Definition 9.1 (Lie group). A Lie group is a smooth manifold \(G\) that is equipped with a smooth map \(G\times G\to G\,,~(g^\prime ,g)\mapsto g^\prime \,g\) (called group multiplication), an element \(e\in G\) (called identity element) and a smooth map \(G\to G\,,~g\mapsto g^{-1}\) (called inversion). These structures have to satisfy the axioms of a group, i.e.

• (i) Associativity: \((g^{\prime \prime }\,g^\prime )\,g = g^{\prime \prime }\,(g^\prime \,g)\) for all \(g,g^\prime ,g^{\prime \prime }\in G\)

• (ii) Identity property: \(g\,e = g = e\,g\) for all \(g\in G\)

• (iii) Inverse property: \(g^{-1}\,g = e = g\,g^{-1}\) for all \(g\in G\)

A Lie group \(G\) is called Abelian if the group multiplication is commutative, i.e. \(g^\prime \,g = g\,g^\prime \) for all \(g,g^\prime \in G\), and otherwise it is called non-Abelian.

Example 9.2 (The Lie groups \(\U (n)\) and \(\SU (n)\)). Let \(n\geq 1\) be a positive integer. The unitary group of degree \(n\) is defined as the submanifold

that consists of all unitary \(n\times n\)-matrices, together with the group multiplication \(U^\prime \,U\) given by matrix multiplication, the identity element \(e=\mathbf {1}\) given by the identity matrix and the inversion \(U^{-1}\) given by the inverse matrix. (Note that the latter is equal to the adjoint matrix \(U^{-1} = U^\dagger \) because we consider unitary matrices.) Setting \(n=1\), we observe that the circle group \(\mathsf {U}(1) = \{U\in \bbC \,:\,U^\ast \,U = 1\}\subseteq \bbC \) is the unitary group of degree \(1\). The special unitary group of degree \(n\) is defined as the Lie subgroup

that consists of all unitary \(n\times n\)-matrices with determinant \(1\). Since the multiplication of matrices of size \(n\geq 2\) is non-commutative, we observe that \(\mathsf {U}(n)\) and \(\mathsf {SU}(n)\) are non-Abelian Lie groups for \(n\geq 2\). In contrast to this, circle group \(\mathsf {U}(1)\) is an Abelian Lie group.

Definition 9.3 (Lie algebra). A (real) Lie algebra is a real vector space \(\g \) together with a bilinear map \([\,\cdot \,,\,\cdot \,] : \g \times \g \to \g \) (called Lie bracket) that satisfies the following axioms:

• (i) Antisymmetry: \([X,Y] = - [Y,X]\) for all \(X,Y\in \g \)

• (ii) Jacobi identity: \([X,[Y,Z]] + [Y,[Z,X]]+[Z,[X,Y]] =0\) for all \(X,Y,Z\in \g \)

A Lie algebra \(\g \) is called Abelian if the Lie bracket is trivial, i.e. \([X,Y]=0\) for all \(X,Y\in \g \), and otherwise it is called non-Abelian.

Example 9.4 (The Lie algebras \(\mathfrak {u}(n)\) and \(\mathfrak {su}(n)\)). Consider the unitary Lie group \(\U (n)\) from Example 9.2, whose elements are unitary \(n\times n\)-matrices \(U\). In the vicinity of the identity element \(\mathbf {1}\in \U (n)\), we can write such matrices in exponential form \(U = e^{X}\), where \(X\) is some small \(n\times n\)-matrix. A first-order Taylor expansion in \(X\) of the unitarity condition \(U^\dagger \,U = \mathbf {1}\) yields

\begin{flalign} U^\dagger \,U = e^{X^\dagger }\,e^X = \mathbf {1} + X^\dagger + X +\mathcal {O}(X^2) \stackrel {!}{=}\mathbf {1}\quad , \end{flalign} which means that the exponent must be an anti-Hermitian matrix \(X^\dagger = -X\). (Note that the second condition \(U\,U^\dagger = \mathbf {1}\) yields the same anti-Hermiticity condition for \(X\).) This implies that the Lie algebra of \(\U (n)\) is given by the vector space

of anti-Hermitian \(n\times n\)-matrices, on which the canonically defined Lie bracket is given by the matrix commutator

