When are all centralizers in a Lie group connected? Let $G$ be a compact connected Lie group acting on itself by conjugation,
$$ G\times G\to G,\qquad (\sigma,h)\mapsto \sigma h \sigma^{-1}.$$
The fixed point set of a closed subgroup $H\le G$ equals the centralizer $C_G(H)$ of $H$ in $G$.
For a current project I am searching for examples of actions $G\times X\to X$ such that every fixed point set $X^H$ is connected, and I would like to fit the above conjugation action into this framework. Hence my questions:


*

*Are there any conditions on a connected Lie group $G$ which ensure that the centralizer $C_G(H)$ of every closed subgroup $H\le G$ is connected?

*Can someone give examples where this is true? I was thinking the unitary groups $U(n)$ might be candidates, since at least their centre $U(1)$ (which clearly is contained in every centralizer) is connected.


Thanks, and apologies if this is too vague or elementary.
 A: Centralizers of various types of subgroups, and their relationship to invariants like the cohomology of the classifying space $BG$ has a long history of study, at least going back to the papers from the 50s by Borel and others.
EDIT: Changed what's below to a more limited statement.
I don't know a reference to your exact question, but the following more limited statement is true:
Theorem: A compact connected Lie group G has the property (**) "For every elementary abelian $p$-subgroup $H \leq G$, $C_G(H)$ is connected" if and only if $G$ has torsion-free fundamental group and has a finite cover isomorphic to a product of copies of $SU(n)$, $Sp(n)$, $S^1$ for various $n \geq 1$.
In particular, if you in addition assume $G$ to be simply connected, the only examples are products of $SU(n)$ and $Sp(n)$ for various $n$.
Sketch of proof:
1) The condition is necessary: If (**) holds holds for all elementary abelian p-subgroups $H$ for all p, then every elementary abelian p-subgroup has to be contained in a maximal torus. By a result of Borel this implies that the fundamental group is torsion free. One easily checks case-by-case that the exceptional groups and Spin(n), $n \geq 6$ cannot be involved in $G$ since they have non-toral elementary abelian p-subgroup (which remain so even after modding out by a central subgroup). This leaves the cases above.
2) The condition is sufficient: We just have to check that the centralizer of an element of finite order of a group as above will again be a group of the same form, which I think you can do by hand.
--
Further info: The groups in the list above are exactly the ones whose classifying spaces have integral cohomology rings a finitely generated polynomial algebra. In fact any polynomial ring which occur as the cohomology ring of the space has to be of this form ---  this is the so-called Steenrod problem, whose solution is described in Andersen-Grodal: The Steenrod problem of realizing polynomial cohomology rings. J. Topology 1 (2008), 747-760. (DOI: 10.1112/jtopol/jtn021). 
Also, see Andersen-Grodal-Møller-Viruel The classification of p-compact groups for p odd. Annals of Math. 167 (2008), 95--210, in particular section 10+12 for a summary of the relationship between subgroups of $G$, their centralizers, and cohomology. Our setup is more general in that paper (we study p-compact groups), but it contains references to the papers of Borel etc.
A: Here's a simple argument for the case of $U(n)$.  Let $h \subseteq U(n)$ be a unitary matrix.  Then $h$ commutes with an element $g \in U(n)$ if and only if $g$ and $h$ are simultaneously diagonalisable (since both are diagonalisable).  Let $\lambda_1, \dots, \lambda_k$ be the distinct eigenvalues of $h$ with eigenspaces $E_1,\dots,E_k$.  For each $j$, let $s_j \in [-i\pi,i\pi)$ be such that $e^{s_j} = \lambda_j$.  Then the $s_j$ are also distinct.  Define $\alpha_h \colon I \to U(n)$ to be the path such that $\alpha_h(t)$ has eigenvalue $e^{t s_j}$ on the subspace $E_j$.  Then $\alpha_h(0) = I_n$ and $\alpha_h(1) = h$ so $\alpha_h$ is a path from $I_n$ to $h$.  By construction, for $t \ne 0$ then $\alpha_h(t)$ has exactly the same eigenspaces as $h$ and so $\alpha_h(t)$ commutes with exactly the same elements of $U_n$ that $h$ commutes with.
Hence for a subset $A \subseteq U(n)$ then if $h \in Z(A)$, $\alpha_h(t) \in Z(A)$ for all $t \in [0,1]$.  Thus $Z(A)$ is connected.
A: Instead of a condition on $G$, there is a simple condition on $H$.
Proposition. For a compact connected Lie group $G$, the centralizer of any torus $S$ is precisely 
the union of all maximal tori of $G$ that contain $S$; in particular, it is connected. 
Proof: Let $g\in G$ centralize $S$. Then 
$S$ is contained in the connected centralizer $Z_G(g)^0$ of $g$, hence $S$ is contained
in a maximal torus $T$ of $Z_G(g)$; but $T$ is also a maximal torus of $G$ (since we can always find a maximal torus of $G$ containing $g$). q.e.d.
A: You are right about $U(n)$.
The group $K=U(n)$ is the maximal compact subgroup of the connected complex reductive group $G=GL_n({\mathbb C})$. 
Any $x\in K$ is semisimple, hence its centralizer $G_x$ in $G$ is reductive.
Let $H\subset K$ be a subgroup, then its centralizer $G_H$ is the intersection of reductive groups, hence reductive.
Further, the Cartan involution of $G$ preserves $G_H$, hence induces a Cartan-involution on this group, which means that $K_H$ is a maximal compact subgroup of $G_H$.
Therefore, $K_H$ is connected iff $G_H$ is.
The centralizer $M_H$ of $H$ in the set of $n\times n$-matrices is connected, as it is a vector subspace.
But $G_H$ is the non-zero set of a single polynomial (the determinant), so it is connected.
A: I'm not sure whether the questions here concern just compact connected Lie groups, but in general the connectedness of centralizers involves some subtleties and may not be dealt with definitively in books on Lie groups.  But since a compact connected Lie group is semisimple (or trivial) modulo a central torus, a lot of structure theory is available. 
One useful indirect method is to follow the lead of Borel and others in passing back and forth between compact semisimple Lie groups in the usual topology and semisimple linear algebraic groups in the coarser Zariski topology.   In both cases the internal structure of the group behaves similarly, while Zariski-closed or Zariski-connected implies the same in the usual topology.
The ideas are brought out especially in the work of Springer and Steinberg on the delicate question of when centralizers of individual semisimple elements (perhaps of finite order) in a Zariski-connected semisimple algebraic group are Zariski-connected.   A sufficient condition is that the ambient group is simply connected (in the algebraic group sense), with an outline in Steinberg's 1963 Nagoya Math. J. paper, followed by a more detailed discussion in the extensive Springer-Steinberg notes on "Conjugacy classes" in Lect. Notes in Math. 131 (Springer, 1970): see especially E, II, sections 4,5.   They also include quite a few comments on related questions about compact Lie groups.   
In the algebraic group setting, centralizers of some closed subgroups such as tori are automatically Zariski-connected (though centralizers of finite subgroups require more care).   Being simply connected means much the same thing in the algebraic group and compact Lie group settings, so that condition is useful to consider when sorting out examples in your situation.    Note especially the older work of Borel-Serre on Lie groups cited by Springer-Steinberg.  
Concerning the broader question of fixed point sets under actions, it's worth keeping in mind that actions on compact manifolds (such as flag manifolds) often involve interesting fixed point sets which are finite but nontrivial and thus not connected.   This includes fixed points under tori.
