By the Euler's theorem, the set $SO_3$ of all 3x3 orthogonal matrices $M$ such that $det(M)=1$, which is called the special orthogonal group of 3x3 matrices, is the set of all 3x3 rotation matrices. The rotation $\rho$ of $R^3$ can also be represented by spin $(\mu, \theta)$ where $\rho$ fixes a unit vector $\mu$(which is called a pole of $\rho$ and rotates the two dimensional subspace $W$ orthogonal to $\mu$ by $\theta$
I come to think that if $M\in SO_3$ represents the spin $(\mu, \alpha)$, then $M^t$ represents the spin $(\mu, -\alpha)$. Is this correct?
My proposition is originated from the observation that $M$ is similar to
$R=\begin{bmatrix} 1 & & \\ & \cos\alpha & -\sin \alpha \\ & \sin\alpha & \cos \alpha \\ \end{bmatrix}$
with some $B \in SO_3$, that is $M=BRB^t$.
$M^t = BR^tB^t$, and
$R^t=\begin{bmatrix} 1 & & \\ & \cos\alpha & \sin \alpha \\ & -\sin\alpha & \cos \alpha \\ \end{bmatrix} =\begin{bmatrix} 1 & & \\ & \cos\alpha & -\sin (-\alpha) \\ & \sin(-\alpha) & \cos \alpha \\ \end{bmatrix}$.
Here $B$ can be interpreted as an ordered set of orthonormal vectors $(\mu, w_1, w_2)$ where $\mu$ is the pole of this rotation and $\{w_1, w_2\}$ is an ordered basis of the two dimensional subspace of $R^3$ orthogonal to $\mu$.
I want to verify that my proposition, together with the proof I suggested, is correct.
Thank you for reading.
I believe you're right (both your end result and your derivation), but I don't know if the similarity is the most direct way to "see" (or convince a skeptic) that $M^t$ is spin $(\mu, -\theta)$.
The most relevant fact here, it seems to me, is that if $M$ is orthogonal then $M^t = M^{-1}$ (this is actually another way to define or characterize the orthogonal matrices), and $\det(M) = 1$ implies $M(\mu) = \mu$ (where $\mu$ represents the pole vector). That means $$M^t(\mu) = M^t(M(\mu)) = I_3(\mu) = \mu,$$ so $\mu$ is also the pole of $M^t$.
Once you have that $M^t$ also fixes $\mu$, we know because $M^t = M^{-1}$ that it must "reverse" the action of $M$ on vectors in $\Bbb{R}^3$. Specifically, vectors normal to $\mu$ in the plane $W := \operatorname{span}\{ \mu \}^\perp$ get rotated by the angle $\theta$ in $W$ by $M$. So these vectors must be rotated in an equal and opposite direction in $W$ by $M^{-1} = M^t$, which implies $M^t$ = spin $(\mu, -\theta)$.