Wikipedia:Reference desk/Archives/Mathematics/2009 January 25

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January 25

Riemann Surface for Dummies

I have looked at Riemann surface article and I don't have a flying clue on what the hell it is. I cannot even construct a Riemann surface for a simple function like Square Root because I do not know where to start. The article looked like it was written by a maths professor to be read by other maths professors. I understand the concept of Complex Number, Real part and Imaginary part but after that I'm lost.

Is Riemann Surface far too advance for someone who understand basic Complex Number System to understand? 122.107.203.230 (talk) 06:31, 25 January 2009 (UTC)

A Riemann surface for the square root function is two copies of the plane, both cut along the same ray from the origin (conventionally the negative real axis), each side of the cut being stitched to the other side of the other sheet. This reflects the double values of the function (±√z). Does this help at all? Unfortunately many WP math articles consist mostly of highly formal language that's over my head. —Tamfang (talk) 07:00, 25 January 2009 (UTC)

Definitely not too advanced for you to understand - you just need some practice in visualising them. Think of two copies of the complex plane, one above the other. Now imagine twisting and stretching the upper sheet so that the two values of ${\sqrt {z}}$ on the upper sheet line up above $z$ on the lower sheet. 0 just has one square root, so 0 sits above 0 on its own. Every other complex number has two distinct square roots, so above every other point apart from 0 the upper sheet has two separate layers. Visualise what happens to the circle $|z|=1$ on the upper sheet. 1 stays above 1 on the lower sheet; i is twisted around the circle by another 90 degrees so it lies above -1; -1 is twisted around another 180 degrees so it lies above 1; -i is twisted around (by how much ?) so that it lies above -1. Each point on the circle is twisted around by an angle equal to its argument again, so the image of this circle on the upper sheet now goes twice round the origin.

Once you are happy with the ${\sqrt {z}}$ example, try visualising the surface for the cube root of z, or for ${\sqrt {z^{2}-1}}$. The difficult part to visualising a Riemann surface is that you can only visualise it in 3 dimensions by making it intersect itself, or by creating a branch cut where the points on the upper sheet suddenly "jump" from one layer to to another one. The actual Riemann surface does not have any self-intersections or cuts - these are just artefacts of its 3D embedding.

Follow the links from this page at MIT to see some images of various different Riemann surfaces. Or search for "Riemann surface" with Google Images to see dozens of examples. Gandalf61 (talk) 10:06, 25 January 2009 (UTC)

I tried doing what you said. "Think of two copies of the complex plane, one above the other." I took two pieces of paper to represent a complex plane each. But I having severe difficulties deforming a piece of paper. How did Riemann deform his paper models? After thinking about it for a well. I decided that Riemann Surfaces cannot be constructed with paper because paper cannot deform at all. What is needed is rubber because rubber can be stretched while paper cannot be stretched. I think I need to start with a square rubber surface to represent a complex plane. Then cut at the branch point and stretched and twist the rubber to make two revolution. Hmmm....122.107.203.230 (talk) 22:35, 25 January 2009 (UTC)

It can't be done without the paper passing through itself, you need to work in 4 or 5 dimensional space if you want to actually construct it that way. --Tango (talk) 01:59, 26 January 2009 (UTC)

A Google image search found a decent rendering — you can browse the directory it's in for images of other functions. --Tardis (talk) 16:14, 26 January 2009 (UTC)

Have you ever played with a Möbius strip? Trace a line along the centre of the strip, and notice how at one point you end up on the opposite side of the paper to where you started. The Riemann surface for the square root is like that, extended up one dimension. So instead of tracing a line around the middle, you instead trace a circle around the origin. At some point, you find yourself on the "other side" (i.e. the other sheet of the surface), where by some measurements you're back where you started and by others you aren't. A bit further on, and you're definitely back where you started. Confusing Manifestation(Say hi!) 22:48, 26 January 2009 (UTC)

