Applied Representation Theory : 006

Lie groups and their algebras

From the example of $\mathfrak{sl}_{2}$, we develop the idea of a Lie algebra. The correspondence with Lie groups is presented, showing how Lie algebras serve as infinitesimal group transformations.

Lightning review of $\mathsf{SL}_{2}\mathbb{C}$

You might remember that the special linear group is the set of $2\times2$ complex matrices with unit determinant:

$$\mathsf{SL}_{2}\mathbb{C} = \left\{ g \in \mathsf{Hom}_{2,2}\mathbb{C} \;\Big|\; \det g = 1\right\}.$$

By the Jacobi relation,

\begin{equation}\label{relation}\det e^{M} = e^{\mathsf{Tr}M},\end{equation}

we see that the subset of $\mathsf{Hom}_{2,2}\mathbb{C}$

$$\mathfrak{sl}_{2} = \left\{ m \in \mathsf{Hom}_{2,2}\mathbb{C} \;\Big|\; \mathsf{Tr}\,m = 0\right\},$$

parametrizes all of $\mathsf{SL}_{2}\mathbb{C}$.

The set $\mathfrak{sl}_{2}$ is a three-dimensional, complex vector space. A basis for $\mathfrak{sl}_{2}$ is given by the physicists' $\sigma$ matrices.

$$\sigma_{1} = \left(\begin{array}{rr} 0 & 1 \\ 1 & 0\end{array}\right),\quad \sigma_{2} =  \left(\begin{array}{rr} 0 & -i \\ i & 0\end{array}\right),\quad \sigma_{3}= \left(\begin{array}{rr} 1 & 0 \\ 0 & -1\end{array}\right).$$

The vector space $\mathfrak{sl}_{2}$ is not a closed algebra under matrix multiplication, as

$$\left\{\sigma_{i},\sigma_{j}\right\} = 2\delta_{ij},$$

which means that in particular,

$$\sigma_{i}\sigma_{i} = \mathbb{1},$$

which is not traceless. This is good, as it facilitates the Euler relation:

$$e^{i\alpha\sigma_{i}} = \cos\alpha \mathbb{1} + i\sin\alpha\sigma_{i},$$

which makes the parametrization of $\mathsf{SL}_{2}\mathbb{C}$ particularly easy to understand.

The Lie algebra $\mathfrak{sl}_{2}$

The vector space $\mathfrak{sl}_{2}$ is closed under a different kind of multiplication operation. It's closed under commutators:

$$[\sigma_{i},\sigma_{j}] = 2i\epsilon_{ijk}\sigma_{j}.$$

As you might recall, the commutator of two items $A$ and $B$ 

$$[A,B] = AB - BA,$$

quantifies the failure of those objects to commute. Commutators are a special case of a more general notion of a Lie bracket.

Given a vector space $V$, a Lie bracket on the elements of $V$ is a bilinear map\(^{1}\) that satisfies two additional requirements. First, it is alternating:

\begin{equation}\label{alternating}[A,A] = 0,\quad \quad A \in V.\end{equation}

Second, it satisfies the Jacobi identity\(^{2}\):

\begin{equation}\label{identity}[A,[B,C]] + [B,[C,A]] + [C,[A,B]]= 0,\quad \quad A,B,C \in V.\end{equation}

The Lie bracket is not associative, as you can infer from \eqref{identity}. Therefore the order in which the operations are performed matters:

$$[A,[B,C]] \neq [[A,B],C].$$

A vector space that is closed under a Lie bracket is called a Lie algebra. Evidently, $\mathfrak{sl}_{2}$ is a Lie algebra. 

Lie Groups

The group of rotations in the plane, $\mathsf{SO}_{2}$ is isomorphic to the unit circle, $S^{1}$. The isomorphism can be made explicit by using its complex form, $\mathsf{U}_{1}$. Any group that can be similarly modeled by a geometric shape - more precisely, a differentiable manifold - is called a Lie group. 

