Equivalences of Lie Algebras

In the following all equivalences are for complex Lie algebras.

The $so(N) \quad{}$ Lie Algebra

A basis for the Lie algebra, $so(N)$, in the fundamental representation, is given by the following $N(N-1)/2$ antisymmetric matrices:

(1)
\begin{align} \left(M_{mn}\right)_{pq}=-\left(M_{nm}\right)_{pq} = \delta_{mp}\delta_{nq}-\delta_{mq}\delta_{np}\ ,\quad m,n,p,q=1,2,\ldots, N\ . \end{align}

The Lie algebra can be obtained by computing the commutator of two such anti-symmetric matrices. It is given by

(2)
\begin{align} [M_{mn},M_{pq}] = -\delta_{mp} M_{nq} + \delta_{mq} M_{np} +\delta_{np} M_{mq} - \delta_{nq} M_{mp} \end{align}

The $so(3)\ \ {}$ Lie Algebra

Let $T_1=M_{23}$, $T_2=M_{31}$, and $T_3=M_{12}$. They satisfy the Lie algebra

(3)
\begin{align} [T_a,T_b]=-\epsilon_{abc} T_c\ , \quad a,b,c=1,2,3 \end{align}

where $\epsilon_{abc}$ is the Levi-Civita tensor in three dimensions.

Equivalence of Lie algebras

We will consider different Lie groups that have isomorphic Lie algebras. Below, we will (loosely) refer to the defining/fundamental representation of a group as ‘the group’. Thus, by $SO(2)$, we mean its realization in the form of $2\times 2$ special orthogonal matrices. This is in keeping with the physicist's blurring of the difference between the abstract group and its (defining) representation.

In order to show that two different realizations of a Lie algebra are equivalent, we need to first choose suitable bases for the two realizations and then provide a map relating the two bases (that is compatible with the Lie algebraic structure). The map should be be compatible with the Lie bracket on either side. In other words, the structure constants (for the basis elements) must be the same.

$\mathbf{so(2)\sim so(1,1)}$

We know that arbitrary elements of the group $SO(2)$ are of the form

(4)
\begin{align} g(\theta) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} = \exp\left[\theta\begin{pmatrix}0 & 1 \\ -1& 0 \end{pmatrix} \right] \end{align}

By expanding the above matrix for small $\theta$, we see that the Lie algebra is one-dimensional with one generator $T$ satisfying $[T,T]=0$.

We also know that an arbitrary element of $SO(1,1)$ is given by a similar matrix with the trigonometric functions being replaced by hyperbolic functions. This can also be achieved by the replacement (a ‘Wick rotation’) $\theta \rightarrow i \phi$. In other words, an arbitrary boost can be written as $g(i\phi)$. More precisely, one has

(5)
\begin{align} S \cdot g(i\phi) \cdot S^{-1} = \begin{pmatrix} \cosh\phi & \sinh \phi \\ \sinh \phi & \cosh \phi \end{pmatrix} \quad \textrm{with } S=\begin{pmatrix} 1 & 0 \\ 0 & -i\end{pmatrix}\ . \end{align}

This shows that the Lie algebra $so(2)\sim so(1,1)$. Of course, the groups are not isomorphic. The parameter $\theta$ is an angle variable with periodicity $2\pi$ while the rapidity $\phi$ takes values on the real line, i.e., $\phi\in {\mathbb R}$. Further, $so(2)$ and $so(1,1)$ are isomorphic as real Lie algebras.

In fact, one has a larger set of equivalences (that subsumes the above one)

(6)
\begin{align} so(p+q)\sim so(p,q)\ . \end{align}

This equivalence can be seen by carrying out similar Wick rotations on various parameters.

$\mathbf{so(3)\sim su(2)}$

An arbitrary group element of $SO(3)$ can be written as

(7)
\begin{align} \exp\left(\theta \hat{n} \cdot \vec{T}\right)\quad,\quad \textrm{where } \left(T_a\right)_{bc} =\epsilon_{abc}\ . \end{align}

An explicit expression (see assignment 2) for the above matrix is given below and it is easy to see that the periodicity of $\theta$ is $2\pi$.

