# Surface representative of homology class.

1. Mar 20, 2009

### WWGD

Hi, everyone:

I wonder if anyone knows the actual definition of

having a surface represent a homology class. Sorry to bother with

just a definition, but I have not been able to find one , neither in books

that a certain surface represents H_2(M,Z) , the second homology

class of an orientable manifold. All I can see is mention of

the inclusion map (and its pushforward), and the fundamental

class ( a generator of top homology).

Anyone know the def., or can give me a link/reference?

Thanks.

2. Mar 20, 2009

### matt grime

Elements in homology groups are equivalence classes of subobjects of the manifold (in this case). I apologise if I get this the wrong way round: they are the cycles. Two such subobjects represent the same element in the homology group if they are homologous, i.e differ by a boundary. Again I apologise if I have things wrong way round and have just described cohomology.

In the case in question it might help if you say what M is.

3. Mar 20, 2009

### WWGD

Thanks, Matt; sorry, M is a manifold.

4. Mar 20, 2009

### matt grime

I meant did you have a particular manifold in mind so that someone could explain what the elements in H_2 are.

The picture I have for the torus, T, is that the generators of H_1 are loops, the same two loops as the fundamental group, and these are 1-submanifolds of T.

5. Mar 20, 2009

### yyat

Let $$N$$ be a manifold, $$M\subseteq N$$ a compact oriented submanifold without boundary, $$\dim(M)=n$$.

$$M$$ has a fundamental class $$[M]\in H_n(M)$$.

Then the homlogy class representing $$M$$ in $$N$$ is the image ("pushforward") of $$[M]$$ under the map (induced by inclusion) $$H_n(M)\to H_n(N)$$.

It is possible to extend this to the case where M is allowed to have a boundary that is contained in the boundary of N. The only difference is that the homology relative to the boundary is used instead of absolute homology.

6. Mar 20, 2009

### WWGD

Thanks, YYAT !.

7. Mar 20, 2009

### wofsy

Every closed orientable manifold has a top dimensional Z homology cycle called the fundamental cycle. This cycle generates its top dimensional homology group. When the manifold is a subset of another manifold then this fundamental cycle is also a homology cycle in the larger manifold.

The fundamental cycle of a sub-manifold is said to be the homology cycle that it represents. So your surface represents its fundamental cycle inside the larger manifold.

One thing though. The important thing is not that the larger manifold is orientable but that the sub-manifold is orientable. If the sub-manifold is not orientable it will not represent a cycle. It has no fundamental class over Z. If it is orientable then its fundamental class will be a cycle whether or not the larger manifold is orientable.

For example the Klein bottle is not orientable. Its second Z homology class is zero. It has no fundamental class over Z. No Klein bottle embedded in a larger manifold can ever represent a Z-cycle.

Another example is the projective plane. It has no fundamental cycle over Z.

8. Mar 20, 2009

### WWGD

Excellent. Very clear, thanks.

9. Mar 20, 2009

### wofsy

If you use Z/2 instead of Z as homology coefficients then non-orientable manifolds do have a fundamental cycle. They will represent Z/2 homology classes in you larger manifold.

10. Mar 23, 2009

### WWGD

Thanks again, Wofsy, all .

Is there a geometric way of seeing/understanding these cycles?. I am trying

to understand this fundamental class in a more geometric way. Say we

take the example of the circle, so that we get H_1(S^1)=Z . What would

[S^1] look like geometrically?

I imagine that

at least within what may be the simplest homologies, singular and simplicial, cycles

can be seen as maps (maybe identified with their images, i.e., if f: sn -->X is

a cycle, with sn an n-simplex , then f(sn)<X can be seen as a cycle) .

Then, as I understand it , at least in simplicial homology, two cycles in X are

homologous if their (point set) sum is a boundary for X , in the sense that if we

remove f(sn) from X, we get a disconnected space. I am thinking here of the

example of the torus, where a closed curve that is not longitudinal nor latitudinal ,

seen as a closed curve on the surface of the torus is a boundary , and any two

latitudinal or any two longitudianl curves are boundaries, since their point-set

sum ( i.e., the union of their images) is a boundary in the sense above ( i.e.,the re

moval will disconnect the space )

Can this geometric idea of cycles be extended to other homologies? .

Thanks.

11. Mar 23, 2009

### wofsy

There is no simple answer to your question but here are a few thoughts.

You can think of a one dimensional cycle as a finite collection of closed curves. Intuitively it is homologous to zero if it is the curves are the boundary of a two dimensional region.

For instance on the torus ,as you wrote, a curve that loops around or through the ring will not be a boundary of a 2 dimensional region but a small circle that does not loop around will be the boundary of a disk.

However general closed curves can be wild and not be the boundary of any region even if they are homologous to zero. For an extreme example think of a closed space filling curve on the sphere. It is homologous to zero. can you picture it as a boundary?

A good example is the annulus. All closed curves, no matter how wild, wind around the annulus a certain number of times. The number of windings is its homology class.

In complex analysis the number of windings can be measured by taking the line integral of a branch of the logarithm along the curve (assuming that it is piecewise differentiable). No matter what the curve, this integral will be an integer multiple of 2 pi and is the total angle that the curve traces around the center of the annulus. This integer is called the winding number of the curve. If I wind around twice then I integrate twice and get twice the winding number. If I wind backwards I get negative the integral and negative the winding number. Two curves are homologous if the integral of the log is the same. A curve is homologous to zero if the integral of the log is zero.

