Left-invariant vector field of the additive group of real number

[AlbertEi]

Hi,

I would like to understand the left-invariant vector field of the additive group of real number. The left translation are defined by
\begin{equation}
L_a : x \mapsto x + a \; , \;\;\; x,a \in G \subseteq \mathbb{R}.
\end{equation}
The differential map is
\begin{equation}
L_{a*} = \frac{\partial (x + a)}{\partial x} = 1,
\end{equation}
and the left-invariant vector field is
\begin{equation}
X = \frac{\partial}{\partial x}.
\end{equation}
So
\begin{equation}
L_{a*} X|_x = \frac{\partial}{\partial x}|_x.
\end{equation}
But I don't understand why
\begin{equation}
L_{a*} X|_x = X|_{x+a} = \frac{\partial}{\partial x}|_{x+a}.
\end{equation}
This should be true if X is really a left-invariant vector field, right?

Thanks in advance for any help.

 

[Matterwave]

The left translations map the line into itself. All the values on the number line are just moved over to the left a distance a. Therefore, the tangent vectors at each point are still $$\frac{\partial}{\partial x}$$ as they were before. The curve which was tangent to the real line at x (the real line itself) is the same curve which is tangent to the real line at x+a (again, the real line itself). This seems to me a pretty trivial result...I don't know if you were asking something deeper?

 

[AlbertEi]

Oh, I thought that $a$ was a variable rather than a constant :shy:. Thanks for the reply.