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Section 4.1 \(\bmit{\SU(2,\CC')}\)

Subsection 4.1.1 Representations

By analogy with \(\SU(2)\text{,}\) we have

\begin{equation} \SU(2,\CC') = \{M\in\CC'^{2\times2}:M^\dagger M=\one, |M|=1\} .\tag{4.1.1} \end{equation}

As with \(\SU(2)\text{,}\) we expect there to be 3 degrees of freedom.

Our old friend \(M(\phi)\in\SO(2)\) is again in \(\SU(2,\CC')\text{,}\) but we give it yet another new name, writing

\begin{equation} T_y(\alpha) = \begin{pmatrix} \cos\alpha \amp -\sin\alpha \\ \sin\alpha \amp \cos\alpha \end{pmatrix} .\tag{4.1.2} \end{equation}

Other elements are

\begin{equation} T_x(\alpha) = \begin{pmatrix} \cosh\alpha \amp L\sinh\alpha \\ L\sinh\alpha \amp \cosh\alpha \end{pmatrix} , \qquad T_z(\alpha) = \begin{pmatrix} e^{L\alpha} \amp 0 \\ 0 \amp e^{-L\alpha} \end{pmatrix} .\tag{4.1.3} \end{equation}

Subsection 4.1.2 Derivatives

We know what to do next: Differentiate! So consider

\begin{equation} \begin{aligned} t_y \amp= \dot{T}_y = T_y'(0) = \begin{pmatrix} 0 \amp -1 \\ 1 \amp 0 \end{pmatrix} ,\\ t_x \amp= \dot{T}_x = T_x'(0) = \begin{pmatrix} 0 \amp L \\ L \amp 0 \end{pmatrix} ,\\ \quad t_z \amp= \dot{T}_z = T_z'(0) = \begin{pmatrix} L \amp 0 \\ 0 \amp -L \end{pmatrix} .\end{aligned}\tag{4.1.4} \end{equation}

These matrices are linearly independent, and therefore span the tangent space at the identity.

Let's compute the commutators. The result is

\begin{equation} [t_x,t_y] = 2t_z , \qquad [t_y,t_z] = 2t_x , \qquad [t_z,t_x] = -2t_y .\tag{4.1.5} \end{equation}

These commutators are almost, but not quite, those of \(\su(2)=\su(2,\CC)\text{.}\) However, complexifying \(\su(2,\CC')\) would yield the same algebra as complexifying \(\su(2,\CC)\text{;}\) the signs wouldn't then matter, since they can be changed by multiplication with \(i\text{.}\) Thus, \(\su(2,\CC')\) must be a real form of \(\su(2)\text{.}\) But which one?

Noting that \(t_y=s_y\text{,}\) \(t_x=L\sigma_x\text{,}\) and \(t_z=L\sigma_z\) shows that

\begin{equation} \su(2,\CC') \cong \sl(2,\RR) \cong \so(2,1)\tag{4.1.6} \end{equation}

since the factors of \(L\) do not change the commutators. However, we can no longer use Hermiticity to determine which elements are rotations, and which are boosts, since all elements of \(\su(2,\CC')\) are anti-Hermitian.

A more robust mechanism is provided by considering matrix squares. Recall that matrix exponentiation yields ordinary trig functions for matrices that square to minus the identity, and hyperbolic trig functions for matrices that square to the identity. Rotations correspond to ordinary trig, with compact orbits; boosts correspond to hyperbolic trig, with noncompact orbits. The correct generalization of this argument is provided by the Killing form

\begin{equation} B(X,Y) = \tr(XY) ,\tag{4.1.7} \end{equation}

which yields a norm that is negative for rotations, and positive for boosts, that is

\begin{equation} \begin{aligned} B(X,X) \lt 0 \amp\Longleftrightarrow \hbox{$X$ is a rotation} \nonumber ,\\ B(X,X) \gt 0 \amp\Longleftrightarrow \hbox{$X$ is a boost} .\end{aligned}\tag{4.1.8} \end{equation}

Using either the Killing form, or simply looking at the trig functions, we conclude that \(t_y\) is a rotation, but \(t_x\) and \(t_z\) are boosts. Comparison with \(\sl(2,\RR)\) shows that the only difference is that we have multiplied the boosts by \(L\text{,}\) changing their Hermiticity, but nothing else.

