Multiplication by one or more quaternions cannot possibly yield a reflection. As the quaternions approach 1, the mapping becomes the identity. Since the mapping is a continuous function of the quaternions, it cannot suddenly jump from being a proper rotation to an improper one. -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, June 26, 2014 9:17 AM Subject: Re: [math-fun] Robert Smith's vector problem...
Perhaps I'm not understanding this, but left multiplication of points in R^4 (identified with H, the ring of quaternions) by any fixed unit quaternion q
L_q: R^4 -> R^4 via L_q(x) := qx
results in a rotation of R^4. (Same for right multiplication R_q(x) := xq.)
If instead we're talking about a pure unit quaternion u (Re(u) = 0), then identifying R^3 with the pure quaternions
H_0 := {x in H | Re(x) = 0}
results in left multiplication by q
f_q: R^3 -> R^3 via f_q(x) := qx
yielding the cross product of q with x, which of course is a projection of R^3 onto the 2-plane perpendicular to q.
So, I'm not sure in what sense a quaternion multiply computes a reflection.
--Dan
On Jun 26, 2014, at 6:07 AM, Henry Baker <hbaker1@pipeline.com> wrote:
. . . each quaternion multiply only computes a _reflection_, and you need 2 reflections to make a rotation.