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I am looking for a clear reason for following fact:Is there any reference ?

Why a $G$-invariant differential form $\omega$ on a homogeneous $G$-manifold $M=G/H$ is uniquely determined by its value at the initial point $m_0$ and this value can be any $H$-invariant antisymmetric poly-linear form on the tangent space $T_{m_0}M$?

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This question is too elementary for this site. The idea is that we have to define $\omega$ at each point $m=gm_0$ to be $\left(g^{-1}\right)^* \omega_0$. The $H$-invariance ensures that this result does not depend on the choice of a particular representative $g \in G$ that takes $m_0$ to $m$. The fact that $G$ acts transitively ensures that there is a choice of such a $g$. To prove smoothness of $\omega$, you need to prove that $G \to G/H$ is a fiber bundle. (See p. 33, theorem 4.3, Brocker and Tom Dieck, Representations of Compact Lie Groups). All details are proven in many books on Lie groups and Lie algebras, and in particular you might like the AMS memoir: Palais, Richard S. (1957), "A global formulation of the Lie theory of transformation groups", Memoirs of the American Mathematical Society 22: iii+123, ISBN 978-0-8218-1222-8, ISSN 0065-9266, MR0121424

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  • $\begingroup$ Although , it was elementary , but your answer was nice, But here G is just Lie group, so can we say G act transitively $\endgroup$
    – user21574
    Commented Apr 12, 2013 at 18:25
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    $\begingroup$ $G$ acts transitively on $M=G/H$ because every element of $G/H$ is a coset, i.e. has the form $m=gH$ for some $g \in G$. But then $m=gm_0$ where $m_0=H$. In case you are unfamiliar with cosets, an $H$-coset in $G$ s a set of points of $G$ of the form $gH = \{gh|h \in H\}$. The definition of $G/H$ is that it is the set of all $H$-cosets in $G$. $\endgroup$
    – Ben McKay
    Commented Apr 12, 2013 at 21:11
  • $\begingroup$ @Ben , Nice comment $\endgroup$
    – user21574
    Commented Apr 12, 2013 at 21:33

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