What is the significance of the modulus of a Riemann surface?

[theorem4.5.9]

I am confused on what is meant by the modulus of a Riemann surface. I have done some very quick searches through my books but haven't found anything.

For context I will paraphrase a portion of Penrose's road to reality that I'm reading.

In the complex plane construct a parallelogram by connecting cyclically the points $0, 1, 1+p, p$ for some complex number $p$. Orient the region by $0\rightarrow 1, p\rightarrow1+p, 0\rightarrow p, 1\rightarrow 1+p$. Form the quotient space by equating opposite sides. Here's the construction I'm looking at
https://www.dropbox.com/sh/h8ws4v10xn7s5bx/j1jxi603FI#f:modulus.PNG
URL: image

Hence this is a construction of a torus. The text then goes on to say "for differing values of $p$, the resulting surfaces are generally inequivalent to each other; that is to say, it is not possible to transform one into another by means of a holomorphic mapping."

I don't really see this. Suppose I choose a different $p$ that preserves the same orientation (say $p+1$ if $p$ is in the 1st quadrant). The parallelogram is different, but by the Riemann mapping theorem there's clearly a holomorphism between the two regions, and so the resulting Riemann surface should be equivalent, though the text says otherwise.

 

[chiro]

Hey theorem4.5.9.

One thing that I'm wondering is if you can create a non-simple region using your parameters for p, since you are using the argument for the Riemann-mapping theorem as a foundation for your stance. I think though that this is OK.

One question I have is regarding the orientation. The object you have mentioned will always be a parrallelogram but if if p and 1+p lie below the x-axis as depicted in your picture, then does the orientation change? If so, does this change the nature of your object and it's property?

 

[lavinia]

Suppose one lattice forms a square, the other a long rectangle. How are these tori conformal?

 

[theorem4.5.9]

I'm restricting the question to choosing $p$ so that the orientation doesn't change, So really the only changes I'm looking at would be things like how skinny or fat the torus is. Here's another figure from the text, that says there is no holomorphism between the two. Specifically it says "intuitively it seems pretty clear that there can be no conformal equivalence between the two, and indeed there is none."
https://www.dropbox.com/s/6zh0cvj7cf1n8gd/modulus2.PNG
URL: image

Suppose one lattice forms a square, the other a long rectangle. How are these tori conformal?

Working in the complex plane, I can find a conformal mapping from the circle to both the square and the rectangle. Invert one mapping, and compose the two. Then I have a conformal mapping between the square and the rectangle. I'm not thinking in terms of lattices, but in terms of quotient spaces, which is why I'm appealing to the Riemann mapping theorem.Perhaps I should just pick up an actual textbook on the subject. I appreciate suggestions for this as well.EDIT: I thought about this some more and realized I need to think about the boundary more. In the outline I gave above, you can't just associate opposite edges, else you risk breaking conformality at those points. To show there doesn't exist a holomorphism between the two (say the square and rectangle) I would look at the two lattices, and the fact that the only conformal mappings between the complex plane and itself are mobious transformations, and none of those candidates work. Hence the only choices of $p$ that makes the torus conformal are those that are given by mobious transformations (like $p, -p, 1/p$ etc. should all be conformal). Though I don't find this result intuitive by any means.

 

[mathwonk]

it is not enough to map the parallelograms onto each other. In fact it would need to be done by a linear mapping. Another way to think of it is that the whole plane has to map to itself taking one parallelogram to the other. Then you can see that parallelograms with different angles cannot be mapped to each other.

 

[lavinia]

it is not enough to map the parallelograms onto each other. In fact it would need to be done by a linear mapping. Another way to think of it is that the whole plane has to map to itself taking one parallelogram to the other. Then you can see that parallelograms with different angles cannot be mapped to each other.

I see - since the only conformal homeomorphism of the plane is a linear transformation.

Is another way of saying this that the map must not only be conformal on the interior of the rectangle but also on the edges? Then the angles that the edges make would be preserved under the mapping so two paralleograms with different angles could not be equivalent.

But also wouldn't it be true that if two paralleograpms had the same angles but different shapes then any map from one onto the other would have to change the angles between some pair of intersecting lines in the rectangle?

A question: Start with a parallelogram as a fundamental domain for a flat torus then slice off a triangle on one side and paste it to the other to make a rectangle. The identification along one edge makes a circular cylinder then on the other a twisted identification - wierd. Is this a general way to create different conformal structures on surfaces - cut along a non-bounding circle then reattach with a twist?

 

[theorem4.5.9]

I see - since the only conformal homeomorphism of the plane is a linear transformation.

Is another way of saying this that the map must not only be conformal on the interior of the rectangle but also on the edges? Then the angles that the edges make would be preserved under the mapping so two paralleograms with different angles could not be equivalent.

But also wouldn't it be true that if two paralleograpms had the same angles but different shapes then any map from one onto the other would have to change the angles between some pair of intersecting lines in the rectangle?

This is what I was trying to say in my previous post. Associating the opposing edges of a parallelogram as equivalent doesn't preserve angles. In this association the topology changes; edges turn into interior points.

 

[lavinia]

This is what I was trying to say in my previous post. Associating the opposing edges of a parallelogram as equivalent doesn't preserve angles. In this association the topology changes; edges turn into interior points.

I am not sure why you say the topology changes. I don't see your picture.

 

[theorem4.5.9]

I am thinking of taking say a square, and gluing the left and right sides together (so it's like a straw) and then gluing the top and bottom sides together to get a torus. This gets rid of the edges, and it's this object I was ultimately investigating.

 

[mathwonk]

the argument i learned from gunning's lectures on riemann surfaces is this: if there is a holomorphic isomorphism, the it induces a holomorphic isomorphism of the covering spaces, which is the complex plane, thus a linear isomorphism. yata yatta...