Compactifying complex spaces

In this post, I’ll try to explain how Riemann surfaces (or topological spaces in general) are transformed into compact spaces. Compact spaces are, in essence, closed and bounded subsets of some larger space. The larger space is unbounded – or ‘infinite’ if you want (the term ‘infinite’ is less precise – from a mathematical point of view at least).

I am sure you have all seen it: the Euclidean or complex plane gets wrapped around a sphere (the so-called Riemann sphere), and the Riemann surface of a square root function becomes a torus (i.e. a donut-like object). And then the donut becomes a coffee cup (yes: just type ‘donut and coffee cup’ and look at the animation). The sphere and the torus (and the coffee cup of course) are compact spaces indeed – as opposed to the infinite plane, or the infinite Riemann surface representing the domain of a (complex) square root function. But what does it all mean?

Let me, for clarity, start with a note on the symbols that I’ll be using in this post. I’ll use a boldface z for the complex number z = (x, y) = reiθ in this post (unlike what I did in my previous posts, in which I often used standard letters for complex numbers), or for any other complex number, such as w = u + iv. That’s because I want to reserve the non-boldface letter z for the (real) vertical z coordinate in the three-dimensional (Cartesian or Euclidean) coordinate space, i.e. R3. Likewise, non-boldface letters such as x, y or u and v, denote other real numbers. Note that I will also use a boldface and a boldface to denote the set of real numbers and the complex space respectively. That’s just because the WordPress editor has its limits and, among other things, it can’t do blackboard bold (i.e. these double struck symbols which you usually see as symbols for the set of real and complex numbers respectively). OK. Let’s go for it now.

In my previous post, I introduced the concept of a Riemann surface using the multivalued square root function w = z1/2 = √z. The square root function has only two values. If we write z as z = rei θ, then we can write these two values as w1 = √r ei(θ/2) and w2 = √r ei(θ/2 ± π). Now, √r ei(θ/2 ± π) is equal to √r ei(±π)ei(θ/2) =  – √r ei(θ/2) and, hence, the second root is just the opposite of the first one, so w= – w1.

Introducing the concept of a Riemann surface using a ‘simple’ quadratic function may look easy enough but, in fact, this square root function is actually not the easiest one to start with. First, a simple single-valued function, such as w = 1/z (i.e. the function that is associated with the Riemann sphere) for example, would obviously make for a much easier point of departure. Secondly, the fact that we’re working with a limited number of values, as opposed to an infinite number of values (which is the case for the log z function for example) introduces this particularity of a surface turning back into itself which, as I pointed out in my previous post, makes the visualization of the surface somewhat tricky – to the extent it may actually prevent a good understanding of what’s actually going on.

Indeed, in the previous post I explained how the Riemann surface of the square root function can be visualized in the three-dimensional Euclidean space (i.e. R3). However, such representations only show the real part of z1/2, i.e. the vertical distance Re(z1/2) = √r cos(θ/2 + nπ), with n = 0 or ± 1. So these representations, like the one below for example, do not show the imaginary part, i.e.  Im(z1/2) = √r sin(θ/2 + nπ) (n = 0, ± 1).

That’s both good and bad. It’s good because, in a graph like this, you want one point to represent one point only, and so you wouldn’t get that if you would superimpose the plot with the imaginary part of wz1/2 on the plot showing the real part only. But it’s also bad, because one often forgets that we’re only seeing some part of the ‘real’ picture here, namely the real part, and so one often forgets to imagine the imaginary part. 🙂 

sqrt

The thick black polygonal line in the two diagrams in the illustration above shows how, on this Riemann surface (or at least its real part), the argument θ of  z = rei θ will go from 0 to 2π (and further), i.e. we’re making (more than) a full turn around the vertical axis, as the argument Θ of w =  z1/2 = √reiΘ makes half a turn only (i.e. Θ goes from 0 to π only). That’s self-evident because Θ = θ/2. [The first diagram in the illustration above represents the (flat) w plane, while the second one is the Riemann surface of the square root function, so it represents but so we have like two points for every z on the flat z plane: one for each root.]

All these visualizations of Riemann surfaces (and the projections on the z and w plane that come with them) have their limits, however. As mentioned in my previous post, one major drawback is that we cannot distinguish the two distinct roots for all of the complex numbers z on the negative real axis (i.e. all the points z = rei θ for which θ is equal to ±π, ±3π,…). Indeed, the real part of wz1/2, i.e. Re(w), is equal to zero for both roots there, and so, when looking at the plot, you may get the impression that we get the same values for w there, so that the two distinct roots of z (i.e. wand w2) coincide. They don’t: the imaginary part of  wand wis different there, so we need to look at the imaginary part of w too. Just to be clear on this: on the diagram above, it’s where the two sheets of the Riemann surface cross each other, so it’s like there’s an infinite number of branch points, which is not the case: the only branch point is the origin.

So we need to look at the imaginary part too. However, if we look at the imaginary part separately, we will have a similar problem on the positive real axis: the imaginary part of the two roots coincides there, i.e. Im(w) is zero, for both roots, for all the points z = rei θ for which θ = 0, 2π, 4π,… That’s what represented and written in the graph below.

branch point

The graph above is a cross-section, so to say, of the Riemann surface  w = z1/2 that is orthogonal to the z plane. So we’re looking at the x axis from -∞ to +∞ along the y axis so to say. The point at the center of this graph is the origin obviously, which is the branch point of our function w = z1/2, and so the y axis goes through it but we can’t see it because we’re looking along that axis (so the y-axis is perpendicular to the cross-section).

