Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have two problems: Let it $\Omega^{*}_{G}:=(\mathbb{C}[\mathfrak{g}]\otimes\Omega^{*}(M))^{G}$ be the complex of equivariant differential forms on a differential manifold $M$ (in which acts a Lie group $G$ and we suppose that $Lie(G)=\mathfrak{g}$). We can define the equivariant differential $d_{\mathfrak{g}}: (\mathbb{C}[\mathfrak{g}]\otimes\Omega^{*}(M))^{G} \rightarrow (\mathbb{C}[\mathfrak{g}]\otimes\Omega^{*}(M))^{G}$ as $(d_{\mathfrak{g}}\alpha)(X)=d(\alpha(X))-\iota_{X}\alpha(X)$ where $\iota$ denotes the inner product and $X \in \mathfrak{g}$. So I have to proove this assertion: Suppose $\alpha \in \Omega^{*}_{G}(M)$ is $d_{\mathfrak{g}}$-cloded. Let $X \in \mathfrak{g}$ be such that the zero set $M_{0}(X)$ is finite. Then $\alpha(X)_{[n]}$ is $d$-exact on $M\setminus M_{0}(X)$. In order to proove the proposition I define the operator $d_{X}=d-\iota_{X}$. Then define $\theta \in \Omega^{1}(M)$ to be the dual of $X$ in the $G$-invariant metrics $\theta(\xi)=(X,\xi)$ for all $\xi \in \Gamma(M,TM)$. What means this notation? (how can I use the identification induced from the action of $\mathfrak{g}$ on $C^{\infty}(M)$ given by $(X.\phi)(x)=\frac{d}{dt}\phi(e^{-tX}.x)\arrowvert_{t=0}$?). Now denoting as $L(x)$ as Lie derivative we compute $(L(X)\theta)(\xi)=\frac{d}{dt}\phi(e^{tX}.\theta)(\xi)\arrowvert_{t=0}=\frac{d}{dt}\theta(e^{tX}.\xi)\arrowvert_{t=0}=\frac{d}{dt}(e^{-tX}.X,\xi)\arrowvert_{t=0}=0$. (I didn't understood the last two steps) Hence $\theta$ is $X$-invariant. Now how I compute $d_{X}\theta$? And how can I invert this form?

share|improve this question
    
Same question on MO. Please link questions when you post them on multiple sites. –  Martin Feb 2 '13 at 8:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.