# Two definitions of left-invariant vector fields of a Lie group

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 $$X=X$$.

• – glS
May 14, 2021 at 16:34

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 $L_a$).
• hi, so shouldn't it be $dL_a([X,Y]_p)f=[X,Y]_{ap}f$ in the initial equation? as $L_a$ sends $p$ to $ap$. this is confusing me as well. thanks! Aug 30, 2022 at 6:25
• @l4teLearner Although I don't quite understand the role of the subscript $p$ here (and the absence of parentheses), since $(dL_a)$ is supposed to act on vector fields rather than individual vectors, the definition of $(dL_a)$ is to change the field not by looking at the same vector field in a different point, but by replacing is argument of the vector field (a function on the group) by a translation of that argument. In this respect the question is correct. Sep 3, 2022 at 20:40