Basic Confusion: Genus, Homology (Exm of Torus)

[Bacle]

Hi, All:

I'm a bit confused about this: the 1st homology of the torus T(over Z) is Z(+)Z,

so that elements of the form (a,b) =/ (0,0) are non-trivial , meaning these are

cycles (closed curves) that do not bound subsurfaces of the torus, and every cycle

that does bound is

equivalent/homologous to (0,0). Still, I don't see why, e.g., the cycle (1,1) of a meridian

and then a parallel, is non-bounding: if we went once around meridionally and once

longitudinally, we would disconnect the torus. Isn't (1,1) then, a cycle that bounds?

Additionally, the genus of the torus is 1 , meaning that we can remove at most one

SCC (simple-closed curve) from the torus without disconnecting it, so it would seem

like any pair (a,b) would consist of 2 curves ( one going a times meridionally and b

times longitudinally ), so that, since the genus is 1, (a,b) would disconnect the torus,

right? But then (a,b) would be a cycle that bounds, so that it would be homologous

to (0,0), which it is not. What am I missing?

EDIT: I realize the obvious fact that (a,b) is not a SCCurve. Is this the issue, i.e.,

must homology classes (at least for the torus) be represented by SCC's?

 

[quasar987]

Hi, All:

I'm a bit confused about this: the 1st homology of the torus T(over Z) is Z(+)Z,

so that elements of the form (a,b) =/ (0,0) are non-trivial , meaning these are

cycles (closed curves) that do not bound subsurfaces of the torus, and every cycle

that does bound is

equivalent/homologous to (0,0). Still, I don't see why, e.g., the cycle (1,1) of a meridian

and then a parallel, is non-bounding: if we went once around meridionally and once

longitudinally, we would disconnect the torus. Isn't (1,1) then, a cycle that bounds?

Nay. If you cut meridionally, you are left with a cylinder. Then when you cut longitudinally, you obtain a rectangle. rectangle = connected.

Additionally, the genus of the torus is 1 , meaning that we can remove at most one

SCC (simple-closed curve) from the torus without disconnecting it, so it would seem

like any pair (a,b) would consist of 2 curves ( one going a times meridionally and b

times longitudinally ), so that, since the genus is 1, (a,b) would disconnect the torus,

right? But then (a,b) would be a cycle that bounds, so that it would be homologous

to (0,0), which it is not. What am I missing?

EDIT: I realize the obvious fact that (a,b) is not a SCCurve. Is this the issue, i.e.,

must homology classes (at least for the torus) be represented by SCC's?

You can realize (a,b) as a SCC. For instance, (1,1) can be realized as the diagonal of the square in the polygonal presentation of the torus. But hopefully by now, you no longer think that (a,b) disconnects.

 

[Bacle]

Thanks, Quasar, its clear now. I was wrongly focusing on the red herring that we

were prying the space open, confusing that with the real issue of the end-space

(after the cuts ) being disconnected. I guess I am getting a head start on senility.