Given a set of $S$ of $ n\times n$ Hermitian matrices with entries in $\mathbb{C}$, the set of all $n\times n$ unitary matrices that commute with $S$ forms a Lie group $G$. My question is what Lie groups can we get from this process by choice of $n$ and $S$? We can get any products of unitary groups of the form $U(p)\times U(q)\times U(r)...$ by taking S to be a matrix with $p$ eigenvalues equal to $0$, $q$ eigenvalues equal to $1$ and $r$ eigenvalues equal to $2$ and so on. Is that the only possibility? Or can I get a group isomorphic to $U(1)\times Sp(m)$ or something? (We're always going to have a factor of $U(1)$ since the identity matrix commutes with everything. Like literally everything.) I can think of neither a counterexample nor a proof.

If I can get other Lie groups than the natural follow up questions would be: What Lie groups can I get? Is there a nice construction get a set $S$ given a Lie group $G$? What representations of $G$ can I get?

  • $\begingroup$ I think the direct products of $U(p)$ is all you can get. It shouldn't be too difficult to prove either, if I'm not completely mistaken. $\endgroup$ – Thomas Klimpel Aug 29 '11 at 18:13
  • $\begingroup$ @Thomas: That sounds right to me, but I can't seem to make the proof happen. $\endgroup$ – BebopButUnsteady Aug 29 '11 at 19:14

If $A$ is a Hermitian matrix whose distinct eigenvalues $\lambda_1,\ldots,\lambda_p$ have multiplicities $n_1,\ldots,n_p$ ($\sum n_j = n$), then $A=QDQ^\ast$ where $D=(\lambda_1 I_{n_1}) \oplus\ldots\oplus(\lambda_p I_{n_p})$ and $Q$ is some unitary matrix. Therefore, all unitary matrices that commute with $A$ are of the form $QUQ^\ast$, where $U=U_{n_1}\oplus\ldots\oplus U_{n_p}$ with each $U_{n_j}$ being an $n_j\times n_j$ unitary matrix. In other words, if $A$ is the only element in $S$, the Lie group $G$ in question is isomorphic to $U(n_1)\times\ldots\times U(n_p)$.

Edit: I misread the question and thought that $S$ is a matrix. $S$ should be a set of Hermitian matrices. So I have the following elaboration.

If $S$ contains more than one elements, things get more complicated. For example, it may happen that the only matrices that commute with all elements in $S$ are multiples of the identity matrix. Let $J(m) = \{\omega I_m: \omega\in\mathbb{C},\ |\omega|=1\}$. We claim that in general, the Lie group $G$ in question is isomorphic to some $$ U(n_1)\times\ldots\times U(n_p)\times J(m_1)\times\ldots\times J(m_q) $$ with $\sum_i n_i + \sum_j m_j = n$.

Here is a sketch of proof. For convenience, let $T_A:\mathbb{C}^n\mapsto \mathbb{C}^n$ denotes the linear map $x\mapsto Ax$. We say that a vector subspace is an invariant subspace to $S$ if it is invariant to every $A\in S$. Now, among all nonzero invariant subspaces to $S$ (including $\mathbb{C}^n$), choose the one, say $V$, with minimum dimension. Since every $A\in S$ is Hermitian, $V^\perp$ is also invariant to $S$. Again, we choose, among all nonzero subspaces of $V^\perp$ that is invariant to $S$, one that is of minimum dimension. Continue in this manner, we have $\mathbb{C}^n = V\oplus W_1\oplus\ldots\oplus W_q$ where $V$ is the direct sum of a number of 1-dimensional invariant subspaces to $S$ and each $W_i$ is a minimal invariant subspace to $S$ with dimension $\ge2$. Since one-dimensional invariant subspaces are eigenspaces, we may write $V=V_1\oplus\ldots\oplus V_p$ so that $T_A|_{V_j}=\lambda\ {\rm id}$ for some eigenvalue $\lambda$ (that is dependent on $A$ and $V_j$). We write $n_j=\dim(V_j)$ and $m_j=\dim(W_j)$.

Hence, with some unitary matrix $Q$, each $A\in S$ can be written in the form of $QDQ^\ast$, where $D$ is a block diagonal matrix with a common block structure $(\lambda_1 I_{n_1}) \oplus\ldots\oplus(\lambda_p I_{n_p})\oplus A_1\oplus\ldots\oplus A_q$ (each $A_i$ is a $m_i\times m_i$ matrix). For any fixed pair of indices $i\not=j$, we can have $\lambda_i\not=\lambda_j$ for some but not for all $A\in S$. It is this block structure that gives rise to the above-mentioned Lie group.

  • $\begingroup$ $S$ is a set of matrices, not a matrix. $\endgroup$ – Qiaochu Yuan Aug 29 '11 at 18:28
  • $\begingroup$ Uh, right. I read the question wrongly. Will make corrections in a moment. $\endgroup$ – user1551 Aug 29 '11 at 18:33
  • $\begingroup$ This is incomplete as it stands. For example take $A_1 = A_2$ for all matrices in $S$ in your decomposition in the last paragraph . Then the two copies $J(m_1)$ you give are actually are subgroups of a representation of $U(2)$ - $m_1$ copies of the fundamental rep. of $U(2)$ glued together. The obvious extension would be as follows: Call $J(m;l)$ the representation of $U(l)$ consisting of $m$ copies of the fundamental representation. Then any $G$ is a direct sum of $J(m_i,l_i)$ with $\sum_i m_i l_i = n$. But I'm still not entirely sure that this exhausts all the possibilities. $\endgroup$ – BebopButUnsteady Aug 30 '11 at 16:41
  • $\begingroup$ To be a bit more specific and to order my own thoughts, you've shown that we can jointly block diagonalize $S$ so that there are no invariant proper subspaces in each block. The needed two bits would be to show that i) if there are no invariant subspaces than only the multiples of the identity commute ii) if I take the direct sum of such blocks what do I get? Its not just the direct sum of the groups. $\endgroup$ – BebopButUnsteady Aug 30 '11 at 17:48
  • $\begingroup$ You are right. I had in mind $A_1,\ldots,A_q$ of different sizes at first, so I had overlooked the case with equal sized $A_j$'s. After some thoughts, I think what you suggested ($G\simeq$ a direct sum of $J(m_i,l_i)$ with $\sum_i m_il_i=n$ shoud be a complete characterization. The proof is not difficult, but it takes quite a lot of spaces so I don't think it's appropriate to write them down here. By the way, what is the origin of this problem? How does it arise? $\endgroup$ – user1551 Sep 1 '11 at 14:55

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