Geometric Example of Torsion Cycles and Relative Cycles (in Homology)

In summary: T and its "reflection" in B. For points x on the boundary of T, you can view the reflection of T in B as a "whirlpool" with x at the eye of the storm. This suggests a deformation retraction of M-x onto M-B, but I haven't worked out the details. In any case, whether or not x is on the boundary of T, this suggests that c is homologous to something that "looks like" a circle winding twice around x. This is what you should try to make precise.In summary, the conversation discussed torsion and relative cycles in a geometric way, mentioning the role of algebraic torsion in homology and the connection between algebra
  • #1
Bacle
662
1
Hi, Everybody:

I am trying to understand torsion and relative cycles in a more geometric way; I
think I understand some of the machinery behind relative cycles (i.e., the LES, and
the induced maps.), and I understand that by ,e .g., Poincare duality, in order to have
torsion in homology, we must have algebraic torsion ( I don't think this came out too
clearly, tho I hope clearly-enough. Please let me know o.wise).

So let me start with this case that may cover both issues:

Say we have the Mobius band M, with boundary component B, and we consider
H_2(M,B;Z)=Z/2 . I can figure this from the LES, but I am having trouble forming a mental
picture of what the actual torsion cycles would be like. I imagine the connection between
algebra and geometry is the obvious one :that going one around will give us a
nonbounding cycle, but going twice around will give us a boundary.


2) How about an example of relative cycles,in (X,A) both trivial ones and bounding ones,
please.?. I understand a cycle in X is a relative boundary if its boundary is fully
contained in A. Still, I am having trouble considering actual geometric examples.

Thanks in Advance.
 
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  • #2
What is LES?

1) I do not see what a generator of H_2(M,B;Z)=Z/2 would be. Are you sure you don't mean H_2(M,B;Z/2)=Z/2 instead?

2) A relative k-cycle c in (X,A) is a k-chain in X whose boundary lies in A. It is a relative boundary if there is some (k+1)-chain whose boundary is c + (some k-chain in A).

Some relative homology groups that occurs often in algebraic topology are the so-called local homology groups of a manifold. These are the homology group of the pairs (M, M-x) for some point x in M. By excision of "everything except a coordinate nbhd of x", this is canonically isomorphic to the homology of (R^n, R-n - 0).

Consider an n-simplex in R^n with 0 in its interior (so that its boundary is in R^n - 0). Seen as an n-chain, this simplex is a relative cycle.

Here is a little mental exercice for you that should be helpful in understanding relative cycles and boundaries: Show that a 1-chain in R^n is a cycle relative R^n - 0 iff there are exactly as many paths entering 0 as there are leaving it. Moreover, every relative 1-cycle is a relative boundary (so that H_1(R^n, R^n - 0)=0). In fact, H_k(R^n, R^n - 0) = 0 for all k other than n.
 
  • #3
it looks to me like a generator of the integral group would be the center circle. it does not bound in the mobius strip, but does bound if you double it. so it is a torsion element. i.e. the boundary of a mobius strip goes around the circle twice.
 
  • #4
Bacle said:
Hi, Everybody:

I am trying to understand torsion and relative cycles in a more geometric way;

Say we have the Mobius band M, with boundary component B, and we consider
H_2(M,B;Z)=Z/2 .

Relative cycles can often (maybe always) be thought of geometrically as cycles in the quotient space M/B.
 
  • #5
mathwonk said:
it looks to me like a generator of the integral group would be the center circle. it does not bound in the mobius strip, but does bound if you double it. so it is a torsion element. i.e. the boundary of a mobius strip goes around the circle twice.

We're talking about H^2.

lavinia said:
Relative cycles can often (maybe always) be thought of geometrically as cycles in the quotient space M/B.

Good point. This is the case when (X,A) are what Hatcher calls "good pairs", meaning A admits an open nbhd in X that strongly deformation retracts onto A. This is very often the case.
 
  • #6
oops i just read the title which did not say torsion in relative homology, so i took the questions one at a time.
 
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  • #7
Thanks, All.

i)LES is the Long-exact sequence.

Could anyone tell me what the generator is, for H_2(M,B;Z/2).?

Clearly, it is a cycle with boundary in B, but I can't figure out what it would be
otherwise.

It is obviously a cycle
 
  • #8
mathwonk said:
it looks to me like a generator of the integral group would be the center circle. it does not bound in the mobius strip, but does bound if you double it. so it is a torsion element. i.e. the boundary of a mobius strip goes around the circle twice.

yet the projection of the Mobius band onto the equator is a retraction.
 
  • #9
Ah, the generator of H_2(M,B;Z/2) that is different from the generator of H_2(M,B;Z). In fact, I think that H_2(M,B;Z)=0...

