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$\newcommand{\diff}{\mathrm{d}}$ TL;DR

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

  3. 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 $G$;

  4. 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:

    1. Definition of the metric.
    2. $\diff_g(R_h\circ R_k)=\diff_{R_kg}R_h\circ\diff_gR_k=\diff_{gk}R_h\circ\diff_gR_k$ by the chain rule.
    3. $R_h\circ R_k=R_{kh}$.
    4. Set $\ell=kg$ and change variables. $h=k^{-1}\ell$.
    5. 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?
    6. 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)

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  • $\begingroup$ You need to assume $G$ is compact, or else the integral in step 3 will not be finite. $\endgroup$
    – Jack Lee
    Commented Feb 16, 2016 at 21:56
  • $\begingroup$ Whoops I forgot about that. $\endgroup$
    – MickG
    Commented Feb 16, 2016 at 21:57
  • $\begingroup$ @JackLee And compactness ensures finiteness because ...? The integrand is bounded and that, on a compact set, implies integrability, perhaps? How do I prove this boundedness? $\endgroup$
    – MickG
    Commented Feb 16, 2016 at 21:59
  • $\begingroup$ The integrand is continuous. Every continuous function on a compact space is bounded and integrable. $\endgroup$
    – Jack Lee
    Commented Feb 16, 2016 at 22:00
  • $\begingroup$ Your proof in 4. is correct. For left invariance, start with $(d_g L_k u, d_g L_k v)_{kg}$, use the fact that $L_k \circ R_h = R_h \circ L_k$ and the left-invariance of $\langle\,\cdot\,,\,\cdot\,\rangle$. $\endgroup$ Commented Feb 16, 2016 at 22:27

1 Answer 1

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$\newcommand{\diff}{\mathrm{d}}$ 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.

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  • $\begingroup$ please revise your latex it doesn't seem to work (\diff doesn't show up correctly) $\endgroup$
    – user153330
    Commented May 8, 2016 at 14:38
  • $\begingroup$ @user153330 Huh? I see a bunch of perfectly fine $\mathrm{d}$s where I guess you see \diff's… $\endgroup$
    – MickG
    Commented May 8, 2016 at 14:41
  • $\begingroup$ that's weird, can u ask someone else in the chat to check in for verification purposes? $\endgroup$
    – user153330
    Commented May 8, 2016 at 14:44
  • $\begingroup$ @user153330 Yes I can. Might I suggest you take the time to type meaningful edit summaries? I just reviewed two edits of yours, one had a meaningless edit summary made of LaTeX commands and the other one had a paragraph of the edited post (deprived of the links) as the edit summary. That is not really that good :). $\endgroup$
    – MickG
    Commented May 8, 2016 at 14:46
  • $\begingroup$ @user153330 here is the chat where I asked. $\endgroup$
    – MickG
    Commented May 8, 2016 at 14:52

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