I Understanding Quaternion Transformations for 3D Vectors

[Tio Barnabe]

I'd like to show why a 3-vector $v$ transforming using a quartenion $q$ must transform as $v' = q^{-1}vq$.

I tried showing that $v^{\dagger}v = v'^{\ \dagger}v'$ as long as $v'$ is given by the above transformation, whereas $v' = qv$ doesn't transform such that the inner product is invariant. However, this "proof" doesn't work.

 

[fresh_42]

I'd like to show why a 3-vector $v$ transforming using a quartenion $q$ must transform as $v' = q^{-1}vq$.

I tried showing that $v^{\dagger}v = v'^{\ \dagger}v'$ as long as $v'$ is given by the above transformation, whereas $v' = qv$ doesn't transform such that the inner product is invariant. However, this "proof" doesn't work.

What do you mean by "must"? Without any conditions, which enforces this necessity, the vector doesn't have to behave anyhow. You may transform it in whatever and in which way ever you like to. We have a natural identification of the 3-sphere with the unitary quaternions $\mathbb{S}^3 \cong U(1,\mathbb{H})$ which connects the two, but there is still the question: "must" because of "what"?

 

[Tio Barnabe]

What do you mean by "must"?

"must" because of "what"?

Hmm, I thought it must be so, because I have seen it on several papers which deals with such transformations. But, I certainly agree with you in that

Without any conditions, which enforces this necessity, the vector doesn't have to behave anyhow

Thanks for remembering me.

We have a natural identification of the 3-sphere with the unitary quaternions $\mathbb{S}^3 \cong U(1,\mathbb{H})$ which connects the two

Would this relation provide us with a clue of how the vector should transform?

 

[fresh_42]

Would this relation provide us with a clue of how the vector should transform?

Conjugation with group elements on its tangent space at $1$ is the adjoint representation of a Lie group on its Lie algebra. This can be generalized a bit further (principal bundles). But basically it is the map $\operatorname{Ad}\, : \,G \longrightarrow GL(\mathfrak{g})$ defined by $g \mapsto (\,X \mapsto gXg^{-1}\,)$ induced by the conjugation in the group, which is a very natural operation of any group on itself, so of Lie groups as well. Since the tangent space of $\mathbb{S}^3$ is $\mathbb{R}^3$ and $\mathbb{S}^3$ the Lie group of unit quaternions, we have in $\operatorname{Ad}$ the desired operation by conjugation. This can be viewed as a "must" in the sense, that it is part of standard Lie theory.

 

[lpetrich]

I'm not sure what the OP is getting at, but I'm guessing something like this result. For quaternion q and vector v,

(q.{0,v}.q^-1)/(q.q) = {0,R(q).v}
(quaternion multiplication), (inner product), (inner product) {0,v} is append 0 to vector v to make a quaternion.

where R(q) is the 3D rotation matrix composed from quaternion q.