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?

Edited: Part (a) answer

  • 1
    $\begingroup$ 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. $\endgroup$ – Pietro Gheri Nov 29 '18 at 2:13
  • $\begingroup$ 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! $\endgroup$ – Icycarus Nov 29 '18 at 8:35
  • $\begingroup$ Also your arguments in (b) and (c) are circular. I.e. you prove (b) by assuming that they don't overlap in positive measure, but in (c) you prove that they don't overlap in positive measure by using the volume you calculated in (b). $\endgroup$ – jgon Nov 29 '18 at 8:43
  • $\begingroup$ @jgon Oh.. yeah, I was wondering if there are any possible ways to better phrase the answer? $\endgroup$ – Icycarus Nov 29 '18 at 10:39
  • $\begingroup$ Well, one can set up a straightforward integral to compute (b), but I would probably start with (c) and argue that the intersection of two such simplices must lie in a plane, and hence has zero volume. (Or at least be in the boundary of your original simplex. $\endgroup$ – jgon Nov 29 '18 at 16:10

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