Proving continuity/smoothness for a special function on a Lie group. So I asked this question, yesterday, forgetting the compactness requirement. Jack Lee commented shortly afterwards, noting that, if we take an inner product on the Lie algebra $T_eG$ of a Lie group $G$, extend it to a left-invariant metric on $G$, and define:
$$(u,v)_g=\int\limits_G\langle\mathrm{d}_gR_h(u),\mathrm{d}_gR_h(v)\rangle_{gh}d\mu(h),$$
where $\mu$ is the unique left-invariant Haar measure that exists on $G$, the integral might not converge. If $G$ is not compact. So I wondered why compactness ensured convergence, and he said the integrand is continuous, hence integrable on a compact set. Now I can see how continuous would imply bounded. But I'm not too sure I can always guarantee a compact set will have finite measure and hence allow integration of constants. In fact, reading that pdf again, the Haar measure is defined to have this property, and in the construction, the property is evident. But above all, I'm not too sure how to prove continuity. I was thinking I could maybe show the two entries of the metric are smooth vector fields, but then $h\mapsto\mathrm{d}_gR_h(u)$ with $u$ tangent at $g$ is not a vector field because it outputs something not tangent at $h$, but at $gh$. So perhaps I could use the left-invariance of $\langle\cdot,\cdot\rangle$ to get actual vector fields and work on those:
$$\langle\mathrm{d}_gR_h(u),\mathrm{d}_gR_h(v)\rangle = \langle(\mathrm{d}_hL_g)^{-1}\mathrm{d}_gR_h(u),(\mathrm{d}_hL_g)^{-1}\mathrm{d}_gR_h(v)\rangle_h.$$
Now the two entries are vector fields, so if I show they are smooth, since I know $\langle\cdot,\cdot\rangle$ is a metric, I conclude the integrand is smooth, hence continuous. But I'm not sure how to do this.
 A: Right, $\hat{U} \colon h \mapsto \mathrm{d}_g R_h(u)$ is not a vector field, it's a vector field composed with a translation. Define
$$U(k) := \mathrm{d}_g R_{g^{-1}k}(u) = \mathrm{d}_g(R_k \circ R_{g^{-1}})(u) = \mathrm{d}_e R_k (\mathrm{d}_g R_{g^{-1}}(u)).$$
Since multiplication is smooth, the map $k \mapsto \mathrm{d}_e R_k$ is smooth, and so $U$ is a smooth vector field (with $U(g) = u$). And hence $\hat{U} = U \circ L_{g}$ is smooth. The same holds for $v$ and its associated vector field, thus we see that
$$h \mapsto \langle \hat{U}(h),\hat{V}(h)\rangle_{gh}$$
is smooth.
Your idea to use the left-invariance and $(\mathrm{d}_hL_g)^{-1} = \mathrm{d}_{gh}L_{g^{-1}}$ works too, but since it has $h$ appearing in two places, it looks more complicated. At least until we rewrite it using $L_{g^{-1}}\circ R_h = R_h \circ L_{g^{-1}}$ to obtain
$$\mathrm{d}_{gh}L_{g^{-1}}(\mathrm{d}_g R_h(u)) = \mathrm{d}_g(L_{g^{-1}}\circ R_h)(u) = \mathrm{d}_g(R_h\circ L_{g^{-1}})(u) = \mathrm{d}_e R_h(\mathrm{d}_gL_{g^{-1}}(u)),$$
which looks quite similar to $U$ above.
