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I've been looking for a space on the internet for which I cannot write down the homology groups off the top of my head so I came across this:

Compute the homology of $X : = \mathbb{R}^3 - S^1$.

I thought that if I stretch $S^1$ to make it very large then it looks like a line, so $\mathbb{R}^3 - S^1 \simeq (\mathbb{R}^2 - (0,0)) \times \mathbb{R}$. Then squishing down this space and retracting it a bit will make it look like a circle, so $(\mathbb{R}^2 - (0,0)) \times \mathbb{R} \simeq S^1$. Then I compute

$ H_0(X) = \mathbb{Z}$

$ H_1( X) = \mathbb{Z}$

$ H_n(X) = 0 (n > 1)$

Now I suspect something is wrong here because if you follow the link you will see that the OP computes $H_2(X,A) = \mathbb{Z}$. I'm not sure why he computes the relative homologies but if the space is "nice" then the relative homologies should be the same as the absolute ones, so I guess my reasoning above is flawed.

Maybe someone can point out to me what and then also explain to me when $H(X,A) = H(X)$. Thanks for your help!

Edit $\simeq$ here means homotopy equivalent.

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"if I stretch $S^1$ it looks like a line" is true, only your stretching is not a continuous function. So your conclusion $\mathbb R^3 \setminus S^1 \simeq (\mathbb R^2 \setminus (0,0)) \times \mathbb R$ is not justified. Moreover, it's not true, provided $\simeq$ means homeomorphic or homotopy-equivalent. – Ryan Budney Aug 14 '11 at 9:13
$\mathbb R^3 \setminus S^1$ has non-trivial $H_2$, for example. – Ryan Budney Aug 14 '11 at 9:14
It's also entertaining to compute $\pi_2 (\mathbb R^3\setminus S^1)$. – Ryan Budney Aug 14 '11 at 9:14
I suggest not trying to follow the physics forum thread. Write $X$ as the union of two contractible subspaces whose intersection you understand, and apply Mayer-Vietoris. – Ryan Budney Aug 14 '11 at 9:17
@palio: The homology of the complement is independent of the embedding. And Matt is only asking about homology, so although your comment about $\pi_1$ is correct it's not what Matt is asking about. – Ryan Budney Aug 14 '11 at 9:47
up vote 5 down vote accepted

Consider $X = \mathbb{R}^3 \setminus (S^1 \times \{ 0 \}) \subseteq \mathbb{R}^3$, $U = X \setminus z \text{-axis}$ and $V = B(0,1/2) \times \mathbb{R}$, where $B(0, 1/2)$ is the open ball with radius $1/2$ and center in the origin of $\mathbb{R}^2$. It is clear that $\{ U, V \}$ is an open cover of $X$.

Now let's compute the homotopy type of $U$. Consider a deformation $f$ of $((0, +\infty) \times \mathbb{R}) \setminus \{(1,0) \}$ onto the circle of center $(1,0)$ and radius $1/2$. Let revolve $f$ around the $z$-axis: obtain a deformation of $U$ onto the doughnut ($2$-torus) of radii $R=1$ and $r = 1/2$.

Since $V$ is contractible and $U \cap V$ is homotopically equivalent to $S^1$, Mayer-Vietoris sequence gives the solution: $H_0(X) = H_1(X) = H_2(X) = \mathbb{Z}$ and $H_i(X) = 0$ for $i \geq 3$.

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How do you compute $H_1$ and $H_2$? I'm stuck : ( Could you expand on your answer please? – Rudy the Reindeer Aug 16 '11 at 7:54
Have you written down Mayer-Vietoris' sequence? – Andrea Aug 16 '11 at 9:33
Yes, look I posted a follow up question here:… – Rudy the Reindeer Aug 16 '11 at 9:48

Another way...

Consider $S^3$ as Alexandrov's compactification of $\mathbb{R}^3$: $S^3 = \mathbb{R}^3 \cup \{ \infty \}$. The set $X = \mathbb{R}^3 \setminus (S^1 \times \{ 0 \})$ can be seen as the complement in $S^3$ of the union $Y$ of a circle $S^1$ and of a point $P$. Since $S^3$ is homogeneous, we can suppose that $\infty$ is a point of $S^1$, so $$ Y = S^3 \setminus (S^1 \cup \{ P \} ) = \mathbb{R}^3 \setminus ((S^1 \setminus \{ \text{a point} \}) \cup \{ P \}. $$ Therefore $X$ is homeomorphic to the complement $Y$ in $\mathbb{R}^3$ of the union of a line and of a point: $Y = \mathbb{R}^3 \setminus (\{ x=y=0 \} \cup \{(1,0,0) \})$. Probably applying Mayer-Vietoris to $Y$ is easier than to $X$.

However it can be seen that $Y$ is homotopically equivalent to the very simple CW-complex $S^1 \vee S^2$, so computing homology is easy using cellular homology.

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