Representation Theory IV Representation Theory IV 2 Representations of Lie groups and Lie algebras - generalities

1 Linear Lie groups and their Lie algebras

1.1 Linear Lie groups

We fix some notation:

•

$\mathbb{K}$ denotes either the field $\mathbb{R}$ or $\mathbb{C}$ ;
•

$\mathfrak{gl}_{n,\mathbb{K}}=M_{n}(\mathbb{K})$ is the vector space of all $n\times n$ matrices over $\mathbb{K}$ .

Definition 1.1.

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

Remark 1.2.

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 1.1 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; sometimes I might omit the word ‘linear’.

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

•

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;
•

the (real or complex) special linear groups $\operatorname{SL}_{n}(\mathbb{K})$ ;
•

we may see $(\mathbb{R},+)$ as a linear Lie group in two ways: either as $\{\begin{pmatrix}1&x\\ 0&1\end{pmatrix}:x\in\mathbb{R}\}\subset GL_{2}(\mathbb{R})$ or as $\{e^{x}:x\in\mathbb{R}\}=(\mathbb{R}^{+},\times)$ .
•

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

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

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

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

Some special cases follow.
•

The orthogonal and special orthogonal groups

$\mathrm{O}(n)=\{g\in\operatorname{GL}_{n}(\mathbb{R}):g^{T}g=I\}$

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

the unitary and special unitary groups

$\mathrm{U}(n)=\{g\in\operatorname{GL}_{n}(\mathbb{C}):g^{\dagger}g=I\}$

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

the symplectic groups

$\operatorname{Sp}(2n)=\{g\in GL_{2n}(\mathbb{R}):g^{T}Jg=J\}$

where $J=\begin{pmatrix}0&I^{\prime}\\ -I^{\prime}&0\end{pmatrix}$ and $I^{\prime}$ is the $n\times n$ matrix with 1s on the antidiagonal and 0s elsewhere. This corresponds to a nondegenerate alternating bilinear form. Remark: it is also common to use $J=\begin{pmatrix}0&I_{n}\\ -I_{n}&0\end{pmatrix}$ .

We may visualise $\mathrm{U}(1)$ as the unit circle in the complex plane; the group $\mathrm{SO}(2)$ can also be visualised as this circle by identifying rotation by $\theta$ with $e^{i\theta}$ .

The group $\mathrm{SO}(3)$ can be visualised as a unit ball in $\mathbb{R}^{3}$ with opposite points on its surface identified; this is identified in section 1.8 below.

Example 1.3.

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}:x\in\mathbb{R}\right\}\subset\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.

1.2 The exponential map

Recall that $\mathbb{K}$ denotes either $\mathbb{R}$ or $\mathbb{C}$ .

Definition 1.4.

Let $X\in\mathfrak{gl}_{n,\mathbb{K}}$ . We define

\exp(X)=\sum_{k=0}^{\infty}\frac{X^{k}}{k!}.

This series is convergent for all $X\in\mathfrak{gl}_{n,\mathbb{K}}$ . Let $||\cdot||$ be the matrix norm

||X||=\left(\sum_{i,j}|x_{ij}|^{2}\right)^{1/2}.

This satisfies the triangle inequality and also $||XY||\leq||X||||Y||$ — this can be proved using Cauchy–Schwarz. Then for any $X\in\mathfrak{gl}_{n,\mathbb{C}}$ with $||X||\leq C$ , we have

||\exp(X)||\leq\sum_{k=0}^{\infty}\frac{||X||^{k}}{k!}\leq\sum_{k=0}^{\infty}% \frac{C^{k}}{k!}=\exp(C).

By the Weierstrass M-test, the power series for $\exp$ converges uniformly and absolutely on the closed ball $\{X:||X||\leq C\}$ , for any $C>0$ . In particular, $\exp$ is uniformly absolutely convergent on all compact subsets of $\mathfrak{gl}_{n,\mathbb{K}}$ , and hence is continuous.

Lemma 1.5.

We have (for all $X,Y\in\mathfrak{gl}_{n,\mathbb{K}}$ , $s,t\in\mathbb{K}$ ):

1.

$\exp(0)=I$ .
2.

$\exp(X+Y)=\exp(X)\exp(Y)$ if $XY=YX$ . (This is NOT true in general).
3.

$\exp(sX)\exp(tX)=\exp((s+t)X)$ .
4.

$\exp(X)$ is invertible, with inverse $\exp(-X)$ . In particular, $\exp(X)\in\operatorname{GL}_{n}(\mathbb{K})$ .
5.

$g\exp(X)g^{-1}=\exp(gXg^{-1})$ .

Proof.

