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Take a cube $C$ $[-1,1]^3\subset\Bbb{R^3}$. How many rotations are there which take $C$ to itself?

What does this question even mean? If you take any line in $\Bbb{R^3}$, and rotate the cube around it by $2\pi$ rad, you are mapping the cube to itself!

Ans: 24. I don't understand.

algebraically_speaking
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  • Skewer the cube through the centers of two opposite faces. Then you can rotate the cube around that axis, and after a $90^\circ$ rotation, for example, you'll end up with the "same" cube, but the vertices will have moved. Now only 23 more to find! – Nick D. Dec 04 '14 at 06:41
  • The rotations are completely specified by their behaviours on $e_1,e_2,e_3$. – copper.hat Dec 04 '14 at 06:41
  • I count 48 rotations, unless you are only counting proper rotations, in which case there are 24. – copper.hat Dec 04 '14 at 06:52
  • Isn's this the Dihedral group? – Passing By Dec 04 '14 at 06:59
  • @copper.hat- How is what I have written incorrect? Shouldn't rotating the cube around different axes count as different rotations? – algebraically_speaking Dec 04 '14 at 07:02
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    That would be the identity rotation... – copper.hat Dec 04 '14 at 07:05
  • Colour the faces on the cube. Assume it is on a flat surface with one face towards you. There are six faces which can be placed at the bottom. When the colour of the bottom face is fixed, there are four colours which can be turned to the front. $6\times 4 = 24$. That these are all rotations follows from the fact that any two rotations of a sphere with centre at the centre of the cube combine to give a third rotation. – Mark Bennet Dec 04 '14 at 08:29
  • @PassingBy The number $24$ is also obtained by observing that the isometric rotations of a cube act as the symmetric group on the four (body) diagonals of the cube - which is not a dihedral group. – Mark Bennet Dec 04 '14 at 08:31
  • @MarkBennet- Say I fix vertex $1$ at a point. How do I arrive upon the answer from here? How do I show that there are 3 different ways I could orient the cube, keeping one vertex fixed? – algebraically_speaking Dec 04 '14 at 09:44
  • Here is a similar question: http://math.stackexchange.com/q/601285/27978 – copper.hat Dec 04 '14 at 23:59

2 Answers2

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Let $A = \{ \pm e_k \}$, and suppose $Q$ is a rotation that satisfies $QC \subset C$. Then we will show that $QA = A$, which will restrict the possibilities for $Q$.

Since $Q$ is a rotation, we have $\|Qx\|= \|x\|$, and so $Q \partial B(0,1) = \partial B(0,1)$. A rotation is invertible and the inverse is also a rotation, so we have $Q B(0,r) = B(0,r)$ for all $r >0$.

We must have $QC = C$. There are various ways of seeing this. Suppose the containment is strict, then $C$ contains some open ball that does not intersect $QC$, which means that the volumes satisfy $m(QC) < m(C)$, which contradicts $Q$ being a rotation. We also have $Q(\lambda C) = \lambda C$ for all $\lambda$ (where $\lambda C = \{ \lambda c \}_{c \in C}$).

We also have $C^\circ = \cup_n (1-{1 \over n}) C$, and so it follows that $Q C^\circ = C^\circ$. Since $C$ is given by the disjoint union $C = \partial C \cup C^\circ$, we have $Q \partial C = \partial C$.

Note that $A=\partial C \cap \partial B(0,1)$, combining the above shows that $QA = A$.

Now count the possibilities for $Q$:

Since $e_1,e_2,e_3$ form a basis, we need only consider the action of $Q$ on these vectors. $Qe_1$ can have $|A| = 6$ values, $Q e_2$ can have $|A|-2 = 4$ values (since $Q$ is injective).

Finally, since $Q$ is injective there are two remaining possibilities. If $Q$ is proper, then only one possibility remains for which $\det Q = 1$.

Hence there are 48 possible rotations of which only 24 are proper.

Aside: I was curious to know if $Q$ is a rotation and $C$ is such that $QC \subset C$ would imply $QC=C$ in general. The answer is negative, for example, let $Q$ be a rotation by 1 radian, and let $C = \{Q^n (1,0)^T \}_{n=0}^\infty$. Since $\pi$ is irrational, all of these points are distinct and $QC = \{Q^n (1,0)^T \}_{n=1}^\infty \neq C$.

copper.hat
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The word "rotation" here does not mean a particular action of rotating the cube around some axis - it refers to the change in the "orientation" of the cube you get by performing such a rotation. (Note that I'm using not using the word "orientation" with its layman's definition, not its usual mathematical meaning.) So if you rotate $2\pi$ radians around an axis, even though you seem to be doing something to the cube, you have in fact not changed the orientation of the cube at all - all of the vertices are right where they started! Thus we call that the identity rotation. You are looking for the number of different final orientations of the cube you can get by rotating it so that it "looks the same" after the rotation as it did before.

ajd
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