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How do I tell if a vector field on a Lie Group is left-invariant? I have the technical definition. But, I want to understand given a specific vector field what should I do to test if it is left-invariant? For instance, here is a vector field in $\mathbb R^2$:

$V(x, y)=y\partial/\partial x-x\partial/\partial y$

If this vector field (is/is not) left-invariant, can you provide me an example worked out as well of a vector field that (is not/is).

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You just use the definition: see what a translation does to it. There is no magic, really. – Mariano Suárez-Alvarez Mar 15 '13 at 21:33
In this particular case, though, you can notice that your vector field vanishes at one point (the origin): if it were translation invariant, it would have to be identically zero. But this is just a reflection of the fact that your example is not an interesting one: there are many vector fields on the plane which do not vanish anywhere.ç – Mariano Suárez-Alvarez Mar 15 '13 at 21:34
If your Lie group is the unit circle, though, then $V$ is left-invariant. You should probably tell us what your example $V$ is referring to! – Ryan Budney Mar 15 '13 at 21:43
I did Ryan, it is in the Lie Group $\mathbb R^2$ – Leo Spencer Mar 15 '13 at 21:50
up vote 1 down vote accepted

Here is an explicit example .

Suppose your Lie group is $G=GL_n(\mathbb R)$ .
The tangent space at a matrix $\Gamma\in G$ is the vector space $T_\Gamma(G)=\oplus_{i,j}\mathbb R\cdot \frac {\partial}{\partial x_{ij}}|_ \Gamma$ .
Any left-invariant vector field on $G$ is then obtained by the following procedure:
choose a matrix $A\in M_n(\mathbb R)$ (beware that $A$ needn't be in $G$ !) and define the left-invariant vector field $\mathcal X_A$ by $$\mathcal X_A (\Gamma)=\sum_{i,j} (\Gamma A)_{ij} \frac {\partial}{\partial x_{ij}}|_\Gamma $$ where $(\Gamma A)_{ij}=\sum_k \Gamma_{ik}A_{kj}$ is the $(i,j)$-entry of the matrix product $\Gamma A$.

This an illustration of the general fact that left-invariant vector fields correspond bijectively to the tangent vector space $T_e G$ of the Lie group $G$ at its identity.
In the example above, where $G=GL_n(\mathbb R)$ and $e=I$ =identity $n\times n$-matrix, we had $T_{I} GL_n(\mathbb R)=M_n(\mathbb R)$.

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