Lie groups have coordinate systems. In particular, these means we can consider continuous paths in them. Such paths afford a notion of distance. Where we can talk about group elements that far apart or infinitesimally close together.

A manifold looks locally like a Euclidean vector space, and a Lie group is no different. What is remarkable about Lie groups is they look locally like Lie algebras. In other words, the Lie bracket can be used to represent an infinitesimal group transformation.

Given an element $A$ of a Lie algebra $\mathfrak{g}$, we can exponentiate $A$ to find

\begin{equation}\label{exp}e^{A} = \sum_{n=0}^{\infty}\frac{1}{n!}A^{n},\end{equation}

which is a member of some Lie group, $G$. At least for finite-dimensional case, $\mathfrak{g}$ completely\(^{3}\) determines $G$.

While not certainly not obvious, it is worth observing that the structure of the Lie bracket - particularly the Jacobi identity - affords the Baker-Campbell-Hausdorff formula:

\begin{equation}\label{bch}e^{A}e^{B} = e^{A + B + \frac{1}{2}[A,B] + \frac{1}{12}[A,[A,B]] + \frac{1}{12}[B,[B,A]]+\cdots},\end{equation}

which can be used to represent the group multiplication in terms of commutators. All possible nested commutators are included in \eqref{bch}.

The group parametrized by $\mathfrak{sl}_{2}$ via the exponential map \eqref{exp} is $\mathsf{SL}_{2}\mathbb{C}$.

Lie Subgroups and Lie subalgebras

Subalgebras of Lie algebras generate subgroups of Lie groups. As an example, let us explore $\mathfrak{su}_{2}$ as a subalgebra of $\mathfrak{sl}_{2}$:

$$\mathfrak{su}_{2} = \mathsf{span}_{\mathbb{R}}\,\left\{ \sigma_{1},\sigma_{2},\sigma_{3}\right\}.$$

That is, $\mathfrak{su}_{2}$ is the restriction of $\mathfrak{sl}_{2}$ to real coefficients. 

We have already seen that the $\sigma$ matrices obey the algebra,

$$[\sigma_{i},\sigma_{j}] = 2i\epsilon_{ijk}\sigma_{k}.$$

That factor of $i$ might cause concern that a Lie bracket operation would take us away from real coefficients. But careful accounting shows that it doesn't - it's related the factor of $i$ in $\sigma_{2}$.

We can remove all reference to $i$ by a slight change of basis. Consider the matrices

$$\sigma_{\pm} = \frac{1}{2}\left(\sigma_{1} \pm i\sigma_{2}\right),$$

and for simplicity, let

$$\sigma = \sigma_{3}.$$

Here

$$\sigma_{+} = \left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right),\quad \sigma_{-} = \left(\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right).$$

The Lie algebra $\mathfrak{su}_{2}$ then can be re-expressed in terms of the commutation relations:

\begin{equation}\label{newsu2}[\sigma,\sigma_{\pm}] = \pm2\sigma_{\pm},\quad [\sigma_{+},\sigma_{-}] = \sigma.\end{equation}

As a Lie algebra, $\mathfrak{su}_{2}$ generates the Lie group, $\mathsf{SU}_{2}$, which is the set of invertible, $2\times 2$ matrices $U$ where

$$U^{\dagger} = U^{-1} \quad \mathrm{and}\quad \det U = 1.$$

It is therefore a subgroup of $\mathsf{SL}_{2}\mathbb{C}$. Notice that, despite the restriction to the real coefficients, it is not the subgroup $\mathsf{SL}_{2}\mathbb{R}$. We shall build that subgroup from a Lie subalgebra of $\mathfrak{sl}_{2}$ in the exercises.

