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I would like to understand the left invariant vector field by using a numerical example. Now we consider a Lie group $G=SE(3)$, and the associated Lie algebra is $\mathfrak{g}=se(3)$. We suppose: $$g=\pmatrix{1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\0 & 0 & 1 & 3\\0 & 0 & 0 & 1}\in G$$ and $$v=\pmatrix{0 & 0 & 0 & 1\\ 0 & 0 & 0 & 2\\0 & 0 & 0 & 0\\0 & 0 & 0 &0}\in\mathfrak{g}$$ $v$ is a vector at the identity element $I$ in a left invariant vector field $X$ of the Lie Group $G$. Then my questions:

1) how to calculate the vector $v_g$ at the point $g$ in the vector field $X$?

2) Now we consider a map: $\phi:G\rightarrow G,x\rightarrow gx$, where $g$ is defined as above. Then

$\quad$ i) how to calculate the vector at the identity element $I$ in the new Lie Group $\phi(G)$? (Is this equal to $v$?)

$\quad$ ii) how to calculate the vector at the point $g=\phi(I)$ in the Lie Group $\phi(G)$?

$\quad$ iii) how to calculate the vector at the point $\phi(g)$ in the Lie Group $\phi(G)$?

Martial
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1 Answers1

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The vector $v_g$ at $g$ is $$v_g=gv=\pmatrix{0 & 0 & 0 & 1\\ 0 & 0 & 0 & 2\\0 & 0 & 0 & 0\\0 & 0 & 0 &0}$$ which is an element in $T_gG$.

To see this, start with the left-multiplication map $\phi(x)=gx$ and compute its differential $d\phi(x, dx)=gdx$ which is the push-forward map of the vector $dx\in T_xG$ to $T_{gx}G$. The push-forward of the vector $v\in T_IG$ to (the tangent spaces of) $I$, $g$ and $g^2$ is $v$, $gv$ and $g^2v$ respectively.

The last example can be spelled out as $d\phi(g,dg)=gdg=g(gv)=g^2v$, which is the push-forward map iterated twice.

Let me also take the opportunity to mention that if you compose the left multiplication push-forward with the inverse right multiplication push-forward you get a map $$Ad_g(v)=gvg^{-1}$$ which maps the Lie algebra back to itself in a nontrivial way. This is the very important adjoint action of the Lie group on its Lie algebra and has many fruitful applications.

Jesper Göransson
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  • Thank you. You mentioned $Ad_g(v)$, what is the relationship between $Ad_g(v)$ and $v$? I mean what is the exact meaning of $Ad_g(v)$. I hope u understand me. for example, we can calculate the $Ad_g(v)$, but what does it represent? – Martial Feb 08 '15 at 10:42
  • @Martial. You're welcome. Well, the adjoint map is simply the natural way that the group acts on its algebra. It's special in that there is no external vector space involved, it is entirely built-in to the Lie group. – Jesper Göransson Feb 08 '15 at 15:27
  • I have one more question about your answer for question 1). It concerns the extention of $T_eG$ to $X$. How did you get the formula $v_g=gv$ (which means, I think, $T_eL_gX(I)=T_eL_g(v)=gv=L_g(v)$, but not reasonable)? Thanks. – Martial Feb 10 '15 at 16:08
  • Perhaps it can be explained like this: differentiate $\phi=gx$ to obtain $\dot{\phi}=g\dot{x}$. The meaning of $\dot{x}$ is the tangent vector $v$ and the meaning of $\dot{\phi}$ is the resulting vector $v_g$. So $v_g=\dot{\phi}=g\dot{x}=gv$. – Jesper Göransson Feb 10 '15 at 17:24