It is well known that any is a rotation, i.e. there exists , , such that and acts on the 2-plane by a rotation of angle . So we are free to call an element of a rotation. The picture below shows a continuous path in :

**Question: **Suppose we are given two rotations with axes and angles respectively. What are the axis and the rotation angle of their composition ?

After discussing with John Ma, WongTing and Zorn Leung, we have come up with the following

Theorem 1 (John Ma’s theorem, 2011)Suppose are two rotations with axis of rotation and rotation angles (see remarks below) for . Then the new rotation have the axis of rotation in the direction ofand the rotation angle is given by

Now we define more precisely what we mean by a rotation about with angle . This means that and there exists an orthnormal basis with such that and . Note that the choice of is uniquely determined up to : and defines the same rotation.

It turns that it is more convenient to work on the quaternions. (Well, originally I work on and work with instead of (see below), it is equivalent (as both are ) but it turns out that is even easier to work with, perhaps due to more built-in structure of (cross product is natural here!), thanks John Ma for pointing this out!)

Let be the space of quaternions, i.e. with . We denote the quaternions with unit norm by (not to be confused with the real or complex symplectic group ). Clearly is a (compact Lie) group. Let be the three dimensional real subspace of all imaginary quaternions. Naturally can be identified with . We define the action of on by

We will denote this action by , where

The following proposition states that this is really a group action.

Proposition 2is a homomorphism .

*Proof:* Clearly . Also, , i.e. .

Proposition 3defines a homomorphism , where we have identified (under induced inner product from ) with the ordinary .

*Proof:* We have to check preserves both the inner product and the orientation. But . So .

Also, is connected, since it is the (three-dimensional) unit sphere inside , so lies in the connected component of containing identity, which is . ( has precisely two (path) connected components corresponding to . )

Remark 1We can check that actually is the double cover of i.e. (see also the proposition below), and it’s easy to see , so . This also shows that , the three-dimensional projective space. Also is simply connected, as it is the 3-sphere. But we don’t need these facts here.

Proposition 4If is a rotation about with angle . By regarding , then for , (and ).

*Proof:* It suffices to show that and where are defined as above.

Denote the conjugate of a quaternion by , so . Consider

where we have used as it is purely imaginary.

On the other hand, since (see here),

As , we conclude that . Thus .

We are now ready to prove John Ma’s theorem.

*Proof of John Ma’s theorem:*

Suppose , are given as in the theorem. Then define for . By Proposition 4, and by Proposition 3 , so we only have to compute (see here):

where and are the angle of rotation and the rotation axis of respectively, by Proposition 4. Here is the cross product in and is the inner product in and we have used the fomula if and are imaginary i.e. .