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In three-dimensional space it's possible to define rotations using the Euler angles $(\Psi,\Theta,\Phi)$ or quaternions $(i,j,k)$ If we have a hyperspace with more than three coordinates, is it still possible to use quaternions or a generalization of them, to describe rotations? Thanks in advance.

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If your vector space is real the group of special orthogonal matrices describe rotations in that space. If it's complex, special unitaries do. Quaternions have a representation as $SU(2)$, so you may think of $SU(N)$ as a generalization. Given that your vector space is provided with the 2-norm, $||v||_2$. – draks ... Apr 23 '12 at 12:47
up vote 2 down vote accepted

In 4D, it's possible to describe any rotation by a pair of unit quaternions $p,q$, with which we can rotate points considered as quaternions as follows $x \mapsto pxq$. It's easy to see that this is either a 4D rotation or a reflection, since a quaternion $x = a + ib + jc + dk$ has norm $a^2 + b^2 + c^2 + d^2$, and this norm is multiplicative, and so multiplying by two quaternions of length one will preserve lengths, and thus must either gives a rotation or a rotoreflection (combination of a reflection and a rotation). But since there always exists a continuous path from such a rotation to the identity rotation, we know it must be a rotation.

To see that any 4D rotation can be expressed in this way, we note that any rotation can be described as the product of an even number of reflections. Now expressing reflections in terms of quaternions

reflection in the hyperplane with normal $n$ $$ x - 2(x \cdot n)n = x - 2(\frac{1}{2}(x\overline{n} + n\overline{x}))n = x - x\overline{n}n - n\overline{x}n = -n\overline{x}n $$ so the product of an even number of reflections is of the form $x \mapsto pxq$.

With geometric algebra, it's possible to construct something similar for dimensions higher than 3 or 4 (using the same reflection trick), but this algebra has zero divisors, so it isn't as nice as the quaternions, and also, the rotation elements are not isomorphic to the the vector elements you are applying your action on.

If you don't know geometric algebra, then you might want to read up on it, but then you should be able to derive it using the fact that in geometric algebra, the dot product of two vectors can be expressed as $$ x \cdot y = \frac{1}{2}(xy + yx) $$

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