Notation: $\chi(G)$ is the set of smooth vector fields on Lie group $G$, which in fact forms a vector space.

  1. Given a Lie group $G$, show that there exists a smooth vector field $X\in \chi(G)$, such that $X(g)\neq 0, \ \forall \ g\in G.$

  2. Let $G$ be $n-$ dimensional Lie group. Show that there exists smooth vector fields $X_1, X_2,...,X_n$ such that for every $g\in G,$ the vectors $X_1(g), X_2(g),...,X_n(g)$ span the tangent space $T_g(G).$

I have no idea how to get started. Could you provide some hints?


1 Answer 1


Hint: Pick a vector $v\in T_{e}(G)$ - that is, pick a vector tangent to $G$ at the identity $e$. Now define a vector field $V$ on $G$ by the rule:


Here, $L_g:G\to G$ is the diffeomorphism of $G$ induced by left multiplication by $g$: $L_g(x)=gx$ for all $x\in G$. And, $(L_g)_{*}: T_e(G)\to T_{g}(G)$ is the induced differential. (Thus, $V_g\in T_{g}(G)$ for all $g\in G$.)

Exercise 1: Prove that $V$ is a smooth vector field on $G$ - that is, $V\in \chi(G)$.

Exercise 2: Prove that if $v_1,\dots ,v_n$ constitutes a basis of $T_{e}(G)$, then $V_1,\dots ,V_n$ has the property of $2$ in your question.

Exercise 3: Prove that $1$ in your question is a corollary of $2$.

Hope this helps! Please let me know if you need further hints.

  • $\begingroup$ yes, I would appreciate more hints... Ex (1) seems to be proved by the fact that $V_g\in T_g(G), \ \forall g\in G.$ No? $\endgroup$
    – math
    Apr 18, 2014 at 10:40
  • 1
    $\begingroup$ @math (it's funny to address you this way!) The key point of Exercise 1 is that $V$ is a smooth vector field. You're right that it's a vector field - just because $V_g\in T_{g}(G)$ for all $g\in G$. But why is it smooth? (Hint: you need to use smoothness of the multiplication map $m:G\times G\to G$.) I would suggest thinking about this a little bit (i.e., recall the definition of "smoothness" of a vector field and try to apply it to $V$) and please let me know if you're still stuck after a while. $\endgroup$ Apr 18, 2014 at 12:43
  • $\begingroup$ back again. Thanks for your response. I know that if $(U,(x^i))$ is a smooth chart in $G$, then for $g\in G$, $$V_g = \sum_{i=1}^n V^i (g) \frac{\partial}{\partial x_i} |_g,$$ where $V^i : U\longrightarrow \mathbb{R}$ is a component function of $V$ in the chart. It is enough to show that $V^i$ are smooth, right? I am not sure how to use smoothness of multiplication map here. Was this what you were saying in the last comment? Thanks. $\endgroup$
    – math
    Apr 19, 2014 at 12:41
  • $\begingroup$ Yes, that is one way of showing smoothness of the vector field $V$. Another way (which I sometimes prefer because it does not rely on an explicit choice of local coordinates) is to show that for all smooth functions $f:G\to \mathbb{R}$, $V(f):G\to \mathbb{R}$ is also smooth. (Recall that $V(f)$ is defined at each point $g$ as the directional derivative of $f$ at $g$ with respect to $V_g$.) $\endgroup$ Apr 19, 2014 at 12:41
  • $\begingroup$ I'm online right now so we can discuss it if you like - I will try to type out some more in the next few minutes. $\endgroup$ Apr 19, 2014 at 12:42

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