Are there mathematical concepts that exist in the fourth dimension, but not in the third dimension? Of course, mathematical concepts include geometrical concepts, but I don't mean to say geometrical concept exclusively. I am not a mathematician and I am more of a layman, so it would be appreciated if you could tell what the concepts are in your answer so that a layman can understand.

  • 7
    $\begingroup$ Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions. $\endgroup$ – eyeballfrog Sep 4 '19 at 15:41
  • 29
    $\begingroup$ What about quaternions? $\endgroup$ – J. W. Tanner Sep 4 '19 at 15:48
  • 2
    $\begingroup$ I've asked a similar question, these answers might be relevant: math.stackexchange.com/questions/374685/… There's also a book Things to Make and Do in the Fourth Dimension by Matt Parker. $\endgroup$ – Džuris Sep 5 '19 at 10:27
  • 2
    $\begingroup$ This is a bit of nitpicking aimed against popular language, but there is no such thing as the (third or) fourth dimension. There is just one (notion of) dimension (which notion applies to certain mathematical situations) whose value is a (usually natural) number; so that value, the dimension, might be $4$. $\endgroup$ – Marc van Leeuwen Sep 5 '19 at 11:02
  • 3
    $\begingroup$ @MarcvanLeeuwen Indeed, it appears the OP is asking about concepts that exist in four dimensions but not in three dimensions. $\endgroup$ – Monty Harder Sep 5 '19 at 15:39

12 Answers 12


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

  • 6
    $\begingroup$ I would think that the third point doesn't count: surely it's just about the parity of the dimension, because an odd order polynomial must have a real root but even order polynomials can be constructed as products of quadratics which are irreducible over the reals? $\endgroup$ – Peter Taylor Sep 5 '19 at 8:09
  • 2
    $\begingroup$ @PeterTaylor well, the question did not specify “special to $4$”, it said “exists in $4$ but not $3$” so that doesn’t seem to be an issue. The thing I thought that might be objectionable is that the property is negative (“in $4$ dimensions there are transformations that don’t...”) $\endgroup$ – rschwieb Sep 5 '19 at 10:27
  • $\begingroup$ "stretch one direction" -- some kind of clarification that eigenvalues of 1 don't "stretch" might be useful. Not sure how to not make it awkward. $\endgroup$ – Yakk Sep 5 '19 at 18:28
  • $\begingroup$ @Yakk I don't have any problem stretching by a factor of $1$. But anyhow, I switched it to scale to make it slightly better. $\endgroup$ – rschwieb Sep 5 '19 at 19:04

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).


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

  • 18
    $\begingroup$ This is closely related to the thing about 2 planes intersecting at a point, mentioned in rschwieb's answer. $\endgroup$ – PM 2Ring Sep 5 '19 at 10:43

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 $\vec{v}, \vec{v} \times \vec{u_0}=\vec{v}.$

  • 5
    $\begingroup$ A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera. $\endgroup$ – tomasz Sep 5 '19 at 0:47
  • 3
    $\begingroup$ Could someone edit this answer to be understandable by a layman like what OP identifies as? $\endgroup$ – user1717828 Sep 5 '19 at 9:38
  • $\begingroup$ user1717828 see the reference I just introduced in my answer. Quaternions can be understood as an extension of complex numbers under the form $r+iV$, where $r$ is a real and $V$ an 3D vector, with a rule of addition which is $(r+iV)+(r'+iV')=(r+r')+i(V+V')$ and a rule of multiplication reminding the product of complex numbers $(r+iV)(r'+iV')=(rr'-V.V'+i(rV'+r'V+V \times V')$ $\endgroup$ – Jean Marie Sep 5 '19 at 21:17
  • $\begingroup$ Here I give a simple proof that 3-dimensional real division algebras don't exist. $\endgroup$ – Jyrki Lahtonen Sep 6 '19 at 10:50

At the risk of sounding flippant, I'm going to propose the universe as one such example. It is a peculiar, but experimentally well verified fact, that the universe has four dimensions (three spatial and one of time). To the extent that the universe is described by mathematical laws, it is an example of something that exists in four, but not three, dimensions.