For a generator of H_2(M,B;Z/2), consider a rectangular piece of paper... triangulate it. Then glue 2 opposite sides of your rectangle to obtain a Mobius band and a triangulation of the Mobius band. View this triangulation as a 2-chain c on M, and observe that for Z/2 coefficients, c is a cycle relative B. Indeed, each side of the triangles in your triangulation either coincides with the side of exactly one other triangle, or lies in B. Since you're counting mod 2, this means that the boundary of c is a 1-chain in B. That is to say, c is a 2-cycle relative B.

According to the general theory of the homology of manifolds, [c] is the generator of H_2(M,B;Z/2) iff when c is viewed as a 2-cycle in M relative M-x, [c] is a generator of H_2(M,M-x) for all points x in M-B (see Hatcher thm 3.26*). For points x that belong to the interior of a triangle T of c, this is obvious because c is homologous to T mod M-x, and it is well known that T is a generator of H_2(M,M-x). For points x that belong to the edges or vertices of triangles, just choose another triangulation of M with the same number of triangles but such that this time x is now in the interior of a triangle. This new triangulation induces a relative 2-cycle which corresponds to the same homology class as the previous one. (This can be seen using the method of subdivision from the proof of thm 2.10 in Hatcher.) So it works for these points also.

*ah, actually Hatcher does not talk about the case of manifolds with boundaries. For this, it might be necessary to venture into Spanier. See around p.306.
 
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  • #10
quasar987 said:
Ah, the generator of H_2(M,B;Z/2) that is different from the generator of H_2(M,B;Z). In fact, I think that H_2(M,B;Z)=0...

For a generator of H_2(M,B;Z/2), consider a rectangular piece of paper... triangulate it. Then glue 2 opposite sides of your rectangle to obtain a Mobius band and a triangulation of the Mobius band. View this triangulation as a 2-chain c on M, and observe that for Z/2 coefficients, c is a cycle relative B. Indeed, each side of the triangles in your triangulation either coincides with the side of exactly one other triangle, or lies in B. Since you're counting mod 2, this means that the boundary of c is a 1-chain in B. That is to say, c is a 2-cycle relative B.

The equatorial circle is a strong deformation retract of the Mobius band so the Mobius band has the same homology as the circle.

Another way to see this is to let Z act by covering transformations on R^2 by the rule, (x,y) -> (x+1/2, -y). The quotient is an open Mobius band.

I tried the triangulation also and got that the boundary of the 2 chain of triangles that covers the Mobius band is a - b +2c where c is the edge identified to its opposite with a half twist and a and b are the opposite edges that are not identified. Pasting a paper rectangle together with scotch tape verifies this formula.

The conclusion from this, if I did it right, is that the boundary circle of the Mobius band is not a homology boundary over Z. Never thought about this before but ... in the case of a cylinder the two bounding circles - with opposite directions/orientations do form a boundary and with the same orientation they represent the same homology class. Their sum is twice the generator of the first homology of the cylinder.

In the Mobius band these two edges are glued to each other at their end points to form a single circle. As a singular chain this new circle consists of the sum of the two pieces added together with the same orientation. So the resulting circle can not form a boundary but rather just as in the case of the cylinder, must represent twice the generator of the first homology

This bounding cycle projects to twice the equatorial circle. The homology map is multiplication by 2.

If one identifies this boundary circle to a point - then it becomes homologous to zero in the quotient (which BTW is homeomorphic to the projective plane). So the first homology of the quotient is Z/2 since twice the generator of the equatorial circle is now homologous to zero.

While a general theorem shows that the n'th Z cohomology of a non-orientable compact n-manifold is Z/2Z, in this case it follows quickly from the cohomology exact sequence of the pair, (M,M-0) where M is the Mobius band, and M-0 is the Mobius band minus its equator.( Notice that M-0 deforms via a homotopy onto the bounding circle.)

H^1(M) -> H^1(M-0) -> H^2(M,M-0) -> H^2(M)

Substitution gives

Z -> Z -> H^2(M,M-0) -> 0

The first map is multiplication by two and H^2(M,M-0) is the second homology of the quotient, M/M-0 i.e. of the projective plane.
 
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  • #11
Cool, thanks for your contribution lavinia.
 
  • #12
quasar987 said:
Cool, thanks for your contribution lavinia.

I was thinking a little more about this. For the Klein bottle the boundary of the triangulation is 2c since the edges a and b are now identified. This shows that the second Z homology of the Klein bottle is zero but that its second Z/2Z homology is Z/2Z.
 
  • #13
Bacle said:
Hi, Everybody:



Say we have the Mobius band M, with boundary component B, and we consider
H_2(M,B;Z)=Z/2 . I can figure this from the LES, but I am having trouble forming a mental
picture of what the actual torsion cycles would be like.


