1
$\begingroup$

Consider an irreducible non-Abelian subgroup $\mathrm{H}$ of group of unitary matrices $\mathrm{U}_n(\mathbb{C})$, that contains the subgroup of diagonal matrices. Does there exist any result regarding the properties or structures of $\mathrm{H}$? The final motivations is to answer this question.

P.S. The question has been updated after some useful comments.

$\endgroup$
13
  • 3
    $\begingroup$ Can't you take block matrices of the form $\begin{pmatrix} A & 0 \\ 0 & \theta \end{pmatrix}$ (where $A$ is a matrix of dimension one less)? $\endgroup$ Jun 2, 2020 at 11:06
  • 2
    $\begingroup$ The question in the first sentence is extremely broad. One short answer is "yes". The second question is quite trivial anyway (use monomial matrices, assuming the dimension is $\ge 2$). A general natural question is then to classify subgroups containing the diagonal group (I'd suggest to edit accordingly). $\endgroup$
    – YCor
    Jun 2, 2020 at 11:08
  • 4
    $\begingroup$ Every finite group, Abelian or not, is isomorphic to a group of unitary matrices. If $n >1$, there is always a proper non-Abelian subgroup $H$ of the unitary group $U_{n}(\mathbb{C})$ which contains the subgroup $D$ of all diagonal unitary matrices, and has $H/D$ isomorphic to the symmetric group $S_{n}$. $\endgroup$ Jun 2, 2020 at 11:10
  • 1
    $\begingroup$ Again, I'm suggesting to edit your question with this (describing subgroups of $U(n)$ containing the group of diagonal matrices), which is a good question (rather than the current one which is half far too broad and half immediate). $\endgroup$
    – YCor
    Jun 2, 2020 at 11:47
  • 1
    $\begingroup$ Because of comments like those of Najib Idrissi, I'd suggest trying to classify irreducible unitary groups containing all diagonal unitary matrices. Even among these, there are many imprimitive groups to be considered which complicate things. $\endgroup$ Jun 2, 2020 at 11:52

1 Answer 1

3
$\begingroup$

1) First, let $H$ be a closed connected subgroup with this property. Let $D$ the diagonal group in $U(n)$; denote Lie algebras with Gothic letters. Then $$\mathfrak{u}(n)=\mathfrak{d}\oplus \bigoplus_{j<k}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj}).$$

Let $e_j:D\to\mathbf{C}^*$, $d\mapsto d_j$ be the projection (valued in the unit circle). As $D$-module, the above decomposition of $\mathfrak{u}(n)$ is invariant, $\mathfrak{d}$ has weight $e_j$ and $\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$ is $\mathbf{R}$-irreducible with 2-dimensional complexification of weights $\pm e_j-e_k$. Since these are distinct when the pair $(j,k)$ with $j<k$ varies, it follows that any $D$-submodule $M$ of $\mathfrak{u}(n)$ containing $\mathfrak{d}$ has the form $$M=\mathfrak{d}\oplus \bigoplus_{j<k;(j,k)\in W}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$ for some subset $W$ of the set of pairs $(j,k)$ with $j<k$. Let $W'$ be the set of pairs $(j,k)$ such that $(j,k)$ or $(k,j)$ belongs to $W$. So $W'$ is symmetric and $$M=\mathfrak{d}\oplus \sum_{(j,k)\in W'}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$ The condition that $M$ is a Lie subalgebra easily implies that $(j,k),(k,\ell)\in W'$ imply $(j,\ell)\in W'$. Hence $W''$, the union of $W'$ and the diagonal, is an equivalence relation on $\{1,\dot,n\}$. Conversely, for every equivalence relation $W''$ on $\{1,\dots,n\}$, $$\mathfrak{h}_{W''}=\mathfrak{d}\oplus \sum_{j\neq k;(j,k)\in W''}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$ is a Lie subalgebra containing $\mathfrak{d}$. The corresponding group is thus the group of block-diagonal matrices with respect to some partition (possibly permuting indices to make it block-wise).

2) Now let $H$ be a closed subgroup containing $D$, possibly not connected. Then $H^0$ has the preceding form, and $H$ normalizes $H^0$. One can check that the normalizer of $H^0$ is always of finite index over $H^0$: indeed, the blocks are precisely the irreducible components of the $H^0$-actions, and are pairwise non-isomorphic $H^0$-modules, hence they are permuted by $H$. That is, this normalizer is the stabilizer of some direct sum according to some partition of indices, possibly permuting blocks.

3) If one wants irreducibility (as I said in a comment, it just complicates the discussion while restricting the scope): it corresponds to the case where $H/H^0$ acts transitively on the set of blocks (this is possible only if all blocks have the same size)

4) The remaining step is to show that any subgroup $H$ containing $D$ is automatically closed. To start with, the connected component of the closure of $H$ is the component-wise stabilizer of some partition $P$ of $\{1,\dots,n\}$.

Let $x=(x_1,\dots,x_m)$ be an $m$-tuple of $H$. Consider the map $D^m\to U(n)$ mapping $(d_1,\dots,d_m)$ to $\prod x_id_ix_i^{-1}$. Let $r_x$ be its rank (maximum rank of its differential over $D^m$). So for some $y=(y_1,\dots,y_m)$, its rank at $y$ is $r_x$. Hence for $x'=(x_1,\dots,x_m,x_m,\dots,x_1)$, its rank at $(y_1,\dots,y_m,y_m^{-1},\dots,y_1^{-1})$ is $\ge r_x$ and moreover the value is $1$. Thus, we can assume that $x$ is chosen such that $r_x$ is maximal and achieved at a point $(y_1,\dots y_m)$ with value $1$. From the maximality it follows that the tangent image is a Lie subalgebra $\mathfrak{l}$ of $\mathfrak{u}(n)$, and it does not depend on the choice of $x$, and the corresponding immersed Lie subgroup $L$ is contained in $H$ and contains $D$. The preceding results concerns Lie subalgebras containing $\mathfrak{d}$ applies, so $\mathfrak{l}$ is the stabilizer of some partition $Q$ of $\{1,\dots,n\}$ (with $Q\subset P$ since $L\subset \bar{H}^0$). But it it easy to see that if $P\neq Q$ then $\mathfrak{h}_Q$ is not normalized by $\mathfrak{h}_P$. So $P=Q$. Hence $H\supset L=\bar{H}^0$. It follows that $H$ is closed.

$\endgroup$
2
  • $\begingroup$ Thanks. Don't you think that this result can be used to prove the following post? mathoverflow.net/questions/361477/… $\endgroup$
    – Mini
    Jun 3, 2020 at 9:52
  • $\begingroup$ @Mini Sure, it's quite immediate then $\endgroup$
    – YCor
    Jun 3, 2020 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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