Let $G$ is a Lie group and $\mathfrak{g}$ is its Lie algebra and $\mathfrak{g}^{\mathbb{C}}$ be its complexification. Also assume that $\mathfrak{h}\subset \mathfrak{g}^{\mathbb{C}}$ be its polarization and $(X_i)_{i=1}^k$ be a basis of $\mathfrak{g}\cap \mathfrak{h}$ and $\alpha=i(X_1^l,X_2^l,...,,X_k^l)d\mu_G$ be a differential form on $G$ where $d\mu_G$ is the Haar measure and $(X_1^l,X_2^l,...,,X_k^l)$ is the left G-invariant fields. Let $D$ be the lie group of $\mathfrak{g}\cap \mathfrak{h}$ and $d\in D$ and $R_d$ be the right action of $d$ on $G$. Then why we have $$R_d^*\alpha=\left(\det Ad_{{\mathfrak{g}}/{\mathfrak{(g\cap h)}}}(d)\right)^{-1}.\alpha$$?

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    $\begingroup$ Simplify first the set up (and notation): $G$ a Lie group, $H$ a subgroup, $X_1,\ldots,X_k$ a basis of the Lie alg of $\mathfrak{h}$ as well as their extension to left-inv vector fields on $G$, $\mu$ a left-inv volume form on $G$ and $\alpha$ the contraction of $\mu$ by the $X_i$'s. Then $R_h^*\alpha=\left(\det Ad_{{\mathfrak{g}}/{\mathfrak{h}}}(h)\right)^{-1}\alpha$ for all $h\in H$. Tell us if you still cannot do it. $\endgroup$ – Gil Bor Feb 19 '14 at 18:00
  • $\begingroup$ Yes, I still can not do it $\endgroup$ – user61135 Feb 19 '14 at 20:02

See my comment on your question for a simplified reformulation of the problem and the notation I am using.

The main idea: consider the map $X\mapsto X(e)$ that assigns to a vector field on $G$ its value at the identity (en element of the Lie algebra of $G$). Restricted to l.i. (left-invariant) vector fields, this map is a $G$-equivariant isomorphism: $[(R_g)_*X](e)=Ad(g)[X(e)]$.

It follows that the same is true for forms: if $\alpha$ is $k$-form on $G$, than $\alpha\mapsto \alpha(e)$ is $G$-equivariant isomorphism between l.i. $k$-forms on $G$ and $\Lambda^k(\mathfrak g^*)$.

Now take a l.i. volume form $\mu$ on $G$. Then $R_g^*\mu$ is also l.i, hence a multiple of $\mu$ (because $\dim \Lambda^n(\mathfrak g^*)=1$), say $R_g^*\mu=c\mu.$ Evaluate both sides at $e$, use the $G$-equivariance, and get $c=\det Ad(g)$.

Next let $\alpha=i(X_1,\ldots,X_k)\mu,$ where $X_1,\ldots,X_k$ is a basis for $\mathfrak h$, extended to l.i. vector fields on $G$. Then $\alpha$ is l.i., hence $[R_g^*\alpha](e)=Ad(g)^*[\alpha(e)].$

The last bit then reduces to the following linear algebra exercise: let $V$ an $n$-dim vector space, $W\subset V$ a $k$-dim subspace with a basis $v_1,\ldots,v_k$, $\mu\in\ \Lambda^n(V^*)$ and $\alpha=i(v_1,\ldots, v_n)\mu.$ Then if $T:V\to V$ leaves $W$ invariant, with induced maps $T':W\to W$ and $T'':V/W\to V/W$, then $T^*\alpha=\det(T'')\alpha.$

(Hint: $\det(T)=\det(T')\det(T'')$).

Hope you can fill in the details.


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