# Understanding the Lie Algebra of left-invariant vector fields

I’m very new to differential geometry and I am currently studying Lie Groups and now I have some trouble understanding the Lie Algebra of left-invariant vector fields over $$G$$, where $$G$$ is a Lie group.

Let’s start stating what I think I got right:

Given a $$\mathcal{C}^{\infty}$$ manifold $$M$$, $$\mathcal{C}^{\infty}(M)$$ is the set of all functions $$f:M\to\mathbb{R}$$ s.t. for all charts $$\varphi:U\to\mathbb{R}^n$$, the map $$f\circ\varphi^{-1}$$ is infinitely differentiable.

A smooth vector field on a manifold $$M$$ is a linear function $$X:\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$$ s.t. $$X$$ is a derivation, i.e., $$\forall f,g\in\mathcal{C}^{\infty}(M)$$ it holds that $$X(fg)=fX(g)+X(f)g$$. The set of all such functions $$X$$, is denoted by $$\mathfrak{X}(M)$$. If we equip $$\mathfrak{X}(M)$$ with the commutator $$[X,Y] = X\circ Y - Y\circ X$$ we obtain a Lie Algebra.

First of all I want to ask you if what I stated so far is correct, in order to proceed with the main question.

Given a Lie group $$G$$, we define the following Lie Subalgebra of $$\mathfrak{X}(M)$$, called the Lie algebra of left-invariant vector fields over $$G$$:

$$\operatorname{Lie}(G):=\{X\in\mathfrak{X}(G) : \forall g\in G, \,\,\,d(L_g)X=X\}$$

where $$L_g:G\to G$$ associate to each $$g'\in G$$ $$L_g(g')=gg'$$. Now I have some trouble understanding the condition $$d(L_g)X=X$$. From what I thought, given a function $$f:G\to \mathbb{R}$$ the differential of $$f$$ gives the following: $$df(X)=X(f)$$. But now we are differentiating $$L_g$$ which is not a real valued function, so I don’t understand the meaning of the condition $$d(L_g)X=X$$. Since what I wrote about the differential can’t hold, what does $$d(L_g)$$ mean?

I know this might be a really naive question, but I’m new to these topics. Any help is very much appreciated. Thank you, guys.

• Have you heard about the differential of a smooth map $f:M\to N$ between two smooth manifolds? The differential of a smooth function $M\to\mathbf R$ is a special case.
– KCd
Aug 15, 2023 at 14:26
• I think so, but it always involve some point. Given two smooth manifolds $M$,$N$, a smooth map $f$ between those two manifolds and a point $x\in M$ I think it should be a map between $T_x M$ and $T_{f(x)} N$. Is that right? But since here there are no point involved I don't see a way to utilize this definition of differential. Aug 15, 2023 at 14:33
• The linear maps $df_x$ at each point $x$ of $M$ can be packaged together to be a single smooth map between the tangent bundles, $df:TM\to TN$. That is what you should be thinking about.
– KCd
Aug 15, 2023 at 14:37
• Could you please unfold this definition of $df$ you gave? I don't think I heard of that one Aug 15, 2023 at 14:59
• I recommend to add some juicy meat to those dry definitions by explicitly calculating left invariant vector fields of a few popular Lie group examples, say, Heisenberg group or $SU(2)\,.$ A bit simpler (potentially boring) could even be the affine group. Aug 15, 2023 at 17:32

Suppose $$f:M\to N$$ is a smooth map between smooth manifolds. This gives us the tangent map between tangent bundles $$Tf:TM\to TN$$. At a fiberwise level this restricts, for each $$x\in M$$, to a map $$Tf_x:T_xM\to T_{f(x)}N$$. The notation is far from standardizes; although I prefer $$Tf$$ and $$Tf_x$$ (or $$T_xf$$), other notations include $$df$$ and $$df_x$$ (or $$d_xf$$ or $$df(x)$$), or $$Df$$ and $$Df_x$$ (or $$D_xf$$ or $$Df(x)$$), or $$\phi_*$$ and $$\phi_{*,x}$$ etc. Basically, anything which reminds you of a derivative. Names for this include ‘tangent map’ (which is the one I prefer), or ‘differential’, or ‘pushforward’.

