There is a strange product that takes two square roots of unit matrix, say $A$ and $B$, $A^2=I$, $B^2=I$ to a square root again, $$ A\star B=(A+B)^{-1}(A-B+2I), \qquad (A\star B)^2=I$$ Could anybody help me with identifying this structure? Where it comes from?

It was obtained from the Caley transform $C(A)=(1-A)^{-1}(1+A)$, $C(C(A))=A$, by expanding $C(C(A),C(B))$ and using $A^2=B^2=I$. Ones we imposed $A^2=B^2=I$ the inverse transform does not exists anymore, so we are at the singular point of the Caley transform. Somehow it looks like adding some points at infinity and extending the action on them.

  • $\begingroup$ Note that $A+B$ need not be invertible. $\endgroup$ Sep 25, 2012 at 12:25
  • $\begingroup$ so it seems it is an idempotent quasigroup with $I$ as the identity element; interesting.... $\endgroup$
    – Suvrit
    Sep 25, 2012 at 15:13
  • $\begingroup$ see my answer. This set is not really a quasi-group, because the operation makes sense only if the spectra (made of $\pm1$s) of $A$ and $B$ are equal. $\endgroup$ Sep 25, 2012 at 15:36
  • $\begingroup$ Indeed, the set does not seem to satisfy any nice relations, like being a group. However, it comes from the group action via Caley transform. Were not for $A^2=B^2=1$ it would be group structure by inverse Caley. It would be nice to find again something like Caley transform for this degenerate case. $\endgroup$ Sep 25, 2012 at 15:50

1 Answer 1


Let me denote $E_\pm(M)$ the eigenspaces of $M$ associated with the eigenvalues $\pm1$. Let me assume that the characteristic of the scalar field $k$ is not $2$. Then your assumption is that $$k^n=E_+(A)\oplus E_-(A)=E_+(B)\oplus E_-(B).$$ In addition, the assumption that $A+B$ is non-singular means $E_+(A)\cap E_-(B)=(0)$ and $E_+(B)\cap E_-(A)=(0)$, from which it follows that $\dim E_+(A)=\dim E_+(B)$.

Then one verifies easily $$E_+(A\star B)=E_+(B),\qquad E_-(A\star B)=E_-(A).$$ Because the dimensions of $E_-(A)$ and $E_+(B)$ sum up to $n$, one deduces $$k^n= E_+(A\star B)\oplus E_-(A\star B),$$ and therefore $(A\star B)^2= I_n$.

  • $\begingroup$ Thank you! This provides some basement for the operation to exist. It would be great to covert it into a more standard operation... $\endgroup$ Sep 26, 2012 at 9:54
  • $\begingroup$ @Eugene. I don't think there's something more standard. $A\star B$ is just the symmetry whose fixed points are those of $B$ and anti-fixed points are those of $A$. There's nothing more to say. It's just nice that you found an analytic expression. $\endgroup$ Sep 26, 2012 at 10:40
  • $\begingroup$ Sad but possibly true, I would like to view this as 'partial' compactification of the group as points close to $A^2=I$ are mapped to very 'large' group elements, so this action somehow extends the action of the group to some small cells attached at infinity. $\endgroup$ Sep 26, 2012 at 11:37

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