Note that \([X,Y]\in \mathfrak {u}(n)\) is indeed anti-Hermitian for all \(X,Y\in \mathfrak {u}(n)\), which is shown by a short calculation

\begin{flalign} [X,Y]^\dagger = [Y^\dagger ,X^\dagger ] = [Y,X] = -[X,Y]\quad . \end{flalign} For the special unitary Lie group \(\SU (n)\subseteq \U (n)\) there is the extra condition that \(\det U=1\) must be \(1\). Using the relationship \(\det (e^X) = e^{\mathrm {Tr}(X)}\) between determinant and trace, we find that the Lie algebra of \(\SU (n)\) is given by the vector space

of anti-Hermitian and trace-free \(n\times n\)-matrices, on which the canonically defined Lie bracket is given again by the matrix commutator

Note that \([X,Y]\in \mathfrak {su}(n)\) is indeed trace-free for all \(X,Y\in \mathfrak {su}(n)\), which is shown by using the cyclicity property of the trace

\begin{flalign} \mathrm {Tr}\big ([X,Y]\big ) = \mathrm {Tr}\big (X\,Y\big ) -\mathrm {Tr}\big (Y\,X\big ) = \mathrm {Tr}\big (X\,Y\big ) -\mathrm {Tr}\big (X\,Y\big ) =0\quad . \end{flalign} We observe that \(\mathfrak {u}(n)\) and \(\mathfrak {su}(n)\) are non-Abelian Lie algebras for \(n\geq 2\), and that the Lie algebra \(\mathrm {u}(1) = \ii \,\bbR \) of the circle group is Abelian and \(1\)-dimensional.

Remark 9.5 (Structure constants of a Lie algebra). Given any Lie algebra \(\g \), we can pick a basis \(\{X_a\in \g \}\) of its underlying vector space. Since the Lie bracket is bilinear, it is completely characterized by its values \([X_a,X_b]\in \g \) on the basis elements, which can be expanded in the chosen basis as

\begin{flalign} [X_a,X_b] = f_{ab}^c\,X_c\quad . \end{flalign} The expansion coefficients \(f_{ab}^c\in \bbR \) are called the structure constants of the Lie algebra in the chosen basis. At the level of the structure constants, antisymmetry of the Lie bracket reads as \(f_{ab}^c = -f_{ba}^c\) and the Jacobi identity translates to

\begin{flalign} f_{ad}^e\,f_{bc}^d +f_{bd}^e\,f_{ca}^d+f_{cd}^e\,f_{ab}^d =0\quad . \end{flalign} This description of Lie algebras in terms of a basis is what you’ll typically see in physics textbooks. Since using a basis \(X_a\) introduces further indices and thereby complicates notation, we will present most of our constructions in this chapter in a basis independent way.

To get a better feeling for how picking a basis works in practice, let us consider the Lie algebra \(\mathfrak {su}(2)\) of anti-Hermitian and trace-free \(2\times 2\)-matrices. A basis for \(\mathfrak {su}(2)\) is given by the following matrices

\begin{flalign} X_1 = -\frac {\ii }{2}\begin{pmatrix} 0 & 1\\ 1 & 0 \end {pmatrix}~~,\quad X_2 = -\frac {\ii }{2} \begin{pmatrix} 0 & -\ii \\ \ii & 0 \end {pmatrix}~~,\quad X_3 =-\frac {\ii }{2} \begin{pmatrix} 1 & 0\\ 0 & -1 \end {pmatrix}\quad , \end{flalign} which you probably recognize as the Pauli matrices rescaled by a factor of \(-\frac {\ii }{2}\). Computing the commutators, one finds

