I accidentally (re-)discovered the circles of Villarceau c. 1992 when trying to foliate an open subset of 3-space by congruent circles. But in fact Villarceau discovered these in 1848. One way to generate them is to start with the Hopf fibration of the 3-sphere S^3 := {(z,w) in C^2 | |z|^2 + |w|^2 = 1}. (C = the complex plane.) Its fibres are circles defined as the equivalence classes of (z,w) ~ (gz,gw) for any (z,w) in S^3 and where g is any complex number of length 1. (These are all great circles of the 3-sphere.) Then project S^3 - {(0,i)} stereographically onto R^3 via S((a+bi, c+di)) := (1/(1-d))(a,b,c). Since S is conformal, it carries the Hopf fibres to circles (except that the incomplete fibre through (0,i) goes to the z-axis). Now, any Hopf fibre in S^3 will lie on a unique torus of the form T_s := S_r x S_s in C^2 with 0 <= s <= 1, where S_t denotes the circle of radius t about the origin in C. Here r^2 + s^2 = 1 by the definition of S^3. (Actually, note that the "tori" T_1 and T_0 are degenerate, and are in fact great circles of S^3.) It is left as an exercise for the reader, to show that for any Hopf fibre F, the size of its image circle S(F) in R^3 depends only on the s for which F lies on T_s. Thus for each s, 0 <= s < 1, we can modify the stereographic projection (S | T_s) with a post-scaling, (S' | T_s) := f(s)*(S | T_s), so that S' takes each Hopf fibre to a circle of radius 1. It is straightforward to verify that S' : S^3 - T_0 -> R^3 is a diffeomorphism onto the open solid torus U of revolution of all points in R^3 that lie at a distance < 1 from the unit circle S^2 in the xy-plane. Then the images of all the Hopf fibres of S^3 - T_1 are unit circles in the image of S', foliating the open solid torus U by congruent circles. Incidentally, the analogous procedure works for all four Hopf fibrations (based on the normed real division algebras, i,e., on R, C, H, O): S^(2k+1) fibred by great S^k's for k = 0,1,3,7 . . . yielding a foliation of an open set of R^1, R^3, R^7, R^15 by S^0's, S^1's, S^3's, S^7's, respectively. (I am told that Bill Thurston knew of these foliations by the mid-1970s.) Interestingly, there are no other values of k for which a non-empty open set in R^(2k+1) can be foliated by congruent round k-spheres. But for k = 1,3,7 there are infinitely many different ways to foliate an open set in R^(2k+1) with (round) unit k-spheres as the leaves. An amusing fact about these is that there is an upper bound M_k to the volume of such a foliated open set U in R^(2k+1). (One needs to assume the leaf space of the foliation is connected, which for k=1,3,7 is the same as assuming that U is connected (i.e., except for k = 0). It's easy to check that for k = 0 this upper bound is 4. I've conjectured that the Hopf fibration method also gives the largest-volume such foliated open set in the other three cases, k = 1,3,7 but have not so far proven it. The volume in the Hopf cases is always A(k)*V(k+1) where A(k) is the k-dimensional content of the (surface of the) unit sphere in R^(k+1), and V(k+1) is the (k+1)-dimensional content of the unit ball in R^(k+1). So we know M_0 = 4 in R^1, and conjecturally M_1 = 2 pi^2 in R^3, M_3 = pi^4 in R^7, M_7 = pi^8/72 in R^15. --Dan P.S. For completeness I'll add that the above readily implies that there do exist non-empty open sets in R^(2k+1+p), for any p >= 0, that are foliated by congruent round k-spheres for k = 0,1,3,7. And conversely, for any k not in {0,1,3,7}, there does not exist a non-empty open set in R^(2k+1-p), for p >= 0 and 2k+1-p >= 0, that is foliated by congruent round k-spheres. But this leaves unresolved infinitely many cases of k-spheres in n-space. First open case: Can congruent round 2-spheres foliate a non-empty open set in R^5 ??? _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele