Compact Lie group bi-invariant metric

Solution 1:

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.