Killing fields on homogeneous spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:41:21Z http://mathoverflow.net/feeds/question/10609 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10609/killing-fields-on-homogeneous-spaces Killing fields on homogeneous spaces Anirbit 2010-01-03T19:01:59Z 2010-01-04T17:12:32Z <p>Let $G$ be a compact lie group and $H$ a closed subgroup and hence think of $G/H$ as a homogeneous space. </p> <p>Then how are the Killing fields on $G/H$ the projection of the right-invariant vector fields on $G$? </p> <p>In the same vein I would like to know why the following construction works:</p> <p>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 $G/H$.</p> <p>This gives a computation approach to writing down the vielbeins on $G/H$.</p> <p>I am thinking of $G/H$ to have the metric induced on it by the bi-invariant metric on $G$. </p> http://mathoverflow.net/questions/10609/killing-fields-on-homogeneous-spaces/10613#10613 Answer by Anton Petrunin for Killing fields on homogeneous spaces Anton Petrunin 2010-01-03T19:14:31Z 2010-01-03T19:14:31Z <p>(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 $H={e}$.)</p> <p>(2) Yes, it is true --- it is a projection of $G$-action on $G/H$...</p> http://mathoverflow.net/questions/10609/killing-fields-on-homogeneous-spaces/10662#10662 Answer by David Bar Moshe for Killing fields on homogeneous spaces David Bar Moshe 2010-01-04T04:31:57Z 2010-01-04T06:57:09Z <p>First I would like to recommend the appendix of article by : <a href="http://www.sciencedirect.com/science?%5Fob=ArticleURL&amp;%5Fudi=B6TVP-46SXPTS-9B&amp;%5Fuser=10&amp;%5Frdoc=1&amp;%5Ffmt=&amp;%5Forig=search&amp;%5Fsort=d&amp;%5Fdocanchor=&amp;view=c&amp;%5Facct=C000050221&amp;%5Fversion=1&amp;%5FurlVersion=0&amp;%5Fuserid=10&amp;md5=c197169cf67b800d6562b1e5ba57479c" rel="nofollow">Roberto Camporesi</a> for a clear exposition of this material.</p> <p>The explanation of both questions comes from the following two facts:</p> <p>The metric on G/H is induced from the Cartan-Killing form on the tangent space of G is G invariant.</p> <p>The generators of the Lie algebra of H are orthogonal to coset generators of G/H (at the origin) under the Cartan-Killing metric. </p> <p>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</p> <p>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.</p> http://mathoverflow.net/questions/10609/killing-fields-on-homogeneous-spaces/10713#10713 Answer by Gil Bor for Killing fields on homogeneous spaces Gil Bor 2010-01-04T16:13:04Z 2010-01-04T16:34:42Z <p>I think that if you generalize that statement a little it becomes clearer (also the proof). </p> <p>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). </p> <p>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. </p> <p>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).</p> <p>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. </p> <p>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). </p> <p>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. </p> <p>Next pick any vector $v\in Lie(G)$ and extend it to a <strong>right</strong> invariant vector field $X$ on $G$. </p> <p>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 <strong>left</strong> translations on $G$. </p> <p>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.</p> <p>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 $[H]\in G/H$. </p> <p>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). </p> http://mathoverflow.net/questions/10609/killing-fields-on-homogeneous-spaces/10720#10720 Answer by Deane Yang for Killing fields on homogeneous spaces Deane Yang 2010-01-04T17:12:32Z 2010-01-04T17:12:32Z <p>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:</p> <p>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</p> <p>There are corresponding examples for complex projective and hyperbolic spaces.</p> <p>Also, I liked the way this stuff was presented in the book by Cheeger and Ebin.</p>