Mobius Transformations in Hyperbolic Geometry: The Upper Half-Plane and the Poincare Disc Model

So much for the trend of lighthearted titles. Maybe this one is just an ironic deconstruction on what it means to be a lighthearted title.

As long as we are considering geometric interpretations of each type of Mobius Transformation, I feel like using hyperbolic geometry gives a more illuminating correspondence of the geometric implications of T. So, let me introduce some new terms, but before I do, I'll broadly explain what hyperbolic geometry is. Hyperbolic geometry is Non-Euclidean geometry, where Euclidean is the normal, everyday space you're used to working with. An example of a Non-Euclidean geometry is the Earth--if you wanted to get from New York to LA , the fastest way to get there might look something like:

But obviously if we're in R³, the fastest way would just be a straight line, which would mean you'd have to go inside the Earth's crust. What's more is that hyperbolic geometry has a lot richer properties and depth of study than Euclidean geometry. Remember from the first blog post about Symmetry, that there are limited ways you can tile a Euclidean plane. But in hyperbolic geometry, there are infinite ways.

Now, let's define some terms. H is the upper half-plane. This is the set of all complex numbers with a positive imaginary part. dH is the boundary of the upper half-plane, also known as the "circle at infinity." dH, then, is just any number that has imaginary part value 0 or is infinity. If you were to stereographically project dH back down to C², you'd get the real number line, plus infinity. So you can see that topologicallydH is a circle.

Now it's useful to see that if we're in H, our Mobius Transformations can be classified by fixed points still, but more elegantly. If T has 2 fixed points, and both are in dH. If T has one fixed point in dH, and none in H, then T is hyperbolic (note: there are no non-hyperbolic loxodromic Mobius Transformations when in H). If T has no fixed points in dH and one in H, then T is elliptic. 

Define D to be the Poincare Disc Model. This will be useful to imagine our elliptic Mobius Transformations. Similar to H, D, is the disc of all complex numbers z such that |z|<1. The boundary of D,dD, or the circle at infinity, is the set of complex numbers with |z|=1. Using h(z) = (z-i)/(iz-1), we can bijectively map H to D and dH to dD, allowing us to transfer the logic we've established thus far and apply it to D and dD. I won't show how they are bijective or how we calculate distances, as that will be unnecessary to a causal reader. But what we can do, knowing h exists, is conjugate a Mobius Transformation T in H by h to get a Mobius Transformation in D, of the form hTh^-1.

This means all Mobius Transformations in D, for some complex numbers e and f, and their conjugates e-bar and f-bar, are of the form: T(z) = (ez+f)/(fbar z + ebar). The way I figured that out is by using an arbitrary Mobius Transformation in H, (az+b)/(cz+d) and then composing and simplifying hTh^-1. Protip--if you want to check that for yourself, remember that composing two Mobius Transformations is the same as multiplying the matrices of their coefficients.

Within D, the classification of a Mobius Transformation T willlook similar to the way it did in H, but a bit different, so be careful not to just mentally replace "H" in your head with "D". If T has 2 fixed points, and both are in dD, then T is hyperbolic. If T has one fixed point in dD, and none in D, then T is parabolic. If T has no fixed points in dD and one in D, then T is elliptic. 

If we take an e value to be e^iθ/2 and an f value to be 0, our Mobius Transformation in D will be T(z)= e^iθ, and acts by rotating the Poincare disc around 0 by an angle of θ. We know that all rotations have one fixed point--the point about which they're rotating--so now we can just think of elliptic transformations as rotations! But let's not stop there. If we know that rotations are elliptic transformations in D, then what do they look like in H? This time, instead of conjugating hTh^-1, we want to go from D to H, so we're instead conjugating h^-1Th. The resulting Mobius Transformation now looks like T(z) = (cos(θ/2)z + sin(θ/2))/(-sin(θ/2)z + cos(θ/2)). And because we remember that all rotations have a fixed point where they rotate around, we need to find the fixed point. Rearranging, we find that -sin(θ/2)z² + cos(θ/2)z = cos(θ/2)z + sin(θ/2), and cancelling the cos(θ/2)z and then dividing by -sin(θ/2), we get that z² =-1, which means the fixed point is at i! i is in H, not dH, which means T is elliptic in H.

There is a lot more cool stuff to do with Mobius Transformations in hyperbolic geometry, and I really regret that I hadn't worked with them in hyperbolic geometry sooner and that I have to debate at the Tournament of Champions this week, so I can't devote much more time to going deeper into concepts like Fuchsian groups. Oh well, I'll see what I can do. At least I finally classified these Mobius Transformations, and at least I finally feel like I've done some real math.
 

The Classification!

