I am reading these lines from a text which shows why the bracket of two left-invariant vector fields is also a left-invariant vector field. But cannot easily get one of the lines.

Let $L_a$ be the left translation of a Lie group $G$ and let $X,Y$ be to left-invariant vector fields on $G, g, a,p \in G$ and $f$ a smooth function on $G$.

Define $X$ to be left-invariant on $G$ if $X_{ag}=(dL_a)_{g}(X_g)$

Now compute $dL_a[X,Y]_pf$.

$dL_a[X,Y]_pf=[X,Y]_p(f \circ L_a)$ (From the definition of differential $dL_a$)

$=X_p(Y(f \circ La))-Y_p(X(f \circ L_a))$ (From the definition of bracket)

$=X_p(dL_aY)f-Y_p(dL_aX)f$ (From the definition of $dL_a$)

$=X_pY(f)-Y_pX(f)$ (From left-invariant definition, i.e. $X=(dL_a)(X$))

$=[X,Y]_pf$ (From the definition of bracket)

So $dL_a[X,Y]_pf=[X,Y]_pf$ and from left-invariant definition, i.e. $X=(dL_a)(X$) we conclude that the bracket of two left-invariant vector fields is a left-invariant vector field.

My specific question is that why these two definitions of left-invariant vector fields are the same: $X_{ag}=(dL_a)_{g}(X_g)$ and $X=(dL_a)(X)$. Clearly in the former $X_{ag} \neq X_g $ whereas in the latter we have $X=X$ .


The equation $X=(dL_a)(X)$ is one of vector fields on$~G$. Here $dL_a$ is a field of differential (linear) maps, each mapping a tangent space at some $g\in G$ to the tangent space at $L_a(g)=ag$. So when applying to the vector field $X$, it means that each tangent vector $X_g$ is sent to $(dL_a)_g(X_g)$, a tangent vector at $ag$. If this is to give the same fields as $X$ itself, it should match the vector of $X$ at the point $ag$, which is $X_{ag}$. So all in all one requires $X_{ag}=(dL_a)_g(X_g)$ to hold for all $g$ (and this for all left translations $L_a$).

  • $\begingroup$ Thanks. Now clear. I see that with respect to left-invariance one equation deals with the same "vector field" yet the another one deals with "two tangent vectors on the same vector field" $\endgroup$ – user169903 Oct 9 '14 at 10:55

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