# Permutations on vertices of cubes and hence finding volume enclosed by the vertices

Denote $$C$$ to be the cube $$C=\{(x_1,x_2,x_3)|0 \leq x_1,x_2,x_3 \leq 1\}$$ and let $$V=\{ (x_1,x_2,x_3)|x_1,x_2,x_3 \in \{0,1 \} \}$$ be the set of vertices of the cube.

Let $$A=$$convex$$((0,0,0) , (1,0,0) , (1,1,0) , (1,1,1))$$, where convex means it is the volume enclosed within the points.

Let $$\sigma \in S_3$$ act on C by $$\sigma . (x_1,x_2,x_3) = (x_{\phi (1)},x_{\phi (2)},x_{\phi (3)})$$

(a) What are the sizes of the orbits?

(b) Let $$A_\sigma = \{ \sigma . (x_1,x_2,x_3)|(x_1,x_2,x_3) \in A \}$$. Determine the volume of this simplex. Show that $$C = \cup_{\sigma \in S_3} A_\sigma$$

(c) Show that any intersection $$A_\sigma$$ and $$A_\tau$$, $$\sigma \neq \tau$$ cannot have any volume.

What I did:

(a) The possible sizes of the orbits are 1, 3 and 6.

Size of orbits 1: Elements that have the form $$(a,a,a)$$

Size of orbits 3: Elements that have the form $$(a,b,b)$$. In a sense, we are rearranging 3 elements but only 2 distinct.

Size of orbits 6: Elements that have the form $$(a,b,c)$$. We are arranging 3 distinct elements.

In a way the number of distinct elements is the number of ways we can partition the number 3, which is 3.

(b) I explained that the volume of the simplex is $$\frac {1}{6}$$ since $$|S_3|=6$$ elements and that each $$A_\sigma, \sigma \in S_3$$ are of equal sizes because they are just rotations/reflections of other points when acted by $$\sigma$$. Since volume of each simplex is $$\frac 1 6$$ for each $$\sigma \in S_3$$, then the union of all simplexes is 1, which is the volume of the cube.

(c) Since each simplex must be of equal size, then the intersection between $$A_\sigma$$ and $$A_\tau$$ must have no volume for $$\sigma \neq \tau$$, otherwise the union of all $$A_\sigma , \sigma \in S_3$$ will not have volume 1, which happens to be the volume of the cube.

I feel that my argument for (b) and (c) is very weak as I cannot justify my answer. Any suggestions on how I can improve my answer?

• I don't really understand your notation in the solution of point (a). What do you mean by $Orb(1)$? Also, I don't think that all the orbits have size 3. For example I would say that $(0,0,0)$ is a fixed point for all the elements of $S_3$, so that its orbit has size 1. The same is true for every element of the form $(a,a,a)$, $0 \leq a \leq 1$. As for the other points, I think that their orbits have size either $3$ or $6$, depending on if they have or do not have two equal coordinates. – Pietro Gheri Nov 29 '18 at 2:13
• I was thinking that $x_1$ gets maps to $x_{\phi (1)}$ so I want to ensure that 1 gets map to 1. Hence I treated them as permutation of numbers. But I get your point now and I didn’t consider the case where all the elements are of the same value! Thanks! – Icycarus Nov 29 '18 at 8:35