[math-fun] winding points and sphere handles
This material is surely elementary, but I couldn't find it anywhere: maybe someone can cite a reference? A 2-sheet Riemann sphere with 2 winding points is homeomorphic to a torus: for if its tube radius remains fixed while the swept radius approaches zero, the torus approaches a double sphere with winding points at opposite ends of a diameter --- whence they may be deformed to arbitrary locations, and the sphere to a topological sphere. In a similar fashion, s sheets with t winding points are homeomorphic to a sphere with (s t)/4 handles, ie. genus = (s t)/4 , at any rate for s, t even; otherwise it perhaps suffices to take the ceiling ... I would probably have remained unaware of this idea if I had not once been experimenting with a simple Maple graphics demo program to plot a torus, and noticed that what I had casually assumed must be a simple sphere instead appeared mysteriously fuzzy and uneven. On this occasion at any rate, Maple was innocent of blame, and indeed had provided valuable insight! Fred Lunnon
Your 2-sheeted Riemann sphere with 2 "winding points" can be thought of as a map from the unknown surface S to the sphere having 2 points that are each "ramified" of (I presume) index 2 (locally the map looks like z |-> z^2). This can be identified by viewing it as 2 spheres, each of which is missing 2 points, but with 2 of those 4 missing points replaced. This means the Euler characteristic is X(S) = 2(X(S^2) - 2) + 2 = 2*0 + 2 = 2. Checking that it's locally Euclidean, this nails it as a sphere, the only surface S with X(S) = 2. Or am I misunderstanding what a "winding point" is? --Dan On Apr 19, 2014, at 6:42 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
This material is surely elementary, but I couldn't find it anywhere: maybe someone can cite a reference?
A 2-sheet Riemann sphere with 2 winding points is homeomorphic to a torus: for if its tube radius remains fixed while the swept radius approaches zero, the torus approaches a double sphere with winding points at opposite ends of a diameter --- whence they may be deformed to arbitrary locations, and the sphere to a topological sphere.
In a similar fashion, s sheets with t winding points are homeomorphic to a sphere with (s t)/4 handles, ie. genus = (s t)/4 , at any rate for s, t even; otherwise it perhaps suffices to take the ceiling ...
I would probably have remained unaware of this idea if I had not once been experimenting with a simple Maple graphics demo program to plot a torus, and noticed that what I had casually assumed must be a simple sphere instead appeared mysteriously fuzzy and uneven. On this occasion at any rate, Maple was innocent of blame, and indeed had provided valuable insight!
Fred Lunnon
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That could explain why I didn't find this anywhere, I suppose. So where does my purported homeomorphism go belly up? Immersed in 3-space, once the sweep inceeds the tube, a nodal singularity develops where the surface self-intersects. But this doesn't behave like a winding point at all --- one revolution around it leaves me back where I started (quite so). In any case, immersion is irrelevant to the homeomorphism: continuity at the winding points would be lost as soon as I tried to depart from radius zero. And another "proof" bites the dust! Still, nice exercise ... WFL On 4/19/14, Dan Asimov <dasimov@earthlink.net> wrote:
Your 2-sheeted Riemann sphere with 2 "winding points" can be thought of as a map from the unknown surface S to the sphere having 2 points that are each "ramified" of (I presume) index 2 (locally the map looks like z |-> z^2).
This can be identified by viewing it as 2 spheres, each of which is missing 2 points, but with 2 of those 4 missing points replaced. This means the Euler characteristic is X(S) = 2(X(S^2) - 2) + 2 = 2*0 + 2 = 2. Checking that it's locally Euclidean, this nails it as a sphere, the only surface S with X(S) = 2.
Or am I misunderstanding what a "winding point" is?
--Dan
On Apr 19, 2014, at 6:42 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
This material is surely elementary, but I couldn't find it anywhere: maybe someone can cite a reference?
A 2-sheet Riemann sphere with 2 winding points is homeomorphic to a torus: for if its tube radius remains fixed while the swept radius approaches zero, the torus approaches a double sphere with winding points at opposite ends of a diameter --- whence they may be deformed to arbitrary locations, and the sphere to a topological sphere.
In a similar fashion, s sheets with t winding points are homeomorphic to a sphere with (s t)/4 handles, ie. genus = (s t)/4 , at any rate for s, t even; otherwise it perhaps suffices to take the ceiling ...
