# Deg 1 map S^3 -> RP(3) ?

1. Jul 27, 2010

### owlpride

It's easy to construct maps of even degree from the three-sphere to real projective three-space. Do there exist maps of odd degree?

2. Aug 3, 2010

### Eynstone

I think the Borsuk-Ulam theorem doesn't obviate the existence of such a map.

3. Aug 3, 2010

### owlpride

I am concerned about the following:

If there existed a map of degree one from S^3 to RP(3), then I could pre-compose it with a degree 1 map from RP(3) to S^3 (whose existence is obvious) to get a degree one map from RP(3) to RP(3) that annihilates the generator of H_1 (RP(3)) = Z_2 (because the map factors through H_1(S^3) = 0).

That seems problematic, although I don't know a theorem off the top of my head which would prevent that.

4. Aug 3, 2010

### quasar987

I should hope so... for any space X the map induced on holomogy by a constant map X-->X is zero (i.e. annihilates everything).

I am curious though: what is your obvious degree 1 map from RP(3) to S^3?

5. Aug 4, 2010

### owlpride

Now you lost me. Where is the constant map coming from? So far I only see degree 1 maps, which are not homotopically trivial (unlike a constant map). Are you arguing that there exists or does not exist a degree 1 map from S^3 to RP(3)?

In an attempt to get a non-existence proof by contradiction, I was assuming the existence of a deg 1 map from S^3 to RP(3) and constructed a map from RP(3) to RP(3) which induces an isomorphism on H_3 and the zero map on H_1. It looks like there should be a contradiction in there somewhere, but I don't know a theorem that would give me a contradiction right off the bat. (All deg 1 maps from a manifold to itself which I have encountered so far were homotopic to a homeomorphism, which would induce isomorphisms on all homology groups. I would love to see a counterexample if one exists.)

There's an easy degree 1 map from any orientable n-manifold to S^n: take a small open ball in the manifold and collapse its complement to a point. If you are a tiny bit more careful this map is even smooth.

6. Aug 4, 2010

### quasar987

If you find the answer to your question, I'd be interested to hear it.

7. Aug 4, 2010

### simeonsen_bg

So, we have a degree 1 map h from RP(3) to RP(3) that annihilates the
generator of H_1(RP(3)) = Z_2 and are looking for a contradiction?
I suppose it possible to get by considering the cohomology algebra of
RP(3) with coefficients in Z_2 and the induced homomorphism h^*.
As Z_2 is a field, homologies and cohomologies are canonically
isomorphic. Let a be the generator of H^3(RP(3)) = Z_2 and x be the
generator of H^1(RP(3)) = Z_2, then it is known that x^3 = a. But then
h^*(a) = h^*(x^3) = [h^*(x)]^3 = 0^3 = 0,
that contradicts the condition deg(h) = 1.
The same argument works in /one of/ the proof of Borsuk-Ulam theorem
(Spanier's textbook).

8. Aug 4, 2010

### owlpride

Thanks! One last question: how do you get the cup products? Or how do you know that x^3 = a? A priori x^2 or x^3 could be zero, couldn't they?

9. Aug 5, 2010

### quasar987

Me I have a question concerning the sentence

"As Z_2 is a field, homologies and cohomologies are canonically isomorphic."

Do you really mean that? As far as I can tell, for M a manifold and F a field, we have a natural isomorphism of vector spaces $H^i(M;F)\approx \mathrm{Hom}_{F}(H_i(M;F),F)=H_i(M;F)^*$, but there is no canonical isomorphism btw H_i(M;F) and its dual, is there?

10. Oct 22, 2010

### quasar987

That is a well known fact concerning the cohomology ring of RP^n with coefficients in Z_2. Look at Hatcher for instance.

I think I found another a little more elementary proof. It is based simply on the fact that since S³ is simply connected and since q:S³-->RP³ is a covering of RP³ (with q the usual quotient map p(x)= {±x}) , by the theory of covering spaces, every map f from S³ into RP³ lifts to a map F from S³ to S³ such that q o F = f. But here (and whenever the dimension of the projective space of interest is odd), q is a local diffeomorphism which preserves the orientation, so deg(q)=2. Thus, deg(f) = 2deg(F) which is even.

So basically: every map S³-->RP³ factors through q so is of even degree.

11. Oct 27, 2010

### quasar987

How do you make it smooth?

12. Oct 27, 2010

### mathwonk

as with a C^infinity partition of unity. i.e. smooth "bump" function.

13. Oct 27, 2010

### mathwonk

I think there is a problem even defining the degree of a map from an oriented to a non orientable manifold. I.e. degree is usually defined by the induced map on top dimensional homology, which is always zero in that case, or using the orientation on tangent spaces to count inverse images of a general point as either +1 or -1. So the question does not even make sense as first posed, at least not to me.

14. Oct 27, 2010

### quasar987

RP^n is orientable for odd n.

15. Oct 27, 2010

### quasar987

I tried something involving a bump function but it did not remove the problem. What did you have in mind specifically?

16. Oct 27, 2010

### mathwonk

thanks for the news about RP^3 being orientable. I have a kind of mind set that real projective space is non orientable, that stems from the case n=2. I should have reaalized that RP^1 is S^1. Thank you.

Ok, about the bump function, I did not give it much thought either but surely this is not a problem too? I mean, let's see,.... the challenge is to define a smooth function from

oh how about a stereographic projection?

17. Oct 27, 2010

### quasar987

Actually, I was in the process of writing what I tried in details, and all of a sudden I don't see a problem with the construction. I have it all on the blackboard at school so I'll double check tomorrow after a good night of sleep but it might have been a ghost. :p

18. Oct 28, 2010

### lavinia

I think it follows from the homotopy exact sequence of a fibration that the degree must be even.

The 2 fold cover S$$^{3}$$ -> RP$$^{3}$$ is a fibration. The exact sequence of homotopy groups looks like

0 -> $$\pi_{3}$$(S$$^{3}$$) -> $$\pi_{3}$$(RP$$^{3}$$) -> 0

since the second and third homotopy groups of two points, the fiber, are both zero.

This means that a map from S$$^{3}$$ into RP$$^{3}$$ is homotopic to a map that factors through the two fold cover - I think. The homology degrees of these two maps are the same because they are homotopic. The second map has even degree because it factors through the two fold cover and the degree of a composition is the product of the degrees.

Last edited: Oct 28, 2010
19. Oct 28, 2010

### mathwonk

In reference to the orientability of RP^3, I fell asleep last night thinking it related to the fact that the antipodal map of that sphere has even degree, although now I am unsure what the rest of my thought was connecting that to the induced map on the volume form.

Now however it seems the orientability is indeed connected to the fact that the antipodal map is homotopic to the identity via a never zero tangent vector field on the sphere. I.e. by flowing along a vector field from each point to its antipodal, we can extend an orientation.....

....uhhh ok, it seems now that it is precisely the fact that the antipodal map has degree one, and is orientation preserving. Thus modding out by that map leaves the quotient space, namely RP^3, with an induced orientation. This seems to clarify the orientability or not for RP^n in all dimensions.

But I like the covering space argument, which seems easier directly from the lifting criterion via the fundamental group than from the full homotopy sequence, probably because I am more familiar with the former.

20. Oct 28, 2010

### lavinia

BTW: RP$$^{3}$$ is diffeomorphic to the tangent circle bundle of S$$^{2}$$ and so is orientable

Last edited: Oct 28, 2010