Let $G$ be a Lie group and $H \subset G$ a (Lie) subgroup. Denote by $\mathfrak{g}:= T_eG$ the corresponding Lie algebra of $G$ and by $T_eH = \mathfrak{h} \subset \mathfrak{g}$ the Lie subalgebra induced by $H$.
A metric $<-,->$ on $G$ is called left invariant iff for all $a,b \in G, X,Y \in T_bG$
$$<X,Y>_b= <T_b L_a(X),T_b L_a(Y)> $$
where $L_a:G \to G, b \mapsto ab$ and $T_bL_a$ it's differential at $b$.
Analogously one defines right invariant metrics.
Remark 1: Obviously a metric on $G$ induces a scalar product on $\mathfrak{g}$ by restricting to $<-,->_e$ to neutral point.
Futhermore we call the pair $(G,H)$ (more suggestively the quotient $G/H$) reductive iff there exist a splitting $\mathfrak{g}=\mathfrak{h} \oplus \mathfrak{m}$ with $Ad_H\mathfrak{m} \subset \mathfrak{m}$.
Remark 2: $Ad_h := T(L_h \circ R_{h^{-1}})= TL_h \circ TR_{h^{-1}}$.
I have a question about following implication (sorry, I still looking for the original source; I suppose that it's from from S. Kobayashi's "Fondations on Differential Geometry 1 or 2"):
The STATEMENT is: Assume $G$ has a metric $g=<-,->$ which is G-left invariant and H-bi invariant (so $L_a ^* g = g, R_b ^* g=g$ for all $a \in G, b \in H$)
Then the theorem says that then $(G,H)$ is reductive (so $\mathfrak{g}=\mathfrak{h} \oplus \mathfrak{m}$) and the corresponding $\mathfrak{m}$ obtains (suppose by restriction) an $Ad_H$ invariant scalar product.
In the proof one consider $\mathfrak{g}$ with induced scalar product $g_e=<-,->_e$ and sets $\mathfrak{m}:=\mathfrak{h} ^{+}$ the orthogonal complement.
QUESTION: Why do we need the assumption that the metric $g=<-,->$ is $G$-left invariant? Doesn't it suffice that it is H-bi invariant?
In other words where we need the $G$-left invariance?
Indeed, if we set $\mathfrak{m}:=\mathfrak{h} ^{+}$ as the orthogonal complement
then $<-,->$ is for every $h \in H$ $Ad_h$ invariant since it is $H$-biinvariant and $Ad_h= TL_h \circ TR_{h^{-1}}$. Obviously $Ad_h \mathfrak{h} \subset \mathfrak{h}$ since $H$ subgroup and therefore if we take $X \in \mathfrak{m}, Y \in \mathfrak{h}$ then
$$<Ad_h X,Y>= <Ad_h X,Ad_h \circ Ad_{h^{-1}}Y>= <X,Ad_{h^{-1}}Y> =0 $$
since by definition $\mathfrak{m}$ is orthogonal to $\mathfrak{h}$.
Futhermore $\mathfrak{m}$ is obtains by restriction $<-,->_e \vert_{\mathfrak{m}}$ an $Ad_h$ invariant scalar product because $<-,->_e$ (at $\mathfrak{g}$) is $Ad_h$-invaraint.
Best Answer
You need $g$ to be left invariant to define the scalar product induced on the Lie algebra of $G$ as you pointed yourself.