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If $\mathfrak{so}(3)$ is the Lie algebra of $SO(3)$ then why are the matrices of $\mathfrak{so}(3)$ not rotation matrices? They aren't infinitesimal rotations either. The matrices of $\mathfrak{so}(3)$ are skew-symmetric matrices which are the type used to calculate the cross product.

How can $\mathfrak{so}(3)$ be tangent to $SO(3)$ if they're never even in $SO(3)$?

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2 Answers 2

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How can $\mathfrak{so}(3)$ be tangent to $SO(3)$ if they're not even in $SO(3)$?

The same way $(0,\frac12)$ can be a tangent vector to the unit circle at $(1,0)$ even though $(0,\frac12)$ is not on the unit circle.

One (loose and informal!) way to think about it is that an element of $\mathfrak{so}(3)$ is the difference between the matrix of an infinitesimal rotation and the identity matrix, but "scaled up by a factor of infinity" such that the entries of the matrix don't need to be infintesimals themselves.

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  • $\begingroup$ I guess that explains "The matrices in the Lie algebra are not themselves rotations; the skew-symmetric matrices are derivatives". Thanks $\endgroup$
    – R. Emery
    Oct 24, 2017 at 20:29
  • $\begingroup$ Would the derivative be with respect to time? $\endgroup$
    – R. Emery
    Oct 24, 2017 at 20:47
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    $\begingroup$ @R.Emery: You can think of it as time if you want. The formalism doesn't care, of course -- it just says that if you have a differentiable function $f:\mathbb R\to SO(3)$ such that $f(t_0)=I$, then $f'(t_0)$ should be in $\mathfrak{so}(3)$ -- and all elements of $\mathfrak{so}(3)$ are such derivatives. $\endgroup$ Oct 24, 2017 at 20:58
  • $\begingroup$ Ok, that makes sense. I cant vote yet but if I could I would give you a plus one. $\endgroup$
    – R. Emery
    Oct 24, 2017 at 21:19
  • $\begingroup$ I have quoted you here: math.wikia.com/wiki/Infinitesimal_rotations hoping you are okay with that. $\endgroup$
    – R. Emery
    Oct 24, 2017 at 21:47
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$A$ is an infinitesimal rotation means that $e^A$ is a rotation.

Consider $e^{tA}$, you have ${d\over{dt}}\langle e^{tA}x,e^{tA}x\rangle=\langle$ $A(e^{tA}x),e^{tA}x\rangle+\langle e^{tA}x,A(e^{tA}x)\rangle=0$ since $A$ is antisymmetric. You deduce that $\langle e^{tA}x,e^{tA}x\rangle$ is constant ant its value is $\langle e^{0A}x,e^{0A}x\rangle=\langle x,x\rangle$.

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  • $\begingroup$ $ \langle x,x \rangle $ is the inner product? $\endgroup$
    – R. Emery
    Oct 24, 2017 at 20:37
  • $\begingroup$ This: math.wikia.com/wiki/… is all I know about that $\endgroup$
    – R. Emery
    Oct 24, 2017 at 20:40

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