Special properties of 'Orientability' of Manifolds ($R$-orientable but not $\Bbb Z$-orientable)

Question

I'm currently self-studying topology, and this a question regarding the orientability of a (closed) manifold that I've been exploring recently by myself.

Motivation:

We know that an $n$-manifold $M$ is $R$-orientable if and only if $H_n(M; R)\cong R$. Replacing the general ring with identity by $\Bbb Z$ gives us the orientability in the usual sense. With integer coefficient, there are plenty of examples of ($\Bbb Z$-) orientable manifolds (e.g. $S^n, T^n$, $\Bbb CP^n$, etc.) and non-orientable manifolds (e.g. $\Bbb RP^n$, "Klein Bottle", "Möbius Strip", etc.) With $\Bbb Z/2$ coefficient, every manifold admits a unique orientation, which is not so interesting, but what about $\Bbb Z/n$ in general?

Question:

• Is there some examples of (closed) manifolds such that they're $R$-orientable ($R\neq\Bbb Z/2$) but NOT $\Bbb Z$-orientable? What about the other direction ($\Bbb Z$-orientable but not $R$-orientable, $R\neq \Bbb Z/2$)?
• Is there a lower dimensional example? (I think this is unlikely, at least in dimension $2$ and $3$)
• Can we construct one by observing some properties that orientability implies?

Some Observation:

The first thing I noticed is the homology groups and the coefficient groups of the form $\Bbb Z/n$. Suppose $M$ is $\Bbb Z/n$-orientable but not $\Bbb Z$-orientable, then by UCT, we have $$H_n(M; \Bbb Z/n)\cong (H_n(M;\Bbb Z)\otimes\Bbb Z/n)\oplus\operatorname{Tor}(H_{n-1}(M;\Bbb Z);\Bbb Z/n)\cong \operatorname{Tor}(\Bbb Z^b\oplus T;\Bbb Z/n) $$ Because $M$ is not $\Bbb Z$-orientable, the torsion of $H_{n-1}(M;\Bbb Z)$ is $\Bbb Z/2$, we know that by applying the usual free resolution of $\Bbb Z/n$, $$H_n(M; \Bbb Z/n)\cong \operatorname{Tor}(\Bbb Z/2,\Bbb Z/n)\cong\ker(\Bbb Z/n\overset{\times 2}{\to}\Bbb Z/n)\cong \Bbb Z/{\operatorname{gcd}(2,n)}$$ To make the top homology isomorphic to $\Bbb Z/n$, we want $\operatorname{gcd}(2,n)=n$, but it's impossible for $n\ge 2$. So, no such thing exists with $\Bbb Z/n$, $n\ge 2$.

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The next class of coefficient rings are probably polynomial rings $R[x]$ or even things like $\Bbb Z[i]$ or $\Bbb Z[\omega]$ and lots of other crazy things, but I'm not sure how to deal with those rings. Any suggestions?

Thanks in advance. ;-)

Answer 1

"However, I'm not sure about whether this is correct as I acquired all my knowledge through self-study."

This seems irrelevant to me. However it is you study you should be able to think critically about your own arguments. Your argument is correct.

In the following "oriented" means "$\Bbb Z$-oriented".

Hatcher, circa Corollary 3.28. If $M$ is an-oriented $n$-manifold, then $H_n(M; \Bbb Z) \cong \Bbb Z$. Otherwise, $H_n(M; \Bbb Z) = 0$ and $H_{n-1}(M;\Bbb Z) \cong \Bbb Z^a \oplus \Bbb Z/2$ for some integer $a \geq 0$.

(This is not quite the statement given there, but you can assemble it from the given theorems and the classification of finitely generated abelian groups.)

It follows from the calculation you sketch that any abelian group $A$ we have either $H_n(M;A) \cong A$ or $$H_n(M;A) \cong A/2A$$ for any abelian group $A$. So if $R/2R \not\cong R$ we have that $M$ is $R$-orientable iff $M$ is orientable.

A more careful argument using the actual definition of orientability (in terms of the local homology at every point) shows that for any ring $R$ whatsoever one has (i) if $2R = 0$ then every manifold is $R$-orientable (proof is straightforward), or (ii) if $2R \neq 0$ then a manifold is $R$-orientable if and only if it is $\Bbb Z$-orientable.

Thus the notion of $R$-orientability is entirely captured by the group-theoretic properties of the underlying additive group $R_+$ and the orientability of $M$ in the traditional sense.

The notion of an $R$-orientation is more complicated, as a connected orientable manifold has a single $R$-orientation for each element of $R^\times$.