The first point is obvious. Let’s prove (2) from which (3) and (4) follow. By definition,

$\displaystyle\exp(X+Y)$	$\displaystyle=\sum_{k=0}^{\infty}\frac{(X+Y)^{k}}{k!}$
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{\binom{k}{l}X^{l}Y^{k-l}}% {k!}$	(using that $X$ and $Y$ commute!)
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{X^{l}Y^{k-l}}{l!(k-l)!}$
	$\displaystyle=\left(\sum_{l=0}^{\infty}\frac{X^{l}}{l!}\right)\left(\sum_{j=0}% ^{\infty}\frac{Y^{j}}{j!}\right),$	(putting $j=k-l$ )

which is equal to the right hand side. Rearranging the sums is valid by absolute convergence. Finally, (5) follows from $gX^{k}g^{-1}=(gXg^{-1})^{k}$ . ∎

Example 1.6.

Let $X=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ and $Y=\begin{pmatrix}0&0\\ -1&0\end{pmatrix}$ . Then $X^{2}=Y^{2}=0$ so

\exp(X)=I+X=\begin{pmatrix}1&1\\ 0&1\end{pmatrix},\qquad\exp(Y)=I+Y=\begin{pmatrix}1&0\\ -1&1\end{pmatrix}.

But $X+Y=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$ , so $\exp(X+Y)=\begin{pmatrix}\cos 1&\sin 1\\ -\sin 1&\cos 1\end{pmatrix}$ . Hence $\exp(X+Y)\neq\exp(X)\exp(Y)$ ; see Problem 1.

In fact the exponential map is differentiable as a function of $X$ . For this, recall that a function $f:\mathbb{R}^{N}\rightarrow\mathbb{R}^{M}$ is differentiable at a point $p\in\mathbb{R}^{N}$ if there is a (necessarily unique) linear map $D_{p}f:\mathbb{R}^{N}\to\mathbb{R}^{M}$ such that

\lim_{h\to 0}\frac{||f(p+h)-f(p)-D_{p}f(h)||}{||h||}=0,

and in this case $D_{p}f$ is called the derivative of $f$ at $p$ . (This definition is independent of the choice of norms on $\mathbb{R}^{N}$ and $\mathbb{R}^{M}$ ).

Proposition 1.7.

The exponential map is differentiable at the origin (zero matrix), and its derivative at the origin is the identity map from $\mathfrak{gl}_{n,\mathbb{C}}$ to itself.

Proof.

In the above definition we have, $N=M=2n^{2}$ , $f=\exp$ , $p=0$ , and we claim $D_{0}\exp$ is the identity. Thus we need to show

\lim\limits_{||X||\to 0}\frac{||\exp(X)-\exp(0)-X||}{||X||}=\lim\limits_{||X||% \to 0}\frac{||\exp(X)-I-X||}{||X||}=0,

which follows from the definition of the exponential map. Indeed,

\frac{||\exp(X)-I-X||}{||X||}=\frac{||\sum_{k=2}^{\infty}\frac{X^{k}}{k!}||}{|% |X||}\leq||X||\cdot\sum_{k=0}^{\infty}\frac{||X||^{k}}{(k+2)!}<||X||e^{||X||},

which tends to zero as $||X||\to 0$ . ∎

Remark 1.8.

In fact, the exponential function has derivatives to all orders at all points. One way to see this is that the matrix entries of $\exp$ are given by multivariable power series in the $x_{ij}$ , and multivariable power series are differentiable (indeed, smooth) on the interior of their region of convergence. Caution: termwise derivatives of matrix functions cannot be calculated naively by applying the formal derivative of the power series. For example, it does not make sense to say that the derivative of $X^{2}$ is $2X$ , so $\exp$ is not "its own derivative" in any meaningful sense.

By the inverse function theorem, it follows from the remark that

Corollary 1.9.

The exponential map is a “local diffeomorphism” at $0$ : there exist neighbourhoods $U\subset\mathfrak{gl}_{n,\mathbb{K}}$ containing $0$ and $V\subset\operatorname{GL}_{n}(\mathbb{K})$ containing $I$ such that $\exp|_{U}$ is a smooth bijection onto $V$ with smooth inverse.

Remark 1.10.

In fact we can take $V=\{X\in\operatorname{GL}_{n}(\mathbb{C}):||X-I||<1\}$ . The inverse of $\exp$ in this neighbourhood is

\log(X)=\sum_{k=0}^{\infty}(-1)^{k}\frac{(X-I)^{k+1}}{k+1},

which is convergent when $||X-I||<1$ .

Of course, $\exp$ is not injective in general. For example, $\exp(2\pi ik)=1$ for $k\in\mathbb{Z}$ , and $\exp\!\left(\begin{pmatrix}0&2\pi\\ -2\pi&0\end{pmatrix}\right)=I$ .

For the next result it will be useful to know the following facts from linear algebra.

Lemma 1.11.

Let $X\in\operatorname{GL}_{n}(\mathbb{C})$ . Then $X$ is conjugate to a matrix of the form $D U$ where

•

$D$ is diagonal
•

$U$ is upper triangular with ‘1’s on the diagonal
•

$D$ and $U$ commute.

Proof.

