# Equaivalent Statements?

1. Jul 25, 2008

### MathematicalPhysicist

Is the statement:
equivalent to:
where PR^n is the projective real space.

I think the answer is yes, just want to be sure here.

2. Jul 25, 2008

### gel

yes.

3. Dec 22, 2008

### quasar987

What is the reason for this?

Using the fact that the quotient map from S^n to RP^n (x-->[x], where x~y iff y=±x) is a local diffeomorphism and the fact that the antipodal map above is orientation reversing, I can get a contradiction from the hypotheses that RP^n is orientable when n is even. But when n is odd, it's another game. How does that work?

Thanks.

4. Dec 23, 2008

### gel

When n is odd, the quotient map f:S^n->RP^n is locally a diffeomorphism, so it locally maps an orientation on S^n to RP^n and, the fact that x->-x is orientation preserving means that the orientation given by antipodal points agree, giving a globally defined orientation on RP^n.

5. Dec 23, 2008

### quasar987

Yes that's obviously the general idea. But thanks, I figured out a way to work out the details just a moment ago.

6. Dec 23, 2008

### quasar987

It's not very elegant though... I start by considering an atlas of consistently oriented charts of S^n such that
(1) The restriction of p to each coordinate chart is a diffeomorphism onto its image.
(2) if U is a coordinate chart around x and V a coordinate chart around -x, then U and V do not intersect.

I then push the atlas to RP^n via p, and now it remains to show that this atlas of RP^n is too, consistently oriented. So I take some point [x] in RP^n. Then for any chart U'=p(U), V'=p(V) around [x], either
(a) U n V is empty and (W.L.O.G.) x is in U, -x is in V
(b) U and V intersect

In each case, the transition function from U' to V' is shown to have positive determinant. In case (a), the antipodal map comes in and the fact that it itself has positive determinant (for n odd) is the key.

Is there an obvious swifter way I am missing?