which generators of SU(n) generate the elements of its center Zn?

Question

It is easy to show the center of SU(n) is $\mathbb{Z}_n$ from its definition, with the elements being $e^{i2\pi m/n}$. However, for $N>2$, I didn't see anywhere giving the explicit generators of them, i.e., some coefficients $\omega_j$ so that $e^{i\omega_jT^j}=e^{i2\pi/n}$, where $T^j$ are the generators in the fundamental representation. Is there a general expression of such coefficients $\omega_j$?

Answer 1

Center of SU(N) in the fundamental representation

In the fundamental representation, the Lie algebra elements are $N\times N$ traceless Hermitian matrices.

Use $e_{ij} $ to represent the matrix where only the element on ith row and jth column is nonzero, for any unitary matrix \begin{equation} U = \exp( i \sum_{i=1}^N x_i e_{ii} + \sum_{i<j} z_{ij} e_{ij} + \sum_{i > j } \bar{z}_{ij} e_{ij} ) \quad x_i \in \mathbb{R}, \sum_{i=1}^N x_i = 0 \,\, z_{ij} \in \mathbb{C} \end{equation}

Elements in the center commute with any the other elements in the group, i.e. if $c \in Z(\text{SU}(N))$, then \begin{equation} U c = c U \implies c = Uc U^{\dagger} \quad \forall U \in \text{SU(N)} \end{equation}

Now take $\partial_{z_{ij}}$ derivative on this expression and then evaluate the value at $U = I$, we have \begin{equation} 0 = i e_{ij} c - ic e_{ij} \implies [ e_{ij}, c] = 0 \end{equation} The $\partial_{\bar{z}_{ij}}$ derivative extend the range of $i,j$ to all pairs of $i\ne j$. The computation of the commutator for one pair of $i,j$ shows that the ith row, jth row, ith column and jth column of $c$ are zero except for the diagonal element. Moreover the two diagonal elements are the same. Iterating over all pairs of $i,j$ then implies $c = \omega I$, where $\omega$ is a constant.

Being a special unitary group element, c has $\det c = \omega^N = 1$, so $\omega$ is the root of unity. We have therefore constructed the center of the special unitary group: \begin{equation} Z( \text{SU(N)} ) = \left\{ \exp\left( 2\pi i \frac{m}{N} \right) I | m = 1, 2, \cdots, N \right \} = \mathbb{Z}_N \end{equation} and its element can be generated by taking \begin{equation} x_i = \left\lbrace \begin{aligned} &2\pi \frac{m}{N} & \quad 1 \le i \le N-1 \\ &-2\pi \frac{m(N-1)}{N} & \quad i = N\ \end{aligned} \right. \end{equation}