Quantum field theory 1, lecture 02

Algebraic approach to Lie groups and Lie algebras

Lie algebra and generators

Because a Lie group is also a manifold, group elements can be labeled by a (usually multi-dimensional) parameter or coordinate \(\xi=(\xi^1,\ldots, \xi^m)\), i.e. we can write them as \(g(\xi)\). Without loss of generality we can assume that \(\xi=0\) corresponds to the unit element, \(g(0)=\mathbb{1}\). Let us now consider infinitesimal transformations. We can write them as \[g(d \xi) = \mathbb{1} + i d\xi^j T_j + \ldots, %\label{eq:para_rep}\] where we use Einsteins summation convention implying a sum over the index \(j\), and the ellipses stand for terms of quadratic and higher order in the infinitesimal \(d\xi\). Note that we can write \[i T_j = \frac{\partial}{\partial \xi^j} g(\xi){\Big |}_{\xi=0}. \label{eq:lie_group_gen}\] Formally, the objects \(i T_j\) constitute a basis of the tangent space of the Lie group manifold at the position of the unit element \(g(\xi)=\mathbb{1}\), which is at \(\xi=0\). The factor \(i\) is conventional and used by physicists, while mathematicians usually work in a convention without it. The \(T_j\) are also known as the generators of the Lie algebra, to which we turn in a moment. The generators constitute a basis such that any element of the Lie algebra can be written as a linear superposition \(v^j T_j\).

Exponetial map

A very important idea is now that one can compose finite group elements, at least in some region around the unit element, out of very many infinitesimal transformations. In other words one writes \[g(\xi) = \lim_{N\to\infty} \left( \mathbb{1} + \frac{i \xi^j T_j}{N} \right)^N = \exp\left(i\xi^j T_j \right). \label{eq:lie_alg_exp_para}\] One recognizes here that the limit in \(\eqref{eq:lie_alg_exp_para}\) would give the exponential if \(T_j\) were just numbers, and one can essentially use this limit to define also the exponential of Lie algebra elements. Alternatively, the exponential may also be evaluated as the usual power series \[\exp\left(i\xi^j T_j \right) = \mathbb{1} + i\xi^j T_j + \frac{1}{2} \left( i\xi^j T_j \right)^2 + \frac{1}{3!} \left( i\xi^j T_j \right)^3+ \ldots\] Note that for \(\alpha, \beta \in \mathbb{R}\) one can combine \[\exp\left(i \alpha \xi^j T_j \right) \exp\left(i \beta \xi^j T_j \right) = \exp\left(i (\alpha+\beta)\xi^j T_j \right). \label{eq:xialphabeta}\] Such transformations (for fixed \(\xi\)) form a one-parameter subgroup.

Combining transformations

It is more difficult to combine transformations \(\exp\left(i \xi^j T_j \right)\) and \(\exp\left(i \zeta^j T_j \right)\) when \(\xi\) is not parallel to \(\zeta\). The reason is that \(\xi^j T_j\) and \(\zeta^j T_j\) can not be assumed to commute. To combine two transformations, one needs to use the Baker-Campbell-Hausdorff formula \[\exp(X) \exp(Y)= \exp(Z(X,Y)), \label{eq:BCH01}\] with \[Z(X,Y) = X+Y+\frac{1}{2}[X,Y]+ \frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+ \ldots \label{eq:BCH02}\] This shows that it is crucial to know how to calculate commutators between the Lie algebra elements.

Commutator and structure constants

For transformations close to the identity element we can write using \(\eqref{eq:BCH01}\) and \(\eqref{eq:BCH02}\) \[\exp\left(i \xi^j T_j \right) \exp\left(i \zeta^j T_j \right) = \exp\left(i \omega^j T_j \right), \label{eq:multiplicationtwoexp}\] with \[\omega^l = \xi^l + \zeta^l - \frac{1}{2} \xi^j \zeta^k f_{jk}^{\phantom{jk}l} + \ldots,\] where the ellipses stand now for terms of quadratic and higher order in \(\xi\) and \(\zeta\). We are using here the structure constants \(f_{jk}^{\phantom{jk}l}\) of the Lie algebra defined through the commutator \[\begin{split} \left[T_j,T_k \right] = i \, f_{jk}^{\phantom{jk}l} \, T_l. \end{split} \label{eq:lie_alg}\] The structure constants are obviously anti-symmetric, \[f_{jk}^{\phantom{jk}l} = - f_{kj}^{\phantom{jk}l}.\] Equation \(\eqref{eq:lie_alg}\) tells that the commutator of two generators can itself be expressed as a linear combination of generators. Together with eqs. \(\eqref{eq:BCH01}\) and \(\eqref{eq:BCH02}\) this makes sure that the group elements \(\eqref{eq:lie_alg_exp_para}\) can be multiplied and indeed form a group. In other words, if eq. \(\eqref{eq:lie_alg}\) holds, we can multiply group elements as in eq. \(\eqref{eq:multiplicationtwoexp}\) to yield another term of the same structure such that they form a group. On the other side, one could also start from the group multiplication law and demand that the left hand side of \(\eqref{eq:multiplicationtwoexp}\) can be written as on the right hand side. At order \(\sim \xi \zeta\) this implies then a relation of the form \(\eqref{eq:lie_alg}\).

