TITLE: Let $S$ be the set of all circles on the unit sphere in $\mathbf{R}^3$. Give a smooth manifold structure to $S$.
QUESTION [5 upvotes]: I am not sure how to do this. $S$ includes singletons too (radius $0$). I have tried mapping a circle to the pair consisting of its center point and its radius, but it isn't injective.

REPLY [2 votes]: Here is my favorite way to see this. Consider $R^3$ as an affine chart in $RP^3$, the real-projective space. Let $B\subset RP^3$ denotes the closed unit ball, with the boundary sphere $S^2$. For any point $p\in RP^3 - B$ consider the conic which is the union of lines through $p$ tangent to $S^2$. The set of points of tangency is a circle. Conversely, every circle in $S^2$ appears this way. One way to see this is to find a projective transformation preserving $B$ and sending the given circle $C$ to a great circle $C'\subset S^2$. Then for $C'$ the relevant conic is the cylinder through $C$ with the axis $L$ passing through the center of $B$ (plus the point $p$ in $RP^3 - R^3$ represented by the line $L$). Thus, with any reasonable topological structure on the set of circles of positive radius in $S^2$, it is homeomorphic to $RP^3 - B$, which is a manifold, being an open subset of $RP^3$.  Now, if you allow circles of zero radius,you have to add to $RP^3 - B$ the boundary sphere $S^2$: I a sequence of points $p_n\in RP^3 -B$ converges to a point $p\in S^2$, the corresponding circles $C_n\subset S^2$ converge to $p$ as well. Hence, your space $S$ is homeomorphic to $RP^3 - int(B)$, which is a smooth manifold with boundary.