Complement of a Handlebody also a Handlebody?

[Bacle]

I have read that the complement of a handlebody Hg--an n-ball embedded

in R^n with g handles attached to the ball-- is itself a handlebody.

But this seems false, e.g., for a torus H1 ( a 3-ball with one

handle attached to it ) embedded in R^3: the fundamental group

of the complement is Z --the integers--as any loop about the complement

can be shrunk and translated into a generic loop, while the fund. group

of a handlebody is Z^2g. (note: B^1 is the unit interval in R^1 , etc.)

All I can think is that we may be considering the complement in

S^3 instead of in R^3, but I don't see how an extra point (point at

infinity) would make a difference.

I can see how the complement of an n-ball in R^n retracts to an (n-1)-sphere, e.g.,we can

retract R^2-{(0,0)} into S^1, and we can then consider the complement of an attached

handle/torus, then the union of the complements, i.e., R^n-{B^n\/{handle}} by De Morgan

or something like it. Then I guess we could have a reverse, or "sunken" handle in the

complement, which we can push outward into a "standard handle".

Any Ideas?

Thanks.



Any Ideas?

Thanks.

 

[lavinia]

not sure but the complement of a solid torus in S^3 is another solid torus.

 

[Bacle]

Thanks, Lavinia, but I am having trouble seeing it.

I will try to be more succinct and precise this time.

Let's use S^3=R^3\/{oo} (by stereo projection, etc.).

Do you agree that the complement of a solid torus T in R^3 itself is not

a torus ( I guess the extra point at oo may be making a difference)?

For one, it seems like the fundamental group of R^3-T is Z, the integers,

since a "lasso" describes all non-trivial loops about the missing torus.

Now we include the missing point at oo. How does that change things?

Or, do you have an explanation from a different perspective?

 

[lavinia]

Thanks, Lavinia, but I am having trouble seeing it.

I will try to be more succinct and precise this time.

Let's use S^3=R^3\/{oo} (by stereo projection, etc.).

Do you agree that the complement of a solid torus T in R^3 itself is not

a torus ( I guess the extra point at oo may be making a difference)?

For one, it seems like the fundamental group of R^3-T is Z, the integers,

since a "lasso" describes all non-trivial loops about the missing torus.

Now we include the missing point at oo. How does that change things?

Or, do you have an explanation from a different perspective?

I am still working on the direct proof that you are trying.

Here are a couple of different approaches.

- the 3 sphere fibers over the 2 sphere as a circle bundle. this is the Hopf fibration.

Above any open disk on the sphere the fibration is trivial and thus is a disk Cartesian product a circle which is a solid torus. The complement of the open disk on the 2 sphere is another open disk so the complement of the solid torus in the 3 sphere is another solid torus.

- Slice the three sphere with the xyz-plane in 4 space. It falls apart into two solid 3-balls
To reassemble the 3 sphere glue these 2 solid balls back together again (with the correct gluing map) along their bounding 2 spheres.

From each ball core out a solid cylindrical tube and choose these tubes so that their ends are identified in the gluing and glue together to become a solid torus. What is left over is two balls with tubes removed and they are glued back together to give the complement of the solid torus.

These two cored out ball are each half bagels. The gluing slaps them together to make a complete bagel. To see this take one and spread it out at one of the end circles of the core until there is enough room to fold the other circle downward into the widenened space. The result is a half bagel and the curved part of it is the everted interior of the core and the flat part is the piece that is glued to the other half bagel. Sorry this hard to describe.

 

[Bacle]

I am still working on the direct proof that you are trying.

Here are a couple of different approaches.

- the 3 sphere fibers over the 2 sphere as a circle bundle. this is the Hopf fibration.

Above any open disk on the sphere the fibration is trivial and thus is a disk Cartesian product a circle which is a solid torus. The complement of the open disk on the 2 sphere is another open disk so the complement of the solid torus in the 3 sphere is another solid torus.

- Slice the three sphere with the xyz-plane in 4 space. It falls apart into two solid 3-balls
To reassemble the 3 sphere glue these 2 solid balls back together again (with the correct gluing map) along their bounding 2 spheres.

From each ball core out a solid cylindrical tube and choose these tubes so that their ends are identified in the gluing and glue together to become a solid torus. What is left over is two balls with tubes removed and they are glued back together to give the complement of the solid torus.

These two cored out ball are each half bagels. The gluing slaps them together to make a complete bagel. To see this take one and spread it out at one of the end circles of the core until there is enough room to fold the other circle downward into the widenened space. The result is a half bagel and the curved part of it is the everted interior of the core and the flat part is the piece that is glued to the other half bagel. Sorry this hard to describe.

Thanks, Lavinia, that was helpful--as usual.

 

[lavinia]

I guess I should have said that when you fold the small circle inwards you evert the inner part of the tube. The idea her is that a cylinder is really an annulus.

 

[lavinia]

Bacle

I think this works and gives a direct decomposition of the 3 sphere into two solid tori.

Parameterize the 3 sphere as ( texp(ia), (1 - t^2)^.5exp(ib))

where t is between zero and 1.

For each t except 0 and 1, a and b trace out a torus ( a circle of circles). For 0 and 1 they only trace out a circle. So I think what is going on here is that the regions

0 <= t <= k and k <= t <=1 are both solid tori that are glued together along the torus t = k ( 0 < k < 1). Yes?

I have a question about these arguments. It has only been shown that the 3 sphere can be decomposed into two solid tori. But your question seemed more general. That is, given any solid torus in the 3 sphere, is its complement also a solid torus? How does the argument work say when the torus is knotted?

 

[lavinia]

Here is an idea that follows your approach.

A simple circular wire ( a solid torus) containing a steady current generates a magnetic field in all of 3 space outside of the wire and whose field lines are loops that link with the circuit. They circle around the wire. The only point on the central disk that is located in the middle of the wire that is not hit by one of these closed field lines is the center of the disk where the field line is a straight line.

Field lines close to the center are straighter and closer to parallel to the the central straight line. After a while they leave the central region and sweep out into space and take a long path around returning to the central disk. In the limit, they converge to the central straight line plus the point at infinity. In the limit, they seem to break but if one includes the point at infinity - thus passing from 3 space to the 3 sphere - these loops do not break but rather converge to the circle on the 3 sphere traced out by the straight line.

This picture illustrates why the complement of the wire is a solid torus. the exact same construction can be made on a solid torus by choosing a disk that is perpendicular to its boundary and intersects the boundary in a circle. through each point in the disk there passes a circle in the interior of the torus. As the point converges to the center of the disk, the circle converges to the central circle that passes through the center of the disk. This is exactly the same picture as for the current loop. The central circle in the complement of the current loop is the straight line plus the point at infinity.

This idea might apply for knotted tori as well. A knotted wire still generates a magnetic field in space that is divergence free and thus should have field lines that are either closed loops or lines (possibly bent rather than straight). These field lines should converge in a similar way but that central disk needs to be found - if it exists at all.

 

[lavinia]

BTW: You can compute the homology of the complement if the solid torus in S^3 using a Mayer-Vietoris sequence. It will immediately show that the first homology is Z as you said and that the higher dimensional homologies are zero.

For links one removes more than one solid torus. The homology calculations lead to Alexander duality.

 

[Bacle]

Excellent; thanks again, lavinia.

 

[WWGD]

I am still working on the direct proof that you are trying.

Here are a couple of different approaches.

- the 3 sphere fibers over the 2 sphere as a circle bundle. this is the Hopf fibration.

Above any open disk on the sphere the fibration is trivial and thus is a disk Cartesian product a circle which is a solid torus. The complement of the open disk on the 2 sphere is another open disk so the complement of the solid torus in the 3 sphere is another solid torus.

I agree, but there seems to be a dimensional problem: the lift to S^3 of the disk inside of a covering neighborhood D^2xS^1 is a 2-torus, and I believe the two Tori Bacle is referring to are 3-Tori S^1 x S^1 X S^1. Or am I missing something?

 

[lavinia]

I am still working on the direct proof that you are trying.

Here are a couple of different approaches.

- the 3 sphere fibers over the 2 sphere as a circle bundle. this is the Hopf fibration.

Above any open disk on the sphere the fibration is trivial and thus is a disk Cartesian product a circle which is a solid torus. The complement of the open disk on the 2 sphere is another open disk so the complement of the solid torus in the 3 sphere is another solid torus.

I agree, but there seems to be a dimensional problem: the lift to S^3 of the disk inside of a covering neighborhood D^2xS^1 is a 2-torus, and I believe the two Tori Bacle is referring to are 3-Tori S^1 x S^1 X S^1. Or am I missing something?

He is referring to solid tori.solid tori are 3 dimensional manifolds with boundary. the boudaries are identified in the 3 sphere.