Proving left-invariance (and proof-verification for right-invariance) for metric constructed from left-invariant Haar measure $\newcommand{\diff}{\mathrm{d}}$
TL;DR


*

*Having read this I know something about Haar measures, in particular that a left-invariant one exists and is unique on any Lie group $G$.

*I know that defining:
$$\langle x,y\rangle_g=\langle(\mathrm{d}_gL_g)^{-1}(x),(\mathrm{d}_gL_g)^{-1}(y)\rangle,$$
where $\langle\cdot,\cdot\rangle$ is any inner product on the Lie algebra $T_eG$ and $L_g$ is the map $L_g(h)=gh$, one obtains a left-invariant metric for $G$.

*I know that if $G$ is compact we can define the metric:
$$(u,v)_g=\int\limits_G\langle\diff_gR_h(u),\diff_gR_h(v)\rangle_{gh}\diff\mu(h),$$
where $\mu$ is the unique left-invariant Haar measure on $G$;

*I believe the following proves right-invariance for that metric:

\begin{align*}
(\diff_gR_ku,\diff_gR_kv)_{gk}={}&\int\limits_G\langle\diff_{gk}R_h\diff_gR_ku,\diff_{gk}R_h\diff_gR_kv\rangle_{gkh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_g(R_h\circ R_k)u,\diff_g(R_h\circ R_k)v\rangle_{gkh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_gR_{kh}u,\diff_gR_{kh}v\rangle_{gkh}\diff\mu(h)=\int\limits_G\langle\diff_gR_\ell u,\diff_gR_\ell v\rangle_{g\ell}\diff\mu(k^{-1}\ell)={} \\
{}={}&\int\limits_G\langle\diff_gR_\ell u,\diff_gR_\ell v\rangle_{g\ell}\diff\mu(\ell)=(u,v)_g.
\end{align*}
Steps thus justified:
  
  
*
  
*Definition of the metric.
  
*$\diff_g(R_h\circ R_k)=\diff_{R_kg}R_h\circ\diff_gR_k=\diff_{gk}R_h\circ\diff_gR_k$ by the chain rule.
  
*$R_h\circ R_k=R_{kh}$.
  
*Set $\ell=kg$ and change variables. $h=k^{-1}\ell$.
  
*This is where I use the invariance. I used it in the form $\diff\mu(k^{-1}\ell)=\diff\mu(L_{k^{-1}}\ell)=\diff\mu(\ell)$. Is that right?
  
*Definition of the metric again.
  

So my question is: is the proof in 4. correct? And how do I prove left-invariance?
(For those wishing to delve deeper into the history of this post and its earlier exact duplicate, see edit history)
 A: $\newcommand{\diff}{\mathrm{d}}$
Converting @Daniel's comment to an answer to get this question answered.
The proof of 4 is, he says, correct.
As for left-invariance, I merely have to exploit the fact that $L_g\circ R_k=R_k\circ L_g$, and do the following:
\begin{align*}
(\diff_gL_ku,\diff_gL_kv)_{kg}={}&\int\limits_G\langle\diff_{gk}R_h\diff_gL_ku,\diff_{gk}R_h\diff_gL_kv\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_g(R_h\circ L_k)u,\diff_g(R_h\circ L_k)v\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_g(L_k\circ R_h)u,\diff_g(L_k\circ R_h)v\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_{gh}L_k\diff_gR_hu,\diff_{gh}L_k\diff_gR_hv\rangle_{kgh}\diff\mu(h)={} \\
{}={}&\int\limits_G\langle\diff_gR_hu,\diff_gR_hv\rangle_{gh}\diff\mu(h)=(u,v)_g.
\end{align*}
The elimination of $\diff_{gh}L_k$ was done thanks to left-invariance of $\langle\cdot,\cdot,\rangle$. That this be "positive definite" is obvious, since the integrand will be strictly positive on the whole of $G$. As proved here, or with similar arguments, if I plug in smooth fields I get smooth functions in the integrand, and the integral of a smooth function is smooth. So this is indeed a biinvariant metric. Of course, this requires the integral to converge, which is guaranteed by the compactness, but it is not strictly required that the group be compact.
