This is regarding Artin's proof of the following statement in his Algebra:
Let $M$ be the matrix in $SO_3$ that represents the rotation $\rho_{(u,\alpha)}$. Let $B$ be another element of $SO_3$, and let $u' = Bu$. Then $M' = BMB^T$ represents the rotation $\rho_{(u',\alpha)}$.
Here $\rho_{(v,\eta)}$ represents the linear transformation of $\Bbb R^3$ that fixes $v$ and rotates $v^{\perp}$ by the angle $\eta$.
The exact sentence I am (somewhat) confused about appears just at the top of p. 138 in the second edition of the book: the author, by that point, proves that the pole of $M'$ is $u'$, and sets $\alpha'$ to be the angle around which $M'$ rotates $u'^{\perp}$. Now we must show that $\alpha = \alpha'$ for the proof to be completed, and here is where I'm confused:
By Euler's theorem, $B$ also represents a rotation, say with angle $\beta$ about some pole. Since $B$ and $M'$ depend continuously on $\beta$, only one of the two values $\pm\alpha$ for $\alpha'$ can occur. On setting $\beta = 0$, we get that $\alpha' = \alpha$, which hence must be the case for all $\beta$.
My understanding of this is that $\alpha'$ is constrained to be either one of $\pm\alpha$, so as we change $\beta$ around, it can't suddenly change from one to the other.
Is my reasoning correct?
Am I correct in assuming that this proof is properly rigorous, or are there some implicit "details" left to the reader?
(Please retag as appropriate.)
Artin in saying that if we write (using his notation) $M' = BMB^t = \rho_{(u', \gamma)}$, then $\gamma = \gamma(M, B)$ is a continuous function of $M$ and $B$ (in the usual topology on $SO(3) \subset GL_3(\mathbb{R})$). He presumably shows that $\gamma = \pm \alpha$ at some part of the proof. Since $\gamma$ is continuous, it follows (ignoring the trivial case $\alpha = 0$) that either $\gamma\equiv \alpha$ everywhere or $\gamma\equiv -\alpha$ everywhere. (The image under a continuous function of a connected set is still connected.) For $B = 0$, we have $M' = M$ and thus $\gamma(M, B) = \alpha$, forcing $\gamma\equiv \alpha$ everywhere.
The proof is rigorous, although it seems a bit clumsier than it needs to be. I don't have a copy of Artin's book with me right now and so haven't seen the full proof, but it seems from the description here that it leaves an amount to the reader that's appropriate for its audience.