This graph is one I made as I tried to get some better understanding of what a ‘branch point’ actually is. Indeed, the graph makes it perfectly clear – I hope 🙂 – that we really have to choose between one of the two branches of the function when we’re at the origin, i.e. the branch point. Indeed, we can pick either the n = 0 branch or the n = ±1 branch of the function, and then we can go in any direction we want as we’re traveling on that Riemann surface, but so our initial choice has consequences: as Dr. Teleman (whom I’ll introduce later) puts it, “any choice of w, followed continuously around the origin, leads, automatically, to the opposite choice as we turn around it.” For example, if we take the w1 branch (or the ‘positive’ root as I call it – even if complex numbers cannot be grouped into ‘positive’ or ‘negative’ numbers), then we’ll encounter the negative root wafter one loop around the origin. Well… Let me immediately qualify that statement: we will still be traveling on the wbranch but so the value of w1 will be the opposite or negative value of our original was we add 2π to arg z = θ. Mutatis mutandis, we’re in a similar situation if we’d take the w2 branch. Does that make sense?

Perhaps not, but I can’t explain it any better. In any case, the gist of the matter is that we can switch from the wbranch to the wbranch at the origin, and also note that we can only switch like that there, at the branch point itself: we can’t switch anywhere else. So there, at the branch point, we have some kind of ‘discontinuity’, in the sense that we have a genuine choice between two alternatives.

That’s, of course, linked to the fact that one cannot define the value of our function at the origin: 0 is not part of the domain of the (complex) square root function, or of the (complex) logarithmic function in general (remember that our square root function is just a special case of the log function) and, hence, the function is effectively not analytic there. So it’s like what I said about the Riemann surface for the log z function: at the origin, we can ‘take the elevator’ to any other level, so to say, instead of having to walk up and down that spiral ramp to get there. So we can add or subtract ± 2nπ to θ without any sweat.

So here it’s the same. However, because it’s the square root function, we’ll only see two buttons to choose from in that elevator, and our choice will determine whether we get out at level Θ = α (i.e. the wbranch) or at level Θ = α ± π (i.e. the wbranch). Of course, you can try to push both buttons at the same time but then I assume that the elevator will make some kind of random choice for you. 🙂 Also note that the elevator in the log z parking tower will probably have a numpad instead of buttons, because there’s infinitely many levels to choose from. 🙂

OK. Let’s stop joking. The idea I want to convey is that there’s a choice here. The choice made determines whether you’re going to be looking at the ‘positive’ roots of z, i.e. √r(cosΘ+sinΘ) or at the ‘negative’ roots of z, i.e. √r(cos(Θ±π)+isin(Θ±π)), or, equivalently (because Θ = θ/2) if you’re going to be looking at the values of w for θ going from 0 to 2π, or the values of w for θ going from 2π to 4π.

Let’s try to imagine the full picture and think about how we could superimpose the graphs of both the real and imaginary part of w. The illustration below should help us to do so: the blue and red image should be shifted over and across each other until they overlap completely. [I am not doing it here because I’d have to make one surface transparent so you can see the other one behind – and that’s too much trouble now. In addition, it’s good mental exercise for you to imagine the full picture in your head.]  

Real and imaginary sheets

It is important to remember here that the origin of the complex z plane, in both images, is at the center of these cuboids (or ‘rectangular prisms’ if you prefer that term). So that’s what the little red arrow points is pointing at in both images and, hence, the final graph, consisting of the two superimposed surfaces (the imaginary and the real one), should also have one branch point only, i.e. at the origin.

[…]

I guess I am really boring my imaginary reader here by being so lengthy but so there’s a reason: when I first tried to imagine that ‘full picture’, I kept thinking there was some kind of problem along the whole x axis, instead of at the branch point only. Indeed, these two plots suggest that we have two or even four separate sheets here that are ‘joined at the hip’ so to say (or glued or welded or stitched together – whatever you want to call it) along the real axis (i.e. the x axis of the z plane). In such (erroneous) view, we’d have two sheets above the complex z plane (one representing the imaginary values of √z and one the real part) and two below it (again one with the values of the imaginary part of √z and one representing the values of the real part). All of these ‘sheets’ have a sharp fold on the x axis indeed (everywhere else they are smooth), and that’s where they join in this (erroneous) view of things.

Indeed, such thinking is stupid and leads nowhere: the real and imaginary parts should always be considered together, and so there’s no such thing as two or four sheets really: there is only one Riemann surface with two (overlapping) branches. You should also note that where these branches start or end is quite arbitrary, because we can pick any angle α to define the starting point of a branch. There is also only one branch point. So there is no ‘line’ separating the Riemann surface into two separate pieces. There is only that branch point at the origin, and there we decide what branch of the function we’re going to look at: the n = 0 branch (i.e. we consider arg w = Θ to be equal to θ/2) or the n = ±1 branch (i.e. we take the Θ = θ/2 ± π equation to calculate the values for wz1/2).

OK. Enough of these visualizations which, as I told you above already, are helpful only to some extent. Is there any other way of approaching the topic?

Of course there is. When trying to understand these Riemann surfaces (which is not easy when you read Penrose because he immediately jumps to Riemann surfaces involving three or more branch points, which makes things a lot more complicated), I found it useful to look for a more formal mathematical definition of a Riemann surface. I found such more formal definition in a series of lectures of a certain Dr. C. Teleman (Berkeley, Lectures on Riemann surfaces, 2003). He defines them as graphs too, or surfaces indeed, just like Penrose and others, but, in contrast, he makes it very clear, right from the outset, that it’s really the association (i.e. the relation) between z and w which counts, not these rather silly attempts to plot all these surfaces in three-dimensional space.

Indeed, according to Dr. Teleman’s introduction to the topic, a Riemann surface S is, quite simply, a set of ‘points’ (z, w) in the two-dimensional complex space C= C x(so they’re not your typical points in the complex plane but points with two complex dimensions), such that w and z are related with each other by a holomorphic function w = f(z), which itself defines the Riemann surface. The same author also usefully points out that this holomorphic function is usually written in its implicit form, i.e. as P(z, w) = 0 (in case of a polynomial function) or, more generally, as F(z, w) = 0.