I find that misleading: Riemann surfaces are orientable; Möbius bands are not. Michael Hardy (talk) 23:58, 26 January 2009 (UTC)

simple quadrilateral from distance matrix

Say I have a Euclidian distance matrix for four points (i.e. six pairwise distances). Unless I'm mistaken, this matrix defines exactly one simple (non-self-intersecting) quadrilateral in the plane, up to rotations/translations/reflections. Is there some easy way to figure out its dimensions? That is, to figure out which four of the six pairwise distances are on the perimeter, and which two are the diagonals? I can't seem to figure anything out, though I'm guessing there's some, probably elementary, identity that I'm missing. --Delirium (talk) 09:42, 25 January 2009 (UTC)

If the quadrilateral is convex then the diagonals will cross. If it is concave then they won't. It may be a trapezoid/ium in shape, in which case the diagonal won't be the longest length. I can't answer the problem (at present) but some empirical thoughts: the six lines form four triangles, each length is part of exactly two trangles; the remaining three lengths can be "attached" to the three triangle corners in 6 ways (ABC -> xyz, xzy, yxz, yzx, zxy, zyx). Chose one length and work your way through the ten combinations of two the remaining five until you get (a) a valid triangle that (b) has three corner centred circles meeting at a point. That is at most (10-1)*6 = 54 evaluations. -- SGBailey (talk) 10:22, 25 January 2009 (UTC)

If you have a triangle plus a point in the centre then you don't have four points on the perimeter. Otherwise you might be interested in the gift wrapping algorithm which is one of the Convex hull algorithms which will order the points round the quadrilateral so finding the diagonals is easy. I didn't read properly Dmcq (talk) 10:35, 25 January 2009 (UTC)

I assume this is what Dmcq meant, but I'll rephrase it a bit. If you have a triangle plus a point in the middle, there are three possible quadrilaterals, not one. So the premise of your question is false. In this case none of the quadrilaterals is convex. If on the other hand none of the points is inside the triangle formed by the other three, then the quadrilateral is convex, and there's only one. (This analysis ignores cases in which three of the points are aligned.) Joeldl (talk) 10:56, 25 January 2009 (UTC)

Ah yeah, that's true; thanks. And yeah, in my case no three points are collinear. It's also relatively easy to figure out cases where one of the points is inside the triangle formed by the other three. In the remaining convex case, though, is there a way to figure out which two distances are the diagonals? Dmcq's suggestion to use the gift wrapping algorithm to order the points is interesting, but seems to require that I have the actual Cartesian coordinates of the points, rather than only the distance matrix. Of course I can first compute an embedding of the distance matrix in the plane and then go from there (in which case I could also look for which two line segments cross, which would be the diagonal), but I was hoping to avoid that step and do something directly from the distance matrix (this is part of a larger computation). --Delirium (talk) 11:05, 25 January 2009 (UTC)

Well here I am again. Tackling the actual question. Take a triangle of lines say AB BC CA, each of the sides say AB defines two half planes, the half plane not containing C and the half plane containing C. The intersections of these half planes define 7 regions (not 8 as it can't always not be on the side containing th other point). And these 7 regions define the shape of the quadrilateral as far as figuring out which lines are the diagonals or if there isn't any. You can find if D is on the same side of AB as C by using the triangle cosine law on ABC and ABD, this gives two possible angles for DAC, these gives possible values for the lengths of DC and if it is the shorter possibility you're on the same side. Anyway I think that'll work, proof left as an exercise! Dmcq (talk) 11:10, 25 January 2009 (UTC)

Here's a way to find out which case you're in, and which segments are your diagonals, using only the lengths of the six segments. First note that Heron's formula gives you a way of computing the area of a triangle from its sides. Your points A, B, C, D form four triangles, depending on which of the four points you leave out. Compute their areas, and order them α ≥ β ≥ γ ≥ δ In the first case described above, the area of the outside triangle is the largest one α, and you have α = β + γ + δ. In the other case, depending on how you divide your convex quadrilateral, you get that two of the triangles together have the same area as the other two together. In this case, α + δ = β + γ. So you can tell which case you're in and which triangles are which. Joeldl (talk) 11:17, 25 January 2009 (UTC)

Very nice. Now why didn't I think of that, duh no don't answer that! Dmcq (talk) 11:27, 25 January 2009 (UTC)

Thanks. actually, I'm realizing it's not clear it works when some of the areas are equal in the convex case. I'll have to think about that. Joeldl (talk) 11:35, 25 January 2009 (UTC)

