Subject: Re: question abt ratl homology spheres
Date: Tue, 15 Oct 2002 15:08:54 -0400
From: taylor.2@nd.edu
To: <dmd1@lehigh.edu>

Hajlasz's remarks and question 2 suggests a complete answer
to the question.

There exists a non-zero degree map $f:S^n\to M^n$
if and only if the universal cover of $M$ is a rational
homology sphere.

pf: If $M$ has infinite fundamental group, several people
pointed out the degree of $f$ is $0$, so $\pi_1(M)$ is finite.
The map $f$ factors through the universal cover and must have
non-zero degree so as Hajlasz observed it's a rational homology
sphere. But conversely, Hajlasz also observed that there is a
map of non-zero degree to a simply-connected rational homology
sphere and then the composition with the covering projection will
still have non-zero degree.

How common or rare is it to have a universal cover a rational
homology sphere? The order of the deck transformation group
must divide the Euler characteristic, so if $n$ is even, the
group has order $2$ and a Lefchetz Fixed point argument shows
that $M$ is non-orientable and therefore has no notion of degree.

Hence in the case $n$ even, $M$ has a non-zero degree map from
a sphere if and only if $M$ is a simply-connected rational homology
sphere.

In the odd dimensional case there are typically lots of non-simply
connected manifolds with non-zero degree maps from a sphere.
Take any spherical space form and connect-sum on a simply-connected
rational homology sphere. As Hatcher suggests, there are not likely
to be examples in dimension 3 except the spherical space forms, but
by dimension five there are lots of simply-connected rational homology
spheres. Pardon, Math. Z. 171 (1980) 247-268, constructs lots of examples
with $\pi_1$ not a spherical space form group.

As several people pointed out, Hajlasz's Question 2 has a negative answer
as soon as $\pi_1(M)$ is infinite, but the answer is still negative in
general even if $\pi_1(M)$ is finite, but not $0$. In every dimension
$n>=6$ and for any finite group $G$, there exist rational homology spheres
of dimension $n$ with $\pi_1=G$ whose universal cover is not a rational
homology sphere.

To construct some examples, take a finite 2-complex, $K'$ with $\pi_1=G$.
We always have $\pi_2(K') \otimes Q\to H_2(K';Q)$ onto since $G$ is finite.
Attach 3-cells carefully to get $K$ with $\pi_1(K)=G$ and
$H_i(K;Q)=0$ for $i>0$, ie. $K$ is a rational disk.
Embed $K$ in $R^{2n+1}$, $n>=3$, and take a regular neighborhood $W$.
Sometimes we take $K$ to be a 2-complex (eg $G$ cyclic) and then  we can
take $n=2$. $W$ is rational disk and $\partial W$ is a rational homology
sphere with $\pi_1(\partial W)=G$.

An Euler characteristic argument shows the universal cover of $\partial W$
is not a rational homology sphere. Indeed, no even dimensional rational
homology sphere with non-trivial fundamental group has a universal
cover which is a rational homology sphere.

To construct odd dimensional examples, take the double of $W$, $D(W)$.
By Mayer-Vietoris, $D(W)$ is a rational homology sphere. Its universal
cover is the double of $U$, the universal cover of $W$. By an Euler
characteristic argument, $U$ is not a rational homology disk and the
cohomology of $U$ is a summand of the cohomology of $D(U)$ so the universal
cover of $D(W)$ is not a rational homology sphere.