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

  1. Apr 6, 2014 #1
    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.
     
  2. jcsd
  3. Apr 8, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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? o_O
     
  4. Apr 8, 2014 #3
    Oh, I thought that $a$ was a variable rather than a constant :shy:. Thanks for the reply.
     
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