Lately I've been looking at the special orthogonal group $SO_\mathbb{R}(3)=\{M\in O(3),\, \det M =1\}$, where $O_{\mathbb{R}}(3)$ is the group of real matrices such that $AA^\top=I$. I'm looking to get a intuitive understanding of a "simple fact".
To my understanding its elements can be seen as rotations around a line through the origin (ie a subspace of dimension 1). $ SO_\mathbb{R}(3)$ forms a group with respect to matrix multiplication, so the product of two elements is still an element of the group. That is to say, if one composes two rotations, around two possibly different lines (through the origin), the resulting automorphism of $\mathbb{R^3}$ is another rotation around some line through 0. Why should that be? Algebraically it checks out, but intuitively I can't see why it should be the case that combining rotations around different axes would have to be another rotation. Why is there always one line left unmoved, fixed by the composition of rotations?
Anyway, if anyone has a good mental image or intuitive perspective they'd like to share, I'd appreciate.
Best Answer
It feels intuitive to me that any orientation preserving isometry of the sphere has at least a fixed point. I mean think of pulling one point of the sphere to another in a (more or less) straight line. The points orthogonal to the plane spanned by the path will stay fixed.
In case you know something about algebraic topology (ie. $H_2(\Bbb S^2)=\Bbb Z$) we can give a lovely proof of this fact.
First note that the composition of two rotations induces an isometry $\Bbb S^2 \rightarrow \Bbb S^2$ and that it suffices to show that this map has a point satisfying $f(x_0)=\pm x_0$. Being an isometry we will then have $f(-x_0)=\mp x_0$, as well so the span of ${x_0}$ is an axis of rotation. Technically there could be an inflection involved as well, but with some work (an inflection would change the orientation given by the fundamental class $[\Bbb S^2] := 1 \in H_2(\Bbb S^2)$ to $(-1)^3=-1=-[\Bbb S^2]$, but rotations preserve it and so do two rotations) we can indeed show $f(x_0)=x_0$.
Now we are left to show that any map $f:\Bbb S^2 \rightarrow \Bbb S^2$ has a point $f(x_0)=\pm x_0$. Assuming otherwise we can state explicit homotopies $\operatorname{id} \sim f \sim -\operatorname{id}$. But then the action of $\operatorname{id},-\operatorname{id}$ on the fundamental class would result in $1=(-1)^3=-1$, a contradiction.
I feel like I will have to defend this answer, so let me say this: What I like of this proof is that it is really conceptual, if one believes in homology, which is basically just a framework allowing you to speak about orientation (among other things). In my opinion this proof better mimics the intuition than algebraic manipulation of matrices does.