Why is the group of sphere rotations isomorphic to $SO(3)$? If we consider the sphere in $\mathbb{R}^3$, the set of all possible rotations of it forms a group isomorphic to $SO(3)$. That is, rotations of the sphere are exactly the orthogonal linear transformations from $\mathbb{R}^3$ to $\mathbb{R}^3$ with determinant $1$.
I understand why rotations of the sphere are linear transformations, and I also understand that since they all preserve length they must be orthogonal. What I don't understand is why are they necessarily of determinant $1$?
My intuition is that if we allow for negative determinant that would correspond to some reflection of the sphere, which would reverse its 'orientation' and that's impossible for rotation. I don't know how to make this intuition precise and follow up on it though.
Is there a way to precisely define the 'orientation' of a space and 'orientation-preserving transformations' and then conclude such transformations have non-negative determinants? If not, is there some other geometric notion that would explain why sphere rotations can't have determinant $-1$?
 A: Here's a good reason why they all have determinant 1. 
Consider a rotation about $v$ by some angle $s$, $Rot(v, s)$; you can connect this rotation to the "identity" rotation by 
$$
c(t) = Rot(v, s(1-t))
$$
where $t$ goes from $0$ to $1$. 
This gives a continuous path in the space of rotations. 
Thus
$$
det \circ c : [0, 1] \to \Bbb R: t \mapsto \det(c(t))
$$
is also continuous. And since the determinant must be $\pm 1$, and the set $\{+1, -1\}$ is disconnected, this function must be a constant function. (Any continuous function to a discrete space is constant). Hence, since the determinant of the identity is $+1$, so is the determinant of any rotation. 
I know this isn't exactly what you asked, but the answer to your question about "Is there a precise way to define orientation" is "Sure, and it's done in many multivariable calculus and linear algebra books, and other places as well...and usually involves determinants." So repeating here what all those books say seems silly. [One such book, although I don't actually recommend it for this topic: Spivak's "Calculus on Manifolds".]
Hidden in the argument above is the claim that every non-identity rotation is actually rotation about some vector by some (nonzero) angle; that requires proof, but it's proved elsewhere as well, so I'm just using it as a lemma here. [See for instance the section on "Log map" in the Wikipedia entry on axis-angle representations.]
A: If $V$ is an $n$-dimensional vector space, then as stated here an orientation of $V$ is a choice of connected component of $\bigwedge^n(V) - \{ 0 \}$ (we have two connected components to choose from because the top exterior power is just a line). 
But we can also formulate the determinant of a linear transformation $A$ as follows. Consider the linear map $\bigwedge^n A : \bigwedge^n V \rightarrow \bigwedge^n V$ given by $v_1 \wedge \cdots \wedge v_n \mapsto Av_1 \wedge \cdots \wedge Av_n$. Then we have
$$ \left( \bigwedge^n A \right) (v_1 \wedge \cdots \wedge v_n) = \det(A) \cdot (v_1 \wedge \cdots \wedge v_n).$$
For the statement of this, see here. You can convince yourself of this easily in the $n=2$ case, by just applying an arbitrary $2 \times 2$ matrix to the wedge $v_1 \wedge v_2$ and noting that the determinant drops out. Note here that I'm using $\{ v_1, \dots, v_n \}$ as a basis of $V$. So if we want our rotation, say $A$, to preserve orientation, we must have $\det(A)=1$ above.
A: Orientation: count how many vectors point in the opposite direction
We can define orientation in $\mathbb{R}^3$ in terms of coordinate axes that become flipped. If a linear transformation flips an odd number of axes, then it inverts the orientation of $\mathbb{R}^3$. If a linear transformation flips an even number of axes, then it preserves the orientation of $\mathbb{R}^3$. 
To see that this definition agrees with our intuitive idea of orientation, see the following transformations involving 0, 1, 2, or 3 coordinate axes being flipped:
$$U_0 :\langle x, y, z\rangle \mapsto\langle x,y,z\rangle \qquad \text{preserved}$$
$$U_1:\langle x, y, z\rangle \mapsto\langle x,y,-z\rangle \qquad \text{inverted}$$
$$U_2:\langle x, y, z\rangle \mapsto\langle -x,-y,z\rangle \qquad \text{preserved}$$
$$U_3:\langle x, y, z\rangle \mapsto\langle -x,-y,-z\rangle \qquad \text{inverted}$$
You can also check using the right-hand rule. (Also note: the orientation-preserving transformations are actually rotations!)

