma$ around $0$ . Then

\left(\gamma\left(\frac{t}{k}\right)\right)^{k}=\exp\left(tX+O\left(\frac{1}{k% }\right)\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 1.3.7.

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

We now collect some properties of $\mathfrak{g}$ .

Proposition 1.3.8.

Let $\mathfrak{g}$ correspond to the (linear) Lie group $G$ . Then

(i)

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

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

For $X,Y\in\mathfrak{g}$ we have $XY-YX\in\mathfrak{g}.$

Proof.

(i)

This is left as an exercise (Problem 1.3.3).
(ii)

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

(iii)

We know by part (ii) 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 $\mathfrak{gl}_{n,k}$ , and so must itself be an element of $\mathfrak{g}$ .

∎

1.3.3. Exercises

Problem 6. Provide a proof of Proposition 1.3.8(i).

1.3. Lecture 3

1.3.1. Linear Lie groups

Definition 1.3.1.

Remark 1.3.2.

Example 1.3.3.

Example 1.3.4.

1.3.2. The set 𝔤\mathfrak{g}

Definition 1.3.5.

Theorem 1.3.6.

Proof.

Remark 1.3.7.

Proposition 1.3.8.

Proof.

1.3.3. Exercises

1.3.2. The set $\mathfrak{g}$