# Does the notion of “rotation” depend on a choice of metric?

Consider the statement:

The Euclidean metric on $\mathbb{R}^n$ is rotationally invariant.

I interpret this to mean (is this interpretation correct?):

The Euclidean metric on $\mathbb{R}^n$ is invariant under the action of the orthogonal group $O(n)$.

However, the orthogonal group $O(n)$ is defined in terms of the Euclidean metric (as the group of all self-maps $\mathbb{R}^n \to \mathbb{R}^n$ which preserve Euclidean distance and fix the origin).

This suggests that we are implicitly using the following definition of "rotation":

Rotations are the set of all (orientation-preserving) isometries of $\mathbb{R}^n$ which fix the origin.

Question: Why is the first claim "the Euclidean metric on $\mathbb{R}^n$ is rotationally invariant" noteworthy/not trivial if we are implicitly using this definition/notion of rotation?

(I.e., of course the metric is preserved by a group of isometries.)

When we define "rotations", how are we not implicitly choosing a preferred metric on $\mathbb{R}^n$?
/Question

Clarifying example: In contrast,

The taxicab metric on $\mathbb{R}^n$ is not rotationally invariant.

In other words,

The taxicab metric on $\mathbb{R}^n$ is not invariant under the action of $O(n)$.

But what if we consider, instead of $O(n)$, what I will call $T(n)$ ("taxicab orthogonal group") of all self-maps $\mathbb{R}^n \to \mathbb{R}^n$ which preserve taxicab distance and fix the origin?

It seems fairly clear that we have:

The taxicab metric on $\mathbb{R}^n$ is invariant under $T(n)$.

or in other words

The taxicab metric on $\mathbb{R}^n$ is "taxicab-rotationally invariant".

Note: This is a very dumb question, so if you have any suggestions for how it could be improved, or if it should just be deleted, please say so (nicely).

• If you saw my previous comment, then you can ignore it, since I've misunderstood the premise of your question. – Wojowu Mar 28 '17 at 20:42
• To answer your question ("When we define "rotations""), to the best of my knowledge $O(n)$ is usually defined in terms of matrices, with no references to metric nor topology. – Wojowu Mar 28 '17 at 20:50
• Two comments: (1) $O(n)$ could be defined as the set of matrices such that $A^T=A^{-1}$, and then the definition is not circular. (2) For a given distance, one could define the set of transformations $T$ that preserve the distance. Then, the distance is invariant under the transformations in $T$, by the definition of $T$, but this is not circular because you don't need $T$ to define the distance function. In addition, one could, for example, choose a group of transformations and (if the group is chosen well), find a distance function that is invariant under the transformations. – Michael Burr Mar 28 '17 at 20:51
• One nontrivial point here is that the Euclidean metric has a lot of symmetries: that is, $O(n)$ (or perhaps better $SO(n)$, since the article says "rotationally invariant" rather than "rotationally and reflectionally invariant") is a Lie group of positive (and rather large) dimension. In contrast, your group $T(n)$ is finite, and some other norm might not have any non-identity symmetries at all. If you want to capture some physical notation of "rotation-invariance" which is clearly continuous in nature, the Euclidean metric is a lot better bet than the taxicab metric. – Micah Mar 28 '17 at 20:53
• I think it's fair to say that 'rotation' is not an entirely fixed technical term; while it has a roughly-canonical definition, it's not uncommon at all to see things a bit more general described as rotations. For instance, I've seen Lorentz boosts (non-rotational elements of the restricted Lorentz group) referred to as hyperbolic rotations or generalized rotations on various occasions. – Steven Stadnicki Mar 28 '17 at 21:00

Most mathematical concepts can be defined in many different ways. In (most) books, great care is taken to ensure there are no clashes among definitions, resulting in a streamlined and concise presentation. But yes, if you put a bunch of definitions and statements in the same bag, you will end up with circularities/trivialities.

Now, one definition of the rotation group $SO_n$ which I like is: $$\{A \in \mathbb{R}^{n\times n} \,|\, A^T A = I_n,\, \det A = 1\}.$$ It is easy to see that the euclidean metric is invariant under the action of $SO_n$.

So the statement 'the euclidean metric is rotationally invariant' is, in this case, a gentle reminder of one geometric property of $SO_n$.

Note. As noted by Michael, there is no circularity in the examples you provide, only redundancy and triviality.

• Can you explain the motivation for that definition? As far as I can tell, it seems concocted to encode "orientation preserving isometries which fix the origin", thus not really answering my question, just avoiding it. Also the answer doesn't address the question in the title. – Chill2Macht Mar 30 '17 at 9:15
• I answered the "question" that was asked in body. More generally, in any euclidean space, you can consider auto-adjoint transforms that preserve orientation (things depend on a choice of scalar product - angles and length - there). In a metric space, it does not make much sense to talk about rotations. – Olivier Mar 30 '17 at 12:17
• I think I answered the question in the body. Basically, there is no circularity in saying that the euclidean metric is rotationally invariant, since that's a statement and not a definition. At worst, its a trivial statement. There's not much more to say, even though I added some meat to the bone. – Olivier Mar 30 '17 at 12:24
• In the comment above, it is stated that "In a metric space, it does not make much sense to talk about rotations". However, it is stated in the question body how this, i.e. "talking about rotations in a metric space", might make sense. An explicit example is given with the taxicab metric. An answer to this question which is based on the implicit claim that "in a metric space, it does not make much sense to talk about rotations" should ideally justify that claim in order to be worthy of acceptance, especially when the original question already provided an explicit argument against the claim. – Chill2Macht Mar 30 '17 at 13:36
• You are right, it deserves more justification. If part of the question is "what is a rotation, generally?", that would deserve another post (its an interesting question). I'm not so sure what I can add while staying in the scope of the original question. – Olivier Mar 30 '17 at 13:42