There are two things you should note here. The first one is that this eminent professor suggests that we should not waste too much time by trying to visualize things in the three-dimensional R3 = R x R x R space: Riemann surfaces are complex manifolds and so we should tackle them in their own space, i.e. the complex Cspace. The second thing is linked to the first: we should get away from these visualizations, because these Riemann surfaces are usually much and much more complicated than a simple (complex) square root function and, hence, are usually not easy to deal with. That’s quite evident when we consider the general form of the complex-analytical (polynomial) P(z, w) function above, which is P(z, w) = wn + pn-1(z)wn-1 + … + p1(z)w + p0(z), with the pk(z) coefficients here being polynomials in z themselves.

That being said, Dr. Teleman immediately gives a ‘very simple’ example of such function himself, namely w = [(z– 1) + ((zk2)]1/2. Huh? If that’s regarded as ‘very simple’, you may wonder what follows. Well, just look him up I’d say: I only read the first lecture and so there are fourteen more. 🙂

But he’s actually right: this function is not very difficult. In essence, we’ve got our square root function here again (because of the 1/2 exponent), but with four branch points this time, namely ± 1 and ± k (i.e. the positive and negative square roots of 1 and krespectively, cf. the (z– 1)  and (z– k2) terms in the argument of this function), instead of only one (the origin).

Despite the ‘simplicity’ of this function, Dr. Teleman notes that “we cannot identify this shape by projection (or in any other way) with the z-plane or the w-plane”, which confirms the above: Riemann surfaces are usually not simple and, hence, these ‘visualizations’ don’t help all that much. However, while not ‘identifying the shape’ of this particular square root function, Dr. Teleman does make the following drawing of the branch points:

Compactification 1

This is also some kind of cross-section of the Riemann surface, just like the one I made above for the ‘super-simple’ w = √z function: the dotted lines represent the imaginary part of w = [(z– 1) + (z– k2)]1/2, and the non-dotted lines are the real part of the (double-valued) w function. So that’s like ‘my’ graph indeed, except that we’ve got four branch points here, so we can make a choice between one of the two branches of at each of them.

[Note that one obvious difficulty in the interpretation of Dr. Teleman’s little graph above is that we should not assume that the complex numbers k and k are actually lying on the same line as 1 and -1 (i.e. the real line). Indeed, k and k are just standard complex numbers and most complex numbers do not lie on the real line. While that makes the interpretation of that simple graph of Dr. Teleman somewhat tricky, it’s probably less misleading than all these fancy 3D graphs. In order to proceed, we can either assume that this z axis is some polygonal line really, representing line segments between these four branch points or, even better, I think we should just accept the fact we’re looking at the z plane here along the z plane itself, so we can only see it as a line and we shouldn’t bother about where these points k and –k are located. In fact, their absolute value may actually be smaller than 1, in which case we’d probably want to change the order of the branch points in Dr. Teleman’s little graph).]

Dr. Teleman doesn’t dwell too long on this graph and, just like Penrose, immediately proceeds to what’s referred to as the compactification of the Riemann space, so that’s this ‘transformation’ of this complex surface into a donut (or a torus as it’s called in mathematics). So how does one actually go about that?

Well… Dr. Teleman doesn’t waste too many words on that. In fact, he’s quite cryptic, although he actually does provide much more of an explanation than Penrose does (Penrose’s treatment of the matter is really hocus-pocus I feel). So let me start with a small introduction of my own once again.

I guess it all starts with the ‘compactification’ of the real line, which is visualized below: we reduce the notion of infinity to a ‘point’ (this ‘point’ is represented by the symbol ∞ without a plus or minus sign) that bridges the two ‘ends’ of the real line (i.e. the positive and negative real half-line). Like that, we can roll up single lines and, by extension, the whole complex plane (just imagine rolling up the infinite number of lines that make up the plane I’d say :-)). So then we’ve got an infinitely long cylinder.

374px-Real_projective_line

But why would we want to roll up a line, or the whole plane for that matter? Well… I don’t know, but I assume there are some good reasons out there: perhaps we actually do have some thing going round and round, and so then it’s probably better to transform our ‘real line’ domain into a ‘real circle’ domain. The illustration below shows how it works for a finite sheet, and I’d also recommend my imaginary reader to have a look at the Riemann Project website (http://science.larouchepac.com/riemann/page/23), where you’ll find some nice animations (but do download Wolfram’s browser plugin if your Internet connection is slow: downloading the video takes time). One of the animations shows how a torus is, indeed, ideally suited as a space for a phenomenon characterized by two “independent types of periodicity”, not unlike the cylinder, which is the ‘natural space’ for “motion marked by a single periodicity”.

plane to torus

However, as I explain in a note below this post, the more natural way to roll or wrap up a sheet or a plane is to wrap it around a sphere, rather than trying to create a donut. Indeed, if we’d roll the infinite plane up in a donut, we’ll still have a line representing infinity (see below) and so it looks quite ugly: if you’re tying ends up, it’s better you tie all of them up, and so that’s what you do when you’d wrap the plane up around a sphere, instead of a torus.

From plane to torus

OK. Enough on planes. Back to our Riemann surface. Because the square root function w has two values for each z, we cannot make a simple sphere: we have to make a torus. That’s because a sphere has one complex dimension only, just like a plane, and, hence, they are topologically equivalent so to say. In contrast, a double-valued function has two ‘dimensions’ so to say and, hence, we have to transform the Riemann surface into something which accommodates that, and so that’s a torus (or a coffee cup :-)). In topological jargon, a torus has genus one, while the complex plane (and the Riemann sphere) has genus zero.

[Please do note that this will be the case regardless of the number of branch points. Indeed, Penrose gives the example of the function w = (1 – z3)1/2, which has three branch points, namely the three cube roots of the 1 – zexpression (these three roots are obviously equal to the three cube roots of unity). However, ‘his’ Riemann surface is also a Riemann surface of a square root function (albeit one with a more complicated form than the ‘core’ w = z1/2 example) and, hence, he also wraps it up as a donut indeed, instead of a sphere or something else.]

I guess that you, my imaginary reader, have stopped reading all of this nonsense. If you haven’t, you’re probably thinking: why don’t we just do it? How does it work? What’s the secret?

