Homology of Euclidean space $H_n(\mathbb{R}^m)$

Question

I've been computing some singular homology groups of different spaces. In particular, I know how to compute the homology of a cell complex. Now I'm wondering how to compute the homology of $\mathbb{R}^m$. Since homology is a way of counting holes and $\mathbb{R}^m$ doesn't have any I guess $H_n(\mathbb{R}^m) = 0$ for all $n,m$.

But how do I rigorously compute this? Thanks for your help.

Edit: I think I can use that $\mathbb{R}^n$ is contractible and then $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.

Do you already know that homotopic maps yield the same morphisms in homology? If you do, you can calculate the homology groups of $\mathbb{R}^m$ in no time! — Olivier Bégassat, Aug 13 '11 at 15:32
Have you already seen the fact that if two spaces are homotopy equivalent then their homologies agree? — Jason DeVito - on hiatus, Aug 13 '11 at 15:32
Remember that the rank of $H_0(\mathbf R^m)$ is the number of path components of $\mathbf R^m$. — Dylan Moreland, Aug 13 '11 at 15:40
I wonder if the Jordan curve theorem could be used here; I know it is like using a tank to kill a fly, but I wonder if JCT says that every cycle (simple closed curve; given an orientation) bounds. — gary, Aug 13 '11 at 16:20

Rudy the Reindeer · Accepted Answer · 2011-08-14T09:05:54.893

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$\mathbb{R}^n$ is contractible therefore homotopy equivalent to a point and so $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.

$$ H_n(\{ \ast \}) = 0 , n > 0$$ $$ H_n(\{ \ast \}) = \mathbb{Z} , n = 0$$

edited Aug 14 '11 at 09:05

answered Aug 13 '11 at 15:55

Rudy the Reindeer

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What is $k$? Perhaps you mean $n$. – user02138 Aug 14 '11 at 07:20
Yes, I do, thanks. Sorry, I used $n$ here and $k$ on paper : P – Rudy the Reindeer Aug 14 '11 at 09:05

gary · Answer 2 · 2011-08-13T16:43:04.613

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I think this may be an alternative proof; maybe overkill, but I think it may work:

I think cycles C on X can be geenralized as injective continuous , injective maps

$S^n\rightarrow X$. JCT states that (the image of ) C separates the plane into two

connected regions with C as the boundary; the interior of C can then be seen as being

bounded by the cycle, C, so that we can then say that every cycle C in $\mathbb R^n$

bounds, and so the homology of $\mathbb R^n $is trivial.

Edit: as Grigory pointed out, my idea was too simple and too good to be true ; e.g.,genus-g surfaces cannot be represented as images of spheres. Maybe my argument can be shown that $\pi_n(\mathbb R^n)=0$, but I am not betting on it.

edited Aug 13 '11 at 16:43

answered Aug 13 '11 at 16:33

gary

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Not every homology class can be realized as an image of a sphere (think, say, of $H_2$ of surfaces with $g>1$)... – Grigory M Aug 13 '11 at 16:37
But, would Sg , I guess as a subbspace of $\mathbb R^n $ be considered a cycle? – gary Aug 13 '11 at 17:48
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Yes, of course. (But beware: in general, not every cycle can be realized by manifold — let alone by embeded manifold.) – Grigory M Aug 13 '11 at 17:54
What is then the general definition of a cycle? – gary Aug 13 '11 at 18:00
Ditto for you on that. – gary Aug 13 '11 at 18:54
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Actually, the idea of yours works nicely in the opposite direction: canonical proof of higher-dimensional JCT uses vanishing of $H(\mathbb R^n)$ explicitly (and ordinary JCT, one might argue, implicitly); in fact, it's an instance of Alexander duality. – Grigory M Aug 13 '11 at 19:41
Grigory: I guess my question is somewhat imprecise, in that the definition of cycle may depend on the choice of homology theory used. And, yes, I should have been more careful before posting, but sometimes posters are aggresive and they use one's observation to post their own answers. – gary Aug 13 '11 at 21:13
@gary: thanks anyway, for trying! – Rudy the Reindeer Aug 14 '11 at 06:55

Homology of Euclidean space $H_n(\mathbb{R}^m)$

2 Answers2