Representation Theory IV 5 Linear Lie groups and their Lie algebras 7 SL₂

6 Representations of Lie groups and Lie algebras - generalities

6.1 Basics

Definition 6.1.

A finite-dimensional (complex) representation $(\rho,V)$ of a Lie group $G$ is a Lie group homomorphism

\rho:G\to\operatorname{GL}(V),

where $V$ is a finite-dimensional complex vector space.

Remark 6.2.

Infinite-dimensional representations are important, but subtle. In general one must equip $V$ with some topology and add topological conditions to all notions which follow. For instance, one might take $V$ to be a Hilbert space.

An example of such a representation arises naturally if you attempt to generalise the regular representation! One must take $V$ to be something like the space of square-integrable functions on the group, rather than the space of arbitrary functions, to get a pleasant theory.

If $\rho$ is a finite-dimensional representation of $G$ as above, then we can take its derivative:

D\rho:\mathfrak{g}\to\mathfrak{gl}(V)=\operatorname{End}(V),

mapping from the Lie algebra $\mathfrak{g}$ of $G$ to the space of endomorphisms of $V$ . Note that

\operatorname{End}(V)

is a Lie algebra with bracket

[S,T]=S\circ T-T\circ S.

The map $D\rho$ is a Lie algebra homomorphism. Often we write, abusively, $\rho$ instead of $D\rho$ .

Note that choosing an isomorphism $V\cong\mathbb{C}^{n}$ induces isomorphisms $\operatorname{GL}(V)\cong\operatorname{GL}_{n}(\mathbb{C})$ and $\mathfrak{gl}(V)\cong\mathfrak{gl}_{n,\mathbb{C}}$ .

Definition 6.3.

A (complex) representation $(\rho,V)$ of a Lie algebra $\mathfrak{g}$ is a Lie algebra homomorphism

\rho:\mathfrak{g}\to\mathfrak{gl}(V),

where $V$ is a complex vector space. That is,

•

$\rho$ is $\mathbb{R}$ -linear;
•

$\rho([X,Y])=[\rho(X),\rho(Y)]$ .

Note that by Theorem 5.37 the differential of a Lie group representation is a Lie algebra representation.

Remark 6.4.

Warning! It is not the case that, if $\rho$ is a Lie algebra representation, then

\rho(XY)=\rho(X)\rho(Y).

Indeed, in general $X Y$ need not be an element of the Lie algebra at all, and even if it is the displayed equation will not usually hold.

The notions of $G$ -homomorphism (or $\mathfrak{g}$ -homomorphism, or intertwiner), isomorphism, subrepresentation, and irreducible representation stay the same as for finite groups. For example, a $\mathfrak{g}$ -homomorphism from $(\rho,V)$ to $(\rho^{\prime},V^{\prime})$ is a linear map $\phi:V\rightarrow V^{\prime}$ such that

\phi(\rho(X)v)=\rho^{\prime}(X)\phi(v)

for all $v\in V$ and $X\in\mathfrak{g}$ .

Definition 6.5.

A $\mathbb{C}$ -linear representation of $\mathfrak{g}$ is a complex representation $\rho$ of $\mathfrak{g}$ such that

\rho(\lambda X)=\lambda\rho(X)

for all $\lambda\in\mathbb{C},X\in\mathfrak{g}$ .

If $G$ is a complex Lie group, then a holomorphic representation of $G$ is a complex representation whose derivative is $\mathbb{C}$ -linear; equivalently, the map $G\to\operatorname{GL}(V)$ is holomorphic.

Theorem 6.6.

Let $G$ be a Lie group, $\mathfrak{g}$ be a Lie algebra.

1.

If $V_{1}$ and $V_{2}$ are irreducible finite-dimensional representations of $G$ or $\mathfrak{g}$ , then

$\dim\operatorname{Hom}_{G/\mathfrak{g}}(V_{1},V_{2})=\begin{cases}1&\text{if $% V_{1}\cong V_{2}$}\\ 0&\text{otherwise}.\end{cases}$

If $V_{1}=V_{2}$ , then any $G$ - or $\mathfrak{g}$ -homomorphism $T:V\rightarrow V$ is scalar.
2.

Any irreducible finite-dimensional representation of an abelian Lie group or Lie algebra is one-dimensional.
3.

