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Assume $G$ is a Lie group. The standard construction of a left invariant metric on $G$ goes as follows:

Take an arbitrary inner product $\langle,\rangle_e$ on $T_eG$ and define $\langle u , v\rangle_x = \langle (dL_{x^{-1}})_xu , (dL_{x^{-1}})_xv\rangle_e $ , where for every $g\in G$ $L_g$:$G\rightarrow G$ is left multiplication by $g$.

How to prove this metric is smooth? (There are several different equivalent criterions,I tried to prove for instance that for any two smooth vector fields X,Y on $G$ $\langle X,Y \rangle$ is a smooth function but failed.

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    $\begingroup$ Can you show that the map $TG\mapsto T_eG$ given by $(x,v)\mapsto (dL_{x^{-1}})v)$ is smooth? $\endgroup$ – Kyle Jun 15 '15 at 22:01
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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.)

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    $\begingroup$ In the expansion in $dF$ this should be $(dR_y)_x(u)$. Also, while I agree your solution is shorter, I do think you are not really avoiding all the part of stating the fact each fiber of a vector bundle is a submanifold, hence the restriction of smooth map whose image is contained in a fiber is smooth. (Your map $(x,v) \mapsto (dL_{x^{-1}})_x (v)$ is indeed smooth, but it is (formally) a smooth map into $TG$. When you put $\langle, \rangle _e$ you are using its smoothness as a map from the cartesian product $T_eg \times T_eg \mapsto $\mathbb{R}$, so you have to say something on that $\endgroup$ – Asaf Shachar Jun 21 '15 at 9:59
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    $\begingroup$ restricting codomain of smooth functions to an embedded submanifold is always smooth. (This is not always the case where immersed submanifolds are considered). Of course, in our situation the fact that each fiber is a submanifold is quite easy to show based on standard theory of submanifolds. (every inverse inage of a regular value, in particular, every inverse image (of a point) of a submersion is an embedded submanifold $\endgroup$ – Asaf Shachar Jun 21 '15 at 10:02
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    $\begingroup$ @AsafShachar: I fully agree with your observations and made edits where appropriate. I should clarify that my aim in providing this argument is to highlight a perspective that avoids some of the bundle-theoretic formality (such as converting from $dL$ to $\widetilde{dL}$ and worrying about the smooth-sections-to-smooth-sections business). Most of the important details are the same and need to be verified in both cases, as you did very carefully in your answer. I think it's really two sides of the same coin. $\endgroup$ – Kyle Jun 22 '15 at 6:05
  • $\begingroup$ I agree. All the passage to the pullback bundle is cumbersome and unnecessary, and your solution is indeed more elegant. Many thanks for your efforts. $\endgroup$ – Asaf Shachar Jun 22 '15 at 7:27
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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 $\mathbb{R}$.

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    $\begingroup$ The motivation for my supplemental answer is that I don't know whether it's obvious to the OP why $(u,v)\mapsto (d(L_{x^{-1}})u,d(L_{x^{-1}})v)$ is smooth (where $x=\pi(u,v)$). Each coordinate "looks" smooth, but the fact that $L_{*}$ itself is varying means that $d(L_{x^{-1}})_x$ isn't the differential of a single smooth function; it's the differential of a different function at each point $x\in G$. That's why it's useful to embed $G$ into a space where the map is the restriction of a more clearly smooth function. $\endgroup$ – Kyle Jun 16 '15 at 3:51
  • $\begingroup$ Yes, you're right, those precisions are useful. $\endgroup$ – Sylvain L. Jun 16 '15 at 8:05
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    $\begingroup$ @squirrel: You were right. It's not obvious to me why $(u,v)\mapsto (d(L_{x^{-1}})u,d(L_{x^{-1}})v)$ is smooth. In fact the smoothness of your map $\phi$ is also not clear to me. What is clear to me is that when you fix $x$ you get a smooth map from the fibered product to itself. I do not see the smoothness of the "whole" map considered as a map from the prodcut. If you could elaborate on that point I would be happy. (Anyway, I have found another way of showing smoothness, you can see my answer) $\endgroup$ – Asaf Shachar Jun 18 '15 at 11:39
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    $\begingroup$ @AsafShachar: I think my intuition for these types of things is that you can usually make things clearer by embedding them in larger spaces, but you're right that the extension of Sylvain L.'s map isn't clearly smooth. As in my original comment, the best thing is to show that $(x,v)\mapsto (dL_{x^{-1}})_x(v)$ is smooth. Your approach provides an argument for showing this, but I think it can be simplified to avoid some of the bundle-theoretic stuff. I tried to convey that in the edits to my answer. A similar argument could have been used directly for the map $G \times( TG \times_G TG)$. $\endgroup$ – Kyle Jun 18 '15 at 15:17
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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.

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