Applied Representation Theory : 003

The Euler Relation

The complex numbers enjoy a polar representation that gives rise to the famous Euler relation. The angular component is parametrized by the unit circle in \(\mathbb{C}\). The quaternions - being a normed vector space - also enjoy a similar representation, although angular component is now parametrized by the three-sphere.

The Commutator

The quaternions are a vector space equipped with a multiplication operation on vectors. Such a space is called an algebra. The multiplication of quaternions does not commute. Instead, basis elements \(i\), \(j\) and \(k\), satisfy:

\begin{equation}\label{commutators}ij - ji = 2k,\quad jk - kj = 2i,\quad ki - ik = 2j.\end{equation}

The lack of commutation is formally captured by the commutator, which for two objects \(A\) and \(B\) is written

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

The idea is that if \([A,B] = 0\), \(A\) and \(B\) commute. The commutator of \(i\) and \(j\) is

$$[i,j] = 2k.$$

Last time we saw how the quaternions could be modeled by traceless, \(2\times 2\) complex matrices, the \(\sigma\) matrices:

$$ I = i\sigma_{1},\quad J = i\sigma_{2}, \quad K = i\sigma_{3}.$$

This construction of \(\mathbb{H}\) also satisfies the commutation relations \eqref{commutators}:

$$[I,J] = 2K, \quad [J,K] = 2I,\quad [K,I] = 2J.$$

Because the abstract basis of quaternions: \(i,j,k\) satisfy the same commutation relations as the concrete, matrix basis: \(I,J,K\)

we say that the two formulations are algebraically isomorphic. That is to say, they are essentially equivalent constructions. By contrast, \(\mathbb{R}^{4}\) and \(\mathbb{H}\) are isomorphic as vector spaces, but not as algebras\(^{1}\).

The commutation relations that define the matrix representation of \(\mathbb{H}\) can be expressed compactly in terms of the \(\sigma\) matrices:

\begin{equation}\label{cr}[\sigma_{m},\sigma_{n}] = 2i\epsilon_{mn\ell}\sigma_{\ell}.\end{equation}

Here \(m,n,\ell\) runs through \(1,2,3\) and the \(\epsilon\) symbol is the totally antisymmetric, three index object where

$$\epsilon_{123} = \epsilon_{231} = \epsilon_{312} =  1, \quad \epsilon_{132} = \epsilon_{321} = \epsilon_{213} = -1.$$

While intricate, this notation is compact and extremely common.

The \(\sigma\) matrices also satisfy set of relations, the so-called anticommutation relations:

\begin{equation}\label{acr}\left\{ \sigma_{m} , \sigma_{n}\right\} = \sigma_{m}\sigma_{n} +  \sigma_{n}\sigma_{m} = 2\delta_{mn}\mathbb{1},\end{equation}

where \(\delta_{mn}\) is the symmetric object where \(\delta_{mn}\) is \(1\) if \(m=n\) and \(0\) otherwise. It immediately follows that the unit imaginary quaternions satisfy

\begin{equation}\label{acc}\left\{ i,j\right\} = 0,\quad \left\{ j,k\right\} = 0 , \quad \left\{ k,i\right\} = 0,\end{equation}

which is to say, they anticommute.

This fact proves useful in evaluating products:

$$(bi + cj + dk)^{2} = -(b^{2} + c^{2} + d^{2}) + bc(ij + ji) + bd(ik + ki) + cd(jk + kj),$$

so that \eqref{acc} implies the cross terms vanish

$$(bi + cj + dk)^{2} = -(b^{2} + c^{2} + d^{2}).$$

The Euler Relation

Let \(\mathsf{U}_{1}\) be the set of complex numbers of unit magnitude:

$$\mathsf{U}_{1} = \left\{ z \in \mathbb{C} \;\Big|\; |z| = 1\right\}.$$ 

Complex numbers have a polar decomposition:

$$z = Re^{i\varphi},\quad R = |R|, \;\varphi \in [0,2\pi).$$

Evidently \(|z| = |R|\), so that the unit complex numbers can alternatively be parametrized by an angle \(\theta\). 

$$\mathsf{U}_{1} = \left\{ e^{i\theta} \in \mathbb{C} \;\Big|\; \theta \in [0,2\pi)\right\},$$

which is equivalent as a set to the unit circle, \(S^{1}\).

