Killing fields on homogeneous spaces - MathOverflow

Killing fields on homogeneous spaces

2010-01-03T19:01:59Z

Let $G$ be a compact lie group and $H$ a closed subgroup and hence think of $G/H$ as a homogeneous space.

Then how are the Killing fields on $G/H$ the projection of the right-invariant vector fields on $G$?

In the same vein I would like to know why the following construction works:

If one looks at the tangent vectors at identity on $G$ which are "transverse" to $H$ and then exponentiate it down and flow along it and project it down to $G/H$ then on $G/H$ you would be flowing along the integral curves of the vielbeins on $G/H$.

This gives a computation approach to writing down the vielbeins on $G/H$.

I am thinking of $G/H$ to have the metric induced on it by the bi-invariant metric on $G$.

Answer by Anton Petrunin for Killing fields on homogeneous spaces

2010-01-03T19:14:31Z

(1) Projections of right-invariant vector fields are Killing fields. The converse hold say if $G$ is the group of isometries on the space (in general it is not true, take $G=S^3$ and $H={e}$.)

(2) Yes, it is true --- it is a projection of $G$-action on $G/H$...

Answer by David Bar Moshe for Killing fields on homogeneous spaces

2010-01-04T04:31:57Z

First I would like to recommend the appendix of article by : Roberto Camporesi for a clear exposition of this material.

The explanation of both questions comes from the following two facts:

The metric on G/H is induced from the Cartan-Killing form on the tangent space of G is G invariant.

The generators of the Lie algebra of H are orthogonal to coset generators of G/H (at the origin) under the Cartan-Killing metric.

H is compact therefore its action on the tangent space of G/H (at the origin) will be unitary. In fact this action will be by means of some orthogonal rotation

As a consequence, the action of G on G/H will translate the tangent space to the tangent space of the translated point accompanied by some H-rotation which doesn't change inner products.

Answer by Gil Bor for Killing fields on homogeneous spaces

2010-01-04T16:13:04Z

I think that if you generalize that statement a little it becomes clearer (also the proof).

Let $G$ be any Lie group (not necessarily compact) with a closed subgroup $H$ and a metric (not necessarily positive definite) on $G$ which is $G$-left-invariant and $H$-right-invariant (not necessarily bi-invariant).

These conditions are equivalent to picking a metric (quadratic form) at $Lie(G)$ (the lie algebra of $G$, thought of as the tangent space at the identity) which is invariant under the Adjoint representation of $G$ restricted to $H$. You extend this metric from the identity to all of $G$ by left translations.

Example: $G=SL(2,R)$, $H=SO(2)$, with the Killing metric on $G$ (bi-invariant but not positive definite). In this case $G/H$ is the hyperbolic plane. Also any semi-simple $G$ with the Cartan-Killing metric and a maximal compact $H$ (then $G/H$ is called a symmetric space).

Another example is $G=SO(3)$, $H=SO(2)$ (standard embedding) with left-invariant metric which is not necessarily right-invariant, but $H$-right-invariant. This is a model for a rigid body motion whose ellipsoid of inertia is axially symmetric.

From these conditions you get that the metric descends to $G/H$ ($G$ modulo right traslations by $H$), and that left translations by $G$, which by definition act by isometries on $G$, descend to isometries on $G/H$ (since left and right translations commute, by associativity).

If you want the metric on $G/H$ to be riemannian (ie positive definite) then you need to ask that $Lie(G)/Lie(H)$ is positive definite. This holds in the examples above.

Next pick any vector $v\in Lie(G)$ and extend it to a right invariant vector field $X$ on $G$.

Exercise: the flow of $X$ is given by the action of the 1-parameter subgroup of $G$ generated by $v$, $g_t=exp(tv)$, acting by left translations on $G$.

Since left translations are isometries of $G$ it follows that $X$ is Killing. Since $X$ is right invariant it descends to a vector field $\tilde X$ on $G/H$ and the left translations by $g_t$ descend to the flow of $\tilde X$, which is by isometries, so $\tilde X$ is Killing.

Note that $v\in Lie(G)$ doesn't have to be transverse to $Lie(H)$. Picking $v\in Lie (H)$ generates Killing fields $\tilde X$ with fixed point $[H]\in G/H$.

Another comment is that this construction doesn't generate in general all the Killing fields on $G/H$. Take for example $G$ compact with bi-invariant metric and $H$ trivial. The construction misses all the left-invariant vector fields on $G$ (generating right translations).

Answer by Deane Yang for Killing fields on homogeneous spaces

2010-01-04T17:12:32Z

I do not have a copy of Camporesi's paper, so I cannot comment on that. However, when I was learning this stuff, I found it much easier to understand if I assumed that G is a matrix group (i.e, a subgroup of GL(n)). Key examples include:

H = O(n), G = group of rigid motions (embedded in GL(n+1)), G/H = Euclidean space H = O(n), G = O(n+1), G/H = sphere H = O(n), G = O(n,1), G/H = hyperbolic space

There are corresponding examples for complex projective and hyperbolic spaces.

Also, I liked the way this stuff was presented in the book by Cheeger and Ebin.