You may ask: What is the mathematical explanation for this peculiar fact? Many physicists have asked themselves the same question. Some answers have been proposed, none of which are yet considered the final word:

  • The most scientifically "modest" proposal is that it has to do with "running of coupling constants under renormalization". In layman's terms, the idea here is that in dimensions other than four, the fundamental fields that make up the universe are strongly coupled, so that it becomes impossible for simple objects like atoms to exist. Without atoms, there is no life, and thus such a universe cannot have people in it to ask the question "why does my universe have this many dimensions?" (This is called an "anthropic argument", and has a number of criticisms, some mentioned in the linked article.)
  • Other anthropic arguments come from general relativity, which suggests that the universe may be unstable or inhospitable for life in various ways if it had higher or lower dimension.
  • String theory predicts that the universe is part of a bigger multiverse that has (depending on the flavor of string theory) something like, ten, eleven, or twenty-six dimensions. Most of these dimensions get curled up so that they become very small (this is called "compactification"). Some physicists have gone looking for extra spatial dimensions, but all such searches have turned up empty so far.
  • Some people speculate that the reason why the universe is 4D has to do with the peculiar topological properties of four dimensional space. As far as I know, no one has a concrete proposal for how this should work--just the aesthetic observation that four dimensions seems to be special in many ways (e.g. exotic spheres and exotic $\mathbb{R}^4$, not that these in themselves explain anything).

Of course, no one has totally ruled out the possibility that there are other universes with different numbers of dimensions, but given that four dimensions seems very stable in our experience, it is interesting to ask whether this is mathematically required.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – quid Sep 7 '19 at 22:06

If it counts: The $4$-dimensional unit sphere $S^4$ in $\mathbb R^5$ has the biggest volume of all unit spheres.

  • 2
    $\begingroup$ NB to people who didn't notice. This is 5-dimensional space, not 4. But the 3-sphere in 4d space also has a bigger volume than the normal, 3d sphere $\endgroup$ – Jam Sep 7 '19 at 13:29
  • 1
    $\begingroup$ I would argue that the sphere is a perfect example of what is not new to four dimensions because it is a straight foward generalization of what we have in two and three dimensions. $\endgroup$ – M. Winter Sep 11 '19 at 9:20

You have neat non-orientable surfaces which embed in $\mathbb R^4$ that do not embed in $\mathbb R^3$, such as the Klein bottle and the real projective plane. For instance, see: A non orientable closed surface cannot be embedded into $\mathbb{R}^3$ or this MathOverflow question on the Klein bottle.

In other words, you have these interesting homogeneous closed surfaces on which there's no consistent way to define counter/clockwise in $\mathbb R^4$, but you can't actually construct them in the 3-d space we live in.

Note: in $\mathbb R^3$, you can embed the non-orientable Mobius strip, but the Mobius strip is a bit different as it is a surface with boundary so is not homogeneous. You can think of the Mobius strip as a slice of a Klein bottle or the real projective plane.

  • $\begingroup$ Right. I personally don't find that a very remarkable property about $\mathbb{R}^4$ though – it's merely that it's “big enough”, allowing enough movement for embedding such manifolds. The manifolds themselves are only 2-dimensional. $\endgroup$ – leftaroundabout Sep 5 '19 at 11:26
  • $\begingroup$ @leftaroundabout Well, I think the interesting thing is that these surfaces don't embed into $\mathbb R^3$. Yes, the properties of the manifolds are simply 2-d properties, but if you restrict to $n$-dim Euclidean spaces (I expect the OP was thinking about Euclidean spaces), the question of what types of submanifolds you have is interesting. $\endgroup$ – Kimball Sep 5 '19 at 17:50
  • $\begingroup$ I would guess that there are non-orientable n-manifolds that can be embedded in R(n+2) but not R(n+1), for any n≥2. $\endgroup$ – Anton Sherwood Sep 6 '19 at 17:30

If you look at real algebras (roughly speaking: arithmetic over vectors of real numbers), there is a 4-dimensional real division algebra, but no 3 dimensional real division algebra. This means that you can construct a 4-dimensional definition for addition, subtraction, multiplication, and division which follows the rules we intuitively expect those operations to follow, but you actually can't do it in 3 dimensions. If you try, you run into inconsistencies with how you and I normally think division should behave.

Real division algebras exist in 1 dimension (real numbers), 2 dimensions (complex numbers), 4 dimensions (quaternions), 8 dimensions (octonions), and 16,32,64... dimensions (sedions). You can go look up rules for how to construct a meaningful concept of division in each of them. The sedions are considered to be mostly just a mathematical peculiarity -- its really hard to find real life applications of them. Octonions are really hard to use in real life (there's an effort to make a grand unified theory(GUT) with them in physics, though its not the most popular GUT candidate).

