## Inversion w.r.t. a sphere: Operator

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 $a$. The inverse transformation is given as $\hat{I}(r)=\frac{a^2}{r}$...We are trying to see if the set of all inversions form a group.

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

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

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

What do you say?

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