Once we have this notion, given any vector field $$X$$ (recall, one way of defining this is as a smooth map $$X:M\to TM$$ such that $$\pi_M\circ X=\text{id}_M$$, i.e a section of the tangent bundle), if we have a diffeomorphism $$f:M\to N$$, then we can define a new vector field $$f_*X$$ on $$N$$, called the pushforward of $$X$$ by/under/via $$f$$. The definition is $$f_*X:= Tf\circ X\circ f^{-1}$$ (draw a commutative diagram so you know where everything is… this is a ‘left-up-right’ movement of the arrows). If you write this out pointwise, then for each $$y\in N$$, we are assigning the vector \begin{align} (f_*X)_y:=Tf_{f^{-1}(y)}\left(X_{f^{-1}(y)}\right)\in T_yN, \end{align} or equivalently (since $$f$$ is a diffeomorphism), for each $$x\in M$$, $$(f_*Y)_{f(x)}=Tf_x\left(X_x\right)\in T_{f(x)}N$$.

Just extra FYI: given a diffeomorphism $$f:M\to N$$ and a smooth vector field $$Y$$ on $$N$$, we can define a vector field $$f^*Y$$ on $$M$$, called the pullback of $$Y$$ under $$f$$, \begin{align} f^*Y:=(f^{-1})_*Y=T(f^{-1})\circ Y\circ f=(Tf)^{-1}\circ Y\circ f. \end{align}

The condition for left-invariance of a vector field $$X$$ on a Lie group $$G$$ is then that for each $$g\in G$$, we require $$(L_g)_*X=X$$. It is equivalent to say that for all $$g\in G$$, $$(L_g)^*X=X$$.

For right-invariance, simply replace $$L_g$$ with $$R_g$$, the right-multiplication by $$g$$.

• Oh, turns out I once answered a related question here. Aug 16, 2023 at 0:07
• Thank you very much, this helped me a lot. Let me just ask something to confirm I got it right. The tangent bundle of $M$ is $TM=\{(x,y) : x\in M, y\in T_xM\}$, given a smooth map $f:M\to N$, we define $Tf: TM\to TN$ such that $\pi_N\circ TF=f\circ\pi_M$. Now given your definitions of vector fields and of pushforward, where $L_g$ plays the role of the diffeomorphism $f$ between G and itself, the condition $(L_g)_*X=X$ is equal to require that $TL_g\circ X\circ L_g^{-1} = X$. Where $TL_g:TG\to TG$ is the map such that $\pi_G\circ TL_g=L_g\circ\pi_G$. Aug 16, 2023 at 8:54
• @cento18 correct. Aug 16, 2023 at 8:56
• I'm sorry to bother almost a week later but I have a very related question. We said above that $(L_g)_*$ is an application that takes a vector field $X:G\to TG$ and gives as output a vector field of the same kind: $Y:G\to TG$. In order to prove that $Lie(G)$ is a subalgebra of $\mathfrak{X}(G)$ (both defined as in the main question), I have to prove that $(L_g)_*[X,Y]=[(L_g)_*X,(L_g)_*Y]$, so that by invariance we get $(L_g)_*[X,Y]=[X,Y]$. Aug 21, 2023 at 9:27
• My problem is that the Lie Bracket is defined as the commutator and uses the definition of vector fields as derivations on $C^{\infty}(G)$, while $(L_g)_*$ uses the defintion of vector fields as maps from $G$ to $TG$. So when trying to prove $(L_g)_*[X,Y]=[(L_g)_*X,(L_g)_*Y]$ the two different definitions of vector field are involved. Is there a way to define the Lie Bracket on $\mathfrak{X}(G)$ according to the latter definition of vector field (maps $G\to TG$)? Otherwise, how do I prove $(L_g)_*[X,Y]=[(L_g)_*X,(L_g)_*Y]$? Aug 21, 2023 at 9:32