Action on the Quaternions

Recall that $\mathbb{H}$ can be modeled by the $\sigma$ matrices. For a generic $q = a + ib + jc + kd$ in $\mathbb{H}$ we have a corresponding matrix:

$$Q = a\mathbb{1} + b i\sigma_{1} + c i \sigma_{2} + d i \sigma_{3},\quad a,b,c,d\in \mathbb{R}.$$

The Lie group $\mathsf{SL}_{2}$ acts on $\mathbb{H}$ by a modified sort of conjugation,

\begin{equation}\label{slaction}\mathsf{SL}_{2}: Q \mapsto M^{\dagger} Q M ,\end{equation}

so that $|Q| = \det Q$ is preserved. That is,

\begin{equation}\label{maction}\det Q \mapsto  \det M^{\dagger} \det Q \det M = |\det M|^{2} \det Q = \det Q,\end{equation}

as $\det M = 1$ by hypothesis.

Notice that the subgroup $\mathsf{SU}_{2}$ does more than preserve $|Q|$, it also keeps the real part of $Q$ fixed:

$$\mathsf{SU}_{2} : a\mathbb{1}\mapsto U^{\dagger} a\mathbb{1} U = a\mathbb{1}.$$

A generic matrix $M$ in $\mathsf{SL}_{2}$ does not have this property. The action of a generic $M$ as in \eqref{maction} rotates amongst all four unit quaternions. It corresponds to six independent rotations. 

The action of the subgroup $\mathsf{SU}_{2}$ on $\mathbb{H}$ restricts to rotations of the imaginary quaternions. It corresponds to three independent rotations.

Exercises

6.1a : Abelian Lie algebras

The unit complex numbers form a Lie group, $\mathsf{U}_{1}$. This group has an action on $\mathbb{C}$ that causes a phase rotation:

$$e^{i\theta}\cdot z = Re^{i(\phi+\theta)},\quad z \in \mathbb{C}.$$

Show that the Lie algebra of $\mathsf{U}_{1}$ is one dimensional, and can be generated by the matrix $J = i\sigma_{2}$. Hence argue that the Baker-Campbell-Hausdorff formula \eqref{bch} defines the group multiplication explicitly. 

Define the $n$-fold cover of $\mathsf{U}_{1}$ by the complex numbers,

$$\mathsf{U}_{1,n} = \left\{ e^{in\theta}\;\Big|\; n\in \mathbb{N}\right\},$$

and show for each $n$, it has the same Lie algebra as $\mathsf{U}_{1}$.

6.1b : The Universal Cover

Show that $\mathbb{R}$ is a Lie group, and argue that it has the same Lie algebra as $\mathsf{U}_{1}$. What is the relationship between $\mathsf{U}_{1,n}$ and $\mathbb{R}$?

6.1c : The $n$-torus

Argue that $\mathsf{R}^{n}$ is a Lie group, where $g$ in $\mathbb{R}^{n}$ acts as $n$ independent copies of $\mathbb{R}$. What is the equivalent construction to $\mathbb{U}_{1}$ in this case? What does the Lie algebra for $\mathsf{U}_{1,n}\times\mathsf{U}_{1,m}$ look like?

6.2 : The Heisenberg Algebra

Consider the Lie algebra $\mathfrak{h}$ generated by three elements, $A$, $B$ and $C$, where

$$[A,B] = C,\quad [C,A] = [C,B] = 0.$$

Use the Baker-Campbell-Hausdorff formula \eqref{bch} to define the group multiplication for the Lie group $H$ generated by $\mathfrak{h}$. Show that a generic element $h$ in $\mathfrak{h}$ can be written as a matrix,

$$h = \left(\begin{array}{cc} a & c \\ 0 & b\end{array}\right),\quad a,b,c \in \mathbb{R}.$$

What is the structure of the corresponding Lie group? What happens if $a,b,c$ are complex numbers?

6.3 : Diffeomorphisms of the Circle

Let $f$ and $g$ be smooth functions from the unit circle to itself:

$$f,g: S^{1} \rightarrow S^{1}.$$

Argue that such functions form a vector space. Show that the bracket

$$[f,g] = \frac{df}{d\theta}g - f\frac{dg}{d\theta},$$

makes this vector space a Lie algebra. What is the corresponding Lie group?

