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I recently heard the following puzzle: There are three nails in the wall, and you want to hang a picture by wrapping a wire attached to the picture around the nails so that if any one nail is removed the picture still stays but if any two nails are removed then the picture falls down. An answer, schematically, is given by abca-1b-1c-1 (i.e., wrap it clockwise around each of the three nails in some order, then counterclockwise around each of the three nails in the same order).

This reminded me very much of the Borromean rings, where three rings are linked but when any one ring is removed the other two become unlinked. So I was trying to figure out if there might be some way to transform one situation into the other. My first instinct was to put the rings in S3 and have one of them pass through ∞, but that isn't really right. What seems to be tripping me up is that with the picture there's an extra object (the wire) that doesn't show up with the Borromean rings, but I have a vague idea that perhaps we could change the former situation by saying that we make the loop in the complement of three unlinked rings, and then perhaps "pulling really hard on the wire" would somehow thread the rings together. Maybe my issue is that what's really going on with the picture is just that we're making a loop in the complement of three points in plane, and I'm confounding phenomena of different dimensions...

Does anyone have any ideas?

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(A totally unhelpful comment -- apologies in advance -- but) why on earth would you want to hang a picture that way? –  Pete L. Clark Jan 6 '10 at 23:44
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@Pete: To paraphrase Oscar Wilde: "to lose one nail may be regarded as a misfortune; to lose both looks like carelessness." :) –  José Figueroa-O'Farrill Jan 7 '10 at 0:04

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up vote 6 down vote accepted

This question is related to this post and its answers: Collapsible group words

You may think of the three nails as giving a 3-punctured plane, which has fundamental group a rank 3 free group. An element of this group may be thought of as pushing a point around in the surface. If one takes the trace of this motion in time, you get a braid with 4 strands, three strands of which are straight. This represents an element of the pure braid group, an example of the Birman exact sequence. Closing up the braid, you get a four component link. Your condition implies that removing any two of the last three strands gives the trivial link, so it is a kind of second order Brunnian condition.

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@Agol: The fundamental group of a 3-punctured sphere is free on 2 generators, right? –  Pete L. Clark Mar 18 '10 at 5:37
    
@ Clark: yes. –  Ian Agol Mar 18 '10 at 6:07
    
That sounds awesome. What do you mean by "take the trace of this motion in time", though? I'd imagine that it means "follow the path traced out by the moving point", except that I don't see how one would choose a particular element of <pi><sub>1</sub>, so that's probably not right... –  Aaron Mazel-Gee Mar 19 '10 at 6:13
    
think of the path of the point in space-time (which is 3-dimensional, since the surface is 2-dimensional). This gives a braid, since in each time slice there is a single point. –  Ian Agol Mar 19 '10 at 16:31

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