TITLE: How to prove that a left-invariant metric on a Lie group is smooth?
QUESTION [7 upvotes]: Let $G$ be a Lie group. Recall the standard construction of a left invariant metric on $G$:
Let $\langle,\rangle_e$ be an (arbitrary) inner product on $T_eG$ and define, for $u,v 
\in T_xG$,
$\langle u , v\rangle_x = \langle (dL_{x^{-1}})_xu , (dL_{x^{-1}})_xv\rangle_e $ , where $L_g$:$G\rightarrow G$ denotes left multiplication by $g$.
How to prove that this construction produces a smooth metric?

REPLY [3 votes]: Here is a way I have found to show smoothness.
Look at the multiplication map $L:G \times G \rightarrow G, L(g,g')=gg'$.
We have the differential map $dL:T(G\times G)\rightarrow TG$, where
$dL_{(g,g')}:T_gG\oplus T_{g'}G \rightarrow T_{gg'}G$.
In order for this map the become a map of bundles over $G \times G$ we use the pullback bundle $L^*(TG)$ which is a vector bundle over $G\times G$. We get an induced map: $\widetilde {dL}:T(G\times G)\rightarrow L^*(TG)$
, $\widetilde {dL}_{(g,g')}:T_gG\oplus T_{g'}G = T_{(g,g')}(G \times G) \rightarrow  L^*(TG)_{(g,g')}=T_{gg'}G$. 
$\widetilde {dL}_{(g,g')}(v,w) = {dL}_{(g,g')}(v,w)$ .[The essential action is the same, the only difference is above which point we consider our fiber to be].
Lemma: Let $v\in T_{g'}G$. Then $dL_{(g,g')}(0,v)=d(L_g)_{g'}(v)$.
Proof:
Take a path $\widetilde \alpha:I \rightarrow G$ such that $\tilde \alpha (0) = g', \dot {\widetilde \alpha}(0)=v$.
Define a path $\alpha :I\rightarrow G\times G, \alpha(t)= (g,{\widetilde \alpha}(t)), \alpha(0)=(g,g'), \dot \alpha(0)=(0,v)$.
Note that $L(\alpha(t))=g \cdot \widetilde \alpha(t) = L_g(\widetilde \alpha(t))$. By the chain rule: $ dL_{(g,g')}(0,v) = \frac {d}{dt} (L(\alpha(t)) = \frac {d}{dt} [L_g(\widetilde \alpha(t))] = d(L_g)_{g'}(v)$.

Now let $X\in \Gamma(TG)$. Define a vector field $\tilde X\in \Gamma(T(G \times G)),  \tilde X(g,g')=(0,X(g'))$.
Since $\widetilde {dL}$ is a smooth bundle map of vector bundles over the same fiber space, it maps smooth sections into smooth sections. By the lemma: $\widetilde {dL}(\tilde X)(g,g')= {dL}_{(g,g')}(\tilde X(g,g'))={dL}_{(g,g')}(0,X(g'))= d(L_g)_{g'}(X(g'))$.
So, we get a smooth map (section), $\widetilde {dL}(\tilde X)= \sigma:G\times G\rightarrow L^*(TG)$. Now precomposing with the smooth map $\beta: G \rightarrow G \times G, \beta(x)=(x^{-1},x)$ we eventually get a smooth map: $h = \sigma \circ \beta: G \rightarrow L^*(TG)$.
$ h(x)=\sigma (x^{-1},x) = d(L_{x^{-1}})_{x}(X(x))$. Now we note that Image($h$) $\subseteq T_eG$. Since the fibers of a vector bundle are embedded submanifolds (inverse inage of a regular value), the restriction of the codomain of $h$ to the fiber $T_eG$ is still smooth. 
Now take two vector fields $X_1,X_2 \in \Gamma(TG) $. We obtain two smooth maps $h_i: G \rightarrow T_eG, h_i(x)=d(L_{x^{-1}})_{x}(X_i(x))$, which rise to the smooth map: $H:G \rightarrow T_eG \times T_eG, H(x)=(h_1(x),h_2(x))$. 
Composing $H$ with the inner product $\langle , \rangle _e$ on $T_eG$ (which is smooth) we get a smooth function $G \rightarrow \mathbb{R} , x\rightarrow \langle (dL_{x^{-1}})_x(X_1(x)) , (dL_{x^{-1}})_x(X_2(x))\rangle_e = \langle X_1(x),X_2(x) \rangle _x$.
This shows the constructed metric is indeed smooth.