# Compact Lie group bi-invariant metric

Let $G$ be a compact Lie group and $\left\langle ,\right\rangle$ be a left invariant metric on $G$; $\omega$ be a positive differential $n$-form on $G$ which is left invariant. Consider the metric $(,)$ on $G$ given by: $$\left(u,v\right)=\int_{G}\left\langle (dR_{x})_{y}u,(dR_{x})_{y}v\right\rangle _{yx}\omega$$ It's not too hard to show that this is left-invariant but I'm wondering how to show that $\left(,\right)$ is right-invariant?

Given $u, v \in T_yG$ arbitrarily, $$(d(R_g)_yu, d(R_g)_yv)_{yg} = \int_G \left\langle d(R_x)_{yg}(d(R_g)_yu),~ d(R_x)_{yg}(d(R_g)_yv) \right\rangle_{ygx}~\omega,$$ by definition. The Chain Rule implies that $d(R_x)_{yg}d(R_g)_y = d(R_x \circ R_g)_y = d(R_{gx})_y$, so the above expression becomes $$\int_G \left\langle d(R_{gx})_yu,~ d(R_{gx})_yv \right\rangle_{y(gx)}~\omega.$$ Now, if you define $f: G \rightarrow \mathbb{R}$ by $f(x) = \langle d(R_x)_yu,~d(R_x)_yv \rangle_{yx}$, what we have is $$\int_G f(gx) ~\omega = \int_G L_g^*(f\omega) = \int_G fw,$$ where the first equality holds because $\omega$ is left-invariant and the last equality is the Change of Variables Theorem.
• I think it would be easier to read if you omit all the pointwise evaluations. Moreover integrating with respect to a left-invariant volume form gives a left invariant integral: Write $\int_G f(g) dg = \int_G f\omega$, then by change of variables: $\int_G f(hg) dg = \int_G f(g) dg$. So eventually for vectorfields $X,Y$ you have $\int_G \langle dR_{hg} X,dR_{hg} Y\rangle dg = \int_G \langle dR_{g} X,dR_{g} Y\rangle dg$ (which is essentially the last line). – T'x Feb 13 '17 at 12:56
There is a problem of the above answer: $$\int_G <d(R_x)_{yg}d(R_g)_y u, d(R_x)_{yg}d(R_g)_y v>_{ygx} \omega(gx)=\int_G f(gx)\omega(gx)\neq \int_G {L_{g}^{*}} f(x) L_{g}^{*}\omega (x)=$$ Because $$L_{g}^{*}f(x)=<d(L_g)_{yx}d(R_{x})_{y}u,d(L_g)_{yx}d(R_x)_{y}v>_{gyx}$$ But $$f(gx)=<d(R_{gx})_{y} u, d(R_{gx})_{y} v)>_{ygx}$$