(nonexaminable) This follows from Jordan normal form. Here’s a direct proof. Firstly write $\mathbb{C}^{n}$ as a direct sum of generalised eigenspaces for $X$ : if $\lambda$ is an eigenvalue of $X$ then we can write the characteristic polynomial $P(T)=(T-\lambda)^{a}Q(T)$ where $Q(T)$ does not have $\lambda$ as a root and $a\geq 1$ is an integer. Then the image of $Q(X)$ on $\mathbb{C}^{n}$ is the generalised eigenspace of $\lambda$ . The kernel of $Q(X)$ is preserved by $X$ and $X$ does not have an eigenvalue equal to $\lambda$ since $\lambda$ is not a root of $Q(X)$ , which must be the characteristic polynomial of $X$ acting on $\ker Q(X)$ . Thus

\operatorname{im}Q(X)\cap\ker Q(X)=\{0\}

and by the rank-nullity theorem

\mathbb{C}^{n}=\operatorname{im}Q(X)\oplus\ker Q(X)

is a decomposition of $\mathbb{C}^{n}$ as a direct sum of the $\lambda$ generalised eigenspace and a subspace preserved by $X$ . Repeating for each eigenvalue gives the required decomposition of $\mathbb{C}^{n}$ . This reduces the proof of the statement to the case where $X$ has only one eigenvalue $\lambda$ . In this case, we can inductively choose a basis $v_{1},\ldots,v_{n}$ of $\mathbb{C}^{n}$ such that, for $1\leq i\leq n$ , the image of $v_{i}$ in $\mathbb{C}^{n}/\left\langle v_{1},\ldots,v_{i-1}\right\rangle$ is an eigenvector of $X$ with eigenvalue $\lambda$ . With respect to this basis, $X$ is then upper triangular with $\lambda$ ’s on the diagonal, and we get the required decomposition with $D=\lambda I$ . ∎

Lemma 1.12.

We have

\det\exp(X)=\exp\operatorname{tr}(X).

Proof.

Conjugate so that $X$ is an upper triangular matrix with diagonal entries $\lambda_{1},\ldots,\lambda_{n}$ , and then note that $\exp(X)$ is also upper triangular with diagonal entries $\exp(\lambda_{1}),\ldots,\exp(\lambda_{n})$ .

Thus

\det\exp(X)=\prod_{i=1}^{n}\exp(\lambda_{i})=\exp\left(\sum_{i=1}^{n}\lambda_{% i}\right)=\exp\operatorname{tr}(X).

∎

Lemma 1.13.

The exponential function $\exp:\mathfrak{gl}_{n,\mathbb{C}}\to\operatorname{GL}_{n}(\mathbb{C})$ is surjective.

Proof.

Sketch. Details are in Problem 4. First prove it for $D$ and $U$ as in Lemma 1.11. The case of diagonal matrices is easy, and for $U$ note that $U-I$ is nilpotent so the power series for $\log(U)$ is a finite sum.

For general $X$ , by conjugating we can reduce to the case where $X=DU$ as above. If $D=\exp(d)$ and $U=\exp(u)$ , then

DU=\exp(d)\exp(u)=\exp(d+u)

because $d$ and $u$ commute (one must choose $d$ and $u$ carefully since $\exp$ is not injective). ∎

Remark 1.14.

The lemma is not true over $\mathbb{R}$ ; by Lemma 1.12, the determinant of $\exp(X)$ is positive for all real matrices $X$ .

The next proposition will be useful when we discuss Lie algebras of linear Lie groups.

Proposition 1.15.

(Lie product formula) We have

\exp(X+Y)=\lim_{k\to\infty}\left(\exp\left(\frac{X}{k}\right)\exp\left(\frac{Y% }{k}\right)\right)^{k}.

Proof.

Let

F(t)=\log(\exp(tX)\exp(tY)),

noting that for $t$ sufficiently small, $\exp(tX)\exp(tY)$ lies in the neighbourhood of $I$ where $\log$ is defined. Since $\log$ and $\exp$ are differentiable and the product of differentiable functions is differentiable, we apply the chain rule to see that $F^{\prime}(0)=X+Y$ . Therefore

F(t)=t(X+Y)+o(t).

(Here, the notation $o(f)$ , for any function $f$ , denotes some function $g$ such that $g(t)/f(t)\to 0$ as $t$ tends to some limit, usually 0 or $\infty$ . In this case, as $t\to 0$ .)

Setting $t=\frac{1}{k}$ and multiplying by $k$ we get

kF(t)=X+Y+ko(1/k)=X+Y+o(1).

Therefore

\left(\exp\left(\frac{X}{k}\right)\exp\left(\frac{Y}{k}\right)\right)^{k}=\exp% \left(kF\left(\frac{1}{k}\right)\right)=\exp\left(X+Y+o(1)\right).

Taking the limit as $k\to\infty$ gives the result. ∎

1.3 One-parameter subgroups

Lemma 1.16.

The map from $\mathbb{R}$ to $\operatorname{GL}_{n}(\mathbb{C})$ given by

t\mapsto\exp(tX)

is a differentiable group homomorphism.

