Note (2021-10-23, Princeton time): This is from my Summer 2017 internship in Grenoble. I originally applied to work in quaternionic Fourier analysis, but got redirected to a project on embedding surfaces in graphs, which amusingly led to the quaternion group Q_8 via the pi_1 computation below. There may be errors below, e.g. I seem to recall pi_1(Y) is not actually "a semi-direct product". I haven't thought about any of this recently, but thought it might be fun to make publicly available anyways. === Fundamental group of deleted product for RP^2 (sent 2017-08-31, Grenoble time) === Hi all, This might not be useful, but If I'm not mistaken, the homotopy long exact sequence for RP^2 \ * -> Y := RP^2 x RP^2 \ Delta -> RP^2 goes: 0 -> pi_2(RP^2) = Z --x4--> pi_1(RP^2 \ *) = Z -> pi_1(Y) = Q_8 (the quaternion group) -> pi_1(RP^2) = Z/2 -> 0 with pi_1(Y) = , where h represents [the image in pi_1(Y) of] a generator of pi_1(RP^2 \ *) = Z, and v represents the action of pi_1(RP^2) = Z/2 on pi_1(RP^2 \ *) = Z. [This black-boxes Reference [2], https://en.wikipedia.org/wiki/Seifert_fiber_space#Positive_orbifold_Euler_characteristic below. I'm surprised how nice the theory is; for example pi_1(Y) is a semi-direct product and not something more complicated.] --- Sketch of argument: Consider the vertical map of fiber bundles: RP^2 \ * -> Y -> RP^2 RP^1 -> Z -> RP^2, where Z is the bundle induced by fiber projection RP^2 \ * -> RP^1 (here * is varying along the base RP^2, and the copy of RP^1 over * can be identified with the space of projective lines through *, so Z is really a bundle). I'm pretty sure the bundle map Y -> Z is (at least a weak) homotopy-equivalence. Z is not the unit circle tangent bundle U(RP^2), but it's closely related: the vertical map S^1 -> U(RP^2) -> RP^2 RP^1 -> Z -> RP^2 is given by modding out fibers by Z/2 (antipodal map). Indeed, the projection RP^2 \ * -> RP^1 (lines through *) identifies opposite tangent directions at *. Since RP^2 is non-orientable, the unit tangent bundle U(RP^2) is clearly non-orientable. However, Z is actually non-orientable too: the generating cycle [v] = 1 in pi_1(RP^2) = Z/2 still reverses orientation on Z (even though we've modded out the fibers S^1 by the antipodal map). Tracing through the reference [2] https://en.wikipedia.org/wiki/Seifert_fiber_space#Classification it seems we are in the case {b; (n2, 1);} (b integral.) This is the Prism manifold with fundamental group of order 4|b| and first homology group of order 4, except for b=0 when it is a sum of two copies of real projective space, and |b|=1 when it is the lens space with fundamental group of order 4. with fundamental group specified in https://en.wikipedia.org/wiki/Seifert_fiber_space#Fundamental_group (which is what I wrote above in the language of Y -> RP^2). I'm pretty sure Z -> RP^2 is b = 2 and U(RP^2) -> RP^2 is b = 1 (see either reference [4] below or the P.S.). Thanks, Victor References: I found these by googling "classification of circle bundles", which seems to be a thing. [1] https://mathoverflow.net/questions/6142/circle-bundles-over-rp2 [2] https://en.wikipedia.org/wiki/Seifert_fiber_space [3] https://en.wikipedia.org/wiki/Spherical_3-manifold#Dihedral_case_.28prism_manifolds.29 [4] Unit Tangent Bundle over Two-Dimensional Real Projective Space, TATSUO KONNO, https://projecteuclid.org/euclid.nihmj/1273779621 P.S. As a sanity check, the fact that the boundary map is x4 can be directly found using the definition at https://en.wikipedia.org/wiki/Fibration#Long_exact_sequence_of_homotopy_groups and patiently drawing some pictures (what I spent most of Wednesday doing before I googled the general theory). P.P.S. de Longueville and MatouĊĦek both discuss "deleted joins" as a sometimes better-behaved notion than deleted products, but I haven't thought much about that (say, whether it gives something nontrivial for S^2). ===