Orientation-preserving isometry of $R^n$ I am preparing for an exam, and would like to have a rigorous definition of the following:
Orientation-preserving isometry of $R^n$
I know that it is something like the following (feel free to correct my wording): 
When the homomorphism $\pi:M_n \rightarrow O_n$ is applied to the unique representation $t_a\phi$ of an isometry $f$, and $\pi(f)=\phi$, define $\sigma:M_n \rightarrow \pm 1$. This map that sends an isometry of $R^n$ to $1$ is orientation-preserving.
 A: The thing that is orientation-preserving is not the map that sends the isometry to $1$; rather it is the isometry itself, not that map, that is orientation-preserving.
An isometry is a function $f:\mathbb R^n\to\mathbb R^n$ that preserves distances, i.e. for any two points $x,y\in\mathbb R^n$, the distance from $x$ to $y$ is the same as the distance from $f(x)$ to $f(y)$.
To say that $f$ is orientation-preserving means that it won't map a left shoe to a right shoe or a left hand to a right hand, etc.  In some contexts, that is demonstrably equivalent to saying that the determinant of a certain matrix is $1$ rather than $-1$.
A: A common classification of isometries of $\mathbb{R}^n$ says that a map $f:\mathbb{R}^n\to \mathbb{R}^n$ is an isometry if and only if $f(x)=Ax+b$ for some orthogonal matrix $A\in O(n)$ and $b\in\mathbb{R}^n$.
If $\det(A)=1$, then $f$ is orientation-preserving. If $\det(A)=-1$, then $A$ is orientation-reversing. But $\{A\in O(n)\mid \det(A)=1\}$ is precisely the special orthogonal group $SO(n)$. So you could give the following definition:
A map $f:\mathbb{R}^n\to \mathbb{R}^n$ is an orientation-preserving isometry if and only if $f(x)=Ax+b$ for some $A\in SO(n)$ and $b\in\mathbb{R}^n$.
