Homotopy Loop problem

• May 15th 2011, 06:54 AM
EinStone
Homotopy Loop problem
Hi, the following is also known as the picture-nail problem.

Consider X = the euclidean plane with N holes. Explain how to construct a loop L in X that is not null-homotopic in X, but if you remove any hole in X, the loop becomes null-homotopic in the new space.

For N = 1 this is trivial. I am mainly concerned with the cases N = 2 and N = 3. The problem is not about proving anything, simple construction is enough.

I actually found an example for N = 2 (click here), I am only looking for an example of N = 3 now.
• May 16th 2011, 12:57 AM
Opalg
Quote:

Originally Posted by EinStone
Hi, the following is also known as the picture-nail problem.

Consider X = the euclidean plane with N holes. Explain how to construct a loop L in X that is not null-homotopic in X, but if you remove any hole in X, the loop becomes null-homotopic in the new space.

For N = 1 this is trivial. I am mainly concerned with the cases N = 2 and N = 3. The problem is not about proving anything, simple construction is enough.

I actually found an example for N = 2 (click here), I am only looking for an example of N = 3 now.

Given points (or holes) $P_1,\,P_2,\,P_2,\ldots$, write $a_k$ for (the homotopy class of) a clockwise loop around the point $P_k$, and write 1 for the homotopy class of a null-homotopic loop. The loop $L_2$ in the example linked to above is given by $L_2 = a_1a_2a_1^{-1}a_2^{-1}$ (where the inverse denotes the corresponding anticlockwise loop, of course). If you remove one of the points $P_k$ then the corresponding loop $a_k$ becomes equal to 1. So for example if you remove the hole at $P_1$ then the path $L_2$ becomes $1a_21a_2^{-1} = a_2a_2^{-1} = 1$.

Now proceed by induction. Suppose that $L_n$ is a path around n points such that if you remove the hole at any one of those n points then the path becomes null-homotopic. Define $L_{n+1} = L_na_{n+1}L_n^{-1}a_{n+1}^{-1}.$ If you remove one of the points $P_k$ (for $1\leqslant k\leqslant n$) then $L_n$ becomes null-homotopic and so $L_{n+1}$ becomes $a_{n+1}a_{n+1}^{-1} = 1$. And if you remove $P_{n+1}$ then $a_{n+1}$ becomes 1 and so $L_{n+1}$ becomes $L_nL_n^{-1} = 1$. That completes the inductive proof that $L_{n+1}$ has the required property.

Once you have that formula, it's easy in principle to draw a geometric picture of the loop $L_3$ for example. But the complexity of the loops increases alarmingly as the number of holes increases. Even for n=3 it will look fairly messy.