Classification of the orientation preserving isometries of $\mathbb{E}^4$.

Question

I can't find a classification of the orientation preserving isometries of $\mathbb{E}^4$. I found this e somewhere on the internet but of course it also raises the question of there being a general classification of the isometries of $\mathbb{E}^4$. Does anyone know if this exists and willing to prove this as well? Thanks!

Answer 1

Any isometry of $\mathbb{R}^n$ is of the form $$ f(x) = Ax + b $$ where $A$ is an orthogonal matrix, and $b$ is a fixed vector. Orientation-preserving isometries would require that $A$ is special orthogonal (ie. $det(A) = +1$).

However, I don't know if there is a well-defined classification of $SO_4(\mathbb{R})$ like there is for $SO_3$ or $SO_2$

Answer 2

I assume by $\mathbb{E}^4$, you mean Euclidean space of dimension $4$. The orientation-preserving isometries of $\mathbb{E}^n$ are a semidirect product of $SO(n)$ and $\mathbb{R}^n$. Geometrically, this is because any isometry can be uniquely decomposed as a product of a rotation and a translation.

To classify the isometries, first observe that translations and rotations are orientation-preserving isometries. Now consider a general orientation-preserving isometry $f:\mathbb{E}^n\to\mathbb{E}^n$. Choose an arbitrary point $p\in\mathbb{E}^n$ and choose geodesic coordinates so that $p=0$.

We know from Riemannian geometry that $f$ is completely determined by $f(0)$ and $df_0$. Since $f$ is an orientation-preserving isometry, $df_0\in SO(n)$, so $$ f(x) = df_0x + f(0).$$ Thus $f$ can be written as a product of rotation and translation.