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.

  • $\begingroup$ 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. $\endgroup$
    – KCd
    Aug 15, 2023 at 14:26
  • $\begingroup$ 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. $\endgroup$
    – cento18
    Aug 15, 2023 at 14:33
  • $\begingroup$ 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. $\endgroup$
    – KCd
    Aug 15, 2023 at 14:37
  • $\begingroup$ Could you please unfold this definition of $df$ you gave? I don't think I heard of that one $\endgroup$
    – cento18
    Aug 15, 2023 at 14:59
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Aug 15, 2023 at 17:32

1 Answer 1


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 $g$.

  • $\begingroup$ Oh, turns out I once answered a related question here. $\endgroup$
    – peek-a-boo
    Aug 16, 2023 at 0:07
  • $\begingroup$ 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 $\pi_G\circ TL_g=L_g\circ\pi_G$. $\endgroup$
    – cento18
    Aug 16, 2023 at 8:54
  • $\begingroup$ @cento18 correct. $\endgroup$
    – peek-a-boo
    Aug 16, 2023 at 8:56
  • $\begingroup$ 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 $(L_g)_*[X,Y]=[X,Y]$. $\endgroup$
    – cento18
    Aug 21, 2023 at 9:27
  • $\begingroup$ 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 $(L_g)_*[X,Y]=[(L_g)_*X,(L_g)_*Y]$? $\endgroup$
    – cento18
    Aug 21, 2023 at 9:32

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