lie groups - Invariant Vector Fields for Homogenous Spaces

The short answer is that many (most? all?) homogeneous spaces do NOT have such a nice description. In particular, at any point $pin M$, the set of the $G$ or $H$ invariant vectors is a strict subset of $T_pM$. (Here, and below, I'm assuming $G$ and $H$ are compact - it's all I'm really familiar with)

Here's the quick counterexample when looking for $G$ invariance: Take any $Hsubseteq G$ with rank$(G)$ = rank$(H)$. Then the Euler characteristic of $G/H$ is greater than $0$. It follows that EVERY vector field has a 0, and hence, the only $G$-invariant vector field is trivial (since $G$ acts transitively).

For (much more) detail, including how it often fails even with rank$(H)<$rank$(G)$...

Suppose $M = G/H$ (and hence, $M$ is also compact), that is, I'm thinking of $M$ as the RIGHT cosets of $G$. For notation, set $ein G$ as the identity and let $mathfrak{g}$ and $mathfrak{h}$ denote the Lie algebras to $G$ and $H$.

Since $G$ is compact, it has a biinvariant metric. The biinvariant metric on $G$ is equivalent to an $Ad(G)$ invariant inner product on $mathfrak{g}$. This gives an orthogonal splitting (as vector spaces) of $mathfrak{g}$ as $mathfrak{h}oplus mathfrak{m}$. (The vector subspace $mathfrak{m}$ is not typically an algebra)

Now, the tangent space $T_{eH}M$ is naturally be identified with $mathfrak{m}$. In fact, this identification is $H$ invariant (where $H$ acts on $mathfrak{m}$ via $Ad(H)$ and $H$ acts on $M$ by left multiplication: $h * (gH) = (hg)H$).

Now, when you say "invariant vector fields", you must first specify $G$ invariant or $H$ invariant. Lets focus of $G$ first, then $H$.

Typically, there are not enough $G$-invariant vector fields to span the tangent space at any point. This is because not only does $G$ act transitively on $M$, but it acts very multiply transitive (an $H$s worth of elements move any element to any other given element). Thus, a $G$-invariant vector field on $M$ is actually equivalent to an $H$ invariant vector in $T_{eH}M$.

So, for example, viewing $S^{n} = SO(n+1)/SO(n)$, You'd find that $SO(n)$ acts transitively on the unit vectors in $mathfrak{m}$, and hence, there are no $SO(n+1)$ invariant vector fields on $S^{n}$.

Now, on to $H$ invariant vector fields. I haven't thought much about this, but even in this case, there are (often? always?) not enough. For example, viewing $S^2 = SO(3)/SO(2)$, we see that $H$ is a circle. In this case, viewing the north pole of $S^2$ as $eSO(2)$, the only $H$ invariant vector fields are the velocity vectors of rotation through the north-south axis - i.e., the flows are lines of lattitude.

Then we see that at generic points, the set of all vectors tangent to an $H$ invariant vector is a 1 dimensional vector space (and hence, doesn't span the whole tangent space), and the situation at the north and south pole is even worse: the only $H$ invariant vectors are trivial.

Even if you come up with a case where at some point, there were enough $H$ invariant vector fields, you can not translate them around all of $M$, since $H$ does NOT act transitively on $M$.

• Next differential topology - When do two holonomy maps determine flat bundles that are isomorphic as just bundles (w/o regard to the flat connections)?
• Previous Counting knots with fixed number of crossings