The unit complex numbers satisfy the Euler relation:

\begin{equation}\label{euler}e^{i\theta} = \cos\theta + i\sin\theta,\end{equation}

which you can readily verify by explicitly expanding the exponential as a series. Recall from our previous discussions that we can represent a complex numbers with \(2 \times 2\) matrices,

$$ Z = a \mathbb{1} + b J,$$

where $J = i\sigma_{2}$. Specified as a matrix, $Z$ is

$$Z = \left(\begin{array}{rr} a & b \\ -b & a \end{array}\right),$$

so that \eqref{euler} becomes

\begin{equation}\label{rotations}e^{i\theta} = \left(\begin{array}{rr} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array}\right),\end{equation}

which is precisely the matrix of rotations in the plane. Notice that 

$$\det Z = a^{2} + b^{2} = |Z|^{2},$$

so that 

$$\det e^{i\theta} = \cos^{2}\theta + \sin^{2}\theta = 1.$$

The Unit Quaternions

Just as the unit complex numbers sweep out a circle in the plane, the unit quaternions sweep out a 3-sphere in \(\mathbb{R}^{4}\). Let us see how that might be. For a given quaternion \(q\)

$$q = a + bi + cj + dk.$$

The unit quaternions are those that satisfy \(|q| = 1\). Let us denote the set of these unit quaternions by \(\mathbb{H}_{1}\). So,

$$\mathbb{H}_{1} = \left\{q \in \mathbb{H} \;\Big|\; |q| = 1\right\}$$

This condition corresponds to the hypersurface in \(\mathbb{R}^{4}\) defined by the equation

\begin{equation}\label{surface3}a^{2} + b^{2} + c^{2} + d^{2} = 1.\end{equation}

Let us split \(q\) into real and imaginary parts:

$$q = a + \Theta,$$

where \(\Theta = bi + cj + dk\). In the exercises you are asked to verify the Euler relation for imaginary quaternions:

\begin{equation}\label{eulerq}e^{\Theta} = \cos|\Theta| + \frac{\Theta}{|\Theta|} \sin|\Theta|,\end{equation}

using the commutators of abstract basis elements \(i,j\) and \(k\) given by \eqref{commutators}.

There is another way to motivate the Euler relation \eqref{eulerq}, using the matrix representation of \(\mathbb{H}\). 

In the 19th century Jacobi proved that for a complex, square matrix \(M\):

\begin{equation}\label{jacobi}\det e^{M} = e^{{\sf Tr}M}.\end{equation}

Since \(\Theta\) is purely imaginary,

$$\Theta = bI + cJ + dK.$$

Notice that the \(\sigma\) matrices all have vanishing trace. Therefore the matrices \(I,J\) and \(K\) are also traceless, so that \eqref{jacobi} gives us:

$$\det e^{\Theta} = |e^{\Theta} | = 1.$$

Therefore, any imaginary quaternion matrix \(\Theta\) can be exponentiated to give a member of \(\mathbb{H}_{1}\).

Exercises

3.1 : Algebras over \(\mathbb{R}^{4}\)

An algebra is a vector space with a notion of vector multiplication. \(\mathbb{H}\) is an algebra based on \(\mathbb{R}^{4}\), where a generic vector \(v\) in \(\mathbb{R}^{4}\):

$$v = a\hat{e}_{1} + b\hat{e}_{2} + c\hat{e}_{2} + d\hat{e}_{4},$$

can be multiplied by other vectors following the identifications

$$\hat{e}_{1} = 1, \quad \hat{e}_{2} = i,\quad \hat{e}_{3} = j, \quad \hat{e}_{4} = k. $$

\(i,j\) and \(k\) of course satisfy the commutation relations \eqref{commutators}.

