orientability of the universal bundle

[philippe]

Does anyone know for which $n$ the manifold {(x,D) in Rn * G1(Rn)
such that x in D } is orientable AS A MANIFOLD (and not as a vector
bundle) (here, G1(Rn) is the set of all 1-subspaces of Rn, i.e. the
projective space of Rn) ???

P.S. For n=2, it is not orientable, because it is the moebius strip.

[W. Dale Hall]

I'll use the notation RP^(n-1) for G1(Rn), and refer to it as "real
projective (n-1) space".

Your space, which I'll denote X, is orientable iff n is odd.

To prove this, note that the sphere bundle of the canonical line
bundle over RP^(n-1) has total space equal to the (n-1) dimensional
sphere, S^(n-1), and its projection is the canonical double cover
f: S^(n-1) --> RP^(n-1). Further, if this projection is used to
attach an n-cell D^n to RP^(n-1), yielding the complex:

D^n \cup_f RP^(n-1)

then the result is RP^n, alias real projective n-space. The normal
bundle of RP^(n-1) in RP^n is the canonical line bundle over RP^(n-1),
so the space X is homeomorphic to a tubular neighborhood of RP^(n-1)
in RP^n. I'll identify X with that tubular neighborhood.

If RP^n is orientable, then X (being an open submanifold) must also be,
and if X is orientable, then RP^n must be orientable, since RP^n \ X is
contractible (it's a disc), .

Finally, RP^n is easily shown to be orientable iff n is odd: RP^n is
the quotient of S^n by the antipodal map, and that map is orientation-
preserving (allowing one to produce an orientation on the quotient) iff
the map x |--> -x is orientation-preserving on R^(n+1).

Dale