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Given a Lie group $G$.

For a left-invariant vector field it holds: $$\mathrm{d}l_gV=V\circ l_g:\quad V_g=\mathrm{d}l_gV_e$$

Conversely rough vector fields are smooth: $$V_g:=\mathrm{d}l_gv:\quad V\in\Gamma_G(\mathrm{T}G)$$ How to prove this in a clever way?

C-star-W-star
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    It's immediate from the fact that group multiplication is a smooth map on $G\times G$. – Ted Shifrin Jan 30 '15 at 14:24
  • How do I see this clearly? – C-star-W-star Jan 30 '15 at 14:34
  • Well, we know that if $f\colon X\times Y\to Z$ is a smooth map of manifolds, then $df\colon TX\times TY\to TZ$ is a smooth map as well. – Ted Shifrin Jan 30 '15 at 14:41
  • @TedShifrin: Aaah so easy. ^^ Why does Lee prove it then explicitely? (My professor does so as well though he did prove smoothness of the differential before.) – C-star-W-star Jan 30 '15 at 14:48
  • Not sure, Freeze. Try writing it out carefully from what I suggested, but in my head I don't see a problem ... yet. :) – Ted Shifrin Jan 30 '15 at 14:49
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    @Ted: Really? A left-invariant vector field $X$ is a map from $G$ to $TG$. The differential of the multiplication map is a map $dm\colon TG\oplus TG \to TG$ (note that the domain is the Whitney sum $TG\oplus TG$, not the product manifold $TG\times TG$). How does smoothness of $X$ follow immediately from that of $dm$? – Jack Lee Jan 30 '15 at 15:21
  • Since @JackLee always catches my errors, I'm going to have to think this through carefully. Good thing I'm retiring :) – Ted Shifrin Jan 30 '15 at 15:45
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    @Ted, I just caught an error of my own: I was wrong about the Whitney sum. The differential $dm$ is a smooth map from $TG\times TG$ to $TG$, as you implied. But I still don't see how to use this to conclude immediately that $X$ is smooth. – Jack Lee Jan 30 '15 at 15:50
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    I'm about to go teach, but it seems to me that if we take $f(g,h) = L_g(h)$, then we want to consider $df((g,0),(e,v))$, with $(g,0)\in TG$ (i.e., $0\in T_gG$) and $(e,v)\in TG$ (i.e., $v\in T_eG$). – Ted Shifrin Jan 30 '15 at 15:58
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    WEll, that seems to work. I'll be darned. I wonder why I went to so much trouble in my proof? :-( – Jack Lee Jan 30 '15 at 19:56

2 Answers2

2

Problematic

Vector field and differential: $$V:G\to\mathrm{T}G:g\mapsto\mathrm{d}l_gv$$ $$\mathrm{d}l_g:\mathrm{T}G\to\mathrm{T}G:v\mapsto\mathrm{d}l_gv$$ (Note parameter and variable!)

Differential

Consider a differential: $$\mathrm{d}F:\mathrm{T}M\to\mathrm{T}N$$

Its coordinate expression: $$\widehat{\mathrm{d}F}(x,v)=(\hat{F}(x),\mathrm{D}\hat{F}(x)v)$$

Its directional derivatives: $$\partial_x\widehat{\mathrm{d}F}(x,v)=(\partial_x\hat{F}(x),\partial_x\mathrm{D}\hat{F}(x)v)$$ $$\partial_v\widehat{\mathrm{d}F}(x,v)=(0,\mathrm{D}\hat{F}(x)\partial_vv)$$

So the differential is smooth!

Vector Field

Regard the map: $$\chi_v:G\to\mathrm{T}(G\times G):g\mapsto[(g,\alpha)]:\quad\hat{\chi}_v(x)=(x,e;0,\hat{v})$$

So the rough vector field writes: $$V_g=\mathrm{d}l_g[\alpha]=[l_g\circ\alpha]=[\mu(g,\alpha)]=\mathrm{d}\mu[(g,\alpha)]=\mathrm{d}\mu(\chi_v(g))$$

Thus it was a smooth!

C-star-W-star
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For $v \in T_e G$, we can find a path $\gamma : I \to G$ such that $\gamma (0) = e$ and $\gamma '(0) = v$.

Then $V_g = d_e \; l_g \;\gamma '(0) = \frac{d}{dt}|_{t=0} \;(l_g \circ \gamma ) = \frac{d}{dt}|_{t=0} \;(g \gamma )$.

The composite of applications $G\times I \to G\times G \to G :(g,t) \mapsto (g,\gamma (t))\mapsto \mu(g,\gamma(t)) = g \gamma(t)$ is smooth,where $\mu$ is the multiplication of $G$. Briefly, $(g,t) \mapsto g\gamma (t)$ is smooth, . It follows that $(g,t) \mapsto \frac{\partial (g\gamma)}{\partial t}$ is smooth, since the partial derivative of a smooth function is still smooth. In particular, $g\mapsto \frac{d}{dt}|_{t=0} \;(g \gamma )=V_g$ is smooth, because it is the composite of $g\mapsto (g,0)\mapsto \frac{d}{dt}|_{t=0} \;(g \gamma )$.

Shunan Sun
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