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Let $G$ be a Lie group, $L_g$ the left-translation on this group with differential $d L_g$. A vector field $X$ on $G$ is called left-invariant if

$$ X \circ L_g = d L_g \circ X \quad \forall g \in G$$

i.e.

$$ X_{gh} = (d L_g)_h (X_h) \quad \forall g,h \in G. $$

Now, this definition seems so natural to me that I cannot come up with a non-trivial counterexample for a vector field that is $\textit{not}$ left-invariant. In my mind, pushing forward on the tangent space is basically always the same as the group action...

Could you provide me with such a counterexample that helps understand the notion of left-invariance?

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    $\begingroup$ Take any non-zero vector in any tangent space. Show that it extends uniquely to a left-invariant vector field. Conclude that the space of left-invariant vector fields has dimension $\dim{G}$, whereas the dimension of the space of vector fields is infinite. Take any non-zero vector field, multiply it by a smooth bump function and get a non-invariant vector field (unless you get zero). Now if you think about how to construct a non-zero vector field on a manifold you should be able to find plenty of non-invariant examples... $\endgroup$ – t.b. Aug 6 '12 at 10:24
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You could start taking the Lie group $(\mathbb R^n,+),$ and considering what does it means for a vector field $X$ on $\mathbb R^n$ to be left-invariant.

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  • $\begingroup$ Thanks! Do you know another example, where the differential $(dL_g)_h$ is not the identity? $\endgroup$ – madison54 Aug 6 '12 at 10:11

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