If $(\rho,V)$ is an irreducible finite-dimensional representation of $G$ (or $\mathfrak{g}$ ) and $Z$ (or $\mathfrak{z}$ ) is the center of $G$ (or $\mathfrak{g}$ ) then there is a homomorphism $\chi:Z\rightarrow\mathbb{C}^{\times}$ (or $\chi:Z\rightarrow\mathbb{C}$ ) such that

$\rho(z)v=\chi(z)v$

for all $z\in Z$ (or $\mathfrak{z}$ ) and $v\in V$ . We call this the central character.

Proof.

The proofs are all the same as in the finite group case! ∎

Proposition 6.7.

Let $(\rho,V)$ be a finite-dimensional representation of a Lie group $G$ . Let $D\rho$ be its derivative.

1.

If $W\subset V$ is invariant under $\rho(G)$ , then $W$ is invariant under $D\rho(\mathfrak{g})$ .
2.

If $D\rho$ is irreducible, then $\rho$ is irreducible.
3.

If $\rho$ is unitary, that is, there is a basis for $V$ such that $\rho(g)\in\operatorname{U}(n)$ for all $g\in G$ , then $D\rho$ is skew-Hermitian, that is, $D\rho(X)\in\mathfrak{u}(n)$ for all $X\in\mathfrak{g}$ (using the same basis for $V$ ).
4.

Let $(\rho^{\prime},V^{\prime})$ be another finite-dimensional representation of $G$ . If $\rho\cong\rho^{\prime}$ , then $D\rho\cong D\rho^{\prime}$ .

If $G$ is connected, then the converses to these statements hold.

So, for connected Lie groups, we can test irreducibility and isomorphism at the level of Lie algebras.

Proof.

For (1), we know that $\rho(\exp(tX))(w)\in W$ for any $X\in\mathfrak{g}$ and $w\in W$ . Taking the derivative at $t=0$ , it follows that $D\rho(X)(w)\in W$ as required. Part (2) follows from (1).

For (3), if $\rho$ is unitary, then after choosing a basis appropriately it is a Lie group homomorphism $\rho:G\to U(n)$ . The derived homomorphism therefore lands in the Lie algebra $\mathfrak{u}(n)$ of $\operatorname{U}(n)$ .

For part (4), let $T$ be a $G$ -isomorphism, so that in particular,

T\rho_{1}(\exp(tX))T^{-1}=\rho_{2}(\exp(tX))

for all $X\in\mathfrak{g}$ and $t\in\mathbb{R}$ . Taking the derivative at $t=0$ gives

TD\rho_{1}(X)T^{-1}=D\rho_{2}(X)

so that $T$ is a $\mathfrak{g}$ -isomorphism as required.

If $G$ is connected, then $G$ is generated by $\exp(\mathfrak{g})$ . Hence all proofs above can be reversed. For example, for (1), suppose that $W$ is preserved by $D\rho(\mathfrak{g})$ . If $w\in W$ and $X\in\mathfrak{g}$ , then

\rho(\exp(X))w=\exp(D\rho(X))w=\sum_{n=0}^{\infty}\frac{(D\rho(X))^{n}}{n!}w\in W

as $W$ is preserved by $D\rho(X)$ and also closed. Since every $g\in G$ can be written as a finite product of $\exp(X_{i})$ for $X_{i}\in\mathfrak{g}$ , we see that $W$ is preserved by $\rho(G)$ as required. ∎

6.2 Standard constructions for representations

We give a list of various constructions with representations of Lie groups, and the analogous constructions for their derivatives.

The standard representation of a linear Lie group $G\subseteq\operatorname{GL}_{n}(\mathbb{C})$ comes from its action on $\mathbb{C}^{n}$ :

	$\displaystyle\rho(g)$	$\displaystyle=g$
	$\displaystyle D\rho(X)$	$\displaystyle=X.$

The direct sum of representations $(\rho_{1},V_{1})$ , $(\rho_{2},V_{2})$ is $(\rho_{1}\oplus\rho_{2},V_{1}\oplus V_{2})$ with derivative

D(\rho_{1}\oplus\rho_{2})=D\rho_{1}\oplus D\rho_{2}.

The determinant representation of $G\subset\operatorname{GL}_{n}(\mathbb{C})$ is $\det:G\to\mathbb{C}^{*}$ which sends $g$ to $\det(g)$ . We have

D\det(X)=\operatorname{tr}(X),

which follows from $\det\exp(tX)=e^{t\operatorname{tr}(X)}$ .