Sorry about the delay in posting--this post is going to be long, meticulous, and introducing new topics, so it took me a long time to write out.

In my last post, I talked about how we know every Mobius Transformation T has at least 1 and potentially 2 fixed points. If T has only 1 fixed point, then we know it's conjugate (meaning equivalent) to a translation on the sphere. If T has 2, then it can either be loxodromic or elliptic, based on what kinds of fixed point they are, and how the points spiral. These transformations are conjugate to a transformation T'(z)=kz. A special case of the loxodromic transformation is hyperbolic, when k is real.

This all becomes a bit easier to conceptualize and think about in hyperbolic geometry, which I'll address later in this post. But first, I'll go through the types of transformations in terms we've talked about so far. Pretty much the logic I'm doing here is given a certain Mobius Transformation T, we conjugate it by another transformation S and then look at its action and see T'(z)=STS^-1(z). S isn't any arbitrary transformation though; it is determined by the fixed points of T. The underlying motivation behind creating S is wanting T' to send ∞ to ∞. This means we need to map the fixed point of T to infinity by S. If there is no other fixed point, we can think of the translation around infinity. If there is another fixed point, again, for ease, we map the other fixed point of T to 0 by S. This makes T' a linear mapping because the only places a line T(z)=az is fixed is at 0 and ∞.

For reference later on. If instead there is only one fixed point, P, then S(z) = Az/(z-P).

How do we know if there's one or two fixed points? Well remember Mobius Transformations are of the form T(z)=(az+b)/(cz+d), and its fixed points are of the form z= (az+b)/(cz+d). If we rearrange and expand, we get cz² + (d-a)z -b =0. Of course you all remember your quadratic formula of z = (-b +/- Sqrt(b² -4ac))/ 2a, right? Well when we plug in what we just rearranged for our fixed points, you'll see it's (a-d +/- Sqrt((d-a)² +4cb))/ 2c. What determines how many fixed points there are is the discriminant, (d-a)² +4cb. If the discriminant is 0, then we have only one fixed point, (a-d)/2c. But if it's nonzero, then we'll get 2 fixed points, because of the plus or minus sign.

Let's go a step further. If we expand (d-a)² +4cb, you get d² +a²-2ad+4cd. We want to factor this, so we do the following: d² +a²-2ad -2ad +2ad +4cd and factoring, now, we get (a+d)² -4(ad-bc). Two things to remember for this to be geometrically significant is 1) the trace of a matrix is defined as a+d; and 2) for all Mobius Transformations, ad-bc=1. (If you have any questions on either of those, comment.) Thus, we know the amount of fixed points a Mobius Transformation has is determined by its trace squared minus four: (TrT)² -4.

Off the bat, if the discriminant is 0, or there's only one fixed point, P, you know that your function is parabolic. We create our S as some Az/(z-P), as then S(P) will be mapped to ∞. Then we can conjugate T by S^-1 and now we have our T'(z)= z+b. This may seem a bit fast, but it'll get clearer when we do more work for other transformations. Parabolic ones are probably the simplest--their fixed point is both. An example is T(z) = (9z-2)/(8z+1). Feel free to verify that. Notice how this transformation when normalized has a trace of 2. This is crucial, because it is a necessary condition for a parabolic transformation to have a trace of +/-2. This is because the only way the discriminant will equal 0

So that's what happens if we have one fixed point, but what about for 2? As you can see in the reference picture above, the difference between a loxodromic and an elliptic Mobius Transformation is what the modulus of k is equal to: if |k|=1, then T is elliptic, if not, T is loxodromic. Or, we can more easily see the difference if we look again to the discriminant. If the discriminant is positive, then you'll get a loxodromic transformation, and if it's negative, you'll get an elliptic one. We know what the discriminant will be based on the trace: (TrT)² -4, so we can easily classify loxodromic transformations as those with -∞< TrT<-2 and 2<TrT<∞ and elliptic transformations as -2<TrT<2.

