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My professor said (hesitantly)

$\textit{If it works in 1-D, it is likely that it also works in N-D}$

Hesitantly because, she remarked, it is not true for everything (which is expected).

After some research, I found one such example:

In 1-D and 2-D, if you start at the origin and continually do random unit steps in each of the cardinal directions (north, south, east, west, etc.), you will return to the origin with probability one. This fails for dimensions three and higher, in fact, you return to origin with probability zero in those cases.

What are some other striking examples of this?

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    $\begingroup$ A simple but important property is: $\mathbb{R}$ is a totally ordered vector space that is Archimedean, $\mathbb{R}^2$ is not. $\endgroup$ Apr 30, 2015 at 19:52
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    $\begingroup$ The statement «the dimension of the space is at most 37» is true only in low dimensions. $\endgroup$ Apr 30, 2015 at 20:34
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    $\begingroup$ How can the probability of returning to the origin be zero in dimension 3? Surely the probability that the first two steps of your walk are $(0,0,0) \to (1,0,0)$ and $(1,0,0) \to (0,0,0)$ is $(1/6)^2$? Maybe you mean the probability of returning to the origin infinitely often or something... $\endgroup$
    – Mike F
    Apr 30, 2015 at 20:34
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    $\begingroup$ Previously: Examples of results failing in higher dimensions and on MathOverflow: Results true in a dimension and false for higher dimensions $\endgroup$
    – user856
    Apr 30, 2015 at 20:58
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    $\begingroup$ The probability of returning to the origin isn't zero for dimensions three and higher, but it is true that it is less than one. $\endgroup$
    – Brian Tung
    Apr 30, 2015 at 21:14

6 Answers 6

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A quick example that came to mind: consider $\Bbb R^n$ with the euclidean metric and $\Omega \subset \Bbb R^n$ a non-empty open and connected set. Take $p \in \Omega$. Then $\Omega \setminus \{p\}$ is also connected..... except when $n = 1 $.

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Expanding my comment:

The vector space $\mathbb{R}$ can be equipped with an order relation $\le$ such that $ (\mathbb{R},\le)$ is a total ordered vector space with the Archimedean property. Tis is not possible in $\mathbb{R}^2$: here the lexicographic order is total but not Archimedean.

Another property:

Rotations in $\mathbb{R}^2$ commutes, but in $\mathbb{R}^3$ not, and in $\mathbb{R}$ they reduce to the trivial group.

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All nontrivial rotations in $n$ dimensions are about a single $(n-2)$-dimensional axis. This holds in dimensions 2 and 3, but fails in dimensions 4 and higher. (Of course, in $1$ dimension, there are no nontrivial rotations.)

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Physics provides a lot of examples of this type which can be translated into mathematical statements.

  1. One of my favorites comes from the history of the Ising model. Let us quote Wikipedia:

    In his 1924 PhD thesis, Ising solved the model for the 1D case. In one dimension, the solution admits no phase transition. On the basis of this result, he incorrectly concluded that his model does not exhibit phase behaviour in any dimension.

    Actually this was not even about the specific model, but about the possibility to describe phase transitions within the formalism of statistical mechanics. The issue was definitively settled by the Onsager's solution of the 2D case.

  2. The potential well of arbitrarily small depth leads to at least one bound state in 1D and 2D, but not in 3D. This has to do with localization.

  3. Conformal symmetry is infinite-dimensional in 2D but only finite-dimensional starting from $D\ge 3$.

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$R^n$ for n = 1,2 can be a field but it is impossible for n = 3. For n=4 it is possible a multiplication of field but this is not commutative (the first example of such a non commutative field, the quaternions, necessarily infinite because of theorem of Wedderburn:"all finite field is commutative")

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$A^2 = 0 \implies A = 0$ if $A$ if a $1\times 1$ matrix. (a number!)

It is false if $A$ is a $n \times n$ matrix, $n \ge 2$

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