I want to prove that the tangent space of a Lie group at its identity $e$ is isomorphic to the vector space of left-invariant vector fields.

Given an element $D \in T_e G$ (a derivation), the corresponding left-invariant vector field is $$X : G \rightarrow TG, \; X(g) := L_g^*(D),$$ where $L_g$ is left-multiplication by $D$.

To verify that this is smooth, I take an open subset $U \subseteq G$ and a smooth function $f : U \rightarrow \mathbb{R};$ then I need to see that $Xf$ is smooth, where $$Xf : U \rightarrow \mathbb{R}, \; \; Xf(g) := X_g(f) = (L_g^*)(D)(f) = D(f \circ L_g).$$

I do not know how to show that $Xf$ is smooth.


1 Answer 1


Possible Outline: You know that group multiplication is smooth, implying that $L_g$ is smooth with $L_g (a) = g\cdot a$. Since $f$ is smooth and $L_g$ is smooth, then $f \circ L_g$ is smooth. Moreover, the (directional) derivative of a smooth function is smooth, so $D(f \circ L_g)$ is smooth.

  • 1
    $\begingroup$ This argument isn't clear to me, since $D(f \circ L_g)$ is a function of $g$ and not of whatever variable $f \circ L_g$ (for a fixed $g$) takes $\endgroup$
    – user180027
    Commented Sep 30, 2014 at 20:20
  • 2
    $\begingroup$ If $m : G \times G \to G$ is defined by $m(a,b) = a \cdot b$, then $L_g = m(g, \cdot)$. The function $m$ is assumed to be smooth (Lie group), so $f \circ L_g = f(m(g, \cdot))$; in particular, the map $g \mapsto X_g f$ is the map $g \mapsto D(f \circ m(g, \cdot))$. But perhaps this is more complicated than it needs to be? $\endgroup$
    – Tom
    Commented Sep 30, 2014 at 20:25
  • $\begingroup$ @Tom I think that's precisely as complicated as it needs to be. If I'm not mistaken, smoothness (in x) of X follows from smoothness of right multiplication, and your comment has pretty much the minimum amount of detail for that subtlety to show. $\endgroup$
    – Solveit
    Commented May 2, 2020 at 7:31

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