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Let $V$ be oriented two-dimensional Euclidean space. Then we can define an oriented angle $\phi$ between two nonzero vectors $u,v\in X$ by formulas: $ \phi=\arccos \frac{\langle u,v\rangle}{\|u\| \|v\|}$ if $(u,v)$ is a positive basis or $u,v$ are linearly dependent, and $\phi=2\pi-\arccos \frac{\langle u,v\rangle}{\|u\|\|v\|}$ if $(u,v)$ is a negative basis in $X$..

Let $X$ be a three dimensional Euclidean space. Let's consider a rotation different from identity $f\in SO(3)$. Then $f$ has a two dimensional invariant subspace $V$. Then its orthogonal complement $V'$ is a one dimensional invariant subspace. Let $V'$ be also oriented. Then orientation on $X$ and $V'$ induce an orientation on $V$: we fix a positive basis $e_1,e_2,e_3$ in $X$ and a positive basis $h$ in $V'$ and we call a basis $f_1,f_2$ in $V$ a positive iff $f_1,f_2,h$ is positive in $X$.

How to determined oriented angle of rotation, that is oriente angle between $u$ and $f(u)$, for $u\in X$ ?

Thanks

P.S. I know how to determine nonoriented angle $\psi$: $Tr f=2\cos \psi+1$.

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There's no need to write $<\bullet>$. One can write $\langle\bullet\rangle$. Angle brackets are coded as \langle and \rangle. I edited your question accordingly. –  Michael Hardy Nov 24 '12 at 19:05
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1 Answer 1

Since $V$ is the orthogonal complement to $V'$, you can find a positively oriented orthormal basis $h_1,h_2,h_3$ with $h_1, h_2,h_3$ such that $h_1,h_2 \in V$ and $h_3:=h/||h|| \in V''$. Now there is a $\psi \in [0,2\pi)$ such that $f(h_1)=\cos(\psi)h_1+\sin(\psi)h_2$. This $\psi$ is the oriented angle of $f$.

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Thanks. I have another idea which not used basis in $W$ but I don't know is it correct. Namely,the nonoriented angle between $u, f(u)$ we can determined by $Trace f=2\cos \psi +1$. We next take arbitrary $u\in W$, $v\neq 0$ and $f(u)$. If $u\times f(u)=c\cdot h$ with $c>0$ then $\phi=\psi$ and if $u\times f(u)=c\cdot h$ with $c<0$ then $\phi=2\pi-\psi$. –  Richard Nov 25 '12 at 20:44
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