So how do I calculate fundamental groups?

**DJDorianGray** · October 21st 2009, 01:34 AM

Hi all,
So I have read the theory of fundamental groups on a couple of introductory textbooks, but I still don't have the slightest idea of how I would go about finding the fundamental group of a path connected space. For example:

"Find the fundamental group of $\mathbb{R}^3 \backslash l$ , where $l$ is a line".

Could anyone help me through this? I guess I should try to show that the space is homotpy equivalent to something whose group is well-known, such as $S^1$ right? The truth is I don't know where to start. Any help is really appreciated. Thanks.

**HallsofIvy** · October 21st 2009, 03:50 AM

Draw closed curves in the space and see which ones are "equivalent"- that is, which ones can be continuously deformed into each other. In the plane, all curves can be deformed into each other so the fundamental group is trivial- it consists of a single element. On a torus, all closed curves are of three kinds- those that can be deformed down to a single point, those that go around the "long" circumference of the torus, and those that go around the "short" circumference of the torus. So the fundamental group of the torus has three members and is isomorphic to the rotation group of a triangle.

In the case of $R^3/l$ , where l is a line, there are two "equivalence classes" of closed curves- those that go around the line and those that don't.

**aliceinwonderland** · October 21st 2009, 06:57 AM

Originally Posted by DJDorianGray

Hi all,
So I have read the theory of fundamental groups on a couple of introductory textbooks, but I still don't have the slightest idea of how I would go about finding the fundamental group of a path connected space. For example:

"Find the fundamental group of $\mathbb{R}^3 \backslash l$ , where $l$ is a line".

Could anyone help me through this? I guess I should try to show that the space is homotpy equivalent to something whose group is well-known, such as $S^1$ right? The truth is I don't know where to start. Any help is really appreciated. Thanks.

$\mathbb{R}^3 \backslash l$ deformation retracts to the $\mathbb{Re}^2$ with a point removed, which is homotopy equivalent to the $S^1$ like your guess.

**DJDorianGray** · October 21st 2009, 09:32 AM

Thanks guys, now I see what the idea is. I'll write down an homotopy between my set and $\mathbb{R}^2$ minus the origin as alice suggested.

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