No symmetries, or $\mathbb{Z}$?

Question

I was thinking about this question from the other day, and some good answers were provided for geometric objects with symmetry group $\mathbb{Z}$. It just occurred to me, though: why isn't $\mathbb{Z}$ the symmetry group of, say, a capital letter "R"?

I usually think of that object as having no symmetries, however, a rotation through $2\pi$ certainly works. Taking that as a generator, we get the group $\mathbb{Z}$. Is there anything wrong with this reasoning? It doesn't seem that the object is entirely relevant in this case, so I wonder if I'm not actually talking about a symmetry of the ambient space, somehow. I'm not really sure how to think about this.

Thanks in advance for any insights.

Answer 1

+1 good question. This is a point of vagueness in a lot of heuristic descriptions of groups and symmetries. I think one confusing point is when you talk of rotation - do you mean it dynammicaly (here I'm rotating the square around, taking some time up to do so), or all at once (via the function that does the rotation).

The difference between the two is the difference between, one the one hand, a path in $SO_2(\mathbb{R})$ starting at the identity matrix and ending at a rotation matrix $M$, and, on the other hand, that rotation matrix $M$.

It's true that you can find paths of transformations in $SO(2)$ that bring the letter $R$ back to itself. However, these paths start and stop at the same point, namely the identity matrix.

When we talk about symmetries of subset $S \subset \mathbb{R}^2$, we generally interpret it the second way - as the subgroup $GL_2(\mathbb{R})$ (or the affine group if you are allowed translations) consisting of the matrices that fix the set $S$ and preserve all distances. (The choice between linear and affine symmetries also reminds me to tell you that you generally need to specify the symmetries of your object as belonging to some subgroup of the group of bijections, depending on the properties you want those symmetries to preserve. When people talk about symmetries in Euclidean space, they typically have in mind isometries.)

(Note that here there are two possible choices - do we want matrices $M$ so that as sets $MS = S$ or do we want to require that for all $s \in S$, $Ms = s$. When we talk about the symmetries of physical shape in Euclidean space we generally mean the former. The latter is much more restrictive. However, there are situations when we mean the latter (pointwise) option. For example, automorphisms of a field extension of $E$ over $F$ are taken to mean (among other requirements) that they fix $F$ pointwise.)

On the other hand, $SO(2)$ is a circle, and the path you describe (besides being set theoretically non-trivial), is also homotpically non-trivial, and indeed generates the fundamental group of $SO(2)$, which is $\mathbb{Z}$.

Answer 2

If you're willing to keep track of the number of times you've rotated "R" counterclockwise then this set will naturally form a group isomorphic to $\mathbb{Z}$. Usually you're thinking of the letter "R" as a subset of $\mathbb{R}^2$ and don't carry this extra information. You're really looking for the set of isometries of the plane which take "R" to itself and of course there is only one.

So what you're doing is considering the symmetry group of a space which maps onto the R. This is basically the same kind of mistake as thinking the integer $4$ is the same thing as the element $0$ in the integers mod $2$.

Answer 3

When you rotate by $2\pi$, that's just the identity operation. That's because, when we think of it as a function, it sends each point to itself.

What does it mean, then, for an operation to be different than the identity? Consider instead rotating the letter H by $\pi$. Now each point gets sent to a different point - for example, the bottom left point gets sent to the top right. So even though rotation by $\pi$ fixes the overall shape of the H (this is what it means to be a symmetry), the image of each point under rotation by $\pi$ is different than its image under rotation by $2\pi$.