We have

\frac{d}{dt}\exp(tX)=X\exp(tX)=\exp(tX)X.

In particular,

\left.\frac{d}{dt}\exp(tX)\right|_{t=0}=X.

Proof.

The given map is a group homomorphism by Lemma 1.5 part 4.

By definition,

\exp(tX)=\sum_{k=0}^{\infty}\frac{X^{k}}{k!}t^{k}.

As this power series (and its termwise derivative) are uniformly convergent on any compact subset, we can compute its derivative by differentiating termwise, which gives

\frac{d}{dt}\exp(tX)=\sum_{k=1}^{\infty}\frac{X^{k}}{(k-1)!}t^{k-1}=X\exp(tX).\qed

Definition 1.17.

A one-parameter subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ is a differentiable group homomorphism $f:\mathbb{R}\to\operatorname{GL}_{n}(\mathbb{C})$ . That is, a differentiable map such that

f(s+t)=f(s)f(t)

for all $s,t\in\mathbb{R}$ .

The infinitesimal generator of a one-parameter subgroup $f$ is the element $f^{\prime}(0)\in\mathfrak{gl}_{n,\mathbb{C}}$ .

By the previous lemma, for every $X\in\mathfrak{gl}_{n,\mathbb{C}}$ the map $t\mapsto\exp(tX)$ is a one-parameter subgroup, with infinitesimal generator $X$ .

Example 1.18.

Let $f:\mathbb{R}\to\mathrm{SO}(3)$ be rotation about the $z$ -axis:

f(t)=\begin{pmatrix}\cos t&-\sin t&0\\ \sin t&\cos t&0\\ 0&0&1\end{pmatrix}.

Then $f$ is a one-parameter subgroup, and its infinitesimal generator is

f^{\prime}(0)=\begin{pmatrix}0&-1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}.

In fact $f(t)=\exp(tf^{\prime}(0))$ . Problem 5 asks you to generalize this to rotation about an arbitrary axis.

Remark 1.19.

(non-examinable) For a one-parameter subgroup $f$ , it actually suffices to require that $f$ is continuous. Differentiability then comes for free.

Indeed, if $f$ is continuous, the integral $\int_{0}^{a}f(t)dt$ exists. Moreover,

f(s)\int_{0}^{a}f(t)dt=\int_{0}^{a}f(s+t)dt=\int_{s}^{s+a}f(t)dt.

The RHS is differentiable with respect to $s$ by the fundamental theorem of calculus. Therefore, to prove that $f(s)$ is differentiable, we only need to show that there is an $a>0$ such that $\int_{0}^{a}f(t)dt$ is an invertible matrix. Now consider the function

F(a)=\frac{1}{a}\int_{0}^{a}f(t)dt.

It is well-defined for $a\neq 0$ and $\lim\limits_{a\to 0}F(a)=I$ . Hence, for $0<a\ll 1$ , $F(a)$ is invertible, and therefore so is $aF(a)=\int_{0}^{a}f(t)dt$ .

The following is a very important property of one-parameter subgroups: that they all come from the exponential map.

Proposition 1.20.

Let $f:\mathbb{R}\to\operatorname{GL}_{n}(\mathbb{C})$ be a one-parameter subgroup with infinitesimal generator $X$ .

Then

f(t)=\exp(tX)

for all $t\in\mathbb{R}$ . That is, all one-parameter subgroups arise from the exponential function.

Proof.

From the definition of one-parameter subgroups, we have

f^{\prime}(t)=\lim_{s\to 0}\frac{f(s+t)-f(t)}{s}=f(t)\lim_{s\to 0}\frac{f(s)-f% (0)}{s}=f(t)f^{\prime}(0)=f(t)X.

Now consider the differential equation

g^{\prime}(t)=g(t)X.

We know that both $f(t)$ and $\exp(tX)$ are both solutions with the same initial condition that $g(0)=I$ . Therefore they must be equal. ∎

1.4 Lie algebras

Definition 1.21.

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 $G$ . We write $\operatorname{Lie}(G)$ for this Lie algebra.

Remark 1.22.

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?

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 1.23.

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 differentiable 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 $X$ lies in the right hand side (i.e. is a tangent vector) 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 $X$ is a tangent vector at $I$ .

Now assume that $X$ is a tangent vector at $I$ . Then there is a differentiable map $\gamma:[-a,a]\to G$ , for some $a>0$ , such that $\gamma(0)=I$ and $X=\gamma^{\prime}(0)$ . By the local inverse property of $\exp$ , there is a neighbourhood of $I$ on which $\log$ is defined and differentiable. For $|t|$ small, set

A(t)=\log(\gamma(t)).

Then $A(0)=0$ , and by the chain rule (using that the derivative of $\log$ at $I$ is the identity map) we have $A^{\prime}(0)=X$ . Hence

A(t)=tX+o(t)

as $t\to 0$ .