2) How about an example of relative cycles,in (X,A) both trivial ones and bounding ones,
please.?. I understand a cycle in X is a relative boundary if its boundary is fully
contained in A. Still, I am having trouble considering actual geometric examples.

Thanks in Advance.

In M/B the torsion 1 cycle is just the equatorial circle of the Mobius band. This is also a cycle in M but in M is is not torsion. The boundary circle is homologous to twice the equatorial circle. If it is identified to a point then the equatorial circle becomes a torsion cycle in the quotient space M/B.
 
  • #14
Bacle said:
2) How about an example of relative cycles,in (X,A) both trivial ones and bounding ones, please.?. I understand a cycle in X is a relative boundary if its boundary is fully contained in A. Still, I am having trouble considering actual geometric examples.

For an example of a relative cycle that is a relative boundary, consider twice the equatorial circle in (M,B). Triangulate the rectangular piece of paper like so: 2 triangles above the equatorial line (which is to become the equatorial circle after gluing), and 2 below. There is a way to orient those triangles so that after gluing, the resulting 2-cycle has boundary "twice the equatorial circle" + the bounding circle. So "twice the equatorial circle" is the relative boundary of that triangulation/2-cycle.
 
  • #15
Bacle said:
I am trying to understand torsion and relative cycles in a more geometric way

One of the first examples of a relative cycle is a generator of the n+1'st homology of (V,V-0) where V is an n dimensional vector space. One can think of this cycle as the homology of the quotient space of V with all of the vectors outside of an open ball around the origin identified to a point. This is an n+1-sphere. It is a geometric realization of this relative cycle.
 
  • #16
Lavinia, is my mistake thinking that the boundary curve "bounds" when actually the thing it is the boundary of (the mobius band itself) is not oriented?
 
  • #17
mathwonk said:
Lavinia, is my mistake thinking that the boundary curve "bounds" when actually the thing it is the boundary of (the mobius band itself) is not oriented?

I don't think you made a mistake. The analysis here used Z coefficients for homology. If one uses Z/2 then everything works. 2c = 0 and -b = b so b + a, the chain representing the bounding circle of the Mobius band, is a boundary.

This difference using Z and Z/2 coefficients can be seen in the exact homology sequence of the pair.

0 -> H_2(M,M-0) -> H_1(M-0) -> H_1(M) -> H_1(M,M-0) ->0

Over Z the sequence is

0 -> 0 -> Z ->Z -> Z/2Z -> 0

over Z/2 the sequence is

0 -> Z/2Z -> Z/2Z -> Z/2Z -> Z/2Z ->0

This reminds me of the case of an unorientable n manifold with no boundary. Over the integers there are no n-cycles even though the manifold has no boundary. One always gets twice some n-1-simplex in the boundary of an n-chain that covers the manifold with simplices that intersect only along n-1 - faces. Over Z/2 though these doubled simplices add to zero and the manifold has a fundamental cycle mod 2.

I am not 100% sure of all of this but the arguments seem right.
 
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  • #18
lavinia said:
One of the first examples of a relative cycle is a generator of the n+1'st homology of (V,V-0) where V is an n dimensional vector space. One can think of this cycle as the homology of the quotient space of V with all of the vectors outside of an open ball around the origin identified to a point. This is an n+1-sphere. It is a geometric realization of this relative cycle.

That is a nice way to look at it!
 

1. What is torsion in homology?

Torsion in homology refers to the phenomenon where homology cycles, which are geometric cycles that can be continuously deformed into each other, are considered equal in homology theory even if they differ by a torsion element. This torsion element is essentially a "twist" or "spiral" in the cycle that cannot be undone by continuous deformations.

2. What is a geometric example of torsion cycles?

A geometric example of torsion cycles can be seen in the Mobius strip. This is a non-orientable surface that has a single boundary curve, which can be continuously deformed into itself but with a twist. This twist is a torsion element in the homology group of the Mobius strip.

3. How are torsion cycles and relative cycles related?

Torsion cycles and relative cycles are related in that both are used to differentiate between homology cycles. While torsion cycles involve a "twist" in the cycle itself, relative cycles involve a "hole" or "gap" in the cycle. These relative cycles are used to define relative homology groups, which are important in studying the properties of manifolds.

4. What is the significance of studying torsion and relative cycles?

Studying torsion and relative cycles is important in understanding the topological properties of spaces. These cycles help to distinguish between different types of geometric structures, and can be used to classify spaces into different homology classes. Additionally, understanding torsion and relative cycles can provide insight into the behavior of more complex homology groups.

5. How is the concept of torsion cycles applied in other areas of mathematics?

The concept of torsion cycles is not limited to homology theory, but can also be applied in other areas of mathematics such as algebraic topology and knot theory. In algebraic topology, torsion elements can be used to study the fundamental group of a space, while in knot theory, torsion elements are used to classify different types of knots.

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