4d, 5d, 6d ... 11d equivalent of cube? 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. 
 A: Occasionally a $4$-dimensional cube is called a tesseract, but mostly an $n$-dimensional cube is just called an $n$-cube.
A: 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.
A: 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 $$(V_n,E_n)$$
