I would like to know if for $A,B\in SO(3)$ the inequality $$ \|AB-BA\|_F\leq \|A-I\|_F\|B-I\|_F $$ holds, where $\|\cdot\|_F$ denotes the Frobenius norm and $I$ the identity matrix. Using the identity $$ AB-BA=(A-I)(B-I)-(B-I)(A-I) $$ one can show the inequality with a factor $2$. However the inequality seems to be true also without this factor.

I encountered this problem when I tried to estimate the error of permuting two small rotations.

  • 6
    $\begingroup$ The inequality holds with constant $\sqrt{2}$ for arbitrary $3\times 3$ matrices, see Bottcher, Wenzel, The Frobenius norm of the commutator, 2008. $\endgroup$ – Christian Remling Nov 9 '16 at 22:21

I confirm Peter's suggestion that the best constant is $\frac1{\sqrt2}$. For let $\omega$ be this best constant. Then the inequality amounts to writing $$3-{\rm Tr}(A^tB^tAB)\le2\omega^2(3-{\rm Tr}\,A)(3-{\rm Tr}\,B).$$ Let me parametrize $A$ by the angle of rotation $\theta$ and the axis of rotation $u$, a unit vector. Likewise $B$ has an angle of rotation $\alpha$ and an axis $v$. The left-hand side above is $8\omega^2(1-c)(1-c')$, where $c=\cos\theta$ and $c'=\cos\alpha$.

Now $B^tAB$ is the rotation of same angle $\theta$ about $w=B^{-1}u$, and $A^t$ the rotation of angle $-\theta$ about $u$. It is an instructive calculus that $${\rm Tr}(A^tB^tAB)=2c+c^2+(1-c)^2(w\cdot u)^2+2s^2w\cdot u.$$ Because of $w\cdot u\ge c'$, we find that $\omega$ is the best constant in $$3-2c-c^2+(1-c)^2c'^2-2s^2c'\le8\omega^2(1-c)(1-c').$$ The left-hand side factorizes as $(1-c)(1-c')(3+c+c'-cc')$, and there remains the inequality $$3+c+c'-cc'\le8\omega^2.$$ Therefore $\omega^2$ is the supremum of $\frac18(3+c+c'-cc')$ over $[-1,1]^2$. Because this quantity if affine in both $c$ and $c'$, the supremum is achieved at a vertex, and its value is $4$. Whence $\omega^2=\frac48=\frac12$.


Numerical experiments suggest that actually $\|AB-BA\|_F\leq\frac{1}{\sqrt{2}} \|A-I\|_F\|B-I\|_F$ could be true, and that this bound is sharp.

I don't know if that helps, but this sharpened inequality is equivalent to \begin{equation} 3-\text{trace}(ABA^tB^t)\le(3-\text{trace}(A))(3-\text{trace}(B)). \end{equation}

  • $\begingroup$ Maybe the last inequality translates to an easier statement for ${\rm SU}(2,\mathbb{C}).$ $\endgroup$ – Geoff Robinson Nov 10 '16 at 13:04
  • 2
    $\begingroup$ I think in fact for $n\times n$ matrices, we could get $1/\sqrt{n-1}$ as the multiplicative factor...does that match your simulations? $\endgroup$ – Suvrit Nov 10 '16 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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