6.4 : Isometries of the Quaternions

What are the six independent rotations of the $\mathsf{SL}_{2}\mathbb{C}$ action on $\mathbb{H}$? Use the Lie algebra $\mathfrak{sl}_{2}$ to parametrize them. Verify the claim that $\mathsf{SU}_{2}$ has only three independent rotations, and that those rotations exist amongst $\mathsf{Im}\,\mathbb{H}$ by explicitly mapping them to a restriction of the aforementioned six.

6.5 : $\mathsf{SL}_{2}\mathbb{R}$

$\mathsf{SL}_{2}\mathbb{R}$ are the $2\times 2$ invertible, real matrices with unit determinant. Prove that $\mathsf{SL}_{2}\mathbb{R}$  is a subgroup of $\mathsf{SL}_{2}\mathbb{C}$. Show that the Lie subalgebra of $\mathfrak{sl}_{2}$ given by

$$\mathfrak{sl}_{2\mathbb{R}} = \mathsf{span}_{\mathbb{R}}\left\{ i\sigma_{1} , \sigma_{2}, \sigma_{3}\right\},$$

generates $\mathsf{SL}_{2}\mathbb{R}$.

Find a change of basis for $\mathfrak{sl}_{2\mathbb{R}}$ given by 

$$\mathfrak{sl}_{2\mathbb{R}} = \mathsf{span}_{\mathbb{R}}\left\{ \tau , \tau_{+},\tau_{-}\right\},$$

so that 

$$[\tau,\tau_{\pm}] = \pm \tau_{\pm}, \quad [\tau_{+},\tau_{-}] = - 2\tau.$$

compare with \eqref{newsu2}.

6.6 : Subgroups of $\mathsf{SL}_{2}\mathbb{C}$

Show that $g$ in $\mathsf{SL}_{2}\mathbb{R}$, $g^{\dagger}\sigma_{1}g = \sigma_{1}.$ Hence argue that $\mathsf{SL}_{2}\mathbb{R}$ action on $\mathbb{H}$ induced by \eqref{slaction} leaves a different a different one-dimensional subspace of $\mathbb{H}$ invariant. Use the Lie algebras to define the corresponding groups that leave the other two, one-dimensional subspaces of $\mathbb{H}$ invariant.

Argue that any subgroup of $\mathsf{SL}_{2}\mathbb{C}$ generated by a three (real) dimensional subalgebra of $\mathfrak{sl}_{2}$ has a basis 

$$\left\{\sigma,\sigma_{+},\sigma_{-}\right\}$$

which satisfies either

$$[\sigma,\sigma_{\pm}] = \pm \sigma_{\pm},\quad [\sigma_{+},\sigma_{-}] = 2\sigma$$

or 

$$[\sigma,\sigma_{\pm}] = \pm \sigma_{\pm},\quad [\sigma_{+},\sigma_{-}] = -2\sigma.$$

That is to say, prove that any such Lie subalgebra of $\mathfrak{sl}_{2}$ is Lie algebra homomorphic\(^{4}\) to either $\mathfrak{su}_{2}$ or $\mathfrak{sl}_{2\mathbb{R}}$. Hence argue that all such groups are homomorphic to $\mathsf{SU}_{2}$ or $\mathsf{SL}_{2}\mathbb{R}$.

\(^{1}\) That is, it's a linear map for both factors.

\(^{2}\) Not to be confused with the Jacobi relation, \eqref{relation}.

\(^{3}\) There are subtleties, of course. There is a unique, simply connected group $G$. There may be homomorphic groups that factorize into products that do not share this technical property. A physical example of this phenomena involves Dirac vs Weyl neutrinos: a fermion in four-dimensions can only interact as both a left or right handed particle if it has a mass. Otherwise the two behave as distinct particles. We shall return to this discussion in due course.

\(^{4}\) Actually, Lie algebra isomorphic, in this case

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