Fix any $t\in\mathbb{R}$ . For $k$ large, $t/k\in(-a,a)$ and $\gamma(t/k)$ lies in the domain of $\log$ , so

\left(\gamma\left(\frac{t}{k}\right)\right)^{k}=\exp\left(kA\left(\frac{t}{k}% \right)\right)=\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),

with $\gamma(t/k)\in G$ for all $k$ , we conclude that $\exp(tX)\in G$ . ∎

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 1.24.

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

\begin{array}[]{lll}G&\mathfrak{g}=\ldots&\dim(\mathfrak{g})\\ \hline\cr\operatorname{GL}_{n}(\mathbb{K})&\mathfrak{gl}_{n,\mathbb{K}}&\dim_{% \mathbb{K}}\mathfrak{gl}_{n,\mathbb{K}}=n^{2}\\ \operatorname{SL}_{n}(\mathbb{K})&\mathfrak{sl}_{n,\mathbb{K}}=\{X\in\mathfrak% {gl}_{n,\mathbb{K}}:\operatorname{tr}(X)=0\}&\dim_{\mathbb{K}}\mathfrak{sl}_{n% ,\mathbb{K}}=n^{2}-1\\ \operatorname{O}(n),\mathrm{SO}(n)&\mathfrak{o}_{n}=\mathfrak{so}_{n}=\{X\in% \mathfrak{gl}_{n,\mathbb{R}}:X+X^{T}=0\}&\dim_{\mathbb{R}}\mathfrak{so}_{n}=% \frac{n(n-1)}{2}\\ \operatorname{U}(n)&\mathfrak{u}_{n}=\{X\in\mathfrak{gl}_{n,\mathbb{C}}:X+X^{% \dagger}=0\}&\dim_{\mathbb{R}}\mathfrak{u}_{n}=n^{2}\\ \mathrm{SU}(n)&\mathfrak{su}_{n}=\{X\in\mathfrak{u}_{n}:\operatorname{tr}(X)=0% \}&\dim_{\mathbb{R}}\mathfrak{su}_{n}=n^{2}-1\\ \operatorname{Sp}(2n)&\mathfrak{sp}_{2n}=\{X\in\mathfrak{gl}_{2n,\mathbb{R}}:X% ^{T}J+JX=0\}&\dim_{\mathbb{R}}\mathfrak{sp}_{2n}=n(2n+1)\end{array}

Here $\mathfrak{so}_{n}$ is the space of skew-symmetric real matrices and $\mathfrak{u}_{n}$ is the space of skew-Hermitian matrices.

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

(1.1)

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

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

Evaluating at $t=0$ , we get

X+X^{T}=0.

Thus $\mathfrak{o}_{n}\subset\{X\in\mathfrak{gl}_{n,\mathbb{C}}:X+X^{T}=0\}$ .

Conversely, if $X+X^{T}=0$ , then equation (1.1) holds because

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

Thus $\{X\in\mathfrak{gl}_{n,\mathbb{C}}:X+X^{T}=0\}\subset\mathfrak{o}_{n}$ and we have equality by the opposite containment above.

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 there are $\frac{n(n-1)}{2}$ entries strictly above the diagonal, that is the dimension of $\mathfrak{o}_{n}$ .

Finally, it is clear that $\mathfrak{so}_{n}\subset\mathfrak{o}_{n}$ . If $X\in\mathfrak{o}_{n}$ then $X+X^{T}=0$ and so $\operatorname{tr}(X)=0$ . Thus $\exp(tX)\in\operatorname{SL}_{n}(\mathbb{C})\cap\operatorname{O}(n)=\mathrm{SO% }(n)$ for all $t$ and so $X\in\mathfrak{so}_{n}$ , giving the claimed equality.

The unitary and symplectic Lie algebras can be computed in a similar way — homework! ∎

Proposition 1.25.

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 multplication, we reparametrise: suppose $X\in\mathfrak{g}$ and $\lambda\in\mathbb{R}$ . Then $t\mapsto\exp(t\lambda X)$ lies in $G$ (as $\lambda t$ is still a real number) and so $\lambda X\in\mathfrak{g}$ .

For addition, let $X,Y\in\mathfrak{g}$ . Then $\exp(tX),\exp(tY)\in G$ for all $t$ , so the product

\gamma(t)=\exp(tX)\exp(tY)

is a differentiable path in $G$ with $\gamma(0)=I$ and $\gamma^{\prime}(0)=X+Y$ . Hence $X+Y\in\mathfrak{g}$ by Theorem 1.23.

Part (2) follows from $\exp(t(gXg^{-1}))=g\exp(tX)g^{-1}\in G$ and 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 1.26.

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 1.27.

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 1.28.

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 1.29.

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 1.30.

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

Definition 1.31.

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 $\mathfrak{g}$ .

1.5 Lie group and Lie algebra homomorphisms

Definition 1.32.

A Lie group homomorphism $\phi:G\to G^{\prime}$ between two linear Lie groups $G$ and $G^{\prime}$ is a continuous group homomorphism.