If $(\rho,V)$ is a representation of $G$ , the dual representation $(\rho^{*},V^{*})$ of $(\rho,V)$ is defined by

\left(\rho^{*}(g)(\lambda)\right)(v)=\lambda\left(\rho(g^{-1})(v)\right),

for $\lambda\in V^{*}$ a linear functional on $V$ and $g\in G$ . It has derivative

D\rho^{*}(X)(\lambda)(v)=-\lambda(D\rho(X)v).

Given a basis of $V$ , then the matrix of $\rho^{*}$ with respect to the dual basis is

\rho^{*}(g)=\rho(g)^{-T},

which differentiates to

(D\rho^{*})(X)=-D\rho(X)^{T}.

If $(\rho_{1},V_{1})$ and $(\rho_{2},V_{2})$ are representations of $G$ , then as before the tensor product representation $\rho_{1}\otimes\rho_{2}$ is the representation on $V_{1}\otimes V_{2}$ defined by

(\rho_{1}\otimes\rho_{2})(g)=\rho_{1}(g)\otimes\rho_{2}(g).

Then using the product rule one sees

D(\rho_{1}\otimes\rho_{2})=D\rho_{1}\otimes\mathrm{id}_{V_{2}}+\mathrm{id}_{V_% {1}}\otimes D\rho_{2}.

The symmetric square and alternating square are also as for finite groups. If $(\rho,V)$ is a representation of $G$ then $\operatorname{Sym}^{2}(V)$ has a representation $\operatorname{Sym}^{2}\rho$ :

\operatorname{Sym}^{2}\rho(g)(v_{1}\cdot v_{2})=(\rho(g)(v_{1}))\cdot(\rho(g)(% v_{2})),

and

D\operatorname{Sym}^{2}\rho(X)(v_{1}\cdot v_{2})=D\rho(X)(v_{1})\cdot v_{2}+v_% {1}\cdot D\rho(X)(v_{2}).

Similarly we have a representation $\Lambda^{2}\rho$ on $\Lambda^{2}(V)$ :

\Lambda^{2}\rho(g)(v_{1}\wedge v_{2})=\rho(g)(v_{1})\wedge\rho(g)(v_{2}),

D\Lambda^{2}\rho(X)(v_{1}\wedge v_{2})=D\rho(X)(v_{1})\wedge v_{2}+v_{1}\wedge D% \rho(X)(v_{2}).

We can take tensor/symmetric/alternating products of more than one factor. Suppose $(\rho_{i},V_{i})$ are representations of $G$ .

•

We form the tensor product

$V_{1}\otimes V_{2}\otimes\ldots\otimes V_{l}.$

It is generated by symbols $v_{1}\otimes\ldots\otimes v_{l}$ subject to the multilinear relations, that is, linearity in each slot:

$v_{1}\otimes\ldots\otimes(av_{i}+bv_{i}^{\prime})\otimes\ldots\otimes v_{l}=a(% v_{1}\otimes\ldots\otimes v_{i}\otimes\ldots\otimes v_{l})+b(v_{1}\otimes% \ldots\otimes v_{i}^{\prime}\otimes\ldots\otimes v_{l}).$

One has

$\dim\left(V_{1}\otimes V_{2}\otimes\ldots\otimes V_{l}\right)=\prod_{i=1}^{l}% \dim V_{i}.$

The action of $G$ is as before: for $g\in G$ , $v_{i}\in V_{i}$ ,

$g(v_{1}\otimes\ldots\otimes v_{l})=(gv_{1})\otimes\ldots\otimes(gv_{l}).$

The derivative is, for $X\in\mathfrak{g}$ and $v_{i}\in V_{i}$ ,

$\displaystyle X(v_{1}\otimes\ldots\otimes v_{l})=$ $\displaystyle(Xv_{1})\otimes v_{2}\otimes\ldots\otimes v_{l}$

$\displaystyle+v_{1}\otimes(Xv_{2})\ldots\otimes v_{l}$

$\displaystyle+\ldots$

$\displaystyle+v_{1}\otimes v_{2}\ldots\otimes(Xv_{l}).$

We also write

$V^{\otimes l}=V\otimes\cdots\otimes V.$
•

The $l$ th symmetric power is the space $\operatorname{Sym}^{l}(V)$ generated by symbols $v_{1}\cdots v_{l}$ with linearity in each slot and any permutation of the vectors giving the same element. We have

$\dim\operatorname{Sym}^{l}(V)=\binom{n+l-1}{l}=\binom{n+l-1}{n-1}$

where $\dim V=n$ . Indeed, if $e_{1},\ldots,e_{n}$ is a basis for $V$ then a basis for $\operatorname{Sym}^{l}(V)$ is

$\{e_{i_{1}}\ldots e_{i_{l}}:1\leq i_{1}\leq i_{2}\leq\ldots\leq i_{l}\leq n\}$

from which finding the dimension is a simple counting problem.

