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3D rotation group
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===Using quaternions of unit norm === {{main|Quaternions and spatial rotation}} The group {{math|SU(2)}} is [[Group isomorphism|isomorphic]] to the [[quaternion]]s of unit norm via a map given by<ref>{{harvnb|Rossmann|2002}} p. 95.</ref> <math display="block">q = a\mathbf{1} + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} = \alpha + \beta \mathbf{j} \leftrightarrow \begin{bmatrix}\alpha & \beta \\ -\overline\beta & \overline \alpha\end{bmatrix} = U</math> restricted to <math display="inline">a^2+ b^2 + c^2 + d^2 = |\alpha|^2 +|\beta|^2 = 1</math> where <math display="inline"> q \in \mathbb{H}</math>, <math display="inline">a, b, c, d \in \R</math>, <math display="inline"> U \in \operatorname{SU}(2)</math>, and <math>\alpha = a+bi \in\mathbb{C}</math>, <math>\beta = c+di \in \mathbb{C}</math>. Let us now identify <math>\R^3</math> with the span of <math>\mathbf{i},\mathbf{j},\mathbf{k}</math>. One can then verify that if <math>v</math> is in <math>\R^3</math> and <math>q</math> is a unit quaternion, then <math display="block">qvq^{-1}\in \R^3.</math> Furthermore, the map <math>v\mapsto qvq^{-1}</math> is a rotation of <math>\R^3.</math> Moreover, <math>(-q)v(-q)^{-1}</math> is the same as <math>qvq^{-1}</math>. This means that there is a {{math|2:1}} homomorphism from quaternions of unit norm to the 3D rotation group {{math|SO(3)}}. One can work this homomorphism out explicitly: the unit quaternion, {{mvar|q}}, with <math display="block">\begin{align} q &= w + x\mathbf{i} + y\mathbf{j} + z\mathbf{k} , \\ 1 &= w^2 + x^2 + y^2 + z^2 , \end{align}</math> is mapped to the rotation matrix <math display="block"> Q = \begin{bmatrix} 1 - 2 y^2 - 2 z^2 & 2 x y - 2 z w & 2 x z + 2 y w \\ 2 x y + 2 z w & 1 - 2 x^2 - 2 z^2 & 2 y z - 2 x w \\ 2 x z - 2 y w & 2 y z + 2 x w & 1 - 2 x^2 - 2 y^2 \end{bmatrix}. </math> This is a rotation around the vector {{math|(''x'', ''y'', ''z'')}} by an angle {{math|2''θ''}}, where {{math|1=cos ''θ'' = ''w''}} and {{math|1={{!}}sin ''θ''{{!}} = {{norm|(''x'', ''y'', ''z'')}}}}. The proper sign for {{math|sin ''θ''}} is implied, once the signs of the axis components are fixed. The {{nowrap|{{math|2:1}}-nature}} is apparent since both {{math|''q''}} and {{math|−''q''}} map to the same {{math|''Q''}}.
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