Does this decomposition hold in every dimension for the $n$-cube? I have the need to cut up a cube ($n$-dimensional cube in fact, but let's stick to the dimension 3 for the moment) in a "wise" way. I come from a totally different field so I know basically nothing about simplices, but I found that maybe a good decomposition in dimension 3 would be the so called Schlafli orthoscheme. Here's what I mean: https://www.youtube.com/watch?v=ffnVCEAcOns&ab_channel=stebulus.
In general, I would need to cut a $n$-cube in a number of pieces such that the orthogonal sides of such pieces all point to linearly independent directions, forming a base of the space. For example, in the 2 dimensional case you cut a square in two right triangles and each of them has the catheti pointing to orthogonal directions. In $\mathbb{R}^3$ each of the tetrahedra in the decomposition above has three sides parallel to the coordinate axis. I already have a hard time picturing the situation in $\mathbb{R}^3$, do you have an idea if in $\mathbb{R}^n$ there's a similar decomposition for the $n$-cube? My intuition says "yes of course you can do it with a $n$-cube"; but I don't know what to look for on the internet.
EDIT: Also, any suggestions on how to visualize/write these things in higher dimensions? Already in dimension 3 I am struggling with my below-average drawing capabilities, I guess there must be a simplerand more clever way, but as I said I am completely out of this field. Thanks!
 A: Ok I thought about this and found an easy solution (for the $n-$cube). Assume you have a unitary $n$-cube $Q$ with a vertex in the origin. It's obvious you can partition your cube in sets like these:
$$ H_{\alpha} = \{ (x_1, \dots, x_n) \in Q | x_{\alpha_1} \geq x_{\alpha_2} \geq \dots \geq x_{\alpha_n} \},$$
where $\alpha= (\alpha_1, \alpha_2 , \dots, \alpha_n)$ is some permutation of $(1,2, \dots, n)$. I want to prove that each of these sets is a tetrahedron (for me, a tetrahedron in $\mathbb{R}^n$ will be the convex hull of a set of $n+1$ affinely independent points). Choose the verteces which you want to do the convex hull of in the following way (I put them in order just for convenience):

*

*First, the vertex $V_1=e_{\alpha_1} =(0,0, \dots, 1, \dots,0)$ ;


*Second, the vertex $V_2 =e_{\alpha_1} + e_{\alpha_2}$ with $1$ in the positions $\alpha_1$ and $\alpha_2$;


*Keep on until you consider $V_n = \sum_{i=1} ^n e_{\alpha_i} = (1,1, \dots, 1)$ as the $n-$th point;


*At last, choose $V_{n+1}$ as the origin.
Then the convex hull of these points will be:
$$ T_\alpha= \{q_1 V_1 + q_2 V_2 + \dots + q_n V_n \text{ such that } q_i \geq 0 \; \forall i \text{  and  } \sum_{i=1} ^n q_i \leq 1 \}. $$
By construction the $\alpha_1 -$th coordinate of points in $T_\alpha$ will be $\sum_{i=1} ^n q_i$, the $\alpha_2 -$th component will be $\sum_{i=2} ^n q_i$ and so on, so it's easy to see that $H_\alpha = T_\alpha$.
