6
$\begingroup$

I apologise if this is a rather basic question- I am relatively new to group theory and geometry/topology etc. I was reading about the Euclidean Group, which Wikipedia defines as being...

the symmetry group of n-dimensional Euclidean space. Its elements, the isometries associated with the Euclidean metric, are called Euclidean motions.

From what I am aware, isometries include translations, rotations and reflections. However later on in the Wiki article it talks about dimensionality and degrees of freedom:

Dimensionality

The number of degrees of freedom for E(n) is n(n + 1)/2, which gives 3 in case n = 2, and 6 for n = 3. Of these, n can be attributed to available translational symmetry, and the remaining n(n − 1)/2 to rotational symmetry.

I was confused as to why reflections did not factor in here, although intuitively I didn't see how a reflection could be a degree of freedom either. There just seems to be something different about reflections compared with translations and rotations, although I'm not sure what (not very mathematical, I know)

After reading about Classification of Euclidean plane isometries on Stack exchange, one of the answers described how you could describe any isometry by a translation, rotation and reflection by considering three points: use a translation to map one point onto its image, a rotation to map the second point to its image, and finally a reflection to match the third. From this I can see how once the translation and rotation are used to match two points, the reflection used for the third is entirely determined by the isomatry- there is no more freedom in choosing the reflection, if you have used a translation and rotation to specify the mapping of two of the points. For an isometry, the final step is fixed.

This kind of makes sense to me, although I am not sure if it is the best way, or even a correct way, of thinking about this.

Also, the answer also stated that

A non-zero translation and a non-zero rotation together are simply a rotation around some other center. A rotation followed by a reflection is a reflection in some different line. (MvG)

So it seems that all combinations of translations and rotations can be described by a single reflection? So reflections 'hide' all of the degrees of freedom? I think this is related to the final point, but I admit I do not quite understand it. Perhaps some of my confusion regarding transformations and degrees of freedom is that I find myself veering towards thinking of the Euclidean group as consisting of transformations as opposed to isometries, and it seems that these are not the same (in this link, it asks about isometries being a subgroup of the reflection group)

$\endgroup$
  • $\begingroup$ If $M$ is a manifold and $X$ is a finite set then $M\times X$ has the same dimension as $M$. For instance, two disjoint spheres have the same dimension as just one sphere. $\endgroup$ – arctic tern Jun 28 '17 at 12:38
5
$\begingroup$

The group of Euclidean motions contains orientation-preserving and orientation-reversing motions. When calculating the "number of degrees of freedom" a.k.a. dimension of this group, one can ignore the orientation-reversing ones because they form another connected component of the same dimension. Therefore the dimension of the group of Euclidean motions of $\mathbb R^n$ is the sum of the dimension of the translations namely $n$, plus the dimension of the special orthogonal group $SO(n)$. For example, for $n=3$ you get $3+3=6$ degrees of freedom.

$\endgroup$
5
$\begingroup$

The phrase "degrees of freedom" refers to the number of independent real number valued parameters (as does the more formal term "dimension").

So, for example, each orientation preserving isometry of $n$-dimensional Euclidean space $E^n$ is a composition $T \circ O$ of a unique orthogonal transformation $O$ followed by a unique translation $T$, and these two factors are independent of each other. Translations are parameterized by vectors which have $n$ degrees of freedom, and rotations of $E^n$ are parameterized by orthogonal $n \times n$ matrices which have $\frac{(n-1)n}{2}$ degrees of freedom. Since the degrees of freedom of translations and of rotations are independent of each other, together they have $n + \frac{(n-1)n}{2}=\frac{n(n+1)}{2}$ degrees of freedom.

Now let's bring in reflections. Let me use $R_n$ to denote the reflection in the coordinate plane $x_n=0$. A general orientation reversing isometry can be expressed uniquely in the form $T \circ O \circ R_n$ where $T,O$ are independently chosen translation and orthogonal transformation. It follows that a general isometry can be written uniquely in the form $T \circ O \circ (R_n)^e$ where we independently choose three things: the translation $T$ with $n$ degrees of freedom; the orthogonal rotation $O$ with $\frac{(n-1)n}{2}$ degrees of freedom; and the exponent $e \in \{0,1\}$.

The choice of the exponent $e$ is discrete --- either $0$ or $1$ --- and this does not represent a "degree of freedom" because it is not a real number valued parameter.

By the way, what makes the calculations in my answer work correctly is the fact that the decomposition $T \circ O \circ (R_n)^e$ is uniquely determined by the isometry. In your question, where you use the fact that every isometry can be written as a composition of reflections, that composition is not unique, and so using it to count degrees of freedom can lead to errors.

Just as an example, picking any even number $n \ge 2$, every translation of the line can be written as a product of $n$ reflections. Since a reflection of the line has $1$ degree of freedom (namely, the reflection point), this would seem to be a proof that the translations of the line have dimension $n$. Since $n$ is arbitrary, there is an error in this argument... which I will leave you to ponder.

$\endgroup$
4
$\begingroup$

The isometry group of an Euclidean space $E$ of dimension $n$ is the semidirect product $$ Isom(E)=T(n)\rtimes O(n), $$ where $T(n)$ is the abelian subgroup of translations, and $O(n)$ is the orthogonal group, which is indeed generated by reflections, by a Theorem of Cartan. It is a linear Lie group of dimension $\dim T(n)+\dim O(n)=n+(n-1)/2=n(n+1)/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.