I have a Lie group, namely on SO(3), i.e. $SO(3,\mathbb{\mathbb{R}})=\left\{ A\in GL\left(3,\mathbb{R}\right)\mid A^{T}A=\mathbb{1},\,\det\left(A\right)=1\right\}$. I have a Left action $L_g$ and I know that by definition a left invariant vector field is a vector field such that $$(L_g)_{\ast}(X_e)=X_g$$ I worked out few thing with the tangent space at the identity and I arrived at formulating the idea that this should be a left invariant vector field on $SO(3)$: $$L_{\mathbf{1}}=y\frac{\partial}{\partial z}-z\frac{\partial}{\partial x}$$ But now how to I effectively demonstrate that is left invariant. Can anybody show me the computation that once in a lifetime must be seen? Or at least give a refernce that actually shows the computation technique?

  • 1
    $\begingroup$ The field you've written down is given in coordinates $x, y, z$ (presuambly it is supposed to be defined on $\Bbb R^3$), but a left-invariant vector field is defined on the group itself, and you've given $SO(3, \Bbb R)$ as a subset of $GL(3, R) \subset \Bbb R^{3 \times 3}$. $\endgroup$ – Travis Apr 29 '16 at 19:45

This kind of formula only works when you have the action of the lie group $G = SO(3)$ on some manifold $M$, in this case, the euclidean space $\mathbb{R}^3$. Let then $G$ act on $M$ by $(g,x) \mapsto gx$.

A left-invariant vector field $V$ in $G$ induces a vector field on $M$ such that the flow in $M$ is: $${\gamma(t,x) = \exp(tV)x,}$$ where $\exp: \mathfrak{g} \rightarrow G$ is the exponential map. Define the induced vector field $\tilde{V}$ on $M$ by: $${\tilde{V} = \frac{d}{dt}\exp(tV)x|_{t=0}.}$$Then we have defined a map $\mathfrak{g} \rightarrow \mathfrak{X}(M)$ by $V \mapsto \tilde{V}$.

In your case, take an element of $\mathfrak{so}(3)$ (the lie algebra of $SO(3)$, e.g.: the tangent space at the identity), namely:

$$X = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix}$$ (You can see that this is an element of $\mathfrak{so}(3)$ because it is an antissymetric matrix).

Then your induced vector field is $\tilde{X} = -z\frac{\partial}{\partial y} + y \frac{\partial}{\partial x}$.

  • $\begingroup$ Here, the exponential map is only the matrix exponential $e^X = I + X + \frac{1}{2} X^2 +...$ that can be done by hand for the matrix in question. $\endgroup$ – B. Chinaski Apr 29 '16 at 19:41
  • $\begingroup$ Thank you for your explanations, I upvoted the answer, but could you also show that X gives rise to a left invariant vector field? I'm not sure if I have understood... thank you $\endgroup$ – Dac0 Apr 30 '16 at 6:39
  • $\begingroup$ As a group of matrices, $SO(3)$ has a Lie Algebra of matrices. The vector field you want is only an induced vector field by the action of this group (as a group of symmetry, for example) in a manifold. If you want to understand that in physical terms, the left-invariant vector field $X$ has an associated operator $\tilde{X}$ that acts on your space as a differential operator on functions, and therefore, as a geometrical vector field. $\endgroup$ – B. Chinaski Apr 30 '16 at 14:51

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