How to prove that a left-invariant metric on a Lie group is smooth?

Question

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?

Answer 1

A natural approach is to show that the map $f(x,v)= (dL_{x^{-1}})_x(v)$ is smooth: The map $F:G\times G \to G$ given by $F(x,y)=xy$ satisfies $dF((x,u),(y,v))=(dL_{x})_y(v)+(dR_y)_x(u)$. Embed $TG$ in $T(G\times G)\cong TG \times TG$ via $(x,v)\mapsto ((x^{-1},0),(x,v))$. The composition $TG \hookrightarrow T(G \times G) \to TG$ is smooth and gives the desired map $$(x,v)\mapsto ((x^{-1},0),(x,v))\mapsto (dL_{x^{-1}})_x (v).$$

Now, given vector fields $X$ and $Y$, we have a smooth map $G \to \mathbb{R}$ given by $x\mapsto \langle X(x),Y(x)\rangle_x=\langle \phi\circ X(x), \phi \circ Y(x)\rangle_e$. (Note that restricting the codomain of the composition from $TG$ down to $T_e G$ results in a smooth map because $T_e G$ is an embedded submanifold of $TG$; see the comments below.)

Answer 2

I think it is just the composition of smooth maps. Let's try to state this formaly.

Consider the "double-tangent bundle" (this is not an official notation, I'm not sure it have a standard name) $T^2G=\bigsqcup_{x\in G}T_x G \times T_x G$ and $\pi : T^2G \rightarrow G$ the canonical projection.

Then, we can define on $T^2G$ the map (with $\mathfrak{g}=T_eG$ the Lie algebra) : $$ \begin{array}{rccc} L : & T^2G & \longrightarrow & \mathfrak{g} \times \mathfrak{g}\\ & (u,v) & \longmapsto & \left( (dL_{\pi(u,v)^{-1}})_{\pi(u,v)} u, (dL_{\pi(u,v)^{-1}})_{\pi(u,v)} v\right), \end{array}$$

Because $\pi$ and $(x,y) \mapsto L_x(y)$ are smooth, so is $L$, by composition of smooth functions. The left-invariant metric is then just the composition of $L$ with the metric on $\mathfrak{g}$ and thus is also smooth, seen as a map from $T^2G$ to $\mathbb{R}$.

Answer 3

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)$.

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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.