There are many possible algebras that can be built from \(\mathbb{R}^{4}\). The alternating algebra in four-dimensions demands

$$\hat{e}_{i}\hat{e}_{j} = - \hat{e}_{j}\hat{e}_{i},$$ 

while the symmetric algebra requires

$$\hat{e}_{i}\hat{e}_{j} = \hat{e}_{j}\hat{e}_{i}.$$ 

The clifford algebra modifies the alternating algebra somewhat:

$$\hat{e}_{i}\hat{e}_{j} + \hat{e}_{j}\hat{e}_{i} = 2\delta_{ij},$$ 

where \(\delta_{ij} = 1\) if \(I=j\), and vanishes otherwise. We can even augment the clifford algebra to service four-dimensional Minkowski space:

$$\hat{e}_{\mu}\hat{e}_{\nu} + \hat{e}_{\nu}\hat{e}_{\mu} = 2\eta_{\mu\nu},$$ 

where off-diagonal terms vanish and \(\eta_{tt} = 1\) and \(\eta_{xx} = \eta_{yy} = \eta_{zz} = -1\).

Prove that each of these algebras are different  - that is, they are not isomorphic - by considering the space of all linearly independent products of basis vectors for each case.

3.2 : Prove the Euler Relation

For a purely imaginary quaternion:

$$\Theta = ib + jc + kd,$$

verify the Euler relation for quaternions

$$e^{\Theta} = \cos|\Theta| + \frac{\Theta}{|\Theta|} \sin|\Theta|.$$

3.3 : Eigenvalues of \(I,J,K\)

Show that the eigenvalues of \(I,J\) and \(K\) are all \(\pm i\). Hence show that the eigenvalues of 

$$\Theta = ib + jc + kd,$$

are \(\pm i|\Theta|\).

3.4 : Polar Decomposition for \(\mathbb{H}\)

For a given quaternion \(q\):

$$q = a + ib + jc + dk,$$

construct the polar decomposition of \(q\). That is, for any \(q\), find a real number \(q_0\) and a totally imaginary quaternion \(\Phi\) such that

$$q = q_{0}e^{\Phi}.$$

3.5 : The Exponential Map as a homomorphism

A homomorphism ( say, \(f\)) is a map between two groups (say \(G\) and \(H\)) that preserves the structure of group multiplication:

$$f : G \rightarrow H,$$

where

$$f(a\cdot b) = f(a)\cdot f(b).$$

Argue that \(\mathsf{U}_{1}\) is a group under multiplication. Show that it is homomorphic to the additive group of the reals \((\mathbb{R},+)\). Show explicitly that the exponential map is a homomorphism that takes this additive group \((\mathbb{R},+)\)to the multiplicative group \((\mathbb{C},\times)\).

3.6 : Quaternions don’t commute

Argue that \(\mathsf{Im}\,\mathbb{H}\) is a vector subspace of \(\mathbb{H}\), and is therefore an additive group under vector addition. We have already seen that the exponential maps \(\mathsf{Im}\,\mathbb{H}\) to \(\mathbb{H}_{1}\). Why does the exponential map fail to to be a homomorphism of groups in this case? For \(\Theta\) and \(\Phi\) in \(\mathsf{Im}\,\mathbb{H}\), explicitly evaluate both

$$e^{\Theta}e^{\Phi},\quad \mathrm{and}\quad e^{\Theta + \Phi}.$$

Use the Euler relation to compare them explicitly and remark on their difference.

\(^{1}\) Why? Because \(\mathbb{R}^{4}\) has no inherent concept of vector multiplication. It is a technical point, to be sure, but one that we need to use often in other contexts.

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