What is Riemann's approach to classifying 2d surfaces?

[AndreasC]

I was reading Bernhardt Riemann's old foundational text on abelian functions, and I found a part that really confused me.

What he is trying to do is set up an invariant to classify 2d surfaces as simply connected, multiply connected, etc via some kind of "connectivity number". From the text, I understand that the idea is to take closed curves on the boundary of such a surface, and come up with a maximal set of such curves so that it does NOT form the complete boundary of the surface, but adding any other such curves to it makes it complete. At least, that's what I THINK he is doing.

Trouble is, under this understanding, none of his lemmas make sense to me. Instead of describing them, I will just post the relevant part of the text:

So I ask, do I have the right idea? Can anyone give an explicit example of how these claims make sense?

 

[quasar987]

I have not read this text but I think I remember reading about it that Riemann is basically taking about Betti numbers but he does it in this language of "k-connectivity". The basic idea I think is this: take a sphere and take any closed loop a on it and take the scisors to it. This then splits the sphere into two parts so Riemann would call {a} a complete boundary of the sphere and so the sphere is 0-connected (H^1=0). Take now a torus, thought of as a square with opposite sides identified. These identified sides form two curves a₁ and a₂ on the torus. When you take the scisors to these two curves you get the square, minus its sides, which is still connected. But if you take a₁, a₂ and any other closed curve and take the scisors to them you get two disconected parts. So according to the definition in italics, the torus is 2-connected (H^1 = ZxZ).

 

[AndreasC]

I have not read this text but I think I remember reading about it that Riemann is basically taking about Betti numbers but he does it in this language of "k-connectivity". The basic idea I think is this: take a sphere and take any closed loop a on it and take the scisors to it. This then splits the sphere into two parts so Riemann would call {a} a complete boundary of the sphere and so the sphere is 0-connected (H^1=0). Take now a torus, thought of as a square with opposite sides identified. These identified sides form two curves a₁ and a₂ on the torus. When you take the scisors to these two curves you get the square, minus its sides, which is still connected. But if you take a₁, a₂ and any other closed curve and take the scisors to them you get two disconected parts. So according to the definition in italics, the torus is 2-connected (H^1 = ZxZ).

Huh, this does make sense, I think I was confused about what it was that he was talking about in general... I thought he was trying to gauge if a surface is simply connected or multiply connected or whatever. In that case it would seem a bit weird, since for instance in a disk with the hole in the middle any closed curve cuts it in two parts, same as a normal disk without a hole. So I was confused, but now it makes more sense that I understand what it is he was examining.

 

[mathwonk]

You may misunderstand the definition of n-connected, not through your own fault, but because of an error in the translation to English. In the original German, it says that a1,...,an do not form the complete boundary of a part of the surface, and yet when augmented by any other closed curve, the enlarged set of curves does form such a boundary. I hope this makes the lemmas make sense.

In fact the last lemma, proving that all such collections of curves have the same cardinality, is proved there by what is now called the "Steinitz exchange lemma", although Steinitz was not yet born when Riemann gave this argument. It was used much later by Steinitz to prove the cardinality of a vector basis is invariant.

[edit: I now think the following paragraph is wrong, and that quasar is indeed right! see post #7]
Also this connectivity measures, not the Betti numbers, but the genus, i.e. half the Betti number. So the sphere has connectivity indeed zero, but the torus has connectivity one, not two. E.g. a single closed curve on a torus can be drawn that does not bound any portion of the torus, namely take one of the edges of the rectangle in quasar's post, say a1. However when combined with a2 in his post, i.e. the other edge, the two curves together do bound the whole rectangle, i.e. not only a part of the surface but the whole surface, or rather that part of the surface interior to the rectangle. Taking in addition to the curve a1, another curve parallel to a1, i.e. a line segment in the rectangle parallel to a1, gives again two curves that together bound a cylinder on the torus, a part of the torus.

Moreover I am partly to blame for this error, since I was the reviewer of this translated volume and did not notice the error, until now.

 

[quasar987]

Thanks for clarifying. My post was just a guess because I don't know what is meant by "the complete boundary" of a surface or of a part of it.

 

[AndreasC]

Moreover I am partly to blame for this error, since I was the reviewer of this translated volume and did not notice the error, until now.

Wow I certainly did not expect that the reviewer of the translation would answer! Thank you very much.

 

[mathwonk]

@quasar987: I have thought more about it and realize I also do not understand what is meant by that term. You may indeed be right that connectivity is Betti number! And you are certainly correct that it rests on the meaning of "complete boundary of a part of the surface".

E.g. if that refers to a "proper" part of the surface, then it means the cuts do separate the surface into two disconnected pieces, i.e. they disconnect the surface. So curves that form the boundary of the entire connected surface would not count. Moreover it is not even clear just what it means for curves to form the complete boundary of the entire surface.

I.e. I argued that a1 + a2 in your example form the complete boundary of the interior of the rectangle, but they do not quite do so, I think. For if they did, then why would just one of them not form the boundary of the entire surface, i.e. in a sense, any closed curve is the boundary of its complement. But actually the complete boundary of the complement of a closed curve should be that curve counted twice, once in each direction.

My claim that connectivity is the genus might be correct if the closed curves are required to be disjoint. My German is not good enough to know off hand if Riemann actually made this requirement.

Reading Riemann took me about one day per page back when I reviewed this book, and it still demands a lot of thought, which is however richly repaid!

There are illustrated examples in the paper but I have trouble visualizing them. I.e. I can't tell if the example Riemann gives of a surface of connectivity one, is a torus or an annulus. Probably it is an annulus, and then you are right, since removing one closed cut from a torus reduces it to an annulus, and one of his lemmas says that connectivity goes down one after removing a closed cut.

So now I think you are right and I am wrong. My apologies!

Ah yes! Reading further Riemann actually mentions the example of a torus, which he says has connectivity three! So you are right, but we are both off by one from Riemann's convention of assigning connectivity n+1, not n, to a surface which has a maximal set of cardinality n, that do not form a boundary.

So in Riemann's convention a sphere has connectivity one, or is "simply connected", since the maximal cardinality of a non bounding family is zero. And a torus has connectivity three since it has, as you said, a maximal non bounding family of cardinality two.

Hence you are correct that he is measuring the Betti number, but he assigns as "connectivity" = Betti number +1.

OK I think I understand it. And you were correct all along. Sorry for muddying the water.

@AndreasC: by the way, an annulus is separated into two parts by removing any closed curve, but that closed curve is not always the complete boundary of one of those parts if the curve goes around the hole in the middle. so an annulus has connectivity 2 it seems, and results from removing a closed curve from a torus of connectivity three. making a transverse cut on an annulus, from a point on the outer boundary circle to a point on the inner boundary circle, changes it into a disc of connectivity one.