Fundamental group of n connect tori with one point removed

[PsychonautQQ]

Well, for starters, $\pi(T)$, the fundamental group of the torus, is $\pi(S^1)x\pi(S^1)=$ which is in turn isomorphic to the direct product of two infinite cyclic groups. Before I tackle the case of n connect tori with one point removed, I'm trying to just understand a torus with a point removed.

I'm imagining the torus as a donut. With no points removed, given any base point, the two loop equivalence classes are obvious (three counting the class containing trivial loops). Removing a point, we will now have a new class of loops that go around that missing point. We will also still have the class that goes around through the 'hole' of the donut. I'm trying to think about what happens to the path that goes around the outside of the donut when we take away a point... Does this get split into two different equivalence classes or just stay one? The more I think about it and try to visualize I'm pretty sure it stays one.

Cool, so perhaps the fundamental group of a torus with a point removed is isomorphic to $ZxZxZ$?

Now, as for a connected sum of n tori... I think all that would happen is we would pick up another $Z$ for each additional 'donut hole'.
How far off am I on this?

 

[andrewkirk]

perhaps the fundamental group of a torus with a point removed is isomorphic to ZxZxZ?

Now, as for a connected sum of n tori... I think all that would happen is we would pick up another Z for each additional 'donut hole'.
How far off am I on this?

I'm pretty confident that will not be the case because removing a point does not have the same effect on fundamental groups as sewing on an extra handle (torus) does. Sewing on a handle does more than just make a new hole. We can see this by observing that the fundamental group of the figure eight, which has two holes, is the free group on two generators, which is non-abelian, while the fundamental group of the torus, which also has two holes, is $\mathbb Z\oplus\mathbb Z$, which is abelian.

I don't know what the answer is since I haven't worked with fund groups for quite a while, but it might have something to do with sewing on discs, as discussed in this thread. I expect that every point removed introduces an additional non-abelian element to the fundamental group, whereas every handle sewn on introduces an 'additional $\mathbb Z$'.

@lavinia will know the answer.

 

[lavinia]

If one thinks of a torus as a square with opposite edges identified these opposite edges form two circles $a$ and $b$ that intersect in a point. They form a figure 8. By itself as Andrewkirk said, the figure 8 has fundamental group the free group on two generators. But in the torus the commutator $aba^{-1}b^{-1}$ equals the identity. This is the only relation so the fundamental group of the torus is the free group on two generators modulo its commutator subgroup.

To see this, take a small disk in the center of the square. Its boundary circle is null homotopic since it can be shrunk to a point in the disk. This circle can be radially deformed onto the figure 8. The picture is the square and radial lines emanating from the center of the disk out to the boundary of the square. Sliding the circle along these radial lines deforms it by a homotopy onto the boundary of the square. As one goes once around the deformed circle, one traverses the loop $aba^{-1}b^{-1}$.

If one removes the center of the disk, then the entire remainder of the torus can be slid along the radial lines onto the boundary of the square. This means that it has the same fundamental group as the figure 8, the free group on two generators.

So you are right that the circle with a point removed introduces a new homotopy class of loops. But it is not a new path. It just is no longer null homotopic.

 

[WWGD]

I remember being able to visualize how the torus minus a point deformation-retracts into a figure 8 , as a larger circle with a satellite tangent circle attached. You start peeling the layer about the smaller circle in $S^1 \times S^1$ But I am too tired now to write something better.

 

[PsychonautQQ]

By itself as Andrewkirk said, the figure 8 has fundamental group the free group on two generators. But in the torus the commutator $aba^{-1}b^{-1}$ equals the identity. This is the only relation so the fundamental group of the torus is the free group on two generators modulo its commutator subgroup.

Can you help me understand why the free group on two generators modulo it's commutator subgroup is equal to the direct product of two infinite cyclic groups? Modulo the commutator subgroup, which is taking out all the elements that don't commute with each other, which makes sense in a way because the free product is built to not commute where as ZxZ does commute... but still, some piece of my understanding is missing.

 

[PsychonautQQ]

Okay, so looking at the torus as a square with proper identifications made, it makes sense why taking out a point makes the fundamental group equal to the free product on two generators, because you can detract the space to the boundary where it is a figure 8.

However, when I look at the torus as a donut and take out a point, it seems to me that the fundamental group should be Z x Z x Z, although apparently this is simply untrue. I want somebody to tell me what is wrong with the reasoning for my belief:
So on the torus with no points missing, one equivalence class of loops is when it goes around the outside of the donut, and one is when it loops around the inside hole, (the two circles the generator the torus). Now, if we remove a point, I feel like both of the original equivalence classes should go unaffected, and furthermore removing a point adds another equivalence class of paths that go around that missing point, so it seems like now the fundamental group should be the $\pi (S^1)$x$\pi (S^1)$x$\pi (S^1)$. What is wrong with my thinking here?

 

[andrewkirk]

Now, if we remove a point [from a torus], I feel like both of the original equivalence classes should go unaffected,

The reason they will be affected is that some of the loops in those equivalence classes go around the deleted point P. So each existing equivalence class will split into a set of new classes. Further, it makes a difference where in the existing loop it goes around P. An equivalence class that orbits the big hole three times clockwise splits into one new class for each number of times the loop goes around P. Then each of those classes splits again according to on which of the three big laps the loop goes around P. For instance, the class of loops that circle P only once splits into three - where the loop circles P on the first, second or third big lap respectively.

removing a point adds another equivalence class of paths that go around that missing point

It adds at least one. That's correct. And this class interacts with other loops around the torus holes in a non-commutative way.

 

[WWGD]

I don't know how practical this is , but maybe you can write the homotopy equivalence between the torus minus a point and the figure 8 analytically and see how a given class in the figure 8 ( better understood fundamental group) pushes forward under the induced map on fundamental group.

 

[lavinia]

I want somebody to tell me what is wrong with the reasoning for my belief:
Now if we remove a point, I feel like both of the original equivalence classes should go unaffected,

No. Before you remove a point, $aba^{-1}b^{-1}$ is null homotopic. After you remove it, it is not.

and furthermore removing a point adds another equivalence class of paths that go around that missing point, so it seems like now the fundamental group should be the $\pi (S^1)$x$\pi (S^1)$x$\pi (S^1)$. What is wrong with my thinking here?

Why do you think that this new equivalence class is not in the group generated by $a$ and $b$? Can you give a proof?

 

[lavinia]

Can you help me understand why the free group on two generators modulo it's commutator subgroup is equal to the direct product of two infinite cyclic groups? Modulo the commutator subgroup, which is taking out all the elements that don't commute with each other, which makes sense in a way because the free product is built to not commute where as ZxZ does commute... but still, some piece of my understanding is missing.

The group has two infinite cyclic generators $a$ and $b$ but if the commutator $aba^{-1}b^{-1}=id$ then $ab=ba$ so the group is commutative.