Frankly, the illustration in Penrose’s Road to Reality (i.e. Fig. 8.2 on p. 137) is totally useless in terms of understanding how it’s being done really. In contrast, Dr. Teleman is somewhat more explicit and so I’ll follow him here as much as I can while I try to make sense of it (which is not as easy as you might think). 

The short story is the following: Dr. Teleman first makes two ‘cuts’ (or ‘slits’) in the z plane, using the four branch points as the points where these ‘cuts’ start and end. He then uses these cuts to form two cylinders, and then he joins the ends of these cylinders to form that torus. That’s it. The drawings below illustrate the proceedings. 

Cuts

Compactification 3

Huh? OK. You’re right: the short story is not correct. Let’s go for the full story. In order to be fair to Dr. Teleman, I will literally copy all what he writes on what is illustrated above, and add my personal comments and interpretations in square brackets (so when you see those square brackets, that’s [me] :-)). So this is what Dr. Teleman has to say about it:

The function w = [(z– 1) + ((z– k2)]1/2 behaves like the [simple] square root [function] near ± 1 and ± k. The important thing is that there is no continuous single-valued choice of w near these points [shouldn’t he say ‘on’ these points, instead of ‘near’?]: any choice of w, followed continuously round any of the four points, leads to the opposite choice upon return.

[The formulation may sound a bit weird, but it’s the same as what happens on the simple z1/2 surface: when we’re on one of the two branches, the argument of w changes only gradually and, going around the origin, starting from one root of z (let’s say the ‘positive’ root w1), we arrive, after one full loop around the origin on the z plane (i.e. we add 2π to arg z = θ), at the opposite value, i.e. the ‘negative’ root w= -w1.] 

Defining a continuous branch for the function necessitates some cuts. The simplest way is to remove the open line segments joining 1 with k and -1 with –k. On the complement of these segments [read: everywhere else on the z plane], we can make a continuous choice of w, which gives an analytic function (for z ≠ ±1, ±k). The other branch of the graph is obtained by a global change of sign. [Yes. That’s obvious: the two roots are each other’s opposite (w= –w1) and so, yes, the two branches are, quite simply, just each other’s opposite.]

Thus, ignoring the cut intervals for a moment, the graph of w breaks up into two pieces, each of which can be identified, via projection, with the z-plane minus two intervals (see Fig. 1.4 above). [These ‘projections’ are part and parcel of this transformation business it seems. I’ve encountered more of that stuff and so, yes, I am following you, Dr. Teleman!]

Now, over the said intervals [i.e. between the branch points], the function also takes two values, except at the endpoints where those coincide. [That’s true: even if the real parts of the two roots are the same (like on the negative real axis for our  z1/2 s), the imaginary parts are different and, hence, the roots are different for points between the various branch points, and vice versa of course. This is actually one of the reasons why I don’t like Penrose’s illustration on this matter: his illustration suggests that this is not the case.]

To understand how to assemble the two branches of the graph, recall that the value of w jumps to its negative as we cross the cuts. [At first, I did not get this, but so it’s the consequence of Dr. Teleman ‘breaking up the graph into tow pieces’. So he separates the two branches indeed, and he does so at the ‘slits’ he made, so that’s between the branch points. It follows that the value of w will indeed jump to its opposite value as we cross them, because we’re jumping on the other branch there.]

Thus, if we start on the upper sheet and travel that route, we find ourselves exiting on the lower sheet. [That’s the little arrows on these cuts.] Thus, (a) the far edges of the cuts on the top sheet must be identified with the near edges of the cuts on the lower sheet; (b) the near edges of the cuts on the top sheet must be identified with the far edges on the lower sheet; (c) matching endpoints are identified; (d) there are no other identifications. [Point (d) seems to be somewhat silly but I get it: here he’s just saying that we can’t do whatever we want: if we glue or stitch or weld all of these patches of space together (or should I say copies of patches of space?), we need to make sure that the points on the edges of these patches are the same indeed.]

A moment’s thought will convince us that we cannot do all this in in R3, with the sheets positioned as depicted, without introducing spurious crossings. [That’s why Brown and Churchill say it’s ‘physically impossible.’] To rescue something, we flip the bottom sheet about the real axis.  

[Wow! So that’s the trick! That’s the secret – or at least one of them! Flipping the sheet means rotating it by 180 degrees, or multiplying all points twice with i, so that’s i2 = -1 and so then you get the opposite values. Now that’s a smart move!] 

The matching edges of the cuts are now aligned, and we can perform the identifications by stretching each of the surfaces around the cut to pull out a tube. We obtain a picture representing two planes (ignore the boundaries) joined by two tubes (see Fig. 1.5.a above).

[Hey! That’s like the donut-to-coffee-cup animation, isn’t it? Pulling out a tube? Does that preserve angles and all that? Remember it should!]

For another look at this surface, recall that the function z → R2/z identifies the exterior of the circle ¦z¦ < R with the punctured disc z: ¦z¦ < R and z ≠ 0 (it’s a punctured disc so its center is not part of the disc). Using that, we can pull the exteriors of the discs, missing from the picture above, into the picture as punctured discs, and obtain a torus with two missing points as the definitive form of our Riemann surface (see Fig. 1.5.b).

[Dr. Teleman is doing another hocus-pocus thing here. So we have those tubes with an infinite plane hanging on them, and so it’s obvious we just can’t glue these two infinite planes together because it wouldn’t look like a donut 🙂. So we first need to transform them into something more manageable, and so that’s the punctured discs he’s describing. I must admit I don’t quite follow him here, but I can sort of sense – a little bit at least – what’s going on.] 

[…]

Phew! Yeah, I know. My imaginary reader will surely feel that I don’t have a clue of what’s going on, and that I am actually not quite ready for all of this high-brow stuff – or not yet at least. He or she is right: my understanding of it all is rather superficial at the moment and, frankly, I wish either Penrose or Teleman would explain this compactification thing somewhat better. I also would like them to explain why we actually need to do this compactification thing, why it’s relevant for the real world.

Well… I guess I can only try to move forward as good as I can. I’ll keep you/myself posted.

