At the end of a lesson in Differential Geometry, my teacher said:
Fatto, che non dimostriamo, non è difficile ma il tempo scarseggia, se $G$ è compatto possiamo sempre trovare una metrica Riemanniana biinvariante su $G$, d'accordo? Tuttavia non è vero che $\varphi$, $\varphi$ invariante a sinistra non implica, d'accordo, non implica invariante a destra, cioè è possibile, lo faremo la prossima volta ormai, costruire metriche invarianti a sinistra ma non a destra.
That is:
A fact which we won't prove, it's not difficult but time is scarce, if [a Lie group] $G$ is compact we can always find a biinvariant Riemannian metric on $G$, OK? However it is not true that $\varphi$, $\varphi$ left-invariant doesn't imply, OK, doesn't imply right-invariant, that is it is possible, we will do it next time by now, to build left-invariant but not right-invariant metrics.
I must admit I have no clue how to use compactness to construct such a metric. I tried Googling, but I basically only found this, and the exponential map is at the end of the course, so if he says "it is not difficult" in a lesson which is barely halfway through the course it must not be strictly necessary to use that map. I found what $\mathrm{Ad}$ is, so I know that $\mathrm{Ad}(g)$ is the differential at the identity of the conjugation map $\Psi_g(h)=ghg^{-1}$:
$$\mathrm{Ad}(g)=\mathrm{d}_e\Psi_g.$$
So how can I prove this fact? Is it possible not to mention the exponential map?
Update 2
After reading this, I have access to Haar measures. I also know how to construct a left-invariant metric on a Lie group, even a noncompact one. So how do I reformulate the metric given in the suggested duplicate in terms of the Haar measure and prove it is bi-invariant?
Update 3
Following Daniel's comments, I tried proving the metric he suggested:
$$(u,v)_g=\int\limits_G\langle\mathrm{d}_gR_hu,\mathrm{d}_gR_hv\rangle_{gh}\mathrm{d}\mu(h),$$
for any $u,v\in T_gG$, is a bi-invariant metric. I started with left-invariance, and got stuck. Here is my try. Now left-invariance means that for any $k\in G$, $L_k^\ast(\cdot,\cdot)=(\cdot,\cdot)$. In other words, for any $g,k\in G,u,v\in T_gG$, we must have:
$$(u,v)_g=(\mathrm{d}_gL_ku,\mathrm{d}_gL_ku)_{kg}.$$
By the definition of our metric, this means the following two integrals must be equal:
$$\int\limits_G\langle\mathrm{d}_gR_hu,\mathrm{d}_gR_hv\rangle_{gh}\mathrm{d}\mu(h)=\int\limits_G\langle\mathrm{d}_{kg}R_h\mathrm{d}_gL_ku,\mathrm{d}_{kg}R_h\mathrm{d}_gL_kv\rangle_{kgh}\mathrm{d}\mu(h).$$
At this point I should exploit that we built $\langle\cdot,\cdot\rangle$ as a left-invariant metric. So I tried viewing that integrand as a pullback, but I got the pullback via $L_k$ of the pullback via $R_h$ of the metric, evalued at $g$, on the original vectors $u,v$. I wish I could eliminate the pullback via $L_k$, but I'd need it to be on the inside to exploit the left-invariance. I do not know if the pullback via $R_h$ of the metric is still left-invariant. I also tried changing variables to exploit the left-invariance of the measure. So setting $\ell=kg$ I got an integrand of $\langle\mathrm{d}_\ell R_h\mathrm{d}_{k^{-1}\ell}L_ku,\mathrm{d}_\ell R_h\mathrm{d}_gL_{\ell g^{-1}}v\rangle_{\ell h}\mathrm{d}\mu(h)$. Notice how I treated the two things in the metric differently to show two possible ways of making the change, writing $g$ in terms of $k,\ell$ on the left and $k$ in terms of $g,\ell$ on the right. But I have two problems: one, the measure is the same, so I cannot exploit its invariance property, and two, the integrand looks worse than before. So how do I proceed here?
That said, right-invariance can indeed be proved as in that answer. Let me read it carefully.
Update 4
$\newcommand{\diff}{\mathrm{d}}$ I think I have proved right-invariance, which decidedly makes this question a NON-DUPLICATE. Here is what I did:
\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 if the above are correct, the point of the other question is solved, and if they aren't, then the reason why they aren't should be explained here, making this decidedly NOT a duplicate of that one, as it never really was IMHO.
(For those wishing to delve deeper in the history of this post and its later exact duplicate, here are the first update, the link in that update and the note on the duplicate issue, which were removed as non-mathematical clutter)