As for higher tensor powers, the actions of $G$ and $\mathfrak{g}$ are

$g(v_{1}\cdots v_{l})=(gv_{1})\ldots(gv_{l}),$

and

$X(v_{1}\cdots v_{l})=(Xv_{1})v_{2}\ldots v_{l}+\ldots+v_{1}\ldots v_{l-1}(Xv_{% l}).$
•

The $l$ th alternating power is the space $\Lambda^{l}(V)$ generated by symbols $v_{1}\wedge\cdots\wedge v_{l}$ with linearity in each slot and having the alternating property: for any permutation $\sigma\in S_{l}$ , we have

$v_{\sigma(1)}\wedge\cdots\wedge v_{\sigma(l)}=\epsilon(\sigma)(v_{1}\wedge% \cdots\wedge v_{l}).$

In particular, switching the places of two components reverses the sign, while $v_{1}\wedge\cdots\wedge v_{l}=0$ if two of the vectors coincide (more generally, if they are linearly dependent).

We have

$\dim\Lambda^{l}V=\binom{n}{l}$

where $\dim V=n$ . Indeed, if $e_{1},\ldots,e_{n}$ are a basis for $V$ then a basis for $\Lambda^{l}(V)$ is

$\{e_{i_{1}}\wedge\cdots\wedge e_{i_{l}}:1\leq i_{1}<i_{2}<\ldots<i_{l}\leq n\}$

from which finding the dimension is a simple counting problem. In particular, $\Lambda^{n}\mathbb{C}^{n}$ is one-dimensional generated by $e_{1}\wedge\cdots\wedge e_{n}$ .

The representation on $\Lambda^{l}(V)$ is given again as above: for $g\in G$ ,

$g(v_{1}\wedge\ldots\wedge v_{l})=(gv_{1})\wedge\ldots\wedge(gv_{l}),$

while for $X\in\mathfrak{g}$

$X(v_{1}\wedge\cdots\wedge v_{l})=(Xv_{1})\wedge v_{2}\wedge\ldots\wedge v_{l}+% \ldots+v_{1}\wedge\ldots\wedge v_{l-1}\wedge(Xv_{l}).$

To give an example of how to justify the claims about derivatives, we do the case of tensor products. Suppose $V$ , $W$ are vector spaces acted on by $G$ . Let $X\in\mathfrak{g}$ , $v\in V$ , $w\in W$ . We must compute

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

Expanding:

	$\displaystyle\exp(tX)v\otimes\exp(tX)w$	$\displaystyle=(v+tXv+O(t^{2}))\otimes(w+tXw+O(t^{2}))$
		$\displaystyle=v\otimes w+t(Xv\otimes w+v\otimes Xw)+O(t^{2}).$

This gives

	$\displaystyle X(v\otimes w)$	$\displaystyle=\left.\frac{d}{dt}\exp(tX)v\otimes\exp(tX)w\right\|_{t=0}$
		$\displaystyle=Xv\otimes w+v\otimes Xw$

as required.

Remark 6.8.

We defined tensor products (and so on) of representations of Lie groups and then differentiated them. We could also directly make these definitions with Lie algebras. For instance, if $\mathfrak{g}$ is a Lie algebra and $V$ is a representation of $\mathfrak{g}$ , we define the symmetric square representation on $\operatorname{Sym}^{2}(V)$ by

X(vw)=(Xv)w+v(Xw).

6.2.1 Functional constructions

We can construct representations as vector spaces of functions on topological spaces with actions of $G$ . If $G$ acts on a set $X$ , then it also acts on the vector space of functions $X\rightarrow\mathbb{C}$ by $(g\cdot f)(x)=f(g^{-1}x)$ . Usually this will be infinite dimensional, and so out of the scope of our course, but sometimes we can impose conditions allowing us to handle it. For example, $\operatorname{GL}_{n}(\mathbb{C})$ acts on $\mathbb{C}^{n}$ , and hence on the space of polynomial functions in $n$ variables. Imposing a further restriction — to homogeneous polynomials of some fixed degree — gives a finite-dimensional representation. The derivative must be calculated on a case-by-case basis. The general scheme for doing this is:

(Af)(x)=\frac{d}{dt}f(\exp(-tA)x)|_{t=0}

for $A\in\mathfrak{g}$ , $x\in X$ , and $f:X\to\mathbb{C}$ suitably regular (e.g. smooth, or polynomial) such that this formula makes sense. You will work this out in a specific example in the second assignment, and we will also return to it in section 7.5.

