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As we know, a $q$ in the group of unit quaternions can be represented as a $2 \times 2$ matrix. So, if $q = a+ib+jc+kd$, then its matrix representation would be $\begin{bmatrix} a+ib & c+id \\ -c +id & a-ib \\ \end{bmatrix}$.

But, I don't see how $qq_1q^{-1} \in \mathbb{R}^3$, if $q_1 \in \mathbb{R}^3$.

I'm guessing first we convert $(x,y,z)=q_1 \in \mathbb{R}^3$ as a $2 \times 2$ matrix $\begin{bmatrix} x & y \\ z & 0 \\ \end{bmatrix}$. And then we do the matrix multiplication.

Then, convert the matrix $qq_1q^{-1}$ back to the form of $\mathbb{R}^3$.

Is this how we compute $qq_1q^{-1}$?

john
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    You convert $q_1$ to a pure quaternion $ix+jy+kz$ and use the matrix representation of that. – Kurt G. Mar 08 '22 at 17:25
  • @KurtG. Oh, I see. So, $q_1$ can be considered as $\begin{bmatrix} ix & y+iz \ -y+iz & -ix \ \end{bmatrix}$. And, the converting map is clearly bijective, thus for each $v \in \mathbb{R}^3$, we get a unique matrix form. – john Mar 08 '22 at 17:30
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    Yes. What makes this operation interesting is that you can rotate vectors in $\mathbb R^3$ by using quaternions. Please dive into this. It is amazing. Look for $SU(2)$ and $SO(3)$. – Kurt G. Mar 08 '22 at 17:32
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    @KurtG. https://math.stackexchange.com/questions/4399056/conjugation-of-quaternions-induces-an-isomorphism-rho-s3-pm-1-rightar is the post I just made, and I think this is what you are talking about. There are some questions regarding the Theorem that $S^3/ {\pm1}$ and $SO(3)$ are isomorphic. If you don't mind, would you also give me some help there? – john Mar 09 '22 at 00:02

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I define a quaternion to be a sum of a scalar and a 3D vector (from $\mathbb{R}^3$). You can also call the scalar and vector parts the real and imaginary parts, respectively. The real inner product is $\langle x,y\rangle=\mathrm{Re}(x\overline{y})$, which is the usual dot product $x_1y_1+x_2y_2+x_3y_3+x_4y_4$. (There are different ways to build up to this, each with its own pedagogical merits, but I'll just leave it as a given you can check.) As a result, multiplying on the left or right by a unit quaternion is an isometry. In particular, if $p$ is a unit quaternion then $p^{-1}=\overline{p}$, and then

$$ \langle pxp^{-1},1\rangle=\langle px\overline{p},1\rangle=\langle px,p\rangle=\langle x,1\rangle $$

so $pxp^{-1}$ is a vector (aka pure imaginary) if and only if $x$ is.

The representation of quaternions by $2\times2$ complex matrices is a separate thing. You can transport the conjugation action over hear, if you want, though. For this, the unit quaternions become $\mathrm{SU}(2)$, the pure imaginary quaternions (real 3D vectors) become the lie algebra $\mathfrak{su}(2)$, which is the set of complex matrices satisfying $X^\dagger+X=0$. Then $SU(2)$ acts on $\mathfrak{su}(2)$ by conjugation, and it is a very quick exercise to check that if $A^\dagger A=I$ and $X^\dagger+X=0$ then $(AXA^{-1})^\dagger+(AXA^{-1})=0$ (keep in mind $(UV)^\dagger=V^\dagger U^\dagger$ and $A^\dagger A=I$ implies $A^{-1}=A^\dagger$). This is called the adjoint representation in Lie theory.

anon
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