Why the induced metric from the lie algebra of lie group $G$ is left invariant.

Question

we say that a Riemannian metric on $G$ is left invariant if $\langle u,v\rangle_y = \langle d(L_x)_y u,d(L_x)_y v\rangle_{L_x(y)}$ to introduce a metric on $G$, take any arbitrary inner product $\langle , \rangle_e$ on the lie algebra of lie group and define $\langle u,v\rangle_x = \langle d(L_{x^{-1}})_x(u),d(L_{x^{-1}})_x (v)\rangle_e$, where $x\in G, u,v\in T_x G$ So, why it is left invariant?

Answer 1

$\langle u,v\rangle_b = \langle d(L_a)_b u,d(L_a)_b v\rangle_{ab}$

Is this equation true?

By our definition of inner product on arbitrary point, which is defined via the inner product on the tangent plane on the uint element of the Lie group, the only equation we have to verified is

$\langle d(L_{b^{-1}})_b(u),d(L_{b^{-1}})_b (v)\rangle_e = \langle d(L_{(ab)^{-1}})_{ab} \circ d(L_a)_b u , d(L_{(ab)^{-1}})_{ab}\circ d(L_a)_b v \rangle_e$

However, it follows from these two facts:

• $L_{xy} = L_x \circ L_y$
• $d(F\circ G)_p = dF_{G(p)}\circ dG_p$, where $G: M\to N$, $F: N\to P$, and $p\in M$