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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.

G Tony Jacobs
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    Rotation by $2\pi$ is just the identity element. – Ian Aug 19 '17 at 16:00
  • Hmm... I guess if it weren't, then the symmetry group of any polygon would contain $\mathbb{Z}$... – G Tony Jacobs Aug 19 '17 at 16:07
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    Nothing finite (topologically, compact) can have $\mathbb Z$ as an isometry symmetry group, I believe. – Thomas Andrews Aug 19 '17 at 17:02
  • I'm pretty sure I was confusing the isometry symmetry group with something else, something involving homotopy. – G Tony Jacobs Aug 19 '17 at 17:03
  • Basically, the set of all isometries (maps of $n$-dimensional geometry that preserve distances) is a group, and an object in that geometry has symmetries the isometries that send the object to itself. The $2\pi$ rotation is just the identity isometry. – Thomas Andrews Aug 19 '17 at 17:07

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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 $2$.

Geoffrey Sangston
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  • I think this is the answer I'm looking for, but I'm a little lost in the second paragraph. Are you saying that $\mathbb{Z}$ is the symmetry group of something in this situation? Also, how is thinking of $4$ as equivalent to $0$ modulo $2$ a mistake? Do you say "the integer $4$" to highlight that you're talking about $4\in\mathbb{Z}$, not in $\mathbb{Z}/2\mathbb{Z}$? – G Tony Jacobs Aug 19 '17 at 16:20
  • $\mathbb{Z}$ is not the symmetry group of "R" if we consider symmetries to mean isometries of the plane taking "R" to itself. I was trying to point out the mistake you might be making and how that could lead to a valid thought. That mistake, I think, is that you're probably "remembering" how many times "R" has been rotated and then considering the set of these rotated "R" letters as distinct. This gives you a group isomorphic to $\mathbb{Z}$.

    And right I'm highlighting that the integer $4$ is different from the "integer $4$ modulo $2$.

     – Geoffrey Sangston Aug 19 '17 at 16:24
  • Yes, I'm thinking of "remembering" the number of rotations. I guess I'm thinking of concepts such as winding number, and fundamental group definitions, where going all the way around is not the same as just standing still. I'm just a little unclear on what, exactly, $\mathbb{Z}$ is the symmeties of here. It's not the "R"; fine. Is it the "R", plus memory of its rotations? – G Tony Jacobs Aug 19 '17 at 16:28
  • So we can talk about the pairs of the letter "R" and an integer which represents the times it's been rotated counterclockwise a full turn. Each time you rotate a member of this set another time you add one to its memory (or subtract if you go backwards). This forms a set which has the operation ("R", n) * ("R", m) = ("R", m + n) and this set is obviously isomorphic to $\mathbb{Z}$. I may have confused you by writing "symmetry group of $\mathbb{Z}$" – Geoffrey Sangston Aug 19 '17 at 16:54
  • I think that, by now, we've basically described the homotopy group of the circle, which is pretty much what I was thinking of. The "R" itself is only relevant because it keeps us from modding out by any smaller rotation. – G Tony Jacobs Aug 19 '17 at 17:02
  • Yea I think you've got it now. I made a similar mistake once. It's all about being exactly clear about what a symmetry is. Eventually you find there may be multiple valid views of what the morphisms between your collection of objects of interest might be and so it may be difficult to define what an object is without essentially stating what the symmetries are. In this letter case the question asker may be asking an ambiguous question if they don't specify that they're talking about isometries (maybe I live in a weird space where my concept of R can be stretched some weird way into itself!) – Geoffrey Sangston Aug 19 '17 at 17:06
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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 $2\pi$.

Carmeister
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  • +1 because your answer is entirely accurate, but this isn't really what I was trying to ask. I'll allow that my question wasn't as well-phrased as I might have hoped. My problem isn't that I don't understand what an identity function is, but rather that I know a little bit of homotopy theory.... just enough to get myself in trouble, but not enough to get out! – G Tony Jacobs Aug 19 '17 at 16:59
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+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 $\mathbb{Z}$.

Elle Najt
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  • This is it, and this is where we ended up with the other answer, by @Geoffrey Sangston, although we haven't called it by its name. As I mentioned in another comment, this question arose because I know just enough about homotopy to get myself in trouble! Thank you. – G Tony Jacobs Aug 19 '17 at 17:01
  • @GTonyJacobs Okay I'm glad you figured it out :) – Elle Najt Aug 19 '17 at 17:09