1
$\begingroup$

I am working over the paper: Target Enumeration via Euler Characteristic Integrals and in order to follow a proof I need to prove:

If $A$ is compact nonempty subset of $\mathbb{R}^2$, then the singular homology groups of $A$, $H_k(A)$ vanish for $k\geq 2$.

The result seems reasonable for me (there are non obvious counterexamples in dimension greater than two but not in dimension two). But I am having difficulties finding a proof.

What I have tried or thought so far:

  • Using the long exact sequence in homology and it didn't work.
  • Maybe trying luck with de Rham cohomology and using some kind of isomorphism followed by the Universal Coefficients theorem in cohomology....
  • Using some dimension theory but I have no clue how...

Any ideas? Thanks in advance and any help would be appreciated.

  • 3
    It is not really *common sense* as it is false for compact subsets of $\mathbb R^3$... There is a famous example of Barratt and Milnor, which you can find [here](http://www.ams.org/journals/proc/1962-013-02/S0002-9939-1962-0137110-9/S0002-9939-1962-0137110-9.pdf) They give subset of $R^3$ with infinitely many nonzero rational homolog groups, and notce that it has integral homology groups above the expected dimension in some range, and say that it probably also works i infinitely many degrees. – 2017-02-26
  • 0
    Yes, you are right @MarianoSuárez-Álvarez as there is a counterexample for dimension greater than two. But I precisely meant that as there were no counterexamples for dimension two it should be true. However you are right. – 2017-02-26
  • 1
    My point is that it is very likely that whatever you had in mind under the phrase of "common sense" also applies to higher dimensions, and it fails there... So common sense has nothing to do wth this. Referring to common sense in relation to this only serves to confuse matters more. – 2017-02-26
  • 0
    @MarianoSuárez-Álvarez The negation of the statement is: "there exists $k\ge 2$ such that $H_k(A)\neq 0$". It is satisfied by the 2-sphere. So although I agree this is not the meaning of "common sense", we can say "there are obvious counterexamples in dimension $\ge 3$". Second, the "common sense" seems to refer to the fact it should hold in dimension 2, not the parenthesis saying the obvious remark in dimension $\ge 3$. – 2017-02-26
  • 0
    Crossposed to MathOF: http://mathoverflow.net/questions/263178/homology-groups-of-compact-subset-of-mathbbr2 – 2017-02-26
  • 0
    The $2$-sphere is not a subset of $\mathbb R^2$, as far as I can tell... – 2017-02-26
  • 0
    Iobject to the idea that it is common sense that the claim should hold in dimension 3, because I find it very unlikely that any "argument" (which is not a proof) that it should hold in dimension 2 should not also apply to dimension 3. On the other hand, there is nothing obvious about the fact that the claim is false in dimensions $\geq3$: Barratt and Milnor wrote a whole paper to give an example, and it systematically is found surprising by people learning the subject. Moeover, the claim *is* true for other cohomology theories, of course. – 2017-02-26
  • 0
    Sounds like the problem is the phrase "common sense." Seems to me one theme of modern mathematics is that "common sense" doesn't apply to compact, nonempty subsets of $\Bbb{R}^n$ for any $n>0$, so I would suggest OP remove it from the question. – 2017-02-26
  • 0
    @MarianoSuárez-Álvarez yes, it's a subset of the 3-space and shows that the main statement does not hold in $\mathbb{R}^3$ instead of $\mathbb{R}^2$. Other remarks of the same level? – 2017-02-26
  • 0
    The statement that's displayed in the question, @ycor, has two appearances of the number 2, and it should be obvious that the generalized statement corresponding to higher dimensions d is the one in which *both* appearances of 2 are replaced by d. The other statement, the one obtained by replacing the *first* 2 by a different number —and to which the 2-sphere surely does give a counterexample, as you remark— is indeed obviously false. No one was talking about *that* statement, so while true, your remark is quite unconnected with the matter at hand. – 2017-02-26
  • 0
    Yes it's another interpretation. Nevermind. – 2017-02-26

0 Answers 0