An isomorphism is a bijective Lie group homomorphism whose inverse is also continuous.

Remark 1.33.

In fact, a continuous homorphism between linear Lie groups is automatically a smooth map of smooth manifolds, and if it is bijective then the inverse is automatically continuous.

Definition 1.34.

A homomorphism $\varphi:\mathfrak{g}\to\mathfrak{g}^{\prime}$ of Lie algebras is an $\mathbb{R}$ -linear map such that

\varphi([X,Y])=[\varphi(X),\varphi(Y)]

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

An isomorphism is an invertible homomorphism.

Definition 1.35.

Let $\phi:G\to G^{\prime}$ be a Lie group homomorphism. Define the derivative (or derived homomorphism)

D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}

D\phi(X)=\left.\frac{d}{dt}\phi(\exp(tX))\right|_{t=0}

for $X\in\mathfrak{g}$ . ¹¹ 1 We can justify taking the derivative by appealing to Remark 1.19.

Remark 1.36.

In fact, $D\phi$ is the derivative of $\phi$ at the identity in the sense of smooth manifolds; recall that $\mathfrak{g}$ and $\mathfrak{g}^{\prime}$ are the tangent spaces to $G$ and $G^{\prime}$ at the identity.

Theorem 1.37.

Let $\phi:G\rightarrow G^{\prime}$ be a Lie group homomorphism with derivative $D\phi$ . Then

1.

The following diagram commutes:

$\begin{CD}\mathfrak{g}@>{D\phi}>{}>\mathfrak{g}^{\prime}\\ @V{\exp}V{}V@V{}V{\exp}V\\ G@>{}>{\phi}>G^{\prime}.\end{CD}$

That is, for $X\in\mathfrak{g}$ we have

$\phi(\exp(X))=\exp(D\phi(X)).$
2.

For all $g\in G,X\in\mathfrak{g}$ ,

$D\phi(gXg^{-1})=\phi(g)D\phi(X)\phi(g)^{-1}.$
3.

The map $D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}$ is a Lie algebra homomorphism.

Proof.

1.

Consider the one parameter subgroup $f:\mathbb{R}\to G^{\prime}$ defined by $f(t)=\phi(\exp(tX))$ . By construction, $f^{\prime}(0)=D\phi(X)$ . By Proposition 1.20, one parameter subgroups are determined by their derivative at 0, so that we must have

$\phi(\exp(tX))=f(t)=\exp(tD\phi(X)).$

We have

	$\displaystyle D\phi(gXg^{-1})$	$\displaystyle=\left.\frac{d}{dt}\phi(\exp(tgXg^{-1}))\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g\exp(tX)g^{-1})\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g)\phi(\exp(tX))\phi(g^{-1})\right\|_{t=0}$
		$\displaystyle=\phi(g)\left.\frac{d}{dt}\phi(\exp(tX))\right\|_{t=0}\phi(g^{-1})$
		$\displaystyle=\phi(g)D\phi(X)\phi(g)^{-1},$

as claimed.

To show that $D\phi$ is a Lie algebra homomorphism, we need to show that

•

$D\phi$ is $\mathbb{R}$ -linear; and
•

$D\phi([X,Y])=[D\phi(X),D\phi(Y)]$ .

So let $X,Y\in\mathfrak{g}$ and $s\in\mathbb{R}$ . By definition,

	$\displaystyle D\phi(sX)$	$\displaystyle=\left.\frac{d}{dt}\phi(\exp(tsX))\right\|_{t=0}.$
If we now set $\mu=st$ , we can rewrite this as:
	$\displaystyle\left.\frac{d}{dt}\phi(\exp(tsX))\right\|_{t=0}$	$\displaystyle=s\left.\frac{d}{d\mu}\phi(\exp(\mu X))\right\|_{\mu=0}$
		$\displaystyle=sD\phi(X).$

So $D\phi$ commutes with scalar multiplication. For additivity, we have

D\phi(X+Y)=\left.\frac{d}{dt}\phi(\exp(t(X+Y)))\right|_{t=0}.

On the other hand, by Proposition 1.15 and using part (1)

	$\displaystyle\phi(\exp(t(X+Y)))$	$\displaystyle=\lim\limits_{k\to\infty}\phi\left(\left(\exp\left(\frac{t}{k}X% \right)\exp\left(\frac{t}{k}Y\right)\right)^{k}\right)$
		$\displaystyle=\lim\limits_{k\to\infty}\phi\left(\exp\left(\frac{t}{k}X\right)% \right)\phi\left(\exp\left(\frac{t}{k}Y\right)\right)^{k}$
		$\displaystyle=\lim\limits_{k\to\infty}\left(\exp\left(\frac{t}{k}D\phi(X)% \right)\exp\left(\frac{t}{k}D\phi(Y)\right)\right)^{k}$
		$\displaystyle=\exp(t(D\phi(X)+D\phi(Y))).$

Taking the derivative at $t=0$ , we conclude that

D\phi(X+Y)=D\phi(X)+D\phi(Y),

showing additivity.