Generators and stucture constants for unitary groups

For unitary Lie groups where \(g^\dagger = (1-id\xi^j T_j^\dagger)= g^{-1} = (1-id\xi^j T_j)\) the generators must be hermitian, \[T_j = T_j^\dagger.\] Also, in that case the structure constants are real, \[f_{jk}^{\phantom{jk}l} = f_{jk}^{*\phantom{k}l}.\] This follows from \[- i f_{jk}^{*\phantom{k}l} T_l = [T_j, T_k]^\dagger = [T_k, T_j] = i f_{kj}^{\phantom{jk}l} T_l = - i f_{jk}^{\phantom{jk}l} T_l.\]

Lie algebra

The definition \(\eqref{eq:lie_alg}\) also makes sure that linear combinations of generators, which obviously form a vector space, constitute a Lie algebra. To that end, the Lie bracket \([\cdot,\cdot]\) must have the properties

• Bilinearity: \([ \lambda A + \mu B, C ] = \lambda [A, C] + \mu [B, C]\),

• Antisymmetry: \([A, B] = - [B, A]\),

• Jacobi identity: \([ A, [ B, C ]] + [B, [C, A]] + [C, [A, B]] = 0\).

From the Jacobi identity for the generators \[\begin{split} [T_j,[T_k,T_l]] + [T_k,[T_l,T_j]] + [T_l,[T_j,T_k]] = 0, \end{split}\] one infers for the structure constants \[f_{jn}^{~~m}f_{kl}^{~~n}+f_{kn}^{~~m}f_{lj}^{~~n}+f_{ln}^{~~m}f_{jk}^{~~n}=0. \label{eq:JacobiIdStructureConst}\]

Representations

The commutation relation \(\eqref{eq:lie_alg}\), expressed also in terms of the structure constants, define a Lie algebra, similar as the multiplication rules do for a group. One distinguishes between a particular Lie algebra as an abstract entity and a concrete incarnation or representation of it.

Fundamental representation

A first example is the fundamental representation \[(T_j^{(F)})^m_{\phantom{m}n} = (t_j)^m_{\phantom{m}n}.\] For SU\((N)\), the generators in the fundamental representation \(t_j\) are hermitian and traceless \(N\times N\) matrices. This is a real vector space of dimension \(N^2-1\).

For SU\((2)\) we can write \(t_j=\sigma_j/2\) with the three Pauli matrices, \[\sigma_1 = \begin{pmatrix} 0 && 1 \\ 1 && 0 \end{pmatrix}, \quad\quad\quad \sigma_2 = \begin{pmatrix} 0 && -i \\ i && 0 \end{pmatrix}, \quad\quad\quad \sigma_3 = \begin{pmatrix} 1 && 0 \\ 0 && -1 \end{pmatrix}.\] For SU\((3)\) one takes \(t_j=\lambda_j/2\) with the eight Gell-Mann matrices \[\begin{split} \lambda_1 = & \begin{pmatrix} 0 && 1 && 0 \\ 1 && 0 && 0 \\ 0 && 0 && 0 \end{pmatrix}, \quad\quad\quad \lambda_2 = \begin{pmatrix} 0 && -i && 0 \\ i && 0 && 0 \\ 0 && 0 && 0 \end{pmatrix}, \quad\quad\quad \lambda_3 = \begin{pmatrix} 1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && 0 \end{pmatrix}, \\ \lambda_4 = & \begin{pmatrix} 0 && 0 && 1 \\ 0 && 0 && 0 \\ 1 && 0 && 0 \end{pmatrix}, \quad\quad\quad \lambda_5 = \begin{pmatrix} 0 && 0 && -i \\ 0 && 0 && 0 \\ i && 0 && 0 \end{pmatrix}, \\ \lambda_6 = & \begin{pmatrix} 0 && 0 && 0 \\ 0 && 0 && 1 \\ 0 && 1 && 0 \end{pmatrix}, \quad\quad\quad \lambda_7 = \begin{pmatrix} 0 && 0 && 0 \\ 0 && 0 && -i \\ 0 && i && 0 \end{pmatrix}, \quad\quad\quad \lambda_8 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -2 \end{pmatrix}. \end{split}\] The restriction to the purely imaginary matrices \(\lambda_2\), \(\lambda_5\) and \(\lambda_7\) generates the Lie algebra of the orthogonal group SO\((3)\) in its fundamental representation.

