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.
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 $g$.