Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For any two matrices $P,Q \in SU(2)$, with $tr(P)=tr(Q)=0$, does there always exist some $G\in SU(2)$ such that $G P G^{-1} = -P$, and $G Q G^{-1} = -Q\ ?$

share|improve this question
add comment

3 Answers 3

up vote 4 down vote accepted

It's not hard to explicitly construct G using the quaternions, assuming P is not \pm Q, and I think this is worth working out in detail because I really like this picture of SU(2). Identify SU(2) with the unit quaternions by the isomorphism

[a        b      ]
[-\bar{b} \bar{a}]  --> a+bj

and since the trace of such a matrix is 2*Re(a), the traceless matrices in SU(2) correspond exactly to the purely imaginary unit quaternions.

Now if we consider R^3 as the space of all imaginary quaternions, we can describe the SU(2) action on it geometrically: Let v=cos(t)+sin(t)w, where w is purely imaginary and |w|=1. Then the conjugation map x -> vxv^{-1} is a rotation of the plane orthogonal to w -- you can easily check that it's in SO(3), and that it fixes w because vw=wv. It's also not hard to check that the angle of this rotation is 2t.

Consider P and Q as imaginary quaternions, hence also as vectors in R^3. Then PQ = -(P.Q) + PxQ, so let G=Im(PQ)/|Im(PQ)|. Now G is a unit vector orthogonal to both P and Q, and conjugation by G is the same as rotating the plane through P and Q by \pi, so GPG^{-1}=-P and GQG^{-1}=-Q as desired.

Finally, we just go back from the unit quaternions to SU(2): up to scale, G was supposed to be the imaginary part of the quaternion PQ, so the matrix G is some constant times the traceless part of the matrix PQ.

share|improve this answer
add comment

Yes. Since det(P)=1 and tr(P)=0, the eigenvalues of P are \pm i. So one of the logarithms of P is p, a matrix in su(2) with eigenvalues i \pi/2 and - i \pi/2. Here su(2) is the Lie algebra of SU(2). (In terms of quaternions, p is of the form bi+cj+dk with b^2+c^2+d^2=(pi/2)^2.) Similarly, Q has a logarithm q in su(2) which also has eigenvalues \pm i pi/2. Let H be the plane in su(2) spanned by p and q.

Now, SU(2) acts on su(2) by the group SO(3) of orthogonal rotations. Let G be the rotation of H by angle pi. So G p G^{-1}=-p and G q G^{-1} = -q. Exponentiating these equations, G has the desired property.

share|improve this answer
add comment

I think in SU(2), whenever two matrices have the same trace, they are conjugate. In other words, the quotient map SU(2) --> SU(2)/conjugation is identified with the trace map SU(2) --> [-2,2].

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.