(8)
\begin{align} \begin{pmatrix} \cos\theta + n_1^2(1-\cos\theta) & n_1n_2 (1-\cos\theta) + n_3 \sin\theta & -n_2 \sin\theta + n_1n_3(1-\cos\theta) \\ n_1n_2 (1-\cos\theta) - n_3 \sin\theta & \cos\theta + n_2^2(1-\cos\theta) & n_1 \sin\theta + n_2n_3(1-\cos\theta) \\ n_2\sin\theta + n_1n_3(1-\cos\theta) & -n_1\sin\theta + n_2n_3 (1-\cos\theta) & n_3^2(1-\cos\theta) + \cos\theta \end{pmatrix}\ , \end{align}

where $(n_1,n_2,n_3)$ are the components of the unit vector $\hat{n}$.

An arbitrary group element of $SU(2)$ can be written as(see assignment 3)

(9)
\begin{align} \exp\left(\theta \hat{n} \cdot \frac{i\vec{\sigma}}{2}\right)=\cos\tfrac{\theta}2\ I + i\sin \tfrac{\theta}2\ \hat{n} \cdot \vec{\sigma} \ . \end{align}

The angle $\theta$ thus has periodicity $4\pi$ — this is the key difference between the Lie groups $SO(3)$ and $SU(2)$. However, as the Lie algebra sees only the infinitesimal transformations, they can be identical. The Lie algebra isomorphism is obtained by the identification

(10)
\begin{align} -iT_a \longleftrightarrow \frac{\sigma_a}{2}\ ,\quad a=1,2,3\ . \end{align}

Of course, one needs to check that both of them satisfy the same commutation relations:

(11)
\begin{align} [J_a,J_b]=i\epsilon_{abc}\ J_c\ . \end{align}

This explains the factor pre-multiplying the Pauli sigma matrices.

$\mathbf{so(1,3)\sim so(4)\sim su(2)\oplus su(2)}$

The Lie algebra $so(4)$ is defined in terms of the generators $M_{mn}$ that we defined above. Set $T_1=M_{23}$, $T_2=M_{31}$, and $T_3=M_{12}$ as we did earlier. Let $B_{a}=M_{a4}$. The Lie algebra then takes the form

(12)
\begin{align} [T_{a},T_{b}]=-\epsilon_{abc} T_c \quad,\quad [T_{a},B_{b}]=-\epsilon_{abc} B_c \quad, \quad [B_{a},B_{b}]=\epsilon_{abc} T_c\ . \end{align}

Define a new basis

(13)
\begin{align} N^{\pm}_a = \frac12\Big(T_{a}\pm i B_a\Big) \ , \quad a=1,2,3 \end{align}

Exercise: Show that

(14)
\begin{align} [N_a^+,N_b^+]=- \epsilon_{abc} N^+_c\quad,\quad [N_a^-,N_b^-]=- \epsilon_{abc} N^-_c\quad,\quad [N_a^+,N_b^-]=0\quad.\quad \end{align}

Thus $N^+_a$ generate a $su(2)$ Lie algebra and $N^-_a$ generate another $su(2)$ Lie algebra. The third Lie bracket is the one which states that two Lie algebras ‘commute’ and hence form $su(2)\oplus su(2)$.

$\mathbf{so(6)\sim su(4)}$

This is being stated without proof. However, one simple check is that both are vector spaces with the same dimension $15=\tfrac{6\times 5}2=(4^2-1)$.

Note: There is another way to see the equivalence of Lie algebras that a mathematician might prefer. One can associate a (distinct) graph known as a Dynkin diagram to every Lie algebra. So if two Lie algebras have identical Dynkin diagrams, they must be equivalent.

References

Robert Cahn has put postscript1 files of the all chapters of his book Semi-Simple Lie Algebras and their Representations (published originally by Benjamin-Cummings in 1984) on his website for free download. If you are interested in learning more on Lie algebras (from a physics perspective), you can work through this book. Update: He has now put up both pdf and postscript files for the full book at the aforementioned URL. All students must thank him if they use his book.

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