# identify a curious subgroup in $U(n)$

Consider the following element $A$ in $U(n)$: $$\begin{pmatrix} 1/2(1+z) & 1/2(1-z) & \\\\ 1/2(1-z) & 1/2(1+z) & \\\\ & &I_{n-2} \end{pmatrix},$$ where $|z| = 1$.

Now conjugate $A$ by permutation matrices $S$, i.e., $S(i,j) =1$ if $\sigma(i) = j$ for a fixed $\sigma \in S_n$. What group does $S A S^{-1}$ generate? What is its dimension?

Finally let $A' = \begin{pmatrix} i & -i & \\\\ -i & i & \\\\ & & 0_{n-2} \end{pmatrix}$ be the associated Lie algebra element to $A$ (i.e., derivative with respect to $z$ at $z = 1$; notice the condition $|z|=1$). Can one give an explicit basis of the Lie algebra closure generated by $S A S^{-1}$, with each element $B$ in the basis of the form $Ad(Ad(\ldots Ad(Ad(Ad(Ad(S_1,A),Ad(S_2,A))\ldots, Ad(S_k,A)),Ad(S_0,A'))$, where $S_i$, $i=0,1,\ldots, k$ are permutation matrices.

Edit: I now have the following conjecture regarding the Lie algebra: It is simply given by $\{A \in \mathfrak{u}(n): \sum_j A_{ij} = 0, \text{ for all }i \in [n]\}$, so it looks like the dimension should be $n^2 - 2n+1$. If so the group will be $U(n-1)$ acting on $V = \{z_1 + \ldots + z_n = 0\}$. The construction of explicit basis by Adjoint action remains.

• I cannot fix the latex for displaying the matrices properly. Someone in charge please help! – John Jiang Mar 25 '12 at 0:00
• You ned to write \\\\ to get a working \\. – Mariano Suárez-Álvarez Mar 25 '12 at 0:08
• @Mariano: Thanks for the tip! I can't seem to remember this. To help those of us who forget about mathoverflow latex idiosyncracies, wouldn't it make sense to allow us to look at others' latex source? – John Jiang Mar 25 '12 at 0:10
• I see that even when I look at latex source of some matrix, it shows up only double slash, not quadruple slash. So there is no way of showing the actual thing typed in? – John Jiang Mar 25 '12 at 0:14

It is clear from the structure of the generators $S A S^{-1}$ that the resulting group lies in the group $U(n-1) \subset U(n) \subset SO(2n)$, where the last inclusion is through the classical identification of $I$ with $1$ and $J$ with $i$, and $U(n-1)$ acts on the complex vector space $V = \{z_1 + \ldots + z_n = 0\}$.
So to show the generated group $G$ is isomorphic to $U(n-1)$, it suffices to show that the Lie algebra has at least $(n-1)^2$ linearly independent elements. One set of linear independent elements are $Ad(S, A'):= S A' S^{-1}$, for $S$ corresponding to $\sigma \in S_n$. These provide $\binom{n}{2}$ elements.
So we still need $\binom{n-1}{2}$ other elements. Those are provided by $Ad(Ad(S_{ij},A'),Ad(S_{jn}, A'))$, $1\le i < j \le n-1$, where $S_{ij}$ stands for the permutation matrix corresponding to the transposition $(ij)$. So we have identified a basis of $\mathfrak{g}$ using only Adjoint action on $A'$ by the generator elements.