\begin{flalign} [X_a,X_b] = \epsilon _{abc}\,X_c\quad , \end{flalign} i.e. the structure constants \(f_{ab}^c = \epsilon _{abc}\) are given by the epsilon-tensor. The reason why there is no factor \(\ii \) on the right-hand side, as you would expect by comparing this with the \(\mathfrak {su}(2)\) Lie algebra \([L_a,L_b] = \ii \,\epsilon _{abc}\,L_c\) from your quantum mechanics module, is that we have taken different conventions for exponentials. Our convention was to write \(U = e^X\), leading to anti-Hermitian exponents \(X\), while in quantum mechanics people typically work with \(U= e^{-\ii \,L}\), leading to Hermitian exponents \(L\). Note that one can pass from one convention to the other by setting \(L_a = \ii \,X_a\), which takes care of the factor \(\ii \) in the Lie bracket relations.

Definition 9.6 (Representation of a Lie group). A (linear) representation of a Lie group \(G\) on a real or complex vector space \(V\) is a smooth group homomorphism

\begin{flalign} \rho \,:\, G \longrightarrow \mathsf {GL}(V)~,~~g\longmapsto \rho (g)\quad , \end{flalign} i.e. it preserves the group multiplication \(\rho (g^\prime \,g) = \rho (g^\prime )\circ \rho (g)\), for all \(g,g^\prime \in G\). (From this property one can deduce that the identity element \(\rho (e) = \mathrm {id}\) and inverses \(\rho (g^{-1}) = \big (\rho (g)\big )^{-1}\), for all \(g\in G\), are preserved too.) In the case where \(V\) is a complex vector space endowed with a sesquilinear inner product \(\langle \,\cdot \,,\,\cdot \,\rangle : V\times V\to \bbC \), one says that a representation \(\rho \) is unitary if it preserves the inner product in the sense that

\begin{flalign} \big \langle \rho (g)(v), \rho (g)(v^\prime )\big \rangle \,=\,\langle v,v^\prime \rangle \quad , \end{flalign} for all \(g\in G\) and \(v,v^\prime \in V\). The notation \(\rho (g)(v)\in V\) means the application of the linear map \(\rho (g) : V\to V\) on the element \(v\in V\).

Example 9.7 (Important representations of \(\SU (n)\)). We list some basic representations of \(\SU (n)\) that are needed for constructing the standard model of particle physics.

• 1. Trivial representation: The trivial representation on the \(1\)-dimensional complex vector space \(\bbC \) is given by the smooth group homomorphism

  \(\seteqnumber{0}{9.}{13}\)

  \begin{equation} \rho _{\mathrm {triv}} \,:\,\SU (n)\longrightarrow \mathsf {GL}(\bbC )~,~~U\longmapsto \rho _{\mathrm {triv}}(U) = \mathrm {id} \end{equation}

  that sends every \(U\in \SU (n)\) to the identity map. This representation is unitary with respect to the standard inner product \(\langle v,v^\prime \rangle = v^\ast \,v^\prime \) on \(\bbC \).

• 2. Defining representation: Consider the standard \(n\)-dimensional complex vector space \(\bbC ^n\). Recall from linear algebra that \(\mathrm {Lin}(\bbC ^n,\bbC ^n)\cong \mathrm {Mat}_{n\times n}(\bbC )\) can be identified with the \(n\times n\)-matrices with complex entries, hence \(\mathsf {GL}(\bbC ^n)\) can be identified with the Lie group of invertible \(n\times n\)-matrices. Explicitly, each \(n\times n\)-matrix \(A\in \mathrm {Mat}_{n\times n}(\bbC )\) defines via matrix multiplication a linear map \(\bbC ^n\to \bbC ^n\,,~v\mapsto A\,v\), and every linear map can be written uniquely in this way. The defining representation

  \(\seteqnumber{0}{9.}{14}\)

  \begin{equation} \rho _{\mathrm {def}}\,:\,\mathsf {SU}(n) \longrightarrow \mathsf {GL}(\bbC ^n)~,~~U\longmapsto \rho _{\mathrm {def}}(U) = U \end{equation}

  is then simply given by the embedding of the unitary matrices with determinant \(1\) into all invertible matrices. Endowing \(\bbC ^n\) with its standard inner product \(\langle v,v^\prime \rangle := v^\dagger \,v^\prime \), one easily checks that this representation is unitary