Note: As mentioned above, there is more than one way to roll or wrap up the complex plane, and the most natural way of doing this is to do it around a sphere, i.e. the so-called Riemann sphere, which is illustrated below. This particular ‘compactification’ exercise is equivalent to a so-called stereographic projection: it establishes a one-on-one relationship between all points on the sphere and all points of the so-called extended complex plane, which is the complex plane plus the ‘point’ at infinity (see my explanation on the ‘compactification’ of the real line above).

Riemann_sphereStereographic_projection_in_3D

But so Riemann surfaces are associated with complex-analytic functions, right? So what’s the function? Well… The function with which the Riemann sphere is associated is w = 1/z. [1/z is equal to z = z*/¦z¦, with z* = x – iy, i.e. the complex conjugate of z = x + iy, and ¦z¦ the modulus or absolute value of z, and so you’ll recognize the formulas for the stereographic projection here indeed.]

OK. So what? Well… Nothing much. This mapping from the complex z plane to the complex w plane is conformal indeed, i.e. it preserves angles (but not areas) and whatever else that comes with complex analyticity. However, it’s not as straightforward as Penrose suggests. The image below (taken from Brown and Churchill) shows what happens to lines parallel to the x and y axis in the z plane respectively: they become circles in the w plane. So this particular function actually does map circles to circles (which is what holomorphic functions have to do) but only if we think of straight lines as being particular cases of circles, namely circles “of infinite radius”, as Penrose puts it.

inverse of z function

Frankly, it is quite amazing what Penrose expects in terms of mental ‘agility’ of the reader. Brown and Churchill are much more formal in their approach (lots of symbols and equations I mean, and lots of mathematical proofs) but, to be honest, I find their stuff easier to read, even if their textbook is a full-blown graduate level course in complex analysis.

I’ll conclude this post here with two more graphs: they give an idea of how the Cartesian and polar coordinate spaces can be mapped to the Riemann sphere. In both cases, the grid on the plane appears distorted on the sphere: the grid lines are still perpendicular, but the areas of the grid squares shrink as they approach the ‘north pole’.

CartesianStereoProj

PolarStereoProj

The mathematical equations for the stereographic projection, and the illustration above, suggest that the w = 1/z function is basically just another way to transform one coordinate system into another. But then I must admit there is a lot of finer print that I don’t understand – as yet that is. It’s sad that Penrose doesn’t help out very much here.

Riemann surfaces (II)

This is my second post on Riemann surfaces, so they must be important. [At least I hope so, because it takes quite some time to understand them. :-)]

From my first post on this topic, you may or may not remember that a Riemann surface is supposed to solve the problem of multivalued complex functions such as, for instance, the complex logarithmic function (log z = ln r + i(θ + 2nπ) or the complex exponential function (zc = ec log z). [Note that the problem of multivaluedness for the (complex) exponential function is a direct consequence of its definition in terms of the (complex) logarithmic function.]

In that same post, I also wrote that it all looked somewhat fishy to me: we first use the function causing the problem of multivaluedness to construct a Riemann surface, and then we use that very same surface as a domain for the function itself to solve the problem (i.e. to reduce the function to a single-valued (analytic) one). Penrose does not have any issues with that though. In Chapter 8 (yes, that’s where I am right now: I am moving very slowly on his Road to Reality, as it’s been three months of reading now, and there are 34 chapters!), he writes that  “Complex (analytic) functions have a mind of their own, and decide themselves what their domain should be, irrespective of the region of the complex plane which we ourselves may initially have allotted to it. While we may regard the function’s domain to be represented by the Riemann surface associated with the function, the domain is not given ahead of time: it is the explicit form of the function itself that tells us which Riemann surface the domain actually is.” 

Let me retrieve the graph of the Riemannian domain for the log z function once more:

Riemann_surface_log

For each point z in the complex plane (and we can represent z both with rectangular as well as polar coordinates: z = x + iy = reiθ), we have an infinite number of log z values: one for each value of n in the log z = ln r + i(θ + 2nπ) expression (n = 0, ±1, ±2, ±3,…, ±∞). So what we do when we promote this Riemann surface as a domain for the log z function is equivalent to saying that point z is actually not one single point z with modulus r and argument θ + 2nπ, but an infinite collection of points: these points all have the same modulus ¦z¦ = r but we distinguish the various ‘representations’ of z by treating θ, θ ± 2π, θ ±+ 4π, θ ± 6π, etcetera, as separate argument values as we go up or down on that spiral ramp. So that is what is represented by that infinite number of sheets, which are separated from each other by a vertical distance of 2π. These sheets are all connected at or through the origin (at which the log z function is undefined: therefore, the origin is not part of the domain), which is the branch point for this function. Let me copy some formal language on the construction of that surface here:

“We treat the z plane, with the origin deleted, as a thin sheet Rwhich is cut along the positive half of the real axis. On that sheet, let θ range from 0 to 2π. Let a second sheet  Rbe cut in the same way and placed in front of the sheet R0. The lower edge of the slit in Ris then joined to the upper edge of the slit in R1. On  R1, the angle θ ranges from 2π to 4π; so, when z is represented by a point on R1, the imaginary component of log z ranges from 2π to 4π.” And then we repeat the whole thing, of course: “A sheet Ris then cut in the same way and placed in front of R1. The lower edge of the slit in R1. is joined to the upper edge of the slit in this new sheet, and similarly for sheets R3R4… A sheet R-1R-2, R-3,… are constructed in like manner.” (Brown and Churchill, Complex Variables and Applications, 7th edition, p. 335-336)

The key phrase above for me is this “when z is represented by a point on R1“, because that’s what it is really: we have an infinite number of representations of z here, namely one representation of z for each branch of the log z function. So, as n = 0, ±1, , ±2, ±3 etcetera, we have an infinite number of them indeed. You’ll also remember that each branch covers a range from some random angle α to α + 2π. Imagine a continuous curve around the origin on this Riemann surface: as we move around, the angle of z changes from 0 to 2θ on sheet R0, and then from 2π to 4π on sheet Rand so on and so on.