Subsection 4.1.3 Comparison with \(\bmit{\SO(2,1)}\)

We can realize this identification explicitly by considering matrices of the form

\begin{equation} Y = \begin{pmatrix}-x\amp z+Lt\\ z-Lt\amp x\end{pmatrix} .\tag{4.1.9} \end{equation}

The matrix \(Y\) is tracefree and Hermitian, and we can act on \(Y\) with \(\SU(2,\CC')\) via the action

\begin{equation} Y \longmapsto MYM^\dagger\tag{4.1.10} \end{equation}

which preserves both of these conditions. Furthermore, we have

\begin{equation} |Y| = -(x^2+z^2-t^2) ,\tag{4.1.11} \end{equation}

and, since \(|M|=1\text{,}\)

\begin{equation} |MYM^\dagger| = |Y|\tag{4.1.12} \end{equation}

Thus, \(M\) can be identified with an element of \(\SO(2,1)\text{,}\) and we have shown that

\begin{equation} \SU(2,\CC') \cong \Spin(2,1) .\tag{4.1.13} \end{equation}

Subsection 4.1.4 Generalization to \(\bmit{\CC'\otimes\CC}\)

Consider now the algebra \(\CC'\otimes\CC\text{.}\) The tensor product of two algebras consists of linear combinations of formal products. We could write \((1,1)\text{,}\) \((L,1)\text{,}\) \((1,i)\) and \((L,i)\) for these products, then define operations by acting on each algebra separately. However, it is easier to drop the parentheses, effectively just multiplying these expressions out; the elements of \(\CC'\otimes\CC\) are linear combinations of \(1\text{,}\) \(L\text{,}\) \(i\text{,}\) and \(iL\text{.}\) Since

\begin{equation} (L,1)(1,i) = (L,i) = (1,i)(L,1)\tag{4.1.14} \end{equation}

we have

\begin{equation} iL = Li .\tag{4.1.15} \end{equation}

As a vector space, \(\CC'\otimes\CC\) is clearly isomorphic to \(\RR^4\text{.}\) But the multiplication table is different from the other 4-dimensional algebras we have seen, \(\HH\) and \(\HH'\text{.}\)

Conjugation in a tensor product algebra is defined by conjugating both elements, that is

\begin{equation} \bar{(a,b)} = (\bar{a},\bar{b}) .\tag{4.1.16} \end{equation}

Thus, although

\begin{equation} \bar{i} = -i, \qquad \bar{L} = -L ,\tag{4.1.17} \end{equation}

as usual, we have

\begin{equation} \bar{iL} = {iL} .\tag{4.1.18} \end{equation}

So what is \(\su(2,\CC'\otimes\CC)\text{?}\) Reasoning by analogy with \(\su(2,\CC')\text{,}\) it is clear that a basis for \(\su(2,\CC'\otimes\CC)\) is given by

\begin{equation} \{L\sigma_x,L\sigma_y,L\sigma_z,-i\sigma_x,-i\sigma_x,-i\sigma_y\} .\tag{4.1.19} \end{equation}

This basis is the same as the basis for \(\sl(2,\CC)\) apart from the factors of \(L\text{,}\) which again appear only in the boosts. These factors change the Hermiticity, but not the commutators, so we conclude that

\begin{equation} \su(2,\CC'\otimes\CC) \cong \sl(2,\CC) \cong \so(3,1) .\tag{4.1.20} \end{equation}

Thus, the use of \(\CC'\) makes it easy to identify the boosts—they contain \(L\)—and allows boosts to be represented using anti-Hermitian matrices.