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Orientation-preserving isometry of R^n

kalish

Member
Oct 7, 2013
99
Orientation-preserving isometry

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**.
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I'm a little baffled by what you mean, here. What is $M_n$?

As I understand it, any isometry is composed of a "translational part", and a "linear part".

Translation does not affect orientation, and is irrelevant, here. Because isometries preserve distances, it is clear to see the linear part is an orthogonal transformation. Thus the linear part has determinant $\pm 1$. The isometries whose linear parts have determinant 1 ("proper rotations") are the orientation-preserving isometries.

Now, it may be what you are saying is the same as what I am if:

$\sigma = \det \circ \pi$, where $\pi$ is the canonical (group) surjection of the quotient of the isometry group by the normal translational subgroup.

Another way to say this is: an isometry is orentation-preserving if it can be written as a composition of a translation and an element of the special orthogonal group.

Because $\sigma$ (as I have defined it) is a composition of group homomorphisms, it is itself a group homomorphism, and the orientation-preserving isometries can be realized as its kernel.