The illustration above also illustrates the meaning of a branch point. Imagine yourself walking on that surface and approaching the origin, from any direction really. At the origin itself, you can choose what to do: either you take the elevator up or down to some other level or, else, the elevator doesn’t work and so then you have to walk up or down that ramp to get to another level. If you choose to walk along the ramp, the angle θ changes gradually or, to put it in mathematical terms, in a continuous way. However, if you took the elevator and got out at some other level, you’ll find that you’ve literally ‘jumped’ one or more levels. Indeed, remember that log z = ln r + i(θ + 2nπ) and so ln r, the horizontal distance from the origin didn’t change, but you did add some multiple of 2π to the vertical distance, i.e. the imaginary part of the log z value. 

Let us now construct a Riemann surface for some other multiple-valued functions. Let’s keep it simple and start with the square root of z, so c = 1/2, which is nothing else than a specific example of the complex exponential function zc = zc = ec log z: we just take a real number for c here. In fact, we’re taking a very simple rational number value for c: 1/2 = 0.5. Taking the square, cube, fourth or  nth root of a complex number is indeed nothing but a special case of the complex exponential function. The illustration below (taken from Wikipedia) shows us the Riemann surface for the square root function.

Riemann_sqrt

As you can see, the spiraling surface turns back into itself after two turns. So what’s going on here? Well… Our multivalued function here does not have an infinite number of values for each z: it has only two, namely √r ei(θ/2) and √r ei(θ/2 + π). But what’s that? We just said that the log function – of which this function is a special case – had an infinite number of values? Well… To be somewhat more precise:  z1/2 actually does have an infinite number of values for each z (just like any other complex exponential function), but it has only two values that are different from each other. All the others coincide with one of the two principal ones. Indeed, we can write the following:

w = √z = z1/2 =  e(1/2) log z e(1/2)[ln r + i(θ + 2nπ)] = r1/2 ei(θ/2 + nπ) = √r ei(θ/2 + nπ) 

(n = 0, ±1,  ±2,  ±3,…)

For n = 0, this expression reduces to z1/2 = √r eiθ/2. For n = ±1, we have  z1/2 = √r ei(θ/2 + π), which is different than the value we had for n = 0. In fact, it’s easy to see that this second root is the exact opposite of the first root: √r ei(θ/2 + π) = √r eiθ/2eiπ = – √r eiθ/2). However, for n = 2, we have  z1/2 = √r ei(θ/2 + 2π), and so that’s the same value (z1/2 = √r eiθ/2) as for n = 0. Indeed, taking the value of n = 2 amounts to adding 2π to the argument of w and so get the same point as the one we found for n = 0. [As for the plus or minus sign, note that, for n = -1, we have  z1/2 = √r ei(θ/2 -π) = √r ei(θ/2 -π+2π) =  √r ei(θ/2 +π) and, hence, the plus or minus sign for n does not make any difference indeed.]

In short, as mentioned above, we have only two different values for w = √z = z1/2 and so we have to construct two sheets only, instead of an infinite number of them, like we had to do for the log z function. To be more precise, because the sheet for n = ±2 will be the same sheet as for n = 0, we need to construct one sheet for n = 0 and one sheet for n = ±1, and so that’s what shown above: the surface has two sheets (one for each branch of the function) and so if we make two turns around the origin (one on each sheet), we’re back at the same point, which means that, while we have a one-to-two relationship between each point z on the complex plane and the two values z1/2 for this point, we’ve got a one-on-one relationship between every value of z1/2 and each point on this surface.

For ease of reference in future discussions, I will introduce a personal nonsensical convention here: I will refer to (i) the n = 0 case as the ‘positive’ root, or as w1, i.e. the ‘first’ root, and to (ii) the n = ± 1 case as the ‘negative’ root, or w2, i.e. the ‘second’ root. The convention is nonsensical because there is no such thing as positive or negative complex numbers: only their real and imaginary parts (i.e. real numbers) have a sign. Also, these roots also do not have any particular order: there are just two of them, but neither of the two is like the ‘principal’ one or so. However, you can see where it comes from: the two roots are each other’s exact opposite w= u2 + iv= —w= -u1 – iv1. [Note that, of course, we have w1w = w12 = w2w = w2= z, but that the product of the two distinct roots is equal to —z. Indeed, w1w2 = w2w1 = √rei(θ/2)√rei(θ/2 + π) = rei(θ+π) = reiθeiπ = -reiθ = -z.]

What’s the upshot? Well… As I mentioned above already, what’s happening here is that we treat z = rei(θ+2π) as a different ‘point’ than z = reiθ. Why? Well… Because of that square root function. Indeed, we have θ going from 0 to 2π on the first ‘sheet’, and then from 2π0 to 4π on the second ‘sheet’. Then this second sheet turns back into the first sheet and so then we’re back at normal and, hence, while θ going from 0π  to 2π is not the same as θ going from 2π  to 4π, θ going from 4π  to 6π  is the same as θ going from 0 to 2π (in the sense that it does not affect the value of w = z1/2). That’s quite logical indeed because, if we denote w as w = √r eiΘ (with Θ = θ/2 + nπ, and n = 0 or ± 1), then it’s clear that arg w = Θ will range from 0 to 2π if (and only if) arg z = θ ranges from 0 to 4π. So as the argument of w makes one loop around the origin – which is what ‘normal’ complex numbers do – the argument of z makes two loops. However, once we’re back at Θ = 2π, then we’ve got the same complex number w again and so then it’s business as usual.

So that will help you to understand why this Riemann surface is said to have two complex dimensions, as opposed to the plane, which has only one complex dimension.

OK. That should be clear enough. Perhaps one question remains: how do you construct a nice graph like the one above?