6.3 The adjoint representation

Let $G$ be a linear Lie group and $\mathfrak{g}$ be its Lie algebra. Then the adjoint representation $\operatorname{Ad}$ of $G$ is the action on $\mathfrak{g}$ by conjugation. We usually write $\operatorname{Ad}_{g}$ instead of $\operatorname{Ad}(g)$ , so that

\operatorname{Ad}_{g}(Y)=gYg^{-1}

for $g\in G$ and $Y\in\mathfrak{g}$ . By Proposition 5.25 (2), $gYg^{-1}$ is indeed in $\mathfrak{g}$ . Thus the map $g\mapsto\operatorname{Ad}_{g}$ is a Lie group homomorphism:

\operatorname{Ad}:G\longrightarrow\operatorname{GL}(\mathfrak{g}).

Remark 6.9.

If you look at how we proved that $\mathfrak{g}$ was closed under conjugation by $G$ , you will see that we can write

\operatorname{Ad}_{g}(Y)=\frac{d}{dt}g\exp(tY)g^{-1}|_{t=0}.

The derivative of $\operatorname{Ad}$ , denoted by $\operatorname{ad}$ , is called the adjoint representation of the Lie algebra $\mathfrak{g}$ . Thus

\operatorname{ad}=D\operatorname{Ad}.

Again, we write $\operatorname{ad}_{X}$ for $\operatorname{ad}(X)$ , so we have

\operatorname{ad}:X\in\mathfrak{g}\mapsto\operatorname{ad}_{X}\in\operatorname% {End}(\mathfrak{g})=\mathfrak{gl}(\mathfrak{g}).

By Theorem 5.37 we have the formula

\operatorname{Ad}_{\exp(X)}=\exp(\operatorname{ad}_{X})

as elements of $\operatorname{GL}(\mathfrak{g})$ .

Theorem 6.10.

Let $G$ and $\mathfrak{g}$ be as above and let $X,Y\in\mathfrak{g}$ . Then

1.

$\operatorname{ad}_{X}(Y)=[X,Y]=XY-YX.$
2.

The map $\operatorname{ad}$ is a Lie algebra homomorphism, so that

$\operatorname{ad}_{[X,Y]}=[\operatorname{ad}_{X},\operatorname{ad}_{Y}].$

Thus, for all $Z\in\mathfrak{g}$ ,

$\operatorname{ad}_{[X,Y]}(Z)=[\operatorname{ad}_{X},\operatorname{ad}_{Y}](Z)=% \operatorname{ad}_{X}(\operatorname{ad}_{Y}(Z))-\operatorname{ad}_{Y}(% \operatorname{ad}_{X}(Z)).$

Proof.

1.

Since $\operatorname{Ad}_{\exp(tX)}(Y)=\exp(tX)Y\exp(-tX)$ , taking the differential at $t=0$ , we get

$\operatorname{ad}_{X}(Y)=XY-YX=[X,Y].$
2.

The map $\operatorname{ad}$ is a Lie algebra homomorphism because it is the differential of a Lie group homomorphism.∎

Remark 6.11.

This explains the origin of the Jacobi identity:

	$\displaystyle[\operatorname{ad}_{X},\operatorname{ad}_{Y}](Z)$	$\displaystyle=\operatorname{ad}_{X}(\operatorname{ad}_{Y}(Z))-\operatorname{ad% }_{Y}(\operatorname{ad}_{X}(Z))$
		$\displaystyle=[X,[Y,Z]]-[Y,[X,Z]]$
		$\displaystyle=[X,[Y,Z]]+[Y,[Z,X]],$

while

	$\displaystyle\operatorname{ad}_{[X,Y]}(Z)$	$\displaystyle=[[X,Y],Z]$
		$\displaystyle=-[Z,[X,Y]].$

Equating these gives the Jacobi identity.

Remark 6.12.

The first formula, $\operatorname{ad}_{X}(Y)=[X,Y]$ , could have been used to define the adjoint representation for any Lie algebra, without reference to Lie groups. The Jacobi identity is then used to show that $\operatorname{ad}_{X}$ is a Lie algebra homomorphism.

Remark 6.13.

