Anyway, most of the work is done, and my German is now a little more fluent. That is, up from “barely at all”. Most of the language is translated, leaving mostly just notation, plus a couple of technical terms like Asymptotenpuncte (asymptotic point) and Ebenenbüscheln (sheaves). You can see the original and the results here:

Steiner, J. – Über die Flächen dritten Grades [1857] (PDF)
Steiner, J. – Über die Flächen dritten Grades (text, edited OCR, German)
Steiner, J. – On surfaces of the third degree (text, edited and Google-translated OCR, English)

It’s some cool stuff, and it kinda makes me look forward to learning to do Math in German. But holy shit was this a pain in the ass.

Briefly, I’m exploring properties of subsets of the modular group, defined as the set of 2×2 matrices with determinant 1. I work with an eye toward the problem of generating congruence subgroups, which is not adequately solved. It is possible to check whether a given group is a subgroup, but it has shown to be difficult to work in the opposite direction.

This week I’m delving into Topology. Any modular subgroup can be described as a polygon on the hyperbolic plane, which in turn can be turned into a Riemann surface by identifying corresponding edges (see for example Wohlfahrt). Of particular interest is the nature of the covering of that surface. Read beneath the fold for diffeomorphisms.

The following is adapted from Schlichenmaier, pp.13-14:

First, suppose we are given two differentiable manifolds and an isomorphism between them:

Now we take a point in the first manifold, and its image in the second. We take a neighbourhood around each of these two points and define the function that maps each neighbourhood to coordinates in the reals:

Now, we can define a function that takes values in , maps them into , then , then back to , like so:

Now, if the mapping is differentiable for all possible choices of and , is called a differentiable map. If this is true of the inverse function as well, then is a diffeomorphism.

Diffeomorphisms help us classify manifolds. It’s easy to see that we can define an equivalence relation based on diffeomorphisms, and therefore divide manifolds into equivalence classes. The special case occurs when we’re talking about compact two-dimensional orientable manifolds (which includes all 1-dimensional complex manifolds). For these, there is exactly one diffeomorphic class for each genus. This is why we can classify such surfaces by genus, and why we get bad jokes like the following:

An employee at a coffee shop filled a topologist’s order, a coffee and donut. The topologist looked up, distraught, and asked, “But which is which?”

In higher dimensions, this is not true. Schlichenmeier gives the example of the 7-sphere, which has 28 inequivalent structures. Fortunately, I am only dealing with Riemann surfaces (1-dimensional complex) here :)

Finally, when dealing with complex manifolds, if all of our are holomorphic (analytic), and is bijective and holomorphic in both directions, then we call the map an analytic isomorphism.

Show that is even, for any given integer n.

And here’s the proof: we can easily see that . Since either or must be even, and the product of an even number with any other number is even, their product must also be even.

This sort of problem has a property similar to many geometry problems, which is that the original statement is easily comprehensible to people without a mathematical background. In this case, the solution also is widely accessible. There are some more complex proofs, though, and I’ve put one below the fold if you’re interested.

It is impossible to pick two integers a and b greater than two such that divides evenly into .

This is what I mean: I think this is a statement that is easily understood by anyone who remembers any 8th-grade math. Finding a proof of this, though, is trickier. We use an iterative approach, manipulating the expression until we get results.

First of all, for the sake of brevity, I’m going to use a bit of shorthand. When I say what I mean is “x divides evenly into y”. This is the same as saying for some integer value of p. The opposite of this is . Now the proof itself.

Proof. Let a and b be any (not necessarily different) integers greater than 2.

Suppose for the time being that . We are going to accept this assumption and show that it leads to a contradiction. This will prove that the assumption is false.

Now,

Since we’ve assumed that divides evenly into , we must be able to subtract any multiple of the first from the second, and the result will still be divisible by the first (if this is not obvious, try it with some integers and you’ll see why it works). So we can subtract from the above expression, and it should still be divisible by .

Here, we’ve shown that if then .

Now, we have three possible cases:

1. 
   In this case, we have . We can do the same thing as above, where we subtract a multiple of the first number, to get:
   
   
   This means that is a factor of 2, namely either 1 or 2. Thus or . 3 is impossible if is an integer, and 2 is only possible if , but we know that so this is impossible.
2. 
   Since , the second term must be equal to or greater than the first. We can’t have if . So this gives us:
   
   
   Now since , this means and therefore . But we divide the last line by two to get:
   
   Note that this difference cannot be zero, since the exponents are not equal. If both and are greater than 0 (and not equal), then their difference will be a multiple of 2. But it clearly is not. So one of the two must be less than or equal to 0. Which means either or is less than or equal to 1. It can’t be , because we know it’s greater than 2. And it can’t be because we already determined that it’s greater than 2. So we have another impossibility.
3. 
   This is the only possibility left, therefore it must be true (if our initial assumption is true).

Now we have , and we know that , since it’s greater than . Thus, we can plug into this whole process, in place of .

Note that is smaller than since is positive. So when we wind up at this same step, with a result of , we’re dealing with an even smaller number. We keep going, and eventually we get a number that is smaller than or equal to (I’m not going to prove that this is inevitable, but you should be able to convince yourself of this). This leads us to possibility 1 or 2, above, which is a contradiction.

Now the contradiction cascades back through the proof. Since a contradiction is generated in one of those three options, the conclusion that must be false, therefore it cannot be the case that . Thus it cannot be true that and so on, back through the iterations of the proof. Thus it cannot be true that , therefore it cannot be true that . This was our initial assumption, therefore we have proved the opposite.

Now, the final step is to generalize the whole thing. Sure, this may work for some choice of and , but is this always true? Well, the proof does not depend on the values of the variables, except for the restriction that they both be greater than 2. So regardless of what values you put in, the original statement (in blockquote) holds.

One final thing: why do we require that and be greater than 2? Because you can find counterexamples otherwise. For instance, if , then the statement is false. And if then the statement is always false.

Anyway, I hope I haven’t bored you too much. I find this sort of work very interesting, and this is just the basic groundwork of the field of Number Theory. Some of the more complicated theorems to do with, for instance, prime numbers, are very deep. Hopefully I’ll succeed in getting to work on this as a research project over the summer, and I can write more about it.