[Math] 4d, 5d, 6d … 11d equivalent of cube

Question

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.

Answer 1

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 $(V_2,E_2)$.

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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 $(V_3,E_3)$.

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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 $$(V_n,E_n)$$