Inversion w.r.t. a sphere: Operator

neelakash

I am trying to give an answer to the question; please rectify me if I am wrong...Consider points on the radius of a sphere of radius [itex]a[/itex]. The inverse transformation is given as [itex]\hat{I}(r)=\frac{a^2}{r}[/itex]...We are trying to see if the set of all inversions form a group.

Identity: [itex]\hat{I}(a)=\frac{a^2}{a}=a[/itex]...thus identity element [itex]\hat{I}_e[/itex] is there.

Inverse: [itex]\hat{I_1}(r_1)=\frac{a^2}{r_1}[/itex]. We can say inverse of [itex]\hat{I_1}[/itex] exists, say [itex]\hat{I_2}[/itex] if we can show [itex]\hat{I_2}(\frac{a^2}{r_1})=r_1[/itex]; the latter is true, however, as [itex]\hat{I_2}[/itex] is also an inversion operator defined as above. Thus, [itex]\hat{I}_2={\hat{I}_1}^{-1}[/itex]

It appears that the closure and associative rules also be satisifed, if we notice that there is only few members of the group: identity [itex]\hat{I}_e[/itex], [itex]\hat{I}_1[/itex] and its inverse [itex]{\hat{I}_1}^{-1}[/itex]...

What do you say?

-Neel

neelakash

In fact, it looks like a Lie group with [itex]r[/itex] being the continuous parameter. With continuous variation of [itex]r[/itex], apparently all the group elements can be generated starting from the identity...