MHB How Does Burnside's Lemma Calculate the Number of Distinct Colored Cubes?

  • Thread starter Thread starter mathjam0990
  • Start date Start date
  • Tags Tags
    Application
Click For Summary
Burnside's Lemma is used to calculate the number of distinct colored cubes with faces colored red, white, or blue. The total number of configurations is 3^6, representing all possible colorings. Fixed points are determined by analyzing the effects of different rotations on the cube, such as a 90-degree rotation about the axis between the top and bottom faces, yielding 3^3 fixed points for that case. The discussion highlights that there are six such rotations, contributing a term of 6·3^3 to the calculation. Further investigation is needed for other rotation types to complete the application of Burnside's Lemma.
mathjam0990
Messages
28
Reaction score
0
Q: How many different colored cubes are there in which each face is colored either red or white or blue?

I know we have to use Burnside's Lemma to solve this. (1/|G|)* Σ f(g) where f(g) are the number of fixed points of g. I also know that one of the values will be 3^6, but that is where I get stuck. Can someone please give an explanation as to how to find the rest of the equation? Thank you!
 
Physics news on Phys.org
The first thing you need to do is figure out what "$G$" is (in this case, it is $S_4$), and then figure out what $X$ is (the set that $G$ is acting upon). In this case it is the configuration space of all $3^6$ possible colorings of the cube, where we keep track of which face got which color (some of these we would not be able to tell apart under a physical rotation in the real world).

Indeed the number of fixed points of the identity is $3^6$, since every face remains unchanged.

Suppose we rotate about the line between the mid-point of the top face and the mid-point of the bottom face, say by 90 degrees. We can then ask: how may fixed points does this have?

Well, the only way this can happen is if the rotated faces (there are 4 of these) are all the same color, all red, or all white, or all blue. For each of these 3 possibilities, we are free to chose the top and bottom faces any way we choose. Since we have $3^2$ choices for top and bottom faces, this gives $3^3$ fixed points of $X$ for this rotation.

The same is true for any such rotation about two opposite faces. We have six such rotations, so a second term is:

$6\cdot 3^3$.

Your task, now, is to investigate the other three types of rotations (a 180 degree face rotation, about a vertex (120 degrees), and about two opposite edges (180 degrees)).
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

Replies
5
Views
6K
  • · Replies 26 ·
Replies
26
Views
773
Replies
6
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
9K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K