  \(\seteqnumber{0}{9.}{15}\)

  \begin{flalign} \big \langle \rho _{\mathrm {def}}(U)(v) , \rho _{\mathrm {def}}(U)(v^\prime )\big \rangle =\big \langle U\,v , U\, v^\prime \big \rangle = v^\dagger \,U^\dagger \,U\,v^\prime = v^\dagger \,v^\prime = \langle v,v^\prime \rangle \quad . \end{flalign}

• 3. Adjoint representation: Every Lie group \(G\) has a canonical linear representation on its Lie algebra \(\g :=T_e G\), which is called the adjoint representation. For \(\SU (n)\), the adjoint representation

  \(\seteqnumber{1}{9.17}{0}\)

  \begin{equation} \rho _{\mathrm {ad}} \,:\,\mathsf {SU}(n)\longrightarrow \mathsf {GL}(\mathfrak {su}(n)) \end{equation}

  is given element-wise in terms of matrix multiplication by

  \(\seteqnumber{1}{9.17}{1}\)

  \begin{equation} \rho _{\mathrm {ad}}(U)(X)\,:=\, U\,X\,U^{-1} \,=\, U\,X\,U^{\dagger }\quad , \end{equation}

  where in the second step we used that \(U\) is unitary, i.e. \(U^{-1} = U^\dagger \). This is indeed well-defined: One easily checks that the matrix \(\rho _{\mathrm {ad}}(U)(X)\) is anti-Hermitian

  \(\seteqnumber{0}{9.}{17}\)

  \begin{flalign} \big (\rho _{\mathrm {ad}}(U)(X)\big )^\dagger = \big (U\,X\,U^{\dagger }\big )^\dagger = U\,X^\dagger \,U^\dagger = -\rho _{\mathrm {ad}}(U)(X) \end{flalign} and trace-free

  \(\seteqnumber{0}{9.}{18}\)

  \begin{flalign} \mathrm {Tr}\big (\rho _{\mathrm {ad}}(U)(X)\big ) = \mathrm {Tr}\big (U\,X\,U^\dagger \big ) = \mathrm {Tr}\big (U^\dagger \,U\,X\big ) = \mathrm {Tr}\big (X\big ) =0\quad , \end{flalign} hence \(\rho _{\mathrm {ad}}(U)(X)\in \mathfrak {su}(n)\) defines an element of the Lie algebra. One also checks easily that \(\rho _{\mathrm {ad}}\) defines a representation, which follows from the short calculation

  \(\seteqnumber{0}{9.}{19}\)

  \begin{flalign} \rho _{\mathrm {ad}}(U^\prime \,U)(X) = (U^\prime \,U) \,X\,(U^\prime \,U)^\dagger =U^\prime \, \big (U \,X\,U^\dagger \big )\, U^{\prime \,\dagger }= \rho _{\mathrm {ad}}(U^\prime )\big (\rho _{\mathrm {ad}}(U)(X)\big )\quad . \end{flalign}

Note that the representations above also make sense when we replace \(\SU (n)\) by the unitary Lie group \(\U (n)\).

Example 9.8 (Representations of \(\mathsf {U}(1)\)). It is quite easy to write down explicit examples of unitary linear representations of the circle group \(\mathsf {U}(1) = \{U\in \bbC \,:\,U^\ast \,U = 1\}\subseteq \bbC \) on the \(1\)-dimensional complex vector space \(\bbC \). In this case we have that \(\mathsf {GL}(\bbC ) = \{U\in \bbC \,:\,U\neq 0\} = \bbC \setminus \{0\}\) is the complex plane minus the origin. For each integer \(k\in \bbZ \), we can define a unitary representation

by taking \(U\) to the power \(k\). Indeed, one easily checks that \(\rho ^{(k)}(U^\prime \,U) = (U^\prime \,U)^k = {U^\prime }^k\,U^k = \rho ^{(k)}(U^\prime )\,\rho ^{(k)}(U)\), for all \(U,U^\prime \in \mathsf {U}(1)\). With some more efforts, which won’t be discuss here, one can show that these are all irreducible linear representations of \(\mathsf {U}(1)\), i.e. every other linear representation can be build from these \(\rho ^{(k)}\)’s.