Warning! It is very easy to misinterpret some of the formulas concerning the adjoint representation. For example, $\operatorname{ad}_{[X,Y]}=[\operatorname{ad}_{X},\operatorname{ad}_{Y}]$ does not mean that

\operatorname{ad}_{[X,Y]}(Z)=[\operatorname{ad}_{X}(Z),\operatorname{ad}_{Y}(Z% )],

but (as already noted and proved) that

\operatorname{ad}_{[X,Y]}(Z)=[\operatorname{ad}_{X},\operatorname{ad}_{Y}](Z)=% \operatorname{ad}_{X}(\operatorname{ad}_{Y}(Z))-\operatorname{ad}_{Y}(% \operatorname{ad}_{X}(Z)).

Similarly, $\operatorname{Ad}_{\exp(X)}=\exp(\operatorname{ad}_{X})$ does not mean that ${\exp(X)}Y\exp(-X)$ is equal to $\exp([X,Y])$ but rather is the identity

	$\displaystyle{\exp(X)}Y\exp(-X)$	$\displaystyle=\exp(\operatorname{ad}_{X})(Y)$
		$\displaystyle=\sum_{k=0}^{\infty}\frac{(\operatorname{ad}_{X})^{k}}{k!}(Y)$
		$\displaystyle=\sum_{k=0}^{\infty}\frac{[X,[X,\dots,[X,Y]]\cdots]}{k!}.$

Proposition 6.14.

If $G$ is abelian, so is $\mathfrak{g}$ . If, moreover, $G$ is connected, then the converse holds.

Proof.

Suppose that $G$ is abelian. Then, for all $g\in G$ and $Y\in\mathfrak{g}$ ,

g\exp(tY)g^{-1}=\exp(tY).

Taking the derivative at $t=0$ we see that $gYg^{-1}=Y$ . Thus $\operatorname{Ad}$ is trivial. Differentiating, we see $\operatorname{ad}$ is trivial, so $\operatorname{ad}_{X}=0$ for all $X$ . Thus $[X,Y]=0$ for all $X,Y\in\mathfrak{g}$ as required.

Conversely, suppose $G$ is connected and $\mathfrak{g}$ is abelian. Then $\operatorname{ad}$ is trivial and, since $G$ is connected, $\operatorname{Ad}$ is trivial. Thus $gYg^{-1}=Y$ for all $g\in G,Y\in\mathfrak{g}$ . Thus $g\exp(Y)g^{-1}=\exp(Y)$ for all $g\in G$ and all $Y\in\mathfrak{g}$ . Since $\exp(\mathfrak{g})$ generates $G$ , we see that $G$ is commutative. ∎

6.4 The homomorphism $\mathrm{SU}(2)\to\mathrm{SO}(3)$

We now return to the example of $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ from Section 5.8, and explain how the adjoint representation produces the homomorphism $\pi:\mathrm{SU}(2)\to\mathrm{SO}(3)$ .

Consider the adjoint representation $\operatorname{Ad}:\mathrm{SU}(2)\to\operatorname{GL}(\mathfrak{su}_{2})$ . Define an inner product on $\mathfrak{su}_{2}$ by

\left\langle X,Y\right\rangle=-2\operatorname{tr}(XY).

(The factor of $2$ is unimportant, it just makes the basis below orthonormal rather than orthonormal up to scalar.)

Exercise 6.15.

Show that $\left\langle\cdot,\cdot\right\rangle$ is a symmetric, positive definite bilinear form on $\mathfrak{su}_{2}$ .

Show that it is preserved by the adjoint action of $\mathrm{SU}(2)$ , i.e. that

\left\langle gXg^{-1},gYg^{-1}\right\rangle=\left\langle X,Y\right\rangle

for all $g\in\mathrm{SU}(2)$ .

Choosing an orthonormal basis for $\mathfrak{su}_{2}$ with the inner product $\left\langle\cdot,\cdot\right\rangle$ , we therefore have that $\operatorname{Ad}(g)\in\mathrm{SO}(3)$ for all $g\in\mathrm{SU}(2)$ . Thus we obtain a Lie group homomorphism (renaming $\operatorname{Ad}$ to $\pi$ )

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

To be more concrete, there is an orthonormal basis $\mathcal{I},\mathcal{J},\mathcal{K}$ of $\mathfrak{su}_{2}$ given by

\frac{1}{2}\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\frac{1}{2}\begin{pmatrix}0&i\\ i&0\end{pmatrix},\frac{1}{2}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}

respectively. We have $[\mathcal{I},\mathcal{I}]=0$ , $[\mathcal{I},\mathcal{J}]=\mathcal{K}$ and $[\mathcal{I},\mathcal{K}]=-\mathcal{J}$ which gives $\operatorname{ad}_{\mathcal{I}}$ in this basis:

(D\pi)(\mathcal{I})=\operatorname{ad}_{\mathcal{I}}=\begin{pmatrix}0&0&0\\ 0&0&-1\\ 0&1&0\end{pmatrix}.

