14
$\begingroup$

I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry.

Looking at the part regarding the orientation of a topological $n$-manifold $M^n$, at page 341 we find the following well-known result, with its usual proof (Proposition 7.1):

enter image description here

So far, so good. However, after five pages we find what follows:

enter image description here

This makes me confused, for at least two reasons:

Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol ${}_2G$ denotes the $2$-torsion part of the abelian group $G$, so if $G$ is torsion-free (for instance, if $G=\mathbf{Z}$) then ${}_2G=0$. This is clearly very different from the free-product $G \ast \mathbf{Z_2}$ (here $\ast$ seems to denote the free-product, see pages 158-159).

Point 2. In Corollary 7.11, take $A=\{x\}$ and $G=\mathbf{Z}$. Then, when $M$ is not orientable one finds $H_n(M, \, M-\{x\}, \, \mathbf{Z})=0$, and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $H_n(M, \, M-\{x\}, \, \mathbf{Z})= \mathbf{Z}$.

Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?

Improve this question
$\endgroup$
14
  • 3
    $\begingroup$ I guess $*$ might be a typo, it would rather be some sort of $\otimes$. $\endgroup$
    – Dima Pasechnik
    Jan 24, 2019 at 11:42
  • 4
    $\begingroup$ Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along $\{x\}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds. $\endgroup$
    – Denis Nardin
    Jan 24, 2019 at 11:59
  • 9
    $\begingroup$ Also it seems that Bredon indicates with $\ast$ what I would call $\mathrm{Tor}_1$, so in particular $A\ast \mathbb{Z}/n$ is exactly the $n$-torsion of $A$. $\endgroup$
    – Denis Nardin
    Jan 24, 2019 at 12:07
  • 3
    $\begingroup$ Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks! $\endgroup$
    – Francesco Polizzi
    Jan 24, 2019 at 12:09
  • 6
    $\begingroup$ @GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough). $\endgroup$
    – Francesco Polizzi
    Jan 24, 2019 at 14:51

2 Answers 2

Reset to default
35
$\begingroup$

Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := \operatorname{Tor}_{\Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.

Improve this answer
$\endgroup$
7
  • 8
    $\begingroup$ Thanks. I was not aware of this (old) notation. $\endgroup$
    – Francesco Polizzi
    Jan 24, 2019 at 13:34
  • 5
    $\begingroup$ (This notation is also used in Spanier's text, for example.) $\endgroup$
    – Pedro
    Jan 24, 2019 at 13:51
  • 9
    $\begingroup$ @FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :) $\endgroup$
    – Denis Nardin
    Jan 25, 2019 at 12:22
  • 3
    $\begingroup$ Bredon carefully defines A*B on page 278 in a section called More homological algebra and even says that it is often denoted Tor(A,B). $\endgroup$
    – Nicholas Kuhn
    Jan 25, 2019 at 17:31
  • 7
    $\begingroup$ Yes, but again: the book is 550 pages long, one that is just interested in one topic cannot read carefully all of it. Since at page 158 Bredon uses $*$ to indicated to free product, I supposed (maybe ingenuously) that this notation would have been kept in all the book, even because I never met the notation $*$ for $\mathrm{Tor}_1$ before. $\endgroup$
    – Francesco Polizzi
    Jan 25, 2019 at 18:55
9
$\begingroup$

I think that you are missing the definition of 'orientable along $A$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $A$' means that if you move a local orientation of $M$ around a closed path that stays in $A$ then it will come back to the same local orientation. In particular, in the case when $A$ is a single point, then $M$ will always be orientable along $A$, regardless of whether $M$ is orientable or not, so the case that you view as wrong doesn't arise.

I agree with Denis T's interpretation of the notation $A*B$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $\mathrm{Tor}_1(A, \, B)$. $\endgroup$
    – Francesco Polizzi
    Jan 25, 2019 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.