By John Klein


2013-07-29 23:16:35 8 Comments

The wikipedia page on Borel-Moore homology claims to give several definitions of it, all of which are supposed to coincide for those spaces $X$ which are homotopy equivalent to a finite CW complex and which admit a sufficiently nice embedding into a smooth manifold.

My question concerns one of the definitions on that page: it is asserted that if $X$ is a space and $\bar X$ is some compactification, then there is a definition of Borel-Moore homology given by $$ H_*^{\text{BM}}(X) := H_*(\bar X,\bar X \setminus X) . $$ The entry additionally requires that the pair $(\bar X,X)$ should be a CW pair for this to work.

It then goes on to assert that $\bar X = X^+$, the one-point compactification, will suffice for this.

Frankly, I don't see how $(X^+,X)$ will form a CW pair in most instances. For example suppose $X = \Bbb Z$ is the set of integers which is considered as a discrete space. Then in this instance $(X^+,X)$ certainly fails to be a CW pair.

Another example: $X = S^1 \setminus \ast$ with one point compactification $S^1$. Then $(S^1,S^1 \setminus \ast)$ is not a CW pair either.

My Questions:

(1) For what class of spaces $X$ does $H_*(X^+,+)$ coincide with definition of Borel-Moore homology given by locally finite chains?

(2) Does the wikipedia entry contain a mistaken assumption? Perhaps we do not need to assume that $(X^+,X)$ can be given the structure of a CW pair?

Note: if $X = \Bbb Z$ and if we use the above as a definition of Borel-Moore homology then $ H_0^{\text{BM}}(X) $ is a free abelian group whose generators are given by the underlying set of $\pi_0(X) = \Bbb Z$. This is clearly the wrong answer: it should be the countably infinite cartesian product of copies of the integers indexed over $X$ instead (using, say, the definition of Borel-Moore homology given by locally finite chains).

Another Question:

(3) Is there a definition the above kind (using compactifications) which will work (i.e., coincide with the locally finite chain definition) for $X = \Bbb Z$?

(I suspect not, since ordinary singular homology in degree zero is always free abelian.)

Incidentally, later in the page it lists the main variance property: Borel-Moore homology is supposed to be covariant with respect to proper maps. The page gives a proof using the above definition. But since the above definition doesn't work in general, I don't see how this is really supposed to be a proof.

1 comments

@Ben Webster 2013-07-30 04:41:31

In Chriss & Ginzburg, it's asserted that the Borel-Moore homology is given by $H_*(\bar X, \bar X\setminus X)$ if $\bar X$ is the one-point compactification or any compactification such that the inclusion is cellular. They very pointedly do not assume that the inclusion of $X$ into $X^+$ is cellular. They reference Bredon but don't cite a particular chapter.

@John Klein 2013-07-30 12:53:58

Ben: I guess you mean that $\bar X \setminus X \to \bar X$ should be cellular. That's a reasonable condition, provided a cell structure on $\bar X$ is given. It seems then that the wikipedia entry has an error then.

@John Klein 2013-07-30 13:26:04

Ben: I just checked Chriss & Ginzburg. It is clear now that the wikipedia entry has a mistake.

@Ricardo Andrade 2013-07-30 22:56:27

I took the liberty to correct the wikipedia page.

@Michał Kukieła 2013-08-16 15:50:33

Another reference (with Google Books preview available) is "Representation Theories and Algebraic Geometry" books.google.pl/books?id=Q0Nyon7-qDMC , pp. 129-130

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