We write $J_{x}$ for the matrix on the right hand side. Similarly, we have:

	$\displaystyle(D\pi)(\mathcal{J})=(J_{y})$	$\displaystyle=\begin{pmatrix}0&0&1\\ 0&0&0\\ -1&0&0\end{pmatrix}$
	$\displaystyle(D\pi)(\mathcal{K})=(J_{z})$	$\displaystyle=\begin{pmatrix}0&-1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}.$

Giving explicit equations for $\pi$ (rather than $D\pi$ ) is possible but less pleasant, so is left as an exercise.

Exercise 6.16.

Write down explicitly the image of $\begin{pmatrix}a&-\overline{b}\\ b&\overline{a}\end{pmatrix}\in\mathrm{SU}(2)$ under $\pi$ .

Finally, one can show that $\ker(\pi)$ is a cyclic group of order 2.

Exercise 6.17.

Show that, if $g\in\mathrm{SU}(2)$ satisfies $gXg^{-1}=X$ for all $X\in\mathfrak{su}_{2}$ , then $g=\pm I$ . Deduce that

\ker(\pi)=\{\pm I\}.

Since $\mathrm{SU}(2)$ is simply-connected, this witnesses the fact that the fundamental group $\pi_{1}(\mathrm{SO}(3))$ is $C_{2}$ .

6.5 Representations of U(1) and Maschke’s theorem

We now discuss the representation theory of the unitary group $\operatorname{U}(1)=\{e^{it}\,:\,t\in\mathbb{R}\}$ , which is isomorphic to $\mathrm{SO}(2)$ , the circle group. Its Lie algebra is $i\mathbb{R}\subset\mathbb{C}$ with trivial Lie bracket, which is isomorphic to $\mathbb{R}$ .

Theorem 6.18.

All irreducible finite-dimensional representations of $\operatorname{U}(1)=\{z\in\mathbb{C};\,|z|=1\}$ are one-dimensional. They are given by

z=e^{it}\mapsto e^{int}=z^{n}

for $n\in\mathbb{Z}$ .

Proof.

It follows from Schur’s lemma that all irreducible finite-dimensional representations of $\operatorname{U}(1)$ are one-dimensional, so are homomorphisms $\operatorname{U}(1)\rightarrow\mathbb{C}^{\times}$ . Since $\operatorname{U}(1)$ is connected, such a homomorphism is determined by the derivative $\mathfrak{u}_{1}\rightarrow\mathfrak{gl}_{1}(\mathbb{C})=\mathbb{C}$ , which has the form $it\mapsto\lambda t$ for some $\lambda\in\mathbb{C}$ . As in Example 5.54, this exponentiates to a map $\operatorname{U}(1)\rightarrow\mathbb{C}^{\times}$ if and only if $\lambda=in$ for some $n\in\mathbb{Z}$ , giving the homomorphism $z\mapsto z^{n}$ . ∎

Theorem 6.19.

1.

All finite-dimensional representations of $\operatorname{U}(1)$ are unitary.
2.

All finite-dimensional representations of $\operatorname{U}(1)$ are completely reducible, that is, decompose as a direct sum of irreducible representations.

Proof.

1.

The details of this are in Problem 66. The idea is to start with any Hermitian form $(\,,\,)$ on a representation $(\rho,V)$ of $\operatorname{U}(1)$ and replace it by the averaged Hermitian form

$(v,w)_{G}=\frac{1}{\pi}\int_{0}^{2\pi}(\rho(e^{it})v,\rho(e^{it})w)dt$

which is positive definite and $G$ -invariant.
2.