Finally we show that $D\phi$ respects the Lie bracket. Let $X,Y\in\mathfrak{g}$ . By parts (1) and (2) we have

D\phi\left(\exp(tX)Y\exp(-tX)\right)=\exp(tD\phi(X))D\phi(Y)\exp(-tD\phi(X)).

Take the derivative for both sides at $t=0$ : the derivative of the RHS is $[D\phi(X),D\phi(Y)]$ , and the derivative of the LHS is $D\phi([X,Y])$ (as $D\phi$ is linear).∎

1.6 Complex Lie groups and holomorphic homomorphisms

Definition 1.38.

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).

Definition 1.39.

Suppose that $G$ and $G^{\prime}$ are complex Lie groups and $\phi:G\rightarrow G^{\prime}$ is a homomorphism. Then $\phi$ is holomorphic if $D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}$ is $\mathbb{C}$ -linear.

(This implies that $\phi$ is a holomorphic map of complex manifolds.)

Example 1.40.

The map $\det:\operatorname{GL}_{2,\mathbb{C}}\rightarrow\operatorname{GL}_{1,\mathbb{C}}$ is holomorphic. The map $\operatorname{GL}_{2,\mathbb{C}}\rightarrow\operatorname{GL}_{2,\mathbb{C}}$ sending $g$ to $\overline{g}$ is not holomorphic.

1.7 Topological properties

We have shown that the differential $\phi\mapsto D\phi$ gives a map

D:\operatorname{Hom}(G,G^{\prime})\longrightarrow\operatorname{Hom}(\mathfrak{% g},\mathfrak{g}^{\prime}).

This raises two natural questions:

1.

Is the map injective? Does the derivative $D\phi$ uniquely determine the Lie group homomorphism $\phi$ ?
2.

Is the map surjective? Or in other words, does every Lie algebra homomorphism $\varphi$ ’exponentiate’ (or ’lift’) to a Lie group homomorphism $\phi$ such that $D\phi=\varphi$ ? We say ’exponentiate’ since if yes, then $\phi$ would need to satisfy $\phi(\exp(X))=\exp(\varphi(X))$ . So this gives a formula for $\phi$ , at least on the image of $\exp$ . The question is whether this is well-defined (the exponential map is neither injective nor surjective in general) and whether this defines a homomorphism.

The answer to these questions is actually of topological nature, which we discuss in this subsection.

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 1.41.

(closed subgroup theorem) Let $G\subset\operatorname{GL}_{n}(\mathbb{C})$ be a closed subgroup, and let $\mathfrak{g}$ be its Lie algebra. Then for every $g\in G$ there is an open subset $g\in U\subset G$ and an open subset $0\in V\subset\mathfrak{g}$ such that $X\mapsto g\exp(X)$ is a homeomorphism $V\to U$ . See Figure 1.

Remark 1.42.

An open subset is a subset $U\subseteq G$ such that $U=G\cap U^{\prime}$ for some open subset $U^{\prime}\subseteq\operatorname{GL}_{n}(\mathbb{C})$ (which has the usual definition in terms of open balls). Likewise, an open subset of $\mathfrak{g}$ means open in its usual Euclidean topology as a finite-dimensional complex vector space.

A homeomorphism is a continuous bijection with continuous inverse.

Figure 1: Chart $\phi$ in closed subgroup theorem.

Proof.

(Sketch, nonexaminable.) Since, for any $g$ , the map ‘multiply by $g$ ’ is continuous with continuous inverse, it suffices to prove this when $g$ is the identity element. Choose a sufficiently small open neighbourhood $W$ of $0\in\mathfrak{gl}_{n,\mathbb{C}}$ such that $\exp:W\to\exp(W)$ is a homeomorphism onto an open neighbourhood of $I\in\operatorname{GL}_{n}(\mathbb{C})$ (for this, you can use the open mapping theorem). Set $V=W\cap\mathfrak{g}$ and $U=\exp(W)\cap G$ . The tricky point is to show that, if $g\in G$ is sufficiently close to the identity (so that $\log(g)$ is defined), then $\log(g)\in\mathfrak{g}$ ; this ensures that $\exp$ restricts to a homeomorphism $V\rightarrow U$ . ∎

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 (or path) $\gamma:[0,1]\to G$ with $\gamma(0)=x$ and $\gamma(1)=y$ . 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 1.43.

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

Proof.

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)&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 1.44.

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

Proof.

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 1.41 $\exp(V)$ is an open subset around $I\in G$ , and $\exp(V)$ is path-connected since $V$ is. Thus $\exp(V)\subset G^{0}$ 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 $G=G^{0}$ .

Exercise 1.45.

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

Proposition 1.46.

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 1.47.

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.

Proposition 1.48.

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 $t$ . ∎

Theorem 1.49.

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.

Proof.

