Math Forum / Mathematics / General Topics / July 2009

Homology spheres. For example, Poincaré's; see <http://
en.wikipedia.org/wiki/Homology_sphere>.

-- m

                       ^^^^^^^^^
                   [don't they say "homotopy equivalent" any more?]

I don't know where you got this idea.

Suppose X = K(G,1). It does not necessarily hold that the
homology of X is concentrated in dimension 1. For instance,
consider X = any closed oriented surface of positive genus.
Its universal cover is the plane (up to homeomorphism; if
you're thinking Riemann surfaces, you may think of the
universal cover [for genus > 1] as the upper half plane),
and so X = K(G,1) for

  G = <a_1, b_1, ..., a_g, b_g |
      a_1 b_1 a_1^(-1) b_1^(-1) ... a_g b_g a_g^(-1) b_g^(-1) >

In any case, you get H_1 ~ Z(+)...(+)Z = Z^(2g)(2g copies),
and H_2 ~ Z. However, the space Y = K(Z^(2g),1) is the product
of 2g copies of S^1, as can be verified directly, or through
the observation that it can be realized as the quotient of R^(2g)
by the lattice Z^(2g). Further inspection shows that the homology
of Y is considerably different from that of X:

    H_q(Y) ~ Z^binomial(2g,q)

I suspect differences in homology are far more prevalent than
similarities in homology, when considering K(G,1) and K(G/[G,G],1).

You mean spaces that are the same up to homology yet differ in
homotopy?

        S^2 v S^1 v S^1    and S^1 x S^1

If you consider the classical lens spaces

(reference http://en.wikipedia.org/wiki/Lens_space)

    L(p,q) = S^3 / Z_p

where Z_p acts on C^2 via this multiplication [which I'll
denote as @], by

zeta = exp(2 pi i / p):

    zeta @ (z1,z2) |--> ( zeta*z1, zeta^q*z2 )

This action is orthogonal, so preserves S^3 as the unit sphere of C^2,
and so the quotient under the action is well-defined. According to the
following reference (Algebraic Topology: a First Course , by Marvin
Greenberg & John Harper):

http://books.google.com/books?id=X2WJ6ykRMuwC&pg=PA299&lpg=PA299

the lens spaces L(5,1) and L(5,2) are not homotopy-equivalent, yet
have isomorphic homotopy groups, isomorphic homology groups, and
isomorphic cohomology rings.

... oops, they don't differ in homotopy groups, duh.

I guess the moral of the story is that these invariants are not
sufficiently powerful to detect homotopy equivalence all by themselves.

Dale

The higher homology groups (with integer coefficients) of K(G,1) are
not, in general, trivial; in fact, they are isomorphic to the homology
groups of the group G, defined algebraically as in e.g. Lang's book on
cohomology of groups. It is also true that the cohomology groups of K
(G,1) agree with the group cohomology of G.

In general, a group cannot be expected to have the same cohomology as
its abelianization, so do not expect the map from K(G,1) to K(G/[G,G],
1) to induce an isomorphism in homology or cohomology.

What you are perhaps thinking of is the Quillen plus-construction:
given a space X, the space X^+ has homology groups isomorphic to X,
and the fundamental group of X^+ is the abelianization of the
fundamental group of X. The higher homotopy groups of X^+, however,
are not generally related in any simple way to the higher homotopy
groups of X; for example, if R is a commutative ring, then GL(R) is a
group which is the union of GL_n(R) for all positive integers n, so
the homotopy groups of K(GL(R),1) are GL(R) itself in dimension 1, and
trivial in all higher dimensions; however, the homotopy groups of K(GL
(R),1)^+ are precisely the algebraic K-theory groups of R, which are
quite complicated.