Negative eigenvectors point in the opposite direction
Eigenvectors give us a precise way to talk about flipped axes. Eigenvectors with negative eigenvalues get inverted by the transformation, pointing in the opposite direction: ($Ux = -\lambda x$). Eigenvectors with positive eigenvalues get preserved, pointing in the same direction.  To decide whether a linear transfromation inverts orientation or not, we can count whether it has an odd or even number of negative eigenvalues.
Actually, note that the determinant of a linear transformation is equal to the product of all its eigenvalues (counting multiplicities). So, a linear transformation on $\mathbb{R}^3$ inverts orientation if and only if it has an odd number of negative eigenvalues if and only if its determinant is negative.
Conveniently, orthogonal matrices have eigenvalues $\pm 1$ only, which means that they have determinant $\pm 1$ only. Hence they invert orientation if and only if their determinant is -1.
We can prove that orthogonal transformations preserve lengths and that they consequently have eigenvalues of $\pm 1$ only.   The proof is that:
$$||Ux||^2 = (Ux)^T (Ux) = x^T U^T U x = x^T x = ||x||^2\qquad \text{($U$ preserves length)}$$
$$||x||^2 = ||Ux||^2 = ||\lambda x||^2 = |\lambda|^2 ||x||^2 \Longrightarrow |\lambda|^2 = 1\qquad \text{($U$ has eigenvalues $\pm 1$ only)}$$ 

Alternative answer:


*

*A linear transformation is orthogonal if and only if it preserves the length of all vectors
$$||x||^2 = ||Ax||^2  \iff x^Tx = (Ax)^T(Ax) = x^TA^TAx \iff A^TA = I$$

*Geometrically, a length-preserving transformation is a rotation and/or a reflection. As we'll show, the pure rotations have determinant +1, and the pure reflections have determinant -1.

*We can get a handle on "orientation" by looking at eigenvalues. Basically, a unitary transformation can only have $\pm 1$ as an eigenvalue. Eigenvectors with value -1 get "flipped/reflected" by the transformation, since $Ux = -x$. The number of flipped axes determine whether orientation has been preserved or not. 
Specifically, flipping an odd number of axes changes the orientation, while flipping an even number of axes preserves it. (This is an appeal to geometric intuition in $\mathbb{R}^3$.)

*The determinant is equal to the product of all eigenvalues, multiplicities included. Therefore, the following circumstances are all equivalent:


*

*the transformation has determinant -1

*the transformation has -1 as an eigenvalue with odd multiplicity

*the transformation has -1 as an eigenvalue with multiplicity one or three 

*the transformation inverts one or three axes in $\mathbb{R}^3$.

*the transformation inverts the orientation of $\mathbb{R}^3.


*For rotations, which preserve orientation, the equivalent properties are:


*

*the transformation has determinant +1

*the transformation has -1 as an eigenvalue with even multiplicity

*the transformation has -1 as an eigenvalue with multiplicity zero or two 

*the transformation inverts zero or two axes in $\mathbb{R}^3$.

*the transformation preserves the orientation of $\mathbb{R}^3.



P.S. Proof that unitary transformations can only have +1 or -1 as an eigenvalue: If $U$ preserves length, then $||Ux||^2  = ||x||^2$ for all vectors $x$. If $x$ is moreover an eigenvalue with eigenvalue $\lambda$, we conclude that $||\lambda x||^2 = ||x||^2$. But $||\lambda x||^2 = |\lambda|^2 \; ||x||^2$, so $|\lambda|^2 = 1$, $\lambda  = \pm 1$.
A: Any rotation has a "square root" which is also a rotation: A rotation about an axis by an angle $\theta$ is the same as the composition of two rotations about the same axis by an angle $\frac\theta2$.
So if the rotation has matrix $A$, we can write $A = B^2$, where $B$ is another rotation matrix, and therefore $\det(A) = \det(B^2) = \det(B)^2 \ge0$.