Since $\exp(\mathfrak{g})$ contains an open neighbourhood of the identity (by Theorem 1.41), it follows that $\exp(\mathfrak{g})$ generates an open subgroup $H$ of $G^{0}$ , which is then necessarily closed. But since $G^{0}$ is connected it has no proper nonempty open and closed subsets,³³ 3 This argument uses the basic fact that a path-connected topological space is connected. and so $H=G^{0}$ . ∎

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

Proposition 1.50.

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 1.51.

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 1.30).

What goes wrong when $G$ is not connected?

Example 1.52.

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 1.53.

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; $\mathrm{SO}(n)$ 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 ( $g\mapsto 1\in\mathbb{R}^{\times}$ ).

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 1.54.

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 $\mathfrak{gl}_{1,\mathbb{C}}$ 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

$x\mapsto e^{a\log(x)}=x^{a}.$
2.

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

$x\mapsto e^{ia\log(x)}.$
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 1.55.

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 1.56.

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.

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

Remark 1.57.

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 1.46, shows that $\operatorname{SL}_{n}(\mathbb{R})$ and $\mathrm{SO}(n)$ are homotopy equivalent. A similar remark applies 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 1.58.

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 $\pi_{1}(U(1))$ .

1.8 The example of SU(2) and SO(3)

We illustrate the previous section with the example of $\mathrm{SO}(3)$ . According to Table 1.7, the fundamental group of $\mathrm{SO}(3)$ is $C_{2}$ . We can visualize this as follows: An element of $\mathrm{SO}(3)$ is rotation by some angle $\theta\in[0,\pi]$ about some (oriented) axis. We can represent this as a vector in $\mathbb{R}^{3}$ of length $\theta$ in the direction of the axis. Elements of $\mathrm{SO}(3)$ then correspond to points in the closed ball in $\mathbb{R}^{3}$ of radius $\pi$ . However, rotation by $\pi$ about the axis $v$ is the same as rotation by $\pi$ about the axis $-v$ , and so we must identify diametrically opposite points on the boundary of this ball.

Figure 2: Picture of

\mathrm{SO}(3)

Now, the straight line in this three-dimensional sphere from a point on the boundary to its diametrically opposite point is a loop in $\mathrm{SO}(3)$ since the endpoints represent the same rotation. You can convince yourself that this loop cannot be shrunk to a point (proving it rigorously requires some topology). However, if you go around the loop twice, then that can (!) be shrunk to a point. The idea is to move one copy of the loop out to the boundary, then use the ‘opposite point’ identification to move it to the other side, when you get a normal loop inside the ball which may be shrunk. See Figure 3.

A nice physical illustration of this is provided by the “Dirac belt trick”; here is a video of this demonstrated with long hair!

According to the general picture of the previous section, there should be a Lie group homomorphism $\tilde{G}\to\mathrm{SO}(3)$ whose kernel has order 2 and such that $\tilde{G}$ is simply connected, and it turns out that we can take $\tilde{G}=\mathrm{SU}(2)$ . So we study this group for a bit.

Firstly, one can show (see problem 12) that every element of $\mathrm{SU}(2)$ has the form

\begin{pmatrix}a&-\overline{b}\\ b&\overline{a}\end{pmatrix}

for $a,b\in\mathbb{C}$ with $|a|^{2}+|b|^{2}=1$ . It follows that $\mathrm{SU}(2)$ is homeomorphic to the unit sphere $S^{3}$ in $\mathbb{R}^{4}$ , which is simply connected (that is, there is a continuous bijection $\mathrm{SU}(2)\to S^{3}$ with continuous inverse).

In Problem 8 we constructed an isomorphism of Lie algebras $\mathfrak{su}_{2}\xrightarrow{\sim}\mathfrak{so}_{3}$ . Since $\mathrm{SU}(2)$ is simply connected, Theorem 1.56 implies that this exponentiates to a Lie group homomorphism

\pi:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3).

We will construct $\pi$ explicitly later, using the adjoint representation (see Section 2.4).

For now, we record the key facts: $\pi$ is surjective and

\ker(\pi)=\{\pm I\}\cong C_{2}.

In particular $\mathrm{SO}(3)\cong\mathrm{SU}(2)/\{\pm I\}$ , so $\pi$ realises $\mathrm{SU}(2)$ as the universal cover of $\mathrm{SO}(3)$ , and the kernel $\{\pm I\}$ is (canonically) isomorphic to $\pi_{1}(\mathrm{SO}(3))$ .

	$\displaystyle D\phi(gXg^{-1})$	$\displaystyle=\left.\frac{d}{dt}\phi(\exp(tgXg^{-1}))\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g\exp(tX)g^{-1})\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g)\phi(\exp(tX))\phi(g^{-1})\right\|_{t=0}$
		$\displaystyle=\phi(g)\left.\frac{d}{dt}\phi(\exp(tX))\right\|_{t=0}\phi(g^{-1})$
		$\displaystyle=\phi(g)D\phi(X)\phi(g)^{-1},$