Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd be interested in knowing if there are any problems that are easier to solve in a higher dimension, i.e. using solutions in a higher dimension that don't have an equally optimal counterpart in a lower dimension, particularly common (or uncommon) geometry and discrete math problems.

share|improve this question
This is not exactly what you are looking for (this is a 2D problem that is more easily solved by going into 3D then back into 2D, rather than a 3D variant of a 2D problem being easier) but might be of interest anyway: tangent lines of 3 circles have collinear intersection –  angryavian Feb 28 '14 at 2:25
Poincaré conjecture. Smale proved the case for dimensions $\ge 5$ in $1961$, Freedman proved the case for $n=4$ in $1982$, and Perelman proved it for $n=3$ in the period $2002-2003$(approved in $2006$). –  chubakueno Feb 28 '14 at 2:29
There's a well known plane geometry problem known as Desargues' Theorem which is easier to prove in 3D. –  user2345215 Feb 28 '14 at 2:37
Sometimes algebraic propositions can be proved for high characteristics but they fail to be tractable for characteristic three or two... in particular, characteristic two is troublesome. Not exactly dimension, but similar in nature to your general inquiry... –  James S. Cook Feb 28 '14 at 2:57
‘R. H. Bing explained the dimension situation in this way: “Dimension 4 is the most difficult dimension. It is too old to spank, the way we might deal with the little dimensions 1, 2, and 3; but it is also too young to reason with, the way we deal with the grown-up dimensions 5 and higher.”’ Quoted from James W. Cannon's review of Embeddings in Manifolds –  MJD Feb 28 '14 at 4:44

2 Answers 2

up vote 12 down vote accepted

The kissing number problem asks how many unit spheres can simultaneously touch a certain other unit sphere, in $n$ dimensions.

The $n=2$ case is easy; the $n=3$ case was a famous open problem for 300 years; the $n=4$ case was only resolved a few years ago, and the problem is still open for $n>4$… except for $n=8$ and $n=24$. The $n=24$ case is (relatively) simple because of the existence of the 24-dimensional Leech lattice, which owes its existence to the miraculous fact that $$\sum_{i=1}^{\color{red}{24}} i^2 = 70^2 .$$ The Leech lattice has a particularly symmetrical 8-dimensional sublattice, the $E_8$ lattice and this accounts for the problem being solved for $n=8$.

There are a lot of similar kinds of packing problems that are unsolved except in 8 and 24 dimensions, for similar reasons.

share|improve this answer

One such example in PDEs is Kirchhoff's formula for the solution to the initial value problem for the wave equation: \begin{equation} \begin{cases} \partial^2_t u - \Delta u = 0 & x \in \mathbb{R}^n,\ t \in \mathbb{R} \\ u(0,x) = g(x) \\ \partial_t u (0, x) = h(x) \end{cases} \end{equation} In space dimension $n=3$ it is relatively easy to derive the formula \begin{equation}\tag{1} u(t,x)=\frac{1}{4\pi t^2} \int_{\partial B(x;t)} \big[ t\cdot h(y) + g(y) + \nabla g(y)\cdot (y-x) \big]\, dS(y) \end{equation} which expresses the solution in terms of the initial data. $^{[1]}$ The same cannot be said for dimension $n=2$, though. Indeed, the usual method to recover a formula analogous to (1) in the two-dimensional case is called method of descent and it works by embedding the two dimensional equation into a three dimensional space and then using formula (1).

$^{[1]}$ One can either exploit symmetries or use the Fourier transform. For the first method one can consult, among others, Evans, Partial differential equations, chapter 2. For the latter method one can consult, among others, Folland's Real Analysis, chapter "Topics in Fourier analysis".

share|improve this answer
Technically, the wave equation behaves very badly in all even dimensions: you lose causality. There's a reason nature picked an odd dimension! –  Alex R. Jul 8 at 20:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.