Well, look carefully at the shape of it. The vertical distance reflects the real part of √z for n = 0, i.e. √r cos(θ/2). Indeed, the horizontal plane is the the complex z plane and so the horizontal axes are x and y respectively (i.e. the x and y coordinates of z = x + iy). So this vertical distance equals 1 when x = 1 and y = 0 and that’s the highest point on the upper half of the top sheet on this plot (i.e. the ‘high-water mark’ on the right-hand (back-)side of the cuboid (or rectangular prism) in which this graph is being plotted). So the argument of z is zero there (θ = 0). The value on the vertical axis then falls from one to zero as we turn counterclockwise on the surface of this first sheet, and that’s consistent with a value for θ being equal to π there (θ = π), because then we have cos(π/2) = 0. Then we go underneath the z plane and make another half turn, so we add another π radians to the value θ and we arrive at the lowest point on the lower half of the bottom sheet on this plot, right under the point where we started, where θ = 2π and, hence, Re(√z) = √r cos(θ/2) (for n = 0) = cos(2π/2) = cos(2π/2) = -1.

We can then move up again, counterclockwise on the bottom sheet, to arrive once again at the spot where the bottom sheet passes through the top sheet: the value of θ there should be equal to θ = 3π, as we have now made three half turns around the origin from our original point of departure (i.e. we added three times π to our original angle of departure, which was θ = 0) and, hence, we have Re(√z) = √r cos(3θ/2) = 0 again. Finally, another half turn brings us back to our point of departure, i.e. the positive half of the real axis, where θ has now reached the value of θ = 4π, i.e. zero plus two times 2π. At that point, the argument of w (i.e. Θ) will have reached the value of 2π, i.e. 4π/2, and so we’re talking the same w = z1/2 as when we started indeed, where we had Θ = θ/2 = 0.

What about the imaginary part? Well… Nothing special really (as for now at least): a graph of the imaginary part of √z would be equally easy to establish: Im(√z) = √r sin(θ/2) and, hence, rotating this plot 180 degrees around the vertical axis will do the trick.

Hmm… OK. What’s next? Well… The graphs below show the Riemann surfaces for the third and fourth root of z respectively, i.e. z1/3 and z1/4 respectively. It’s easy to see that we now have three and four sheets respectively (instead of two only), and that we have to take three and four full turns respectively to get back at our starting point, where we should find the same values for z1/3 and z1/4 as where we started. That sounds logical, because we always have three cube roots of any (complex) numbers, and four fourth roots, so we’d expect to need the same number of sheets to differentiate between these three or four values respectively.

Riemann_surface_cube_rootRiemann_surface_4th_root

In fact, the table below may help to interpret what’s going on for the cube root function. We have three cube roots of z: w1, wand w3. These three values are symmetrical though, as indicated be the red, green and yellow colors in the table below: for example, the value of w for θ ranging from 4π to 6π for the n = 0 case (i.e. w1) is the same as the value of w for θ ranging from 0 to 2π for the n = 1 case (or the n = -2 case, which is equivalent to the n = 1 case).

Cube roots

So the origin (i.e. the point zero) for all of the above surfaces is referred to as the branch point, and the number of turns one has to make to get back at the same point determines the so-called order of the branch point. So, for w = z1/2, we have a branch point of order 2; for for w = z1/3, we have a branch point of order 3; etcetera. In fact, for the log z function, the branch point does not have a finite order: it is said to have infinite order.

After a very brief discussion of all of this, Penrose then proceeds and transforms a ‘square root Riemann surface’ into a torus (i.e. a donut shape). The correspondence between a ‘square root Riemann surface’ and a torus does not depend on the number of branch points: it depends on the number of sheets, i.e. the order of the branch point. Indeed, Penrose’s example of a square root function is w = (1 – z3)1/2, and so that’s a square root function with three branch points (the three roots of unity), but so these branch points are all of order two and, hence, there are two sheets only and, therefore, the torus is the appropriate shape for this kind of ‘transformation’. I will come back to that in the next post.

OK… But I still don’t quite get why this Riemann surfaces are so important. I must assume it has something to do with the mystery of rolled-up dimensions and all that (so that’s string theory), but I guess I’ll be able to shed some more light on that question only once I’ve gotten through that whole chapter on them (and the chapters following that one).  I’ll keep you posted. 🙂

Post scriptum: On page 138 (Fig. 8.3), Penrose shows us how to construct the spiral ramp for the log z function. He insists on doing this by taking overlapping patches of space, such as the L(z) and Log z branch of the log z function, with θ going from 0 to 2π for the L(z) branch) and from -π to +π for the Log z branch (so we have an overlap here from 0 to +π). Indeed, one cannot glue or staple patches together if the patch surfaces don’t overlap to some extent… unless you use sellotape of course. 🙂 However, continuity requires some overlap and, hence, just joining the edges of patches of space with sellotape, instead of gluing overlapping areas together, is not allowed. 🙂

So, constructing a model of that spiral ramp is not an extraordinary intellectual challenge. However, constructing a model of the Riemann surfaces described above (i.e. z1/2, z1/3, z1/4 or, more in general, constructing a Riemann surface for any rational power of z, i.e. any function w = zn/m, is not all that easy: Brown and Churchill, for example, state that is actually ‘physically impossible’ to model that (see Brown and Churchill, Complex Variables and Applications (7th ed.), p. 337).

Huh? But so we just did that for z1/2, z1/3 and z1/4, didn’t we? Well… Look at that plot for w = z1/2 once again. The problem is that the two sheets cut through each other. They have to do that, of course, because, unlike the sheets of the log z function, they have to join back together again, instead of just spiraling endlessly up or down. So we just let these sheets cross each other. However, at that spot (i.e. the line where the sheets cross each other), we would actually need two representations of z. Indeed, as the top sheet cuts through the bottom sheet (so as we’re moving down on that surface), the value of θ will be equal to π, and so that corresponds to a value for w equal to w = z1/2 = √r eiπ/2 (I am looking at the n = 0 case here). However, when the bottom sheet cuts through the top sheet (so if we’re moving up instead of down on that surface), θ’s value will be equal to 3π (because we’ve made three half-turns now, instead of just one) and, hence, that corresponds to a value for w equal to w = z1/2 = √r e3iπ/2, which is obviously different from √r eiπ/2. I could do the same calculation for the n = ±1 case: just add ±π to the argument of w.