Something else needs to be done when α = β. Any thoughts? Joeldl (talk) 11:41, 25 January 2009 (UTC)

If theres two pairs of equal amounts then you've two parallel sides. There's two sets of possible diagonals, the one with the largest sum is the real pair. If all four amounts are the same then just take the largest and the other diagonal as it is a parallelogram. Dmcq (talk) 12:11, 25 January 2009 (UTC)

In fact the sum of the diagonals will always be greater than the sum of two opposite sides for any convex quadrilateral. So we only really need to do the three sums and see which is he biggest. Dmcq (talk) 12:22, 25 January 2009 (UTC)

Is that so? Are you using an outside result, or is this something you've just found? If the latter, how? Joeldl (talk) 12:28, 25 January 2009 (UTC)

Never mind, I agree with you. Joeldl (talk) 12:30, 25 January 2009 (UTC)

Great answer, that completely solves the convex case. Joeldl (talk) 12:31, 25 January 2009 (UTC)

In the non-convex case I'll just say that pair are the honorary diagonals and that'll solve that too :) I think the triangle area idea really needs to be used to determine convexity. Dmcq (talk) 12:40, 25 January 2009 (UTC)

Everything depends on the sign of α - β - γ (the largest area minus the two middle areas). I really prefer your idea in the convex case, because it would not have been appropriate to set up α = β as a special case. That's because when doing calculations on a computer, you may not know it when two numbers are exactly equal. So we would have needed to set up special criteria when α and β were close, and we would have needed to say what "close" meant. That being said, if α, β, γ, δ have been computed, and it has been determined that α > β, and α - β - γ is negative, then that's already enough to answer the question. Joeldl (talk) 12:53, 25 January 2009 (UTC)

The other good thing about the areas is you can check the points really are planar. Sorry if I'm a bit cynical about accepting that sort of thing for real world data. Dmcq (talk) 13:49, 25 January 2009 (UTC)

Very nice, guys. I wonder if Joeldl's criterium with areas may be put in simpler form after algebraic manipulations. Does one need to compute four times a square root and evaluate Hero's formula? For instance, if we want to get rid of all the square roots in α + δ = β + γ to get a polynomial expression in the six distances, in principle, we need to square both sides three times. Now the resulting polynomial sounds horrible, but, maybe, all the symmetry may produce simplifications... --pma (talk) 17:37, 25 January 2009 (UTC)

<--- I just had a look at the formula for a tetrahedron as the six lines form a degenerate one with zero volume. If you take the determinate form of the volume in that article and differentiate on the length of a line the result should be positive for 4 lines and negative for two for a proper quadrilateral as increasing the length of a side pops the tetrahedron up but you have to decrease the size of a diagonal to do that. When the quadrilateral isn't convex then you get three which are positive and three negative. If q12 is the side d12 squared then differentiating on q12 one gets

$-2q_{34}^{2}-4q_{12}q_{34}+2(q_{23}q_{34}+q_{24}q_{34}+q_{13}q_{34}+q_{14}q_{34})-2(q_{24}q_{14}+q_{23}q_{13})+2(q_{23}q_{14}+q_{24}q_{13})$

I think that's right. With similar expressions for the other five. Not a particularly pretty expression that I can see unfortunately. So you don't need square roots but I'm not sure you gain much. Dmcq (talk) 18:00, 25 January 2009 (UTC)

Thanks folks! I think between Joeldl and Dmcq's answers I should be able to implement something reasonable. --Delirium (talk) 21:58, 25 January 2009 (UTC)

By the way it looks like you'd just have to evaluate that last equation twice for a pair of opposites like 12 and 34 and multiply together. If the result is +ve you've got a convex quadrilateral and if it's -ve one of the points is inside, otherwise there's a point on an edge. Dmcq (talk) 13:28, 26 January 2009 (UTC)

As a bit of a followup to my own question, it looks like there's a field of distance geometry that studies questions of the form "can I determine X just from the distance matrix?". I haven't followed it up beyond that, though. --Delirium (talk) 20:23, 26 January 2009 (UTC)

The formula there for 4 points is the same as the one for the tetrahedron that I differentiated. You can check using it whether the 6 distances all lie on a plane. Always a good idea. Dmcq (talk) 00:07, 27 January 2009 (UTC)

