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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).

Chill2Macht
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    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
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    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
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    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
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    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
  • @Micah So maybe it would have to be a choice of metric for which the resulting group is path-continuous at least? Is the Euclidean metric the only such metric possible on $\mathbb{R}^n$? (I have no idea.) – Chill2Macht Mar 28 '17 at 20:55
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    Note that the distance function where $d(x,y)=1$ if $x\not=y$ is invariant under all of $GL(n)$. This is as connected as $O(n)$ is (since you can have positive or negative determinants). – Michael Burr Mar 28 '17 at 20:59
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    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
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    I think "the group of all self-maps which preserve distance and fix the origin" is a great way to generalize the orthogonal group to arbitrary metric spaces. Note that the orthogonal group can do more than rotating, it can also do reflection. Would there be a way to generalize the special orthogonal group (rotation without reflexion) to arbitrary metric spaces or are metric spaces too abtract to make a distinction between the two? If it's not possible with metric spaces, would it be possible with a less abstract space such as arbitrary normed vector spaces? – Paul Mar 29 '17 at 00:04
  • @RobArthan If it makes life easier, sticking with normable vector spaces would be fine. Ultimately though I would like to know which types of metric spaces, besides normable vector spaces, have a notion of length. (I think it is those for which the induced metric is finite, but I'm not sure which metric spaces this describes -- it might just be normable affine spaces for all I know.) – Chill2Macht Mar 29 '17 at 06:24
  • @Paul These are great questions. In particular, with the taxicab metric, it is unclear to me how one could single out certain taxicab isometries fixing the origin as "reflections" and others as "pure rotations". Seemingly the distinction between reflections and other members of the orthogonal group depends on the orientable (vector space) structure of $\mathbb{R}^n$ -- I am not sure if that initial perception is deceptive however. – Chill2Macht Mar 29 '17 at 06:26
  • @RobArthan a metric is a notion of distance -- distance is between two points. Length belongs to a set of points, properly a path in the metric space. If we are willing to consider paths with infinite length, then all metric spaces have a notion of length. However, if we want any two points to be joinable by a path with finite length (also known as a rectifiable curve), this limits the possible metric spaces considerably. The condition that any two points are joinable by a path with finite length is equivalent to the condition that the intrinsic metric of the metric space is actually a metric – Chill2Macht Mar 29 '17 at 23:42
  • @RobArthan i.e. is finite (does not take any infinite values) https://en.wikipedia.org/wiki/Intrinsic_metric In general, even when the intrinsic metric $d_1$ is finite, it does not necessarily agree with the original metric $d$. Spaces for which the two do agree are called length spaces. Note that one can define something called "length structures" on arbitrary topological spaces, but if such a structure exists, it turns the topological space into a length (metric) space (see: http://math.ucr.edu/~monnot/geometry) Thus it makes sense to say that these spaces are the most general on which – Chill2Macht Mar 29 '17 at 23:49
  • a notion of length exists. This reference also looks interesting http://www.math.psu.edu/petrunin/papers/alexandrov/bbi.pdf -- apparently it is more readable than a similar account by Gromov, who is supposed to be a smart/famous mathematician (I don't really know though firsthand). (As for restrictions, apparently length spaces have to be locally path connected, which seems like a pretty sharp restriction: http://www.math.ucsd.edu/~benchow/math257a/2-1-9.pdf) -- Also, to be fair, I thought distance and length were the same thing one-two months ago too. – Chill2Macht Mar 29 '17 at 23:49
  • @RobArthan The comments have nothing (directly) to do with the question, just with the claim that 'all "types of metric spaces" have a notion of length: that's what the metric is!' It might make sense to move the comments to chat instead of closing the question. – Chill2Macht Mar 30 '17 at 00:04

1 Answers1

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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.

Chill2Macht
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Olivier
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  • 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 09:15
  • 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
  • Obviously it's not literally circular. I meant figuratively, in the sense that it is a trivial statement. Your comments above seem to agree that it is a trivial statement. Also, this answer does not address the entirety of the question as stated in the body, namely this answer does not address the second part of the question. 'When we define "rotations", how are we not implicitly choosing a preferred metric on $\mathbb{R}^n$?' – Chill2Macht Mar 30 '17 at 13:33
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    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
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    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
  • You are right that I was not clear originally about what I was asking, thus making my criticism of your answer unfair. I have edited my post and removed my downvote. I apologize. – Chill2Macht Mar 30 '17 at 13:48