Huh? You’ll probably wonder what I am trying to say here. Well, what I am saying here is that plot of the surface gives us the impression that we do not have two separate roots w1 and won the (negative) real axis. But so that’s not the case: we do have two roots there, but we can’t distinguish them with that plot of the surface because we’re only looking at the real part of w.

So what?

Well… I’d say that shouldn’t worry us all that much. When building a model, we just need to be aware that it’s a model only and, hence, we need to be aware of the limitations of what we’re doing. I actually build a paper model of that surface by taking two paper disks: one for the top sheet, and one for the bottom sheet. Then I cut those two disks along the radius and folded and glued both of them like a Chinese hat (yes, like the one the girl below is wearing). And then I took those two little paper Chinese hats, put one of them upside down, and ‘connected’ them (or should I say, ‘stitched’ or ‘welded’ perhaps? :-)) with the other one along the radius where I had cut into these disks. [I could go through the trouble of taking a digital picture of it but it’s better you try it yourself.]

ChineseHat

Wow! I did not expect to be used as an illustration in a blog on math and physics! 🙂

🙂 OK. Let’s get somewhat more serious again. The point to note is that, while these models (both the plot as well as the two paper Chinese hats :-)) look nice enough, Brown and Churchill are right when they note that ‘the points where two of the edges are joined are distinct from the points where the two other edges are joined’. However, I don’t agree with their conclusion in the next phrase, which states that it is ‘thus physically impossible to build a model of that Riemann surface.’ Again, the plot above and my little paper Chinese hats are OK as a model – as long as we’re aware of how we should interpret that line where the sheets cross each other: that line represents two different sets of points.

Let me go one step further here (in an attempt to fully exhaust the topic) and insert a table here with the values of both the real and imaginary parts of √z for both roots (i.e. the n = 0 and n = ± 1 case). The table shows what is to be expected: the values for the n = ± 1 case are the same as for n = 0 but with the opposite sign. That reflects the fact that the two roots are each other’s opposite indeed, so when you’re plotting the two square roots of a complex number z = reiθ, you’ll see they are on opposite sides on a circle with radius √r. Indeed, rei(θ/2 + π) = rei(θ/2)eiπ = –rei(θ/2). [If the illustration below is too small to read the print, then just click on it and it should expand.]

values of square root of z

The grey and green colors in the table have the same role as the red, green and yellow colors I used to illustrated how the cube roots of z come back periodically. We have the same thing here indeed: the values we get for the n = 0 case are exactly the same as for the n = ± 1 case but with a difference in ‘phase’ I’d say of one turn around the origin, i.e. a ‘phase’ difference of 2π. In other words, the value of √z in the n = 0 case for θ going from 0 to 2π is equal to the value of √z in the n = ± 1 case but for θ going from 2π to 4π and, vice versa, the value of √z in the n = ±1 case for θ going from 0 to 2π is equal to the value of √z in the n = 0 case for θ going from 2π to 4π. Now what’s the meaning of that? 

It’s quite simple really. The two different values of n mark the different branches of the w function, but branches of functions always overlap of course. Indeed, look at the value of the argument of w, i.e. Θ: for the n = 0 case, we have 0 < Θ < 2π, while for the n = ± 1 case, we have -π < Θ < +π. So we’ve got two different branches here indeed, but they overlap for all values Θ between 0 and π and, for these values, where Θ1 = Θ2, we will obviously get the same value for w, even if we’re looking at two different branches (Θ1 is the argument of w1, and Θ2 is the argument of w2). 

OK. I guess that’s all very self-evident and so I should really stop here. However, let me conclude by noting the following: to understand the ‘fully story’ behind the graph, we should actually plot both the surface of the imaginary part of √z as well as the surface of the real part of of √z, and superimpose both. We’d obviously get something that would much more complicated than the ‘two Chinese hats’ picture. I haven’t learned how to master math software (such as Maple for instance), as yet, and so I’ll just copy a plot which I found on the web: it’s a plot of both the real and imaginary part of the function w = z2. That’s obviously not the same as the w = z1/2 function, because w = z2 is a single-valued function and so we don’t have all these complications. However, the graph is illustrative because it shows how two surfaces – one representing the real part and the other the imaginary part of a function value – cut through each other thereby creating four half-lines (or rays) which join at the origin. 

complex parabola w = z^2

So we could have something similar for the w = z1/2 function if we’d have one surface representing the imaginary part of z1/2 and another representing the  real part of z1/2. The sketch below illustrates the point. It is a cross-section of the Riemann surface along the x-axis (so the imaginary part of z is zero there, as the values of θ are limited to 0, π, 2π, 3π, back to 4π = 0), but with both the real as well as the imaginary part of  z1/2 on it. It is obvious that, for the w = z1/2 function, two of the four half-lines marking where the two surfaces are crossing each other coincide with the positive and negative real axis respectively: indeed, Re( z1/2) = 0 for θ = π and 3π (so that’s the negative real axis), and Im(z1/2) = 0 for θ = 0, 2π and 4π (so that’s the positive real axis).

branch point

The other two half-lines are orthogonal to the real axis. They follow a curved line, starting from the origin, whose orthogonal projection on the z plane coincides with the y axis. The shape of these two curved lines (i.e. the place where the two sheets intersect above and under the axis) is given by the values for the real and imaginary parts of the √z function, i.e. the vertical distance from the y axis is equal to ± (√2√r)/2.

Hmm… I guess that, by now, you’re thinking that this is getting way too complicated. In addition, you’ll say that the representation of the Riemann surface by just one number (i.e. either the real or the imaginary part) makes sense, because we want one point to represent one value of w only, don’t we? So we want one point to represent one point only, and that’s not what we’re getting when plotting both the imaginary as well as the real part of w in a combined graph. Well… Yes and no. Insisting that we shouldn’t forget about the imaginary part of the surface makes sense in light of the next post, in which I’ll say a think or two about ‘compactifying’ surfaces (or spaces) like the one above. But so that’s for the next post only and, yes, you’re right: I should stop here.