Unity

The quote by Kronecker that "God created the integers..." greives me somewhat because integers, even natural numbers seem to be just as artifical constructs as complex numbers, or matrices, or groups. My reasoning is this:

A human being may regard a tree and say, "I see one tree, therefore the concept of unity exists in nature." But surely this reasoning is flawed as the tree has many constituent parts- branches, bark, cells, atoms. We could regard the tree as part of a greater whole, such as a forest, or we could regard it as the aggregate of many other things. Wether we regard a tree as one thing or many things is completely arbitrary and human, since nature does not at any point designate the property of "one" to something.

Unless, that is, that there were some structure or object in nature that could not be split or divided into two or more peices. This would show that unity wasn't an artifical construct and that the natural numbers were something that was outside of human experience and NOT artifical.

I would like to know if anyone sees any flaws in considering all mathematics to be the product of human thought, having no true representation in nature, ignoring the issue of whether or not a truly indivisible structure exists in the universe. 82.32.49.186 (talk) 22:51, 25 January 2009 (UTC)

You might find something relevant to your question at Philosophy of mathematics. I go in for some opinions on the refdesk but I shouldn't and as to this I think you're better debating it with a salesman over the price. You never know they might give up and reduce it for you. Dmcq (talk) 23:30, 25 January 2009 (UTC)

I think a common view is that we (whether this means humans or a broader group of animals) evolved to have a concept of numbers because it was well adapted to understanding (and therefore surviving in) our surroundings. "Understanding" can probably be taken here to mean "rendering predictable." Looking at things this way, before there were living beings, the organization of the universe was such that the emergence of a concept of number in any intelligent beings that appeared in it was likely. Joeldl (talk) 04:47, 26 January 2009 (UTC)

I am not sure about the meaning of Kronecker's sentence indeed. He could have said as well "the integers came from the Hell", with analogous meaning. I mean, taken in a weak sense, the sentence means that (at Kronecker times) mathematicians had succeeded in giving a satisfying foundation of almost all parts of mathematics, and that everything was based on the natural numbers (e.g., euclidean geometry may be founded on real linear algebra, linear algebra on reals numbers, reals numbers on rational numbers etc), BUT that a satisfying foundation for natural numbers was somehow still lacking (Peano's axiomatization is 20 or 30 years later). Today, after von Neumann's model for ordinal numbers we could say "God created the empty set, etc", that may be agreeded even by an atheist ;) --pma (talk) 22:12, 26 January 2009 (UTC)

In the sense A implies B I'm quite happy to accept that God created everything :) Dmcq (talk) 22:51, 26 January 2009 (UTC)

But if everything is proven, everybody has to go home .) pma (talk) 22:56, 26 January 2009 (UTC)

Even if there are many patterns to see, that doesn't invalidate those people choose to focus on. Even if the tree is both divisible and part of a larger system, there's still a tree. Here's an interesting thought experiment, though: what's one bowl of water plus one bowl of water? Black Carrot (talk)05:21, 27 January 2009 (UTC)

Two bowls of water I believe. Perhaps you meant one cloud plus one cloud? Or one rabbit plus one rabbit? Dmcq (talk) 09:02, 27 January 2009 (UTC)

One bowl of water plus one bowl of water equals two bowls of water to anyone trained (pretty much everyone) in basic arithmetic at least. But the universe does not apply a property to a bowl of water that says "this is one bowl of water", only the human mind. Like I said, it seems that only the existence of an object that was itself and itself alone could mean that basic mathematical structures existed in nature. Although I like the idea that the null set is something that could exist in nature, since we can define a successor function that constructs natural numbers from empty sets? I'll need to think more on how the empty set can exist in nature. 82.32.49.186 (talk) 12:40, 30 January 2009 (UTC)

Question about probability and ergodicity

Can somebody point me to something that helps motivate the connection between a stationary ergodic process and independence? Along those lines, it's only an infinite sequence of iid random variables that can be thought of as an ergodic process, right? Because for any finite collection of iid random variables, I'm thinking it's possible to construct a set of positive, non-full measure which is preserved under all shift operators. Thanks, Ray (talk) 22:58, 25 January 2009 (UTC)