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}={}&amp;\int\limits_G\langle\diff_{gk}R_h\diff_gR_ku,\diff_{gk}R_h\diff_gR_kv\rangle_{gkh}\diff\mu(h)={} \\
{}={}&amp;\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)={} \\
{}={}&amp;\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)={} \\
{}={}&amp;\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)
$\newcommand{\diff}{\mathrm{d}}$
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}={}&amp;\int\limits_G\langle\diff_{gk}R_h\diff_gL_ku,\diff_{gk}R_h\diff_gL_kv\rangle_{kgh}\diff\mu(h)={} \\
{}={}&amp;\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)={} \\
{}={}&amp;\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)={} \\
{}={}&amp;\int\limits_G\langle\diff_{gh}L_k\diff_gR_hu,\diff_{gh}L_k\diff_gR_hv\rangle_{kgh}\diff\mu(h)={} \\
{}={}&amp;\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.
Finiteness of Haar volume of a compact group : let $U$ be an open neighborhood of identity. The open sets $g\,U$ for $g \in G$ form an open cover of $G$. Since $G$ is compact, this open cover has a fintie subcover. However, since the Haar volume is invariant, all the sets $g\,U$ have the same volume. Thus, we have bounded the volume of $G$ by an integer multiple of the volume of $U$, (the volume of $U$ is finite, of course!) -- Salem