I would probably have remained unaware of this idea if I had not once been experimenting with a simple Maple graphics demo program to plot a torus, and noticed that what I had casually assumed must be a simple sphere instead appeared mysteriously fuzzy and uneven. On this occasion at any rate, Maple was innocent of blame, and indeed had provided valuable insight!
Fred Lunnon
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Fred, I don't know where you went wrong, since that style of reasoning is unfamiliar to me. But the Riemann-Hurwitz formula covers cases like yours: (From < http://en.wikipedia.org/wiki/Riemann–Hurwitz_formula >): ----- For an orientable surface S the Euler characteristic χ(S) is 2-2g, where g is the genus (the number of handles) . . .. In the case of an (unramified) covering map of surfaces pi:S' -> S that is surjective and of degree N, we should have the formula X(S') = N*X(S) That is because each simplex of S should be covered by exactly N in S′ — at least if we use a fine enough triangulation of S, as we are entitled to do since the Euler characteristic is a topological invariant. What the Riemann–Hurwitz formula does is to add in a correction to allow for ramification (sheets coming together). Now assume that S and S′ are Riemann surfaces, and that the map pi is complex analytic. The map pi is said to be ramified at a point P in S′ if there exist analytic coordinates near P and pi(P) such that pi takes the form pi(z) = z^n, and n > 1. An equivalent way of thinking about this is that there exists a small neighborhood U of P such that pi(P) has exactly one preimage in U, but the image of any other point in U has exactly n preimages in U. The number n is called the ramification index at P and also denoted by e_P. In calculating the Euler characteristic of S′ we notice the loss of e_P − 1 copies of P above pi(P) (that is, in the inverse image of pi(P)). Now let us choose triangulations of S and S′ with vertices at the branch and ramification points, respectively, and use these to compute the Euler characteristics. Then S′ will have the same number of d-dimensional faces for d different from zero, but fewer than expected vertices. Therefore we find a "corrected" formula X(S') = N*X(S) - Sum over P in S' of (e_P - 1) (all but finitely many P have e_P = 1, so this is quite safe). This formula is known as the Riemann–Hurwitz formula and also as Hurwitz's theorem. ----- On Apr 19, 2014, at 9:07 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
That could explain why I didn't find this anywhere, I suppose. So where does my purported homeomorphism go belly up?
Immersed in 3-space, once the sweep inceeds the tube, a nodal singularity develops where the surface self-intersects. But this doesn't behave like a winding point at all --- one revolution around it leaves me back where I started (quite so).
In any case, immersion is irrelevant to the homeomorphism: continuity at the winding points would be lost as soon as I tried to depart from radius zero.
And another "proof" bites the dust! Still, nice exercise ... WFL
On 4/19/14, Dan Asimov <dasimov@earthlink.net> wrote:
Your 2-sheeted Riemann sphere with 2 "winding points" can be thought of as a map from the unknown surface S to the sphere having 2 points that are each "ramified" of (I presume) index 2 (locally the map looks like z |-> z^2).
This can be identified by viewing it as 2 spheres, each of which is missing 2 points, but with 2 of those 4 missing points replaced. This means the Euler characteristic is X(S) = 2(X(S^2) - 2) + 2 = 2*0 + 2 = 2. Checking that it's locally Euclidean, this nails it as a sphere, the only surface S with X(S) = 2.
Or am I misunderstanding what a "winding point" is?
--Dan
On Apr 19, 2014, at 6:42 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
This material is surely elementary, but I couldn't find it anywhere: maybe someone can cite a reference?
A 2-sheet Riemann sphere with 2 winding points is homeomorphic to a torus: for if its tube radius remains fixed while the swept radius approaches zero, the torus approaches a double sphere with winding points at opposite ends of a diameter --- whence they may be deformed to arbitrary locations, and the sphere to a topological sphere.
In a similar fashion, s sheets with t winding points are homeomorphic to a sphere with (s t)/4 handles, ie. genus = (s t)/4 , at any rate for s, t even; otherwise it perhaps suffices to take the ceiling ...
I would probably have remained unaware of this idea if I had not once been experimenting with a simple Maple graphics demo program to plot a torus, and noticed that what I had casually assumed must be a simple sphere instead appeared mysteriously fuzzy and uneven. On this occasion at any rate, Maple was innocent of blame, and indeed had provided valuable insight!
Fred Lunnon
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