# Minimal rotational degrees of freedom needed in N-space

How many rotational matrices are needed in N-space to achieve an arbitrary attitude when allowed to be successively applied?

Each N by N rotation matrix will allow only a rotation in a plane, so only two dimensions may be selected for rotation at a time. What is a smallest set of these matrices which, when used in succession, can achieve any desired orientation in N space?

Assume an N dimensional hypercube with distinct features at every vertex is being rotated on a computer screen under control of $$m$$ sense switches. The idea is to minimize the number of switches without sacrificing the ability to orient the object, whose projection appears on the screen, in any desired attitude, eventually, by application of this minimal set of rotations.

So far I have $$m=N-1$$. This is the set comprised of the set of rotations that all include the $$x$$ dimension. Is there a smaller set? Perhaps not. But the real question is:

If we open it up to allow multiple dimensions being rotated per switch, does that change the answer?

• Generally the number of degrees of freedom for N-dimensional rotation matrix is the same as the number of DOF for skew-symmetric N-dimensional matrix so we have $n(n-1)/2$ DOF. – Widawensen Oct 29 '18 at 9:42
• Thank you. So for four dimensions, 6? – John L Winters Oct 29 '18 at 19:15
• But I already proved I need only 3, xy, xz, and xw. – John L Winters Oct 29 '18 at 19:17
• By using these three in sequences I can achieve any final orientation. The question is not about degrees of freedom. It is about the minimal number of rotation matrices to cover the space if orientations when sequences of these rotations are allowed. – John L Winters Oct 29 '18 at 19:19
• Hmm, I'm not sure whether we can omit planes yz,yw,wz - with three mentioned by you it gives 6 planes for 4D. $\ \ \ \ \ \$ In 3 dimensions we could have three (=(3*2)/2) planes xy,yz,zx. – Widawensen Oct 30 '18 at 11:36