This follows from (1) as in the proof of Maschke’s theorem for finite groups. Here is another proof: take $(\rho,V)$ , a finite-dimensional representation of $\operatorname{U}(1)$ of dimension $n$ . Consider its differential $D\rho:\mathfrak{u}(1)=i\mathbb{R}\to\mathfrak{gl}_{n,\mathbb{C}}$ and let $A=D\rho(i)$ . After conjugation, we may write $A=D+N$ with $D$ diagonal with integer entries, $N$ upper triangular with zeros on the diagonal, and $D$ and $N$ commuting. We have that $\rho(e^{it})=\exp(itD)\exp(itN)$ . By Theorem 6.18, the eigenvalues of $D$ are integers, so the eigenvalues of $\exp(itD)$ are complex numbers of absolute value 1. Setting $t=2\pi$ , we have $\exp(itD)=I$ and so, since $\rho(e^{2\pi it})=\rho(1)=I$ , we must have $\exp(2\pi iN)=I$ . But Problem 51 shows that $\exp$ is injective when restricted to the set of strictly upper-triangular matrices, whence $N=0$ .

Thus (after conjugating) $\rho(g)$ is diagonal for all $g\in\operatorname{U}(1)$ , so $V$ is completely reducible.

∎

Remark 6.20.

Complete irreducibility does not hold for representations of a general Lie group. For example, the standard representation of

N=\left\{\begin{pmatrix}1&x\\ 0&1\end{pmatrix}:x\in\mathbb{R}\right\}

does not decompose into a direct sum of two one-dimensional invariant subspaces (otherwise we could diagonalize $\begin{pmatrix}1&x\\ 0&1\end{pmatrix}$ , which is impossible). Furthermore, it is not unitary (as unitary matrices are diagonalizable).

Some of this actually generalizes substantially:

Theorem 6.21.

(Maschke’s theorem for compact Lie groups) Let $G$ be a compact Lie group.

1.

Any finite-dimensional representation of $G$ is unitarizable.
2.

(complete reducibility) Any finite-dimensional representation of $G$ is a direct sum of irreducible representations.

Proof.

The second part is proved exactly as for finite groups using the averaging technique. Let $(\rho,V)$ be a finite-dimensional representation and let $W$ be a subrepresentation. Let $\left\langle\,,\,\right\rangle$ be the $G$ -invariant Hermitian inner product on $V$ guaranteed by the first part. Then the orthogonal complement $W^{\perp}$ is also a subrepresentation, and $V=W\oplus W^{\perp}$ . Iterating, we obtain that $V$ is a direct sum of irreducible representations.

The proof of the first part also uses the same idea as for finite groups. Take $(\,,\,)$ to be any Hermitian inner product on $V$ . Then define

\left\langle v,w\right\rangle=\int_{g\in G}(gv,gw)dg.

This is also a Hermitian inner product, and

$\displaystyle\left\langle hv,hw\right\rangle$	$\displaystyle=\int_{g\in G}(ghv,ghw)dg$
	$\displaystyle=\int_{k\in G}(kv,kw)d(kh^{-1})$	(putting $k=gh$ )
	$\displaystyle=\int_{k\in G}(kv,kw)dk$	(since $dk=d(kh^{-1})$ )
	$\displaystyle=\left\langle v,w\right\rangle.$

The challenge here is to show that there is an appropriate notion of $\int_{g\in G}f(g)dg$ for which the step “ $dk=d(kh^{-1})$ ” is valid — this goes by the name of ‘existence of Haar measure’. For $U(1)$ you can do it by hand, see problem 66. ∎

	$\displaystyle X(v_{1}\otimes\ldots\otimes v_{l})=$	$\displaystyle(Xv_{1})\otimes v_{2}\otimes\ldots\otimes v_{l}$
		$\displaystyle+v_{1}\otimes(Xv_{2})\ldots\otimes v_{l}$
		$\displaystyle+\ldots$
		$\displaystyle+v_{1}\otimes v_{2}\ldots\otimes(Xv_{l}).$

6 Representations of Lie groups and Lie algebras - generalities

6.1 Basics

Definition 6.1.

Remark 6.2.

Definition 6.3.

Remark 6.4.

Definition 6.5.

Theorem 6.6.

Proof.

Proposition 6.7.

Proof.

6.2 Standard constructions for representations

Remark 6.8.

6.2.1 Functional constructions

6.3 The adjoint representation

Remark 6.9.

Theorem 6.10.

Proof.

Remark 6.11.

Remark 6.12.

Remark 6.13.

Proposition 6.14.

Proof.

6.4 The homomorphism SU⁢(2)→SO⁢(3)

Exercise 6.15.

Exercise 6.16.

Exercise 6.17.

6.5 Representations of U(1) and Maschke’s theorem

Theorem 6.18.

Proof.

Theorem 6.19.

Proof.

Remark 6.20.

Theorem 6.21.

Proof.

6.4 The homomorphism $\mathrm{SU}(2)\to\mathrm{SO}(3)$