Is there a way to find a homeomorphism of the torus using a 2x2 integer matrix?

[lavinia]

For each 2x2 integer matrix with determinant +/-1 is there a homeomorphism of the 2 torus so that the the induced map on the first homology group is multiplication by this matrix?

 

[mathwonk]

why not? is it defined by the same matrix, on C/Z^2?

 

[homeomorphic]

Yeah, there is a map from the mapping class group of the torus (the homeomorphism group mod isotopy) to SL(2, Z) that is an isomorphism. You map a guy in the mapping class group to the guy in SL(2, Z) that is the induced map on homology. The surjectivity of this map gives you what you want.

One way to see the surjectivity is that the guy in SL(2, Z) induces a homeomorphism on the universal cover, R^2 (or ℂ), that descends to a homeomorphism downstairs (the deck transformations are integers translations, which commute with the linear maps from SL(2,Z) up to integer translations). A guy (p,q) in the lattice upstairs, with p and q relatively prime corresponds to p, q in homology (homology is the same as the fundamental group, here). So, you can map (1,0) to any such vector upstairs, and map (0,1) to a complementary vector. So, you can get anybody in SL(2,Z) that way.

You can get injectivity from the fact that the torus is an Eilenberg-Maclane space.

 

[homeomorphic]

Oh, I forgot about orientations. Mapping class group is the orientation-preserving homeos, only.

Also, what I said just works for the determinant 1 guys. I think if you include orientation-reversing homeos, you can get the determinant -1 guys. A reflection upstairs of the universal cover should induce a reflection of the torus.

 

[mathwonk]

i am still puzzled why it isn't obviously given by the same matrix. i.e. given an integer matrix with determinant one, it maps the integer lattice in R^2 isomorphically to itself. As a 2by 2 real matrix it aldo maps the plane R^2 isomorphically to itself, hence it maps the quotient space T = R^2/Z^2, namely the two torus, isomorphically and homeomorphically, onto itself.

Since H^1 of the quotient space R^2/Z^2 is the integer lattice Z^2 itself, the map on H^1(T,Z) is given by the original matrix.

isn't this obvious? or is it wrong for some reason? I.e. it seems very elementary and concrete.

 

[lavinia]

Yeah, there is a map from the mapping class group of the torus (the homeomorphism group mod isotopy) to SL(2, Z) that is an isomorphism. You map a guy in the mapping class group to the guy in SL(2, Z) that is the induced map on homology. The surjectivity of this map gives you what you want.

One way to see the surjectivity is that the guy in SL(2, Z) induces a homeomorphism on the universal cover, R^2 (or ℂ), that descends to a homeomorphism downstairs (the deck transformations are integers translations, which commute with the linear maps from SL(2,Z) up to integer translations). A guy (p,q) in the lattice upstairs, with p and q relatively prime corresponds to p, q in homology (homology is the same as the fundamental group, here). So, you can map (1,0) to any such vector upstairs, and map (0,1) to a complementary vector. So, you can get anybody in SL(2,Z) that way.

You can get injectivity from the fact that the torus is an Eilenberg-Maclane space.

Right. So just project the linear map down to the torus.

I was interested in this because I was wondering what closed three manifolds you could get by gluing two solid tori together. Since one of the first homology generators of the torus is null homologous in the solid torus, the resulting manifold's homology would be determined easily from the gluing homeomorphism and Van Kampen's Theorem(meyer vietoris sequence also).

A reversal of first homology generators would yield a simply connected manifold so it is a three sphere. The identity on homology would yield a manifold with first homology group the integers and which has an orientation reversing involution that covers two solid Klein bottles glued together.

The map which takes the null homologous generator to itself plus twice the non null homologous generator would yield a manifold with first homology equal to Z2 so I guess it must be projective space.

And then there are infinitely many others each with distinct homology.

 

[homeomorphic]

i am still puzzled why it isn't obviously given by the same matrix. i.e. given an integer matrix with determinant one, it maps the integer lattice in R^2 isomorphically to itself. As a 2by 2 real matrix it aldo maps the plane R^2 isomorphically to itself, hence it maps the quotient space T = R^2/Z^2, namely the two torus, isomorphically and homeomorphically, onto itself.

Since H^1 of the quotient space R^2/Z^2 is the integer lattice Z^2 itself, the map on H^1(T,Z) is given by the original matrix.

isn't this obvious? or is it wrong for some reason? I.e. it seems very elementary and concrete.

Yeah, pretty much. You could change basis somewhere to make the matrices different, but that's kind of silly.

I was interested in this because I was wondering what closed three manifolds you could get by gluing two solid tori together. Since one of the first homology generators of the torus is null homologous in the solid torus, the resulting manifold's homology would be determined easily from the gluing homeomorphism and Van Kampen's Theorem(meyer vietoris sequence also).

Well, that's a genus one Heegard splitting. Those are known as three dimensional lens spaces, L(p,q), with p and q relatively prime, at least if you use orientation-preserving gluings. I forget why the different lens space definitions are equivalent. I think Hatcher explains it in his 3-manifold notes somewhere. Lens spaces can be classified using Reidemeister torsion.

http://en.wikipedia.org/wiki/Lens_space#Classification_of_3-dimensional_lens_spaces

A reversal of first homology generators would yield a simply connected manifold so it is a three sphere. The identity on homology would yield a manifold with first homology group the integers and which has an orientation reversing involution that covers two solid Klein bottles glued together.

I don't follow. The identity map should also give S^3.

The map which takes the null homologous generator to itself plus twice the non null homologous generator would yield a manifold with first homology equal to Z2 so I guess it must be projective space.

Yes, I think so.

 

[lavinia]

I don't follow. The identity map should also give S^3.

I was thinking that if the null homologous circle on one surface torus was identified with the not null homologous circle on the other - and visa versa, then the gluing would cancel all the first homology out and leave a simply connected manifold.

 

[lavinia]

Here is another one that is a little different.

Divide the surface torus of a solid torus into two topological cylinders that share their two circles as a common boundary. Identify these circles with a half twist so that the boundary of this solid is now two intersecting Klein bottles.

take two of these solids and identify them along their Klein bottle boundaries to form a closed 3 manifold. What are the possible manifolds that result? I know that one of them is the Hanshe-Wendt manifold, a flat manifold that has zero first Betti number.

 

[Bacle2]

Here is another approach:

By functoriality of π₁ , every automorphism of the torus gives rise to

an automorphism ( isomorphism) of Z(+)Z to itself . Now, we need to show that

this functor is onto. We know that Aut(Z(+)Z) is SL(2,Z) .

We also have that SL(2,Z) has a generating set with three elements. These elements

are standard homeomorphisms from the torus to itself. Then, for any matrix in SL(2,Z),

find its expression in terms of these three generators, and you're done.