Remark 9.9 (New representations from old ones). There exist general techniques to construct new representations out of given ones. A detailed discussion of these methods goes beyond the scope of this module, but as a taster I would like to sketch some of the basic ideas.

• (i) Tensor product representations: Let \(G\) be a Lie group and suppose that we have given two vector spaces \(V\) and \(W\) with representations \(\rho _V\) and \(\rho _W\), respectively. Then the tensor product representation is defined by

  \(\seteqnumber{0}{9.}{21}\)

  \begin{flalign} \rho _{V\otimes W} \,:\,G\longrightarrow \mathsf {GL}(V\otimes W) ~,~~g\longmapsto \rho _V(g)\otimes \rho _W(g)\quad . \end{flalign} You probably have seen simple examples of such tensor product representations when you’ve studied spin in quantum mechanics, which are unitary \(\mathsf {SU}(2)\) representations. More concretely, a non-relativistic spin \(\frac {1}{2}\) particle is described by the defining representation of \(\mathsf {SU}(2)\), which in physics is typically written in terms of the states \(\ket {\,{\uparrow }\,},\ket {\,{\downarrow }\,}\in \bbC ^2\). The tensor product of this representation with itself is described by the tensor product states \(\ket {\,{\uparrow \uparrow }\,},\ket {\,{\uparrow \downarrow }\,},\ket {\,{\downarrow \uparrow }\,}, \ket {\,{\downarrow \downarrow }\,} \in \bbC ^2\otimes \bbC ^2\). A standard exercise in quantum mechanics is then to show that this tensor product representation can be decomposed into a spin \(1\) and a spin \(0\) representation by using Clebsch-Gordan coefficients.

• (ii) Dual representations: Let \(G\) be a Lie group and suppose that we have given a representation \(\rho \) on a vector space \(V\). Recall that the dual vector space \(V^\ast :=\mathrm {Lin}(V,\mathbb {K})\) is given by the vector space of linear maps \(f:V\to \mathbb {K}\) from \(V\) to the \(1\)-dimensional vector space \(\mathbb {K}\), i.e. \(\mathbb {K} = \bbR \) in the real case and \(\mathbb {K} = \bbC \) in the complex case. Then the dual representation

  \(\seteqnumber{1}{9.23}{0}\)

  \begin{flalign} \rho ^\ast \,:\,G\longrightarrow \mathsf {GL}(V^\ast ) \end{flalign} is defined element-wise by

  \(\seteqnumber{1}{9.23}{1}\)

  \begin{flalign} \rho ^\ast (g)(f)\,:=\, f\circ \rho (g^{-1})\quad , \end{flalign} for all \(g\in G\) and \(f\in V^\ast \).

Definition 9.10 (Representation of a Lie algebra). A representation of a Lie algebra \(\mathfrak {g}\) on a real or complex vector space \(V\) is a Lie algebra homomorphism

\begin{flalign} \underline {\rho }\,:\, \mathfrak {g} \longrightarrow \mathfrak {gl}(V)~,~~X\longmapsto \underline {\rho }(X)\quad , \end{flalign} i.e. a linear map that preserves the Lie bracket

\begin{flalign} \underline {\rho }\big ([X,Y]\big ) = \big [\underline {\rho }(X),\underline {\rho }(Y)\big ]\quad , \end{flalign} for all \(X,Y\in \mathfrak {g}\). In the case where \(V\) is a complex vector space endowed with a sesquilinear inner product \(\langle \,\cdot \,,\,\cdot \,\rangle : V\times V\to \bbC \), one says that a representation \(\underline {\rho }\) is unitary if it preserves the inner product in the sense that

\begin{flalign} \big \langle \underline {\rho }(X)(v), v^\prime \big \rangle + \big \langle v, \underline {\rho }(X)(v^\prime )\big \rangle \,=\,0\quad , \end{flalign} for all \(X\in \mathfrak {g}\) and \(v,v^\prime \in V\). The notation \(\underline {\rho }(X)(v)\in V\) means the application of the linear map \(\underline {\rho }(X) : V\to V\) on the element \(v\in V\).

