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In do-carmo's Book "Riemannian Geometry" there is an exercise on proving existence of a bi-invariant metric on any compact connected Lie group. (pg 46, question 7).

In the first stage, you are required to prove that every differential form (of top degree) which is left invariant, is also right invariant.

However, I do not see where we are using this property later. The basic method of the proof is this: take any left invariant metric on $G$ , and use integration of right translates of it to produce a bi-invariant metric.

The proof of the bi-invariance of the constructed metric does not use the fact $\omega$ is bi-invariant, but only its left-invariance. (More specifically, the left invariance of the metric is implied by the left invariance of the original metric taken, and the right invariance follows by using left invariance of $\omega$ together with the fact that the integral of a form does not change under pullback via orientation preserving diffeomorphism).

So, I am a bit puzzled by the first stage. Is it really needed for the proof?

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    $\begingroup$ I think sometimes people start with an arbitrary Riemannian metric (rather than one which is left-invariant) and form an integral over $G\times G$. This requires the volume form to be bi-invariant. But I agree with you in this case. $\endgroup$ – Tim kinsella Jun 10 '15 at 14:21
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For existence of biinvariant metric, biinvariant volume form is needed :

Assume that $g$ is a left invariant Riemannian metric on $G$.

If $F:\mathbb{R}^n\rightarrow G $ is local chart on $U$ around $e\in G$, then $\{E_i=dF\ e_i\}$ are coordinate vector field. Define $$ g_{ij}:= g(E_i,E_j) $$

Hence corresponded volume form is $$ V:=\sqrt{{\rm det}\ g_{ij}} \ E_1^\ast \cdots E_n^\ast $$

(1) Left invariance of $V$ : Note that $L_x\circ F $ is chart on $xU$ around $x\in G$. Hence $$ g_{ij}(xp):=g(dL_x\ E_i(p), dL_x\ E_j(p) )=g_{ij}(p)$$ and $$ E_i(xp)=dL_x\ E_i(p) $$ so that $$ V(xp)=\sqrt{{\rm det}\ g_{ij} (p)}\ (dL_x\ E_1(p))^\ast \cdots (dL_x \ E_n(p))^\ast$$

So $$ L_x^\ast \ V(xp)= V(p)$$

(2) In further assume that $g$ is right invariant.

Right invariance of $V$ :

$$ V(x)=\sqrt{{\rm det}\ g_{ij}(e) } \ (dL_x\ E_1(e))^\ast \cdots (dL_x\ E_n(e))^\ast $$

$$ R_p^\ast\ V (xp)= \sqrt{{\rm det}\ g_{ij} ( p )} \ R_p^\ast\ (dL_x\ E_1(p))^\ast \cdots R_p^\ast \ (dL_x\ E_n(p))^\ast $$

Here $$V(x) (dL_x\ E_1(e),\cdots, dL_x\ E_n(e))=\sqrt{{\rm det}\ g_{ij} (e)}$$ And \begin{align*}& \sqrt{{\rm det}\ g_{ij} ( p )} \ R_p^\ast\ (dL_x\ E_1(p))^\ast \cdots R_p^\ast \ (dL_x\ E_n(p))^\ast\\& \bigg( dL_x\ E_1(e),\cdots, dL_x\ E_n(e)dL_x\ E_1(e),\cdots, dL_x\ E_n(e) \bigg)\\&= \sqrt{{\rm det}\ g_{ij} ( p )} \ E_1(p)^\ast \cdots E_n(p)^\ast \ ( dR_p\ E_1(e),\cdots, dR_p\ E_n(e) ) \end{align*}

In further if $E_i(e)$ is orthonomal wrt $g$, then so is $\{dR_p\ E_i(e)\}$.

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I believe that to construct a bi-invariant metric is equivalent to constructing an inner product on the Lie algebra (i.e. $T_eG$) that is invariant under the adjoint representation. To construct such an inner product, the standard trick is to start with any inner product $\langle \cdot , \cdot \rangle$ on $T_eG$ and average over the group: given $X,Y \in T_eG$, define $f(p): = \langle Ad_p X , Ad_pY\rangle$ for $p\in G$, and define $$[X,Y]:=\int_G f(p)dV $$ (sorry that I used the bracket: it is not the Lie bracket but the notation for our new inner product)

This is clearly an inner product. To check it is $Ad$-invariant, we need to show that for any (fixed) $q\in G$: $$[X,Y]=[Ad_q X, Ad_qY] = \int_G \langle Ad_{pq}X , Ad_{pq}Y\rangle dV = \int_G f(pq)dV$$ Now you may wish to use the fact that $G$ is a group and make a change of variable. This is exactly where you need the condition $dV$ is right invariant: to apply a change of variable you are basically doing $$\int_G f(pq)dV=\int_G R_{q^{-1}}^*(f(pq)dV) = \int_G f(p)(R_{q^{-1}}^*\ dV) $$ Apply right invariance of $dV$, and the above integral is just $[X,Y]$. This is what we desire.

For more details, see the textbook of Lee: Riemannian Manifolds - An Introduction to Curvature, Problem 3.11 and 3.12.

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