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Is a homomorphism sending an element

  1. Apr 25, 2005 #1
    How is it that with taking coefficients over a "simpler" ring, we are able to recover more information about the homology of a space? For example, if we take the projective plane as the union of a mobius band and the 2-disk glued along their boundaries, with coefficients in R we only get something in [tex]H_0(P^2)[/tex]. Over R, we get something in [tex]H_0(P^2)[/tex] and [tex]H_1(P^2)[/tex], but it is only in [tex]\mathbb{Z}_2[/tex] do we realise that the projective plane is a "true" 2-dimensional object, not being homotopically equaivalent to an object of lower dimension.

    We have in the Mayer-Vietoris sequence a section that looks like

    [tex]0 \rightarrow H_2(P^2) \rightarrow H_1(S^1) \rightarrow H_1(D^2)\oplus H_1(M) \rightarrow ...[/tex]

    And taking coefficients over [tex]\mathbb{Z}_2[/tex] we get

    [tex]0 \rightarrow H_2(P^2) \rightarrow \mathbb{Z}_2 \rightarrow \mathbb{Z}_2 \rightarrow ... [/tex]

    Where the homomorphism from [tex]\mathbb{Z}_2 \rightarrow \mathbb{Z}_2[/tex] defined by sending 1 to 2, which is due to the circle wrapping twice around the mobius band, which means that the map from [tex] H_2(P^2) \rightarrow \mathbb{Z}_2[/tex] is an isomorphism. I guess my question is that is there any other reason that [tex]\mathbb{Z}_2[/tex] is able to pick out this extra structure in homology than other rings than that there is a homomorphism sending an element d in the ring to 2d? A more geometric interpretation?
  2. jcsd
  3. Apr 26, 2005 #2

    matt grime

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    Well, the reason for this kind of thing is, I suppose, because if we pick R, it has no torsion, and every map between fields is either an isomorphism or is null. So it is less to do with the nature of the space then the algebraic sturcture of the ring of coefficients.

    There are the Universal Coefficients theorem and the Kunneth Formula that also sa similar things. Roughly that coefficients in Z contain all the information. Indeed in the above if you used Z not Z_2 you could work out that the first homology group is not trivial, and is still Z_2, since its abelianization is the fundamental group.

    Perhaps someone else has a better explanation for this, but to me R is the simplest ring since it has no non-trivial endomorphisms.
  4. Apr 26, 2005 #3
    Thanks for the explanation. I guess I was looking at one ring being "simpler" than another the wrong way around.
  5. Apr 26, 2005 #4


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    I assume you mean continuous endomorphisms -- it has field automorphisms up the wazoo!

    As for the original question, I'd only be stabbing in the dark, but I'd say the same thing as matt. Z is sort of universal in the fact it can act nontrivially on every ring. So, even if the "best" ring to use for a particular problem was something else, Z should still be able to tell you something interesting.
  6. Apr 26, 2005 #5

    matt grime

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    No, I meant trivial to exclude automorphism: every ring homomorphism from R to R (ie endomorphism) is an automorphism or is null, in fact the same is true for any field. Which really does make it "lucky" that you've picked Z_2 as the choice of coefficients in your example since that means you can actually get a non-trivial first homology group.

    The geometric reason is, I suppose, becuase the first homology group ought to measure the fundamental group in some way via some universal property. The fundamental group is Z_2, so that's a sort of geometric argument.

    And remember that it is not objects that is important in maths but maps between objects, and maps between maps between objects.
  7. Apr 26, 2005 #6


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    I think Matt put his finger on the key element which is torsion. I.e. the projective has some torsion in its geometry, hence trying to mirror its geometry within algebra works better if the algebraic object also has torsion.

    Consider the double cover of the projective plane by the sphere. and consider the image in the projective plane of an arc of a great circle on the sphere joining two antipodal points. This arc maps to a closed loop on the projective plane, which is not homotopic to a constant in the projective plane. But the full great circle on the sphere maps to twice this loop.

    Since the rgeat circle on the sphere is homotopic to a constant, so is tis image on the projective plane. hence that loop on the projective is an example of geometric torsion.

    since a ring of characteristic zero has not torsion it cannot capture this.

    so as hurkyl suggested, the most universal ring is Z, the integers. homology and cohomology over other rings is derived from that one, and the process, tensoriong, actually loses information, in particular tensoring with Q or R loses torsion.

    on the other hand R has one feature that Z2 does not, ordering - it can distinguish between +1 and -1.

    so for some purposes, such as detecting the non existence of never zero vector fields on a sphere, R works better than Z2.

    i.e. the existence of a never zero vector field on the 2 sphere would imply that 1 = -1 in the cohomology group, which is a contradiction in R, but not in Z2.
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