Example 9.11 (Important representations of \(\mathfrak {su}(n)\)). Before going into the examples, let me sketch the basic idea how one can construct Lie algebra representations from linear Lie group representations: Given any linear representation \(\rho : \SU (n)\to \mathsf {GL}(V)\) of the Lie group \(\SU (n)\), we can consider the linear map \(\rho (e^{X}) : V\to V\) that is associated to \(U=e^X\in \SU (n)\) for a Lie algebra element \(X\in \mathfrak {su}(n)\). Using a first-order Taylor expansion in \(X\), we define

\begin{flalign} \rho (e^{X}) =: \mathrm {id} + \underline {\rho }(X) + \mathcal {O}(X^2)\quad . \end{flalign} From the Baker-Campbell-Hausdorff formula \(e^{X}\,e^Y = e^{X+Y + \frac {1}{2}[X,Y] + \cdots }\) and the property that \(\rho \) is a Lie group representation, one shows that \(\underline {\rho }\) is a Lie algebra representation, i.e.

\begin{flalign} \underline {\rho }\big ([X,Y]\big )\,= \, \big [\underline {\rho }(X),\underline {\rho }(Y)\big ]\quad , \end{flalign} for all \(X,Y\in \mathfrak {su}(n)\). The Lie algebra representations associated with the \(\SU (n)\) representations listed in Example 9.7 read concretely as follows:

• 1. Trivial representation: By definition, we have that

  \(\seteqnumber{0}{9.}{28}\)

  \begin{flalign} \rho _{\mathrm {triv}}(e^X) = \mathrm {id}\quad , \end{flalign} hence \(\underline {\rho _{\mathrm {triv}}}(X) = 0\) for all \(X\in \mathfrak {su}(n)\). This Lie algebra representation is unitary.

• 2. Defining representation: Let us compute

  \(\seteqnumber{0}{9.}{29}\)

  \begin{flalign} \rho _{\mathrm {def}}(e^X) = e^X = \mathrm {id} + X +\mathcal {O}(X^2)\quad , \end{flalign} hence we have that \(\underline {\rho _{\mathrm {def}}}(X) = X\) for all \(X\in \mathfrak {su}(n)\). This Lie algebra representation is unitary when \(\bbC ^n\) is endowed with its standard inner product \(\langle v,v^\prime \rangle := v^\dagger \,v^\prime \). Indeed,

  \(\seteqnumber{0}{9.}{30}\)

  \begin{flalign} \big \langle \underline {\rho _{\mathrm {def}}}(X)(v) , v^\prime \big \rangle + \big \langle v , \underline {\rho _{\mathrm {def}}}(X)(v^\prime )\big \rangle = v^\dagger \, X^\dagger \,v^\prime + v^\dagger \, X\,v^\prime =0\quad , \end{flalign} for all \(X\in \mathfrak {su}(n)\) and \(v,v^\prime \in \bbC ^n\), where in the last step we used that \(X^\dagger = -X\) is anti-Hermitian.

• 3. Adjoint representation: Let us compute

  \(\seteqnumber{0}{9.}{31}\)

  \begin{flalign} \rho _{\mathrm {ad}}(e^X)(Y)\,:=\, e^{X}\,Y\,e^{-X} \,=\,Y + X\,Y - Y\,X +\mathcal {O}(X^2) \,=\, Y + [X,Y] +\mathcal {O}(X^2)\quad , \end{flalign} hence we have that \(\underline {\rho _{\mathrm {ad}}}(X)(Y) = [X,Y]\) is given by the Lie bracket for all \(X,Y\in \mathfrak {su}(n)\).

Note that the representations above also make sense when we replace \(\SU (n)\) by the unitary Lie group \(\U (n)\).