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Here is a rather vague and subjective question: for which $n$ and $m$ are $\mathbb{R}^n$ and $\mathbb{R}^m$ ``essentially similar''? The answer depends partly on what type of mathematician is answering it.

Of course, in a certain technical sense, $\mathbb{R}^n$ and $\mathbb{R}^m$ are different for any $n\neq m$ since they are not homeomorphic (or linearly isomorphic); thus they by definition differ in some describable topological (or algebraic) way. But qualitatively all Euclidean spaces, especially ones of large dimension, seem to be ``similar.'' The question is really: how large does $n$ have to be for Euclidean spaces of dimension $\geq n$ all to behave essentially the same way?

I have included my own discussion of some possible answers in my Real Analysis manuscript at http://wolfweb.unr.edu/homepage/bruceb/Meas.pdf (Section XI.18.3). I invite comments on this discussion.

Bruce Blackadar
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    Although I don't have anything smart to say about this, I do see how the question could make sense: algebraic/differential topology behaves significantly differently in 1, 2, 3, 4, and then $\ge 5$ dimensions. As dimension goes up, spheres become (perhaps surprisingly) much smaller subsets of the circumscribing cube. Etc. – paul garrett Apr 22 '17 at 23:11
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    Couldn't this question be asked about the integers themselves? I.e., how large do integers have to be before they're basically the same? It's hard to imagine a more context dependent question. – Geoffrey Irving Apr 22 '17 at 23:12
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    There's of course the simple answer: as sets, groups, and $\mathbb{Q}$ or $\bar{\mathbb{Q}}$-vector spaces, they are isomorphic. – user44191 Apr 22 '17 at 23:13
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    One could look at this question to find theorems which are true in low dimensions but false in high dimensions. The highest dimension involved seems to be $65$. So $\mathbb R^{65}$ is similar to all higher dimensional spaces in the sense that each property mentioned in that post either holds for all of them or fails for all of them – Oscar Cunningham Apr 22 '17 at 23:44
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    It's not clear to me that $\mathbb{R}^{2n}$ is all that similar to $\mathbb{R}^{2n+1}$. For instance, consider the following property of a vector space $V$: ``There exists a projection $\pi: V \to V$ and an isomorphism $\psi: V \to V$ such that $V = \pi(V) \oplus \psi(\pi(V))$." This characterizes the $\mathbb{R}^n$ for $n$ even. – Danielle Ulrich Apr 23 '17 at 00:04
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    The PDF of the full book seems to be a rather large file. Would it be possible to replace your link with a link to an excerpt? – Yemon Choi Apr 23 '17 at 00:09
  • I took a (very brief) look at the PDF and the discussion within. I applaud your work in trying to come to grips with both the foundations of mathematics and some of the deeper questions beyond the foundations. But it seems to me that your question is meaningless. I am voting to close. – Sam Nead Apr 23 '17 at 07:12
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    @SamNead I am ambivalent at the moment about whether the question should be reopened, but "meaningless" seems too strong. Perhaps "ill-defined" might be a fairer criticism? (Having read some of the OP's work I am inclined to start from the position that he has serious intent and proven experience, to put it mildly.) – Yemon Choi Apr 23 '17 at 07:54
  • One of the infinitely many aspects in which high-dimensional products of the real numbers can be considered to be eventually similar is whether they are suitable for embedding a given class of objects into them (w.r.t. a given class of maps). There is a plethora of examples. Elementary example: w.r.t. smoothly embedding countable simple graphs, $\mathbb{R}^n$ is equivalent to $\mathbb{R}^m$ for any $m,n\geq 3$. Classical example: H. Whitney proved that for smoothly embedding a $d$-dimensional manifold, they are equivalent as soon $m,n\geq 2d$. – Peter Heinig Apr 23 '17 at 15:26
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    Of course $n$ has to be at least $42$, but everybody knows that. – Franz Lemmermeyer Apr 23 '17 at 15:35
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    This was indeed intended to be a serious question, and I don't think it is at all meaningless. It is ill-defined, and intentionally so; $\mathbb{R}^n$ is many things. I asked it to try to provoke the kind of discussion which has begun. As for the manuscript, it is not such a huge file (48MB), but it is long, although I have tried to make it easily searchable (I eventually hope to do better), and the relevant section can be easily found by a hyperlink from the table of contents. – Bruce Blackadar Apr 23 '17 at 15:39
  • The original question was "how large does $n$ have to be for Euclidean spaces of dimension $\geq n$ all to behave essentially the same way?" This question can't mean anything, because we aren't told what "the same way" means or what "essentially" means. What is the notion of sameness that is of interest here? – Sam Nead Apr 24 '17 at 09:46
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    To be clear, I think this question violates the principles listed here: https://mathoverflow.net/help/dont-ask -- the question is open ended, cannot have a "correct" answer, and invites discussion. Note that I do not think the poster is crazy, or the question is dumb, etc... It is just that the question doesn't have (and cannot have) a definite answer. – Sam Nead Apr 24 '17 at 09:48

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The qualitative intuition that you mention — that $\mathbb{R}^n$ begin to be similar in sufficiently high dimension — is the same intuition underlying my question Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimensions? from some time ago.

Specifically, I had asked whether every statement in the language of what I called real-matrix algebra eventually stabilizes in $\mathbb{R}^n$ for sufficiently high dimension. The language has sorts for scalars, matrices, row vectors and column vectors and operations for the natural additions and multiplications. The language does not allow one to make explicit reference to the dimension. I had thought that statements in this language might stabilize in high dimension.

But the answers showed that that was wrong, and that one can find statements whose truth values do not stabilize in high dimension, but rather detect various number-theoretic features of $n$.

Thus, those answers tend to refute the intuition that even for very simple statements, $\mathbb{R}^n$ begins to look alike in all sufficiently large dimension.

Joel David Hamkins
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    This is pretty definitive from a linear algebra standpoint, but the answer might be quite different from a topology or analysis viewpoint. – Bruce Blackadar Apr 23 '17 at 16:25
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There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even.

Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if and only if there exists an $(n+1)\times(n+1)$ Hadamard matrix.

Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.

Wlodek Kuperberg
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I remember there were some speculations by Yurii Manin that our arithmetic is impect because it distinguishes between huge numbers that differ by 1 which should for all practical purposes be the same. I am not sure I follow Manin here but anyway he did make such an argument, though for the moment I can't find it in MathSciNet. Perhaps this did not reach MSN, and perhaps in the context of physics this actually makes sense. At any rate this seems like an intriguing question, though in my view the answer is negative: you can have an infinite prime number $H$ which is unlike $H+1$.

Mikhail Katz
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