I am currently reading a paper about Einstein manifolds.

There is a comment where I don't know exactly the meaning of the words, namely a certain metric has a group of isometries of dimension $4$ which admits a principal orbit of dimension $3$.

What does the principal orbit describe?

Many thanks for your help!


1 Answer 1


These concepts are explained in every introduction to smooth Lie group actions on manifolds, for example chapter $IV$ in Bredons book "Introduction to compact transformation groups" (even though it is a standard reference I personally find this book somewhat hard to read).

To give a short overview: The isometry group $G$ of a complete Riemannian manifold $(M,g)$, or any closed subgroup of it, is a Liegroup acting on $(M,g)$ by isometries. The orbits $G \ast p$ and $G \ast q$ of points $p$ and $q$ are said to be of the same type if the istropy groups are conjugate, that is there exists $g \in G$ with $G_p = gG_qg^{-1}$. $G \ast p$ is of lower type than $G \ast q$ if a conjugate of $G_q$ is a subgroup of $G_p$, intuitively $G_q$ is smaller than $G_p$. It is a well known theorem, that in case $(M,g)$ is compact and connected there exists a maximal orbit type, that is there exists some $q \in M$ such that any other orbit has the same or bigger type than $G \ast q$. An orbit of maximal type is then called a principal orbit. There are many important structure theorems on this, e.g. the set of principal orbits is open, dense and convex in $M$.

Moreover any orbit $G \ast p$ is an embedded submanifold and hence has a dimension. In fact $G \ast p \cong G/G_p$. This also shows, that orbits of bigger type are essentially bigger and principal orbits are the biggest orbits.

  • 4
    $\begingroup$ It's worth noting that the terminology here ("maximal") is referring to the size of the orbit, rather than the size of the stabilizer. This makes the second paragraph above sound a little awkward (everything has bigger type than the maximal orbit?). Bredon does indeed define the maximal orbit to be the one with the smallest stabilizer (hence the largest orbit). $\endgroup$
    – Dan Ramras
    Commented Sep 21, 2014 at 15:39

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