0
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.