Adjoint representation

From the Jacobi identity \(\eqref{eq:JacobiIdStructureConst}\), one can see that the structure constants can actually be used to construct another representation, the so-called adjoint representation. Here one sets the matrices to \[(T_j^{(A)})^{m}_{\phantom{m}l} = i f_{jl}^{\phantom{jn}m}. \label{eq:adjointRepAlgebra}\] Indeed one has for the commutator of two generators \[^{m}_{\phantom{m}l} = - f_{jn}^{\phantom{jn}m} f_{kl}^{\phantom{kl}n} + f_{kn}^{\phantom{kn}m} f_{jl}^{\phantom{jl}n} = - f_{jk}^{\phantom{jk}n} f_{nl}^{\phantom{ln}m} = i f_{jk}^{\phantom{jk}n} (T_n^{(A)})^m_{\phantom{m}l}.\]

The dimension of the adjoint representation equals the number of generators of the Lie algebra. For example, the Lie algebra of SU\((3)\) has 8 generators and accordingly the adjoint representation is given by \(8\times 8\) matrices.

The fundamental and the adjoint representation are the most important representations. The adjoint representation always exists and can be used to classify Lie algebras.

However, there are many more representations of Lie algebras and they all induce corresponding representations of the Lie group through the exponential map \(\eqref{eq:lie_alg_exp_para}\).

Differential geometric approach to Lie groups and Lie algebras

Left translation

Now that we understand already some of the properties of Lie groups and Lie algebras, let us discuss them also from a geometric point of view. It is interesting to consider the group multiplication as a map on the group manifold, \[L_h: G \to G, \quad\quad\quad L_h(g) = h g. \label{eq:leftTranslation}\] This is the so-called left translation.

Tangent space at unit element

Recall that in \(\eqref{eq:lie_group_gen}\) we have introduced the generators \(T_j\) as a basis for the tangent space of the group manifold \(G\) at the identity \(g=\mathbb{1}\) or \(\xi=0\). More formally, one can construct the tangent space of a manifold as a basis for vectors, which are in turn defined through curves. For this construction we first consider a curve in the group manifold parametrized by some parameter \(\alpha\in \mathbb{R}\) and we assume that it goes through the unit element \(g(\alpha_0)=\mathbb{1}\). We can write the curve as \(g(\alpha)\), or in terms of coordinates \(\xi\) on the group manifold as \(\xi(\alpha)\) such that \(g(\alpha) = g(\xi(\alpha))\) and \(\xi(\alpha_0)=0\). Now consider the derivative \[\frac{d}{d\alpha} g(\alpha) {\Big |}_{\alpha=\alpha_0} = \frac{\partial}{\partial \xi^j} g(\xi) {\Big |}_{\xi=0} \frac{d\xi^j}{d\alpha} = iT_j \frac{d\xi^j}{d\alpha}.\] This is now an element of the tangent space \(T_\mathbb{1}(G)\) of the group manifold at the point where \(g(\xi)=\mathbb{1}\). Any element of this vector space can be written as a linear combination of the basis elements \[i T_j = \frac{\partial}{\partial \xi^j} g(\xi){\Big |}_{\xi=0}.\]

