If you are in a Euclidean vector space V of dimension n and you have two totally orthogonal subspaces S and T of dimensions k and n-k, then a rotation s of S and a rotation t of T can be continued to a rotation v of V such that the restriction of v to S is s and the restriction of v to T is t.
As i1 = i, ii = -1, ij = k, ik = -j, multiplication from the left with i means rotation of the (1,i)-flat (as with the complex numbers) and an analogous rotation if the (j,k)-flat. These rotations can be continued to a rotation of four-space around the origin.
Analogous for j and k.
I heard that quaternions are used in geometric optics, but I am not a physicist.
But if you represent i as the left multiplication by i in the space of quaternions that contains i itself, it seems to me that this is not an explanation of i in terms previously known geometrical concepts.
So I cannot be completely satisfied of your correct solution.
there is a vector-field interpretation of quaternions, in which the generators of the algebra of linear vector fields in the 4-dimensional Euclidean space, are tangent to the 3-dimensional sphere.
I think, you are looking for a geometric interpretation of quaternions via rotations of a 3d space. One possible answer is discussed in my article "On a Quaternionic Analogue of the Cross-Ratio" just before Lemma 1. Hope this helps.
Any unitary quaternion can be interpreted as a rotation of 3-space. Namely, if you take q with norm one, it can be written as q=cos\theta + \sin\theta w, where w is a quaternion without real part , that is, w=xi+yj+zk, and norm one. Then q can be interpreted as the rotation of axis w=(x,y,z) and rotation angle \theta/2. The exact assignment is the map S^3 \to SO(3) sending q into the conjugation map x \maps qxq^{-1}, see for instance Chevalley's book on Lie groups, or many other sources. Your question refers to w=i,j,k, then they are rotations of angle \pi/2 around the three coordinate axis.
It seems to me that the angle of the rotation is 2*\theta and not \theta/2.
In every case it seems to me that the map that to each unitary quaternion associate a rotation is not a bijection. Opposite quaternions have the same rotation.
Finally the rotations associated to i j k are of \pi radiants not \pi/2 , I believe.
you are right, of course,it is 2*\theta, sorry. and i,j,k are \pi rotations, indeed.
yes, it is not a bijection, this is called a two-fold covering map, in fact both S^3 (unit quaternions) and SO(3) (rotations) are Lie groups, and the kernel of the map is the center \pm id, this is why changing the sign of q does not affect the rotation.
the interpretation of quaternions as rotations was a great controverse betwen Hamilton (who was wrong) and Olinde Rodrigues (who was right),
I see that you have repeated a number of times your plea for a simple verbal description of the geometrical meaning of the quaternion units i, j and k.
They each correspond to a rotation by pi in the right-hand sense.
I've found no simpler illustration of the workings of the unit quaternions than Chapter 11 of Roger Penrose's book, The Road to Reality.
If you can find Section 11.3 "Geometry of quaternions" in Chapter 11 "Hypercomplex numbers" (p. 204 in my editions - I have two copies), I'm sure you'll find it all perfectly clear, and you'll be initiated into the mysteries of spinors at the same time.
You might even find there's a PDF of the book online...