Are there mathematical concepts that exist in dimension $4$, but not in dimension $3$?

Solution 1:

The one that sticks out for me the most is that there are five regular polytopes (called Platonic solids) in $3$ dimensions, and they all have analogues in $4$ dimensions, but there is another regular polytope in $4$ dimensions: the 24 cell.

The kicker is that in dimensions higher than $4$... there are only three regular polytopes!

Another thing that can happen in $4$ dimensional space but not $3$ is that you can have two planes which only intersect at the origin (and nowhere else.) In $3$ dimensions you'd get at least a line in the intersection.

I don't know if this also counts, but linear transformations in $3$-dimensions always scale one direction (that is, they have a real eigenvector). This means that in all cases, a line in one direction must either stay put or be reversed to lie upon itself. In $4$ dimensions, it's possible to have transformations (even nonsingular ones) that don't have any real eigenvectors, so all lines get shifted.

Also not sure if this counts, but there are no $3$ dimensional asociative algebras over $\mathbb R$ which allow division (they're called division algebras) but there is a unique $4$ dimensional one. (Look up the Frobenius theorem

Solution 2:

The fourth dimension is, in some ways, very peculiar.

The $3$-dimensional Euclidean space has a unique differentiable structure. In fact, the same is true in any dimension not equal to $4$. By contrast, the $4$-dimensional space has a continuum of incompatible differentiable structures. Intuitively, this means that something that looks like the familiar $4$-dimensional space can have a large variety of rather odd geometries.

An at most $3$-dimensional sphere admits a unique differentiable structure (there are no exotic spheres).

It is not known whether there are exotic $4$-dimensional spheres, or whether there are finitely many of them (in contrast, in any other dimension, there are only finitely many differentiable structures on the sphere --- e.g. $28$ in $7$ dimensions).

Solution 3:

Double rotation. In 4 dimensions we can have 2 rotations independent of each other: $$ \begin{pmatrix} \cos\alpha & \sin\alpha & \cr -\sin\alpha & \cos\alpha & \cr && \cos\theta & \sin\theta & \cr && -\sin\theta & \cos\theta & \cr \end{pmatrix}. $$

Solution 4:

The fact that $\mathbb{R}^4$ can be given a multiplication such that with vector's addition, it has a structure of (non commutative) field, i.e., quaternions. See remark 1 below.

Whereas one can prove that it is impossible for $\mathbb{R}^3$ to have a vector multiplication such that, with usual addition of vectors, $\mathbb{R}^3$ is a field.

I just saw now that at the end of his answer @rschwieb has mentionned "division algebras" which in fact is equivalent to (non commutative) or "skew") field (thanks to Tomasz for having done this remark).

Remark 1 : For quaternions, see my answer to what is the relation between quaternions and imaginary numbers?

Remark 2 : Think for example to an unsuccessful candidate, cross product, which is

• not associative : in general $(\vec{u} \times \vec{v}) \times \vec{w} \neq \vec{u} \times (\vec{v} \times \vec{w}).$

• does not possess a neutral element : no vector $\vec{u_0}$ exists such that for all $\vec{v}, \vec{v} \times \vec{u_0}=\vec{v}.$