Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to know what does represent the equivalent of 3d cube in 4d, 5d, 6d .... till 11d??

2d is square (a x a)
3d is cube (a x a x a)
4d is ???
5d is ???
11d is ???

Thanks in advance.

share|cite|improve this question
No idea what you want. – Will Jagy Oct 14 '12 at 21:06
$\underbrace{a \times \cdots \times a}_{n\ \text{times}}$? – EuYu Oct 14 '12 at 21:10
Is you question related to this by any chance? – user39572 Oct 14 '12 at 21:15
4D = HyperCube (Tesseract) 5D = 5-cube 6D = 6-cube etc. Here is some work I did on visualizing the structure of a 4d cube's shadow a while back:… – Albert Renshaw Feb 19 '13 at 21:12
up vote 2 down vote accepted

One way to represent a square in $2D$ is as follows.

Let $$V_2 = \{(0,0),(1,0),(1,1),(0,1)\}$$ For $v_1,v_2 \in V_2$, let $$d(v_1,v_2) = \text{Sum of absolute values of components of }v_1 - v_2$$ For instance, $d_2((0,0),(1,1)) = 1 + 1 = 2$ and $d_2((0,0),(1,0)) = 1$. We want an edge between two vertices whose length is $1$.

Let $$E_2 = \{(v_1,v_2): v_1,v_2 \in V_2 \text{ and }d(v_1,v_2) = 1\}$$

Then the square is nothing but $(V_2,E_2)$.

Similarly in $3D$, define the vertex set as $$V_3 = \{(x_1,x_2,x_3): x_1,x_2,x_3 \in \{0,1\}\}$$ For $v_1,v_2 \in V_3$, let $$d_3(v_1,v_2) = \text{Sum of absolute values of components of }v_1 - v_2$$ As before, define the edge set as $$E_3 = \{(v_1,v_2): v_1,v_2 \in V_3 \text{ and }d_3(v_1,v_2) = 1\}$$

Then the cube is nothing but $(V_3,E_3)$.

The same extension works in $n$-D as well.

Define the vertex set as $$V_n = \{(x_1,x_2,x_3,\ldots,x_n): x_1,x_2,x_3,\ldots,x_n \in \{0,1\}\}$$ For $v_1,v_2 \in V_n$, let $$d_n(v_1,v_2) = \text{Sum of absolute values of components of }v_1 - v_2$$ As before, define the edge set as $$E_n = \{(v_1,v_2): v_1,v_2 \in V_n \text{ and }d_n(v_1,v_2) = 1\}$$

A $n$-dimensional cube is $$(V_n,E_n)$$

share|cite|improve this answer

Occasionally a $4$-dimensional cube is called a tesseract, but mostly an $n$-dimensional cube is just called an $n$-cube.

share|cite|improve this answer

The prefix "hyper" is often used for higher dimensional versions of geometric objects. Look up hyperplanes, hyperspheres, and hypercubes. This does not give any clarity about what dimension is intended though; just that there is the potential for the dimension to be greater than 1, 2, or 3.

share|cite|improve this answer

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.