As for the others, you can use them all the time. Obviously you know the real numbers. Complex numbers get used in all sorts of situations in higher math for many reasons (one is that they're analytic. Another is that they model signals that real life engineers are interested in). Quaternions get used all the time to model rotations.

And that's all of them. There simply isn't a 3-d real division algebra. Its proven that one cannot exist.


Things that are completely new to four dimensions, that is, they appear first in four dimensions, but not in one/two/three dimensions, and which are also not straight forwards generalizations of concepts in one/two/three dimensions:

  • Double rotation (a way to rotate in two independent way simultaneously)
  • duo-prisms and duo-cylinders (geometric objects that "visualize" the concept of double-rotation).
  • A centrally symmetric self-dual polytope other than a polygon (24-cell, which is even regular),
  • More than one way to develope differential geometry on a manifold (e.g. multiple differentiable structures on $\Bbb R^4$ and maybe the 4-sphere)
  • Polytopes that cannot have rational vertex-coordinates, or polytopes which have strange realization spaces (not connected or not simply-connected, ...). For example, there is a (combinatial type of) polytope that can be realized in two ways, but these two ways cannot be continuously deformed into each other without changing the combinatorial structure of the polytope at some point.
  • Polytopes in which there is an edge between any pair of vertices, other than a simplex (so-called neighborly polytopes).
  • A knotted surface, e.g. a sphere embedded into 4-space, in such a way that it is free of self-intersections, but cannot be continuously transformed into the "standard embedding" without creating self-intersections in the process (okay, this is kind of a straight foward generalization of knots from 3D).

It doesn't properly answer the question but maybe it is worth noticing : James-Stein estimator.

For Gaussian vectors in dimension $m\ge3$, the estimator $$ \hat\theta_{\rm J.-S.} = \left(1-\frac{(m-2)\sigma^2}{\|\bar X_n\|_2^2}\right)\bar X_n $$ is uniformly better than the empirical mean $\bar X_n$ with respect to the Mean Squared Error measure $\mathbb{E}\left[\|\hat\theta-\theta\|^2_2\right]$.

It consists in shrinking the vector.


Going in the other direction, (nontrivial) knots exist in three dimensions, but not in four.

  • 1
    $\begingroup$ That's false. You just need to tie up a sphere instead of a circle. $\endgroup$ – John Dvorak Sep 5 '19 at 16:47
  • $\begingroup$ @John, I'm defining a knot as a continuous image of a circle. $\endgroup$ – Gerry Myerson Sep 5 '19 at 22:08

My answer is a much more basic one, so much so that part of it is almost non-mathematical.

Three-dimensional space can be fully visualized by human beings. Four-dimensional space, however, can only be inferred by analogy, or describing analytically, usually by extending theorems in lower dimensions making them definitions in higher dimensions.

For example of the latter, by formula, think of the formula for distance in one, two, and three dimensions. In all three cases, we may regard the situation as being in 3D, where distance is found by the square root of the sum of the squares of the differences of coordinates. We can see that. In 4D, by contrast, we take what was a theorem such as distance in lower dimensions and simply make that the definition. Can you really SEE the distance of point (1,1,1,1) to the point (2,2,2,2), which happens to be 2? Can you take out a ruler to get at least approximate verification of this?

On the other hand, by analogy, think of visualizing a person moving out of just three dimensions and going into the fourth dimension. How do we visualize that? Think of you in three dimensions standing on a plane of two-dimensional entities. They see only line segments representing contact points made by your two feet. Now jump up. The segments shrink and disappear (not necessarily at same exact moment, depending on how you jump). The 2D dwellers may ask: "Where did the 3D man go?" Well, you really did go off the x, y plane by using the z-direction, which 2D people can't visualize.

If there were a 4D man in Brooklyn, New York, by this analogy, and if he moved in the fourth dimension (whether spatial or temporal, doesn't matter), he - or rather his 3D "contact point shadow" - would seem to disappear from Brooklyn as he moved outside our limited view, and then suddenly appeared in, say, Tokyo.

P.S. Perhaps we can get the 4D man to fight Rodan or Godzilla while visiting Tokyo?


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