Quaternion Rotation formula

Question

Can someone explain what exactly what the last part $P^{-1}$ in the formula $PQP^{-1}$ does. I heard it rotates halfway but I'm not understanding that. If it is the inverse, it goes back the exact same as it was?

Answer 1

In the product $PQP^{-1}$, $Q=q_1\mathbf i +q_2 \mathbf j +q_3 \mathbf k$ is a pure imaginary quaternion that represents the vector $\vec q=[q_1,q_2,q_3]^T$ in $\mathbb{R}^3$ and $P$ is a unitary quaternion, i.e. a quaternion $ p_0+ p_1\mathbf i +p_2 \mathbf j +p_3 \mathbf k=p_0+\mathbf p $ such that $||P||=\sqrt{p_0^2+p_1^2+p_2^2+p_3^2}=1$ so that $P$ can be represented in polar form as $P=e^{\theta \mathbf p}$, with $ \cos \theta =\frac{p_0}{||P||}$.

Whan we multiply $PQ$ the result is a quaternion, but not a pure imaginary quaternion, and this means that we cannot identify such product with a vector in $\mathbb{R}^3$ but, if we calculate the product $PQP^{-1}=e^{\theta \mathbf p}Qe^{-\theta \mathbf p}$ than this gives a pure imaginary quaternion, and we can see that it corresponds to a rotation of $\vec q$ by an angle $2 \theta$ around the axis oriented by the vector $\mathbf p$.

So, to give a direct answer to your question; the part $P^{-1}$ has no independent role, but is required, together with P to obtain as a result a vector.

Answer 2

Any quaternion $q=x_0+x_1 i+x_2 j + x_3 k$ (where $x_l$ are real numbers and $ij=k$, $i^2=j^2=k^2=-1$) which is not a real number or equivalently such that $x_1^2+x_2^2+x_3^2\neq 0$ can be uniquely written as

$$q=x_0+Iy_0$$ with $y_0:= \sqrt {x_1^2+x_2^2+x_3^2}$ a positive real number and

$$I=\dfrac{x_1 i+x_2 j + x_3 } {\sqrt{x_1^2+x_2 ^2+x_3^2}}$$ is a (unitary) quaternion such that $I^2=-1$.

Thus, since real numbers form the center of the algebra of quaternions, for any quaternion $p$ different from $0$ one has

$$pqp^{-1}= px_0p^{-1}+p(Iy_0)p^{-1}= x_0+pIp^{-1}y_0$$ and since $J:=pIp^{-1}$ is a (unitary) quaternion and $J^2=-1$, it turns out that the quaternion $pqp^{-1}$ has same real part of the quaternion $q$ and can be obtained from a rotation of $q$ around the real axis.

Answer 3

Just as complex numbers represent a rotation in 2d, one can represent unit quarterions as a clifford rotation in 4d.

Let's start with the geometry CE2, which is ordinary algebraic geometry done with complex numbers. The equation for a line is still y = ax + b, and where b=0, y=ax. Now we introduce a cyclic variable w = cos t + i sin t = cis t.

Now y = ax gives wy = awx. Since at t=1/2 cycle, w = -1, then this shows that every point must rotate in a circle centred on the origin. This is a clifford rotation, or clifford parallels in the S3 space (equidistants, but not in the same plane).

The slope a is a complex number, and we can turn this into a sphere, by supposing the complex plane is in the real plane (r,i, 0), and place the sphere such that the diameter is (0,0,0) to (0,0,1). The points r,i are mapped onto this sphere by drawing the ray (0,0,1) to (r,i,0). Diametrically opposite points are then at right-angles. The angle between two points on this sphere is twice the angle between the parallels in S3.

There are two completely separate spheres. If we suppose the one described is x1,x2,y1,y2, then a rotation w would convert x1 to x2, and y1 to y2. The other rotation rotates x1 to x2, and y2 to y1.

In quarterion multiplication, we transform the point 1,0,0,0 to x,y,z,w by multiplication. If the rotation-value is given first, such as PQ, then the space rotates leftwise (ie x,y,z,w). If we put the multiplier after it, then it uses a right-clifford rotation,

The effect of using PQP'is to rotate the point outwards and inwards, in a way that P rotates both 2-spaces forward and the P' rotates one 2-space backwards, and the second 2-space forwards again. So one 2-space is left unchanged, and the orthogonal 2-space undergoes a double rotation.

Answer 4

Think about quaternions as planes. In G3 a quaternion is a bivector. The general rotation formula is $$ V^\prime = \exp( \mathbf{b} \phi/2) V \exp( - \mathbf{b} \phi/2) $$ where $\mathbf{b}$ is a unit bivector s.d. $\mathbf{b}^2=-1$. Then we have a generalization of the de Moivre formula $$ \exp( \pm \mathbf{b} \phi/2) = \cos(\phi/2) \pm \mathbf{b} \sin( \phi/2) $$