I believe that the idea of homology began with this example because people realized that line integrals of holomorphic functions depend only on the homology of the curve and not on the specific path that it traverses. A function that is holomorphic in some region of the plane (possibly with many holes in it) will have integrals that depend only on the number of times that the curve winds around the holes. This winding count is the homology class of the curve.

These ideas extend naturally to Riemann surfaces and the study of integrals of meromorphic functions on Riemann surfaces reveals that there are more complicated ways that curves can wind around than just winding around holes. But the same principle holds.

For the circle, curves are classified also by the number of times they wind around. A fundamental cycle is any curve that winds around once. This is really the same situation as the annulus.

The reason cycles are thought of as mappings is that the geometric image does not tell you the number of times that the curve winds around and thus does not tell you the homology class.

Last edited: Mar 24, 2009
12. Mar 24, 2009

### yyat

Suppose you have a closed oriented manifold M and a simplical structure on M. A cycle representing the fundamental class is obtained as the sum of all the the top-dimensional simplices, assuming they have the same orientation as the manifold (otherwise there are signs).

13. Mar 26, 2009

### WWGD

I don't mean to beat these questions to death, it is just that these answers really help me
tie down a lot of loose ideas I have (which may be natural to have in one's first year).

uniqueness of smooth structures for dimensions less than or equal to 3. Here are
the refs:

i) E. Moise: "Geometric Topology in Dimensions 2 and 3". Springer-Verlag, N.Y, 1977

ii) J. Munkres " Obstructions to the smoothing of piecewise differentiable
homeomorphisms." Annals of Math., 72:521-554 , 1960.

14. Mar 26, 2009

### WWGD

Sorry that quotations

15. Mar 26, 2009

### WWGD

Re: Surface representative of homology class. Sorry for previous

Wofsy:
"There is no simple answer to your question but here are a few thoughts.

You can think of a one dimensional cycle as a finite collection of closed curves. Intuitively it is homologous to zero if it is the curves are the boundary of a two dimensional region.

I see, so the region is simply-connected iff every closed curve (and therefore every cycle)
is homologous to zero.

For instance on the torus ,as you wrote, a curve that loops around or through the ring will not be a boundary of a 2 dimensional region but a small circle that does not loop around will be the boundary of a disk.

However general closed curves can be wild and not be the boundary of any region even if they are homologous to zero. For an extreme example think of a closed space filling curve on the sphere. It is homologous to zero. can you picture it as a boundary?

A good example is the annulus. All closed curves, no matter how wild, wind around the annulus a certain number of times. The number of windings is its homology class.

In complex analysis the number of windings can be measured by taking the line integral of a branch of the logarithm along the curve (assuming that it is piecewise differentiable). No matter what the curve, this integral will be an integer multiple of 2 pi and is the total angle that the curve traces around the center of the annulus. This integer is called the winding number of the curve. If I wind around twice then I integrate twice and get twice the winding number. If I wind backwards I get negative the integral and negative the winding number. Two curves are homologous if the integral of the log is the same. A curve is homologous to zero if the integral of the log is zero."

So I guess log is defined in the region R whenever the winding number about a point
not in R is zero. Then the integral dz/z is zero , and so it is independent of path, which
I understand to be equivalent to the log being defined/definable, i.e., that the integral
of dz/z around any curve is zero, and so is independent of path.

Wofsy Wrote:

"I believe that the idea of homology began with this example because people realized that line integrals of holomorphic functions depend only on the homology of the curve and not on the specific path that it traverses. A function that is holomorphic in some region of the plane (possibly with many holes in it) will have integrals that depend only on the number of times that the curve winds around the holes. This winding count is the homology class of the curve."

But there seems to be an issue of homotopy here too, right?. If curves C_1, C_2 contained
in R have the same winding number about points in C-R , and (it follows that )C_1 and C_2 are homotopic, and the homotopy takes place entirely in R (I think this property has the weird name of R being "semilocally simply-connected" ), then, for any f holomorphic in the region we have:

Int_C_1 f(z)dz =Int_C_2 f(z)dz

uniqueness of smooth structures for dimensions less than or equal to 3. Here are
the refs:

i) E. Moise: "Geometric Topology in Dimensions 2 and 3". Springer-Verlag, N.Y, 1977

ii) J. Munkres " Obstructions to the smoothing of piecewise differentiable
homeomorphisms." Annals of Math., 72:521-554 , 1960.

16. Mar 27, 2009

### wofsy

Re: Surface representative of homology class. Sorry for previous

I have to apologise. I mispoke about the log. The winding number is defined by the integral of dz/z-a where a is chosen in the center of the annulus. This integral's value is the difference in the two values of the log as the curve winds on its Riemann surface. dz/z-a is holomorphic in the annulus.

Second. Homotopy and homology are not the same. While it is true that a region of the plane is simply connected iff its first Z homology is zero there are spaces where this is not true. Further, in the plane some regions do not have an abelian fundamental group. In these regions number of times a curve winds around the holes does not determine the homotopy class. It is also the order in which it winds around the holes. In homology the order doesn't matter. Notice that the order doesn't matter for line integrals either.