Induced basis for tangent spaces at other points

Interestingly, this basis for \(T_\mathbb{1}(G)\) can be extended to a basis for the tangent spaces at other positions of the group manifold. To that end we can use the left translation \(\eqref{eq:leftTranslation}\). Specifically, from the curve \(g(\alpha)\) we can construct another curve through the left translation \(\eqref{eq:leftTranslation}\) \[\tilde g(\alpha) = L_h (g(\alpha)) = h g(\alpha).\] The derivative at the point \(\alpha_0\) is now \[\frac{d}{d\alpha} \tilde g(\alpha){\Big |}_{\alpha=\alpha_0} = \frac{\partial}{\partial \xi^j} h g(\xi) {\Big |}_{\xi=0} \frac{d\xi}{d\alpha}. %= i \tilde T_j(h) \frac{d\xi}{d\alpha}.\] One observes that a basis for the tangent space \(T_h(G)\) is given by \[i T_j(h) = \frac{\partial}{\partial \xi^j} h g(\xi) {\Big |}_{\xi=0} =h \frac{\partial}{\partial \xi^j} g(\xi) {\Big |}_{\xi=0} = i \, h T_j. \label{eq:mapTangentSpaces}\] In this way we can actually get a basis for the tangent spaces everywhere in the entire group manifold. It is quite non-trivial that the tangent spaces can be parametrized by a single set of basis functions \(iT_j(h)\). One says that the manifold \(G\) is parallelizable.

Vector fields on group manifolds

Formally, the map \(\eqref{eq:mapTangentSpaces}\) between the tangent spaces \(T_\mathbb{1}(G)\) and \(T_h(G)\) is an example for a pushforward, induced by the map \(\eqref{eq:leftTranslation}\) on the manifold itself. One also writes this as \[T_j(h) = L_{h*} T_j(\mathbb{1}) = L_{h*} T_j.\] One may now construct vector fields on the entire manifold as linear combinations, \[V(h) = v^j(h) T_j(h). \label{eq:vectorFieldonG}\]

Left-invariant vector fields

Such a vector field is called left invariant if \[L_{g*} V(h) = V(gh).\] Because the basis \(T_j(h)\) is left-invariant by construction, the vector field \(\eqref{eq:vectorFieldonG}\) is left-invariant when the coefficients \(v^j(h)\) are independent of the position on the manifold, i. e. independent of \(h\).

In summary, we may say that the generators of the Lie algebra \(T_j\) induce actually a left-invariant basis for vector fields on the entire group manifolds. One may even understand the Lie algebra itself as an algebra of left-invariant vector fields. The Lie bracket is then introduced as the Lie derivative of vector fields.

Examples for matrix Lie algebras

Let us end this section with a few examples for Lie algebras corresponding to matrix Lie groups introduced previously.

• \(\mathfrak{su}(n)\) is the Lie algebra corresponding to the group SU\((n)\). We write the group elements as \(U=\exp(i t)\). From \(U^\dagger U = \mathbb{1}\) one infers \(t^\dagger = t\). Writing this in components, the real part is symmetric, \(\text{Re}(t_{nm}) = \text{Re}(t_{mn})\), and the imaginary part is anti-symmetric, \(\text{Im}(t_{nm}) =- \text{Im}(t_{mn})\). Moreover, we have the condition \(\det(U)=1\). The latter can be rewritten as \[\det(U) = \exp(\ln(\det(U))) = \exp(\text{Tr}\{ \ln(U)\}) = \exp(i \text{Tr}\{ t\})=1,\] so that we need \(\text{Tr}\{t\}=0\). These arguments show that the Lie algebra \(\mathfrak{su}(n)\) as a real vector space has \(n^2-1\) linearly independent generators \(T_j\).

• \(\mathfrak{so}(n)\) is the Lie algebra corresponding to the group SO\((n)\). Here we write the group elements as \(R=\exp(it)\) and they are real matrices such that \(R^T R = \mathbb{1}\). For the Lie algebra elements we have again \(t=t^\dagger\). In order for an infinitesimal transformation \(R=\mathbb{1}+i t\) to be real, the components \(t_{mn}\) must be imaginary, and therefore also anti-symmetric. The condition \(\text{Tr}\{ t\}=0\) is then automatically fulfilled. These arguments show that the Lie algebra \(\mathfrak{so}(n)\) has \(n(n-1)/2\) linearly independent generators \(T_j\).

• \(\mathfrak{sp}(2n)\) is the Lie algebra corresponding to the group Sp\((2n)\). The group elements \(R=\exp(it)\) are real matrices that satisfy \(R^T \Omega R=\Omega\) with \(\Omega=-\Omega^T\) given in \(\eqref{eq:OmegaSpn}\). For an infinitesimal transformation \(R=\mathbb{1}+i t\) one finds the condition \[\Omega t + t^T \Omega = \Omega t - t^T \Omega^T = \Omega t - (\Omega t)^T=0.\] In other words, \(\Omega t\) must be symmetric. These arguments show that the Lie algebra \(\mathfrak{sp}(2n)\) has \(n(2n+1)\) linearly independent generators \(T_j\).