# sphere, torus, degree, cohomology

by quasar987
Tags: cohomology, degree, sphere, torus
 PF Patron HW Helper Sci Advisor P: 4,755 I am trying to show that any smooth map F from the 2-sphere to the 2-torus has degree zero. The definition of the degree of F can be taken to be the (integer) degF such that $$\int_{\mathbb{S}^2}F^*\omega=(\deg F)\int_{\mathbb{T}^2}\omega$$ For omega any 2-form on T². Another definition would be $$\deg F= \sum_{x\in F^{-1}(y)}sgn(dF_x)$$ where y is any regular value of F (meaning dF_x is invertible at every x is in F^{-1}(y)) and where the sign of det(dF_x) is +1 if the isomorphism dF_x preserves the orientation and -1 if it reverses it. So! Of course, if it were to happen for some reason that no smooth map from S² to T² can be surjective, then the result would follow because we would only have to select in the second definition a y in T² that has an empty preimage, in which case degF=0 by convention. But I don't see how could argue in this direction. Another trail is this. Consider T² as R²/Z² and let p:R²-->R²/Z² be the projection map. R² | |p | v T²<---F---S² Topologically, there exists a lift G:S²-->R² making the above diagram commutative. If such a smooth lift exists, then noting that H²(S²)=R and H²(R²)=0 and passing to the cohomology in the preceeding diagram, we would end up with the following commutative diagram: H²(R²)=0 ^ | |p* | H²(T²)---F*--->H²(S²)=R (and G*:H²(R²)=0-->H²(S²)=R). We conclude that the pullback F* of F in cohomology is identically 0. That is to say, the usual pullback pulls every 2-form to an exact 2-form (since every 2-forms on the 2-torus is closed). But on the n-sphere, the integral of an n-form vanishes if and only if the form is exact. In particular, taking omega to be an orientation (aka volume) form so that $\int\omega \neq 0$, we could conclude that degF=0. So the question is, does the smooth version of the lifting lemma used above holds??? Thanks.
 PF Patron HW Helper Sci Advisor P: 4,755 If p is a local diffeomorphism, this will undoubtedly imply that the a priori only continuous G is smooth. I'm gonna look into this.
 P: 6 I like your second approach. Since the 2-sphere is simply-connected, every map f:S^2 -> T^2 lifts to the universal cover (i.e., lifts to a continuous map F:S^2 -> R^2.) This lift is unique up to the choice of a point in the fiber above f(1,0,0), where I'm assuming (1,0,0) to be the chosen basepoint of S^2. Since the covering projection R^2 -> T^2 has smooth local sections, the lift F is smooth. Thus, you have the commutative diagram you're after. Quick question: I usually work with singular homology, in which case the degree of a smooth map f:M -> N between manifolds of the same dimension n can be defined to be the integer corresponding to the induced map f_*:H_n(M) -> H_n(N). Relative to this approach, the result (that deg(f) = 0) follows from the existence of the commutative diagram. Does this apply to the (co)homology theory you're using?
HW Helper
P: 9,371

## sphere, torus, degree, cohomology

you don't seem to need any help.
 PF Patron HW Helper Sci Advisor P: 4,755 Indeed, it turns out that p is a local diffeomorphism and approach #2 works. It often happens that the key idea comes to me as I am typing the question or a minute after I'm done.
 P: 707 here's a proof that you might like. The volume element of the torus is the wedge-product of the two closed one forms dual to its generating circles. If you think of the torus as R2/Z2 then these are just the forms dx and dy projected onto the torus. But every closed form on the sphere is exact. Thus the pullbacks of these two one forms on the torus to the sphere are exact and the integral of their wedge product is zero.
 P: 491 Don't homotopic maps induce the same map in De Rham cohomology? I would think that $$\pi_2 T^2 = 0$$ would do the trick.
P: 707
 Quote by zhentil Don't homotopic maps induce the same map in De Rham cohomology? I would think that $$\pi_2 T^2 = 0$$ would do the trick.
that also works.

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