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Let $SO(3)$ be the Lie group of 3D rotations. Rotation about z-axis by an angle $\phi$ is represented in standard basis by this matrix:

$$ \begin{pmatrix} \cos \phi & -\sin\phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Differentiating this matrix at $\phi = 0$ we get an infinitesimal generator of rotation about z-axis: $$ Z = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

In similar fashion we get matrices $X$ and $Y$ corresponding to rotations about x- and y-axis. These matrices are elements of $\mathfrak{so}(3)$, Lie algebra of $SO(3)$ and bracketing gives us:

$$[X, Y] = -Z, [Y, Z] = -X, [Z, X] = -Y$$

I'm not really good at visualizing things, but I'm curious is there a nice visual explanation for this?

ante.ceperic
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1 Answers1

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$[X,Y]$ refers to the rate of change of $Y$ over $X$. So it is equal to $Z$. It appears to be the cross product, as @Will Jagy said. If you want a more sophicated explanation of the whole thing, I will have to start over and take some time.

Troy Woo
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  • Let's imagine you are seeing those notions in my post for the first time. Could one with good visual intuition guess $[X, Y]$ proportional to $Z$ before calculating? If I understand correctly, when you say that Lie bracket refers to the rate of change, you think about the Lie bracket of left-invariant vector fields generated by $X, Y, Z$, right? It's not really obvious to me that $[X,Y]$ should be proportional to $Z$ when I think that way. Is it to you? Could you explain how? – ante.ceperic Aug 30 '14 at 21:29
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    @ante.ceperic Rotate $Y$ about $X$ a very small angle, say $\theta$, and the difference $e^{X\theta}Y-Y$ is almost parallel to $Z$ right? You take the limit of $\lim_{\theta\to 0}(e^{X\theta}Y-Y)/\theta$, and it should be $Z$ right? I abused the notation a little bit, sorry. – Troy Woo Aug 30 '14 at 21:34