Let $T$ be a solid torus, how to calculate the fundamental group $\pi_1(\mathbb R^3- T)$? intuitively, I think it's a free group with one generator. So if it is so, how to obtain it, and what the generator would be?

  • $\begingroup$ Why do you think it's free on one generator? $\endgroup$ Jul 14, 2013 at 16:08
  • $\begingroup$ if it helps, shrink it down to $\mathbb{R}^3 \setminus S^1$ and try using seifert van kampen $\endgroup$ Jul 14, 2013 at 16:32
  • $\begingroup$ @ChrisEagle it is not? $\endgroup$
    – Ronald
    Jul 14, 2013 at 16:43
  • $\begingroup$ I didn't say it is or it isn't. I asked why you think it is. Are you going to answer? $\endgroup$ Jul 14, 2013 at 16:44
  • $\begingroup$ I dont know actually. I just have the sense of that $\endgroup$
    – Ronald
    Jul 14, 2013 at 16:45

1 Answer 1


I'll offer some hints as I think this is an exercise that you should get some practice with to really understand some of the tools being used.

As has already been mentioned a few times, it is probably easier to think of this space $X=\mathbb{R}^3\setminus T$ as the homotopy equivalent space $\mathbb{R}^3\setminus S^1$ where $S^1$ is an unknotted circle embedded in $\mathbb{R}^3$ (It's interesting to note that the qualifier 'unknotted' here is very important. If $S^1$ is knotted, then the fundamental group of its complement in $\mathbb{R}^3$ is known as the knot group of the knot and can be very different from the knot group of the unknot).

There are a few ways to tackle the problem now. Probably the easiest computationally (although perhaps hardest to visualise) is to note that $\mathbb{R}^3\setminus S^1$ is homotopy equivalent to the wedge sum of a $2$-sphere and a circle. That is $\mathbb{R}^3\setminus S^1\simeq S^2\vee S^1$. To see this, deformation retract $\mathbb{R}^3\setminus S^1$ on to the subspace $B\setminus S^1$ where $B$ is a solid ball containing $S^1$. Now, 'push' the negative space of the removed $S^1$ to the boundary of the ball so that we have the space $S^2\cup I$ where $I$ is an interval connecting the north and south pole of the sphere. You can now drag one end of the interval around to the other end so that it becomes a circle, giving $S^2\vee S^1$. The rest is just some simple applications of Seifert-van Kampen and knowing the fundamental groups of usual spaces.

Another approach would be to 'slice' $3$-space along a plane which also cuts the circle, leaving you with two half spaces, each with a removed interval whose ends are on the boundary of the half-spaces. If you 'thicken' these spaces up so that their intersection is open in $\mathbb{R}^3\setminus S^1$, then you can apply Seifert-van Kampen directly to this union. The only tricky part here is working out which elements in the fundamental group of the intersection correspond to the normal subgroup you will factor out of the free product.

One other approach might be to use the space that another user constructed in his answer.

EDIT that answer referenced was deleted so I will just say that it is the product of the space $S^1$ with $H\setminus \{p\}$ where $H$ is an open half plane, $p\in H$ and we view this space as a 'surface of revolution' in $\mathbb{R}^3$, which is homotopy equivalent to $(\mathbb{R}^3\setminus S^1)\setminus l$ where $l$ is an infinite line which passes through $S^1$.

If you then consider that the union of this space with an open tubular neighbourhood of $l$ is exactly the space you are considering, then another application of Seifert-van Kampen should give you the fundamental group of your space readily (I haven't written this down so I'm not sure if there will be any difficulties in this decomposition, but I can't see why there would be).

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    $\begingroup$ As a topology noob, is there a good reference explaining Seifert-van Kampen and how to apply it? The Wikipedia article (en.wikipedia.org/wiki/Seifert%E2%80%93van_Kampen_theorem) unfortunately gets bogged down in extremely intimidating category theory... $\endgroup$
    – user7530
    Jul 15, 2013 at 3:43
  • $\begingroup$ @Daniel Rust, that was helpful, thanks a lot. But I need more explanation please to understand how the dragging of the end points of $I$ gives $S^2\vee S^1$. My imagination of $ S^2\vee S^1$ is $S^1$ joining $S^2$ in just one point! But dragging the line inside $S^2$ makes a circle inside it.. $\endgroup$
    – Ronald
    Jul 15, 2013 at 5:42
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    $\begingroup$ @user7530 I agree there's a lot more category theory on the wiki page than a beginner can hope to handle. Almost every intro book to algebraic topology will have a section on the Seifert-van Kampen theorem, and indeed most online lecture notes. The most often cited, and easiest to access is Allen Hatcher's Algebraic Topology which is kindly made freely available on his website math.cornell.edu/~hatcher/AT/ATpage.html. $\endgroup$
    – Dan Rust
    Jul 15, 2013 at 9:50
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    $\begingroup$ @Danial I suppose it is hard to imagine at first. You need to place a contractible neighbourhood on the sphere which contains both end points of the interval and then shrink that entire neighbourhood to a point. The effect is a homotopy between the two mentioned spaces because both end points of the interval are now connected to the same point on a sphere which has had a contractible subspace shrunk to a point (which is just a sphere again). $\endgroup$
    – Dan Rust
    Jul 15, 2013 at 10:00

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