# What problems are easier to solve in a higher dimension, i.e. 3D vs 2D?

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.

• 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 Feb 28, 2014 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 $2006$). Feb 28, 2014 at 2:29
• There's a well known plane geometry problem known as Desargues' Theorem which is easier to prove in 3D. Feb 28, 2014 at 2:37
• ‘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, 2014 at 4:44
• Similar question recently posted to MO, mathoverflow.net/questions/360924/… May 22, 2020 at 12:46

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.

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. $$^{}$$ The same cannot be done directly for dimension $$n=2$$, though. The usual method to recover a formula analogous to (1) in the two-dimensional case is called method of descent; it works by embedding the two dimensional equation into a three dimensional space and then using (1).

$$^{}$$ One can either exploit symmetries or use the Fourier transform. The first method is known as the “method of spherical means”; see e.g. Evans, Partial differential equations, chapter 2. For the latter method, see e.g. Folland's Real Analysis, chapter "Topics in Fourier analysis".

• Technically, the wave equation behaves very badly in all even dimensions: you lose causality. There's a reason nature picked an odd dimension! Jul 8, 2015 at 20:30

Counting regular convex polytopes

In 2D, there are infinitely many regular polygons.

In 3D, there are five regular convex polyhedra.

In 4D, there are six regular convex polytopes.

In 5D and above, there are only three.

I was quite surprised when I first learned this.

List of regular polytopes and compounds (Wikipedia)