To clarify this and to check our processes, let's look to some examples. First, let T(z) = (2z+5)/(-3z-1). So, -5/2 maps to 0, -1/3 maps to ∞, -6/5 maps to 1, and -1/2 + Sqrt(69)i = P and -1/2 - Sqrt(69)i = Q are our fixed points. Thus, we construct an S(z) = A(z-Q)/(z-P). Now P maps to ∞, Q maps to 0, and 0 maps to AQ/P. So now we have a conjugation of T by S, T'(z)= STS^-1(z), such that T'(z)=kz. Whatever |k| is equal to will tell us if T is elliptic or loxodromic. That can be done by finding |T'(1)|=|k(1)|=|STS^-1(1)|, because |T'(1)|=|k|, but that is really arduous and tedious work, so I'm just going to quickly demonstrate to you this T I've selected is elliptic by showing one |T'(z)|=|z|. Obviously this isn't mathematically rigorous, but given that I'm not going to choose any particular number to ensure it comes out that way, this is useful to a more casual reader for a heuristic understanding, but I still do promise (and please verify it yourself if you justly don't take me at my word) that |T'(1)|=1). I've already said in this paragraph that S(0) is AQ/P, and T maps -5/2 to 0. We can easily figure out S^-1maps A((-5/2)-Q)/((-5/2)-P) to -5/2. So, T'(A((-5/2)-Q)/((-5/2)-P))= AQ/P. Because A is a constant, we can remove it when considering the modulus. Having calculated the modulus of both, I have found that |((-5/2)-Q)/((-5/2)-P)| = |Q/P| = 1. Thus T is elliptic.

Now we should find out if T being elliptic coheres to the set of conditions we've also derived, so let's now check if the discriminant of T is negative, which would indicate T being elliptic. And indeed it is, as TrT = 1/13, and -2<(1/13)-4 <2!

You can verify that T(z)=(4z+5)/(3z+2) is indeed loxodromic, using either of the ways above.

To sum thus far up, if TrT =2 or -2, T is parabolic, if -∞<TrT<-2 or 2<TrT<∞, T is loxodromic, and if -2<TrT<2, then T is elliptic. That is completely exhaustive of all the possibilities of T.

Next, let's turn to the geometric interpretations of the different kinds of Mobius Transformations.
A loxodromic Mobius transformation has two fixed points--one attracting and one repelling. Naturally, points here spiral out from the source (repelling point) and into the sink (attracting point).

These pictures are, again, from Indra's. Notice how everything is flowing out of the fixed point on the left, and flowing into the fixed point on the right.

And, a special kind of loxodromic Mobius Transformation, as I said at the top of this post, is a hyperbolic Mobius Transformation, when |k| and TrT are real. For instance, having conjugated by S just as we did before, T'(z) could end up equaling something like (i+9)z, or T'(z) could end up equaling something like 9z. In the latter case, T would be hyperbolic. Hyperbolic Mobius Transformations don't have points spiral into or out of either fixed point, but instead move in circles through them:

Finally, elliptic transformations don't have any sources or sinks, meaning neither of their fixed points are attracting or repelling, that both of them are neutral. Here, the points simply move in circles around the fixed points:

Actually, I'm going to break the post here for the sake of consistency of post-theme, but this will immediately continue in the next post.

How Attracting

Sorry about the hiatus on posting, I was in San Diego for a bit of debate prep for the upcoming Tournament of Champions. But fear not, the project lived on, even if it wasn't chronicled on blogger. Today I'm going to post about fixed points and their importance in the classification of Mobius Transformations.

A fixed point is pretty simple. For a transformation T(z), a fixed point is any point z such that T(z)=z. See? It's fixed, as if we imagine the transformation moves or changes the domain, that point z stays put. Fixed points are important, as they allow us to understand the transformations around a dynamic focused on the point.

Let's first consider a linear function T(z)=az. If we want to find the fixed points, we just have to find anywhere z=az. The only two places where this happens are 0=0z and ∞ = ∞z. Geometrically, this looks like a line from 0 to ∞, pretty intuitive, ie, it's a linear function, but on the Riemannian Sphere, it looks like a spiral:

From Indra's. Also known as a loxodrome from the Greek "running obliquely." Here, 0 acts as a "source" and ∞ as a "sink." I'll clarify some of the terminology more later.

This gets even more interesting. 0 and ∞ seem like the most special numbers, but on the Riemannian Sphere, because 1/0 and 1/∞ are well defined, they are just like any other numbers. This means we can conjugate by other transformations to have the same spiral on the sphere, just with different sinks and sources! Let's say we wanted to make our sinks and sources -1 and 1 (which are arbitrary, we could make them whatever), then we just need another Mobius Transformation R such that R(z) maps 0 to -1 and ∞ to 1. Fortunately, we know how to do this from the last post, and we create R(z) = (z-1)/(z+1). Check: R(0) = -1/1 = -1 and R(∞) = ∞/∞ = 1. Now because we have to conjugate T to find RTR^-1(z), so we need to find R^-1. That's not difficult if we imagine Mobius Transformations as matrices, and then we know that R^-1 is (z+1)/(-z+1). To check this, we should verify that -1 is sent to 0 and 1 is sent to ∞. That is the case, so we are good.

T^*= RTR^-1(z) has the properties we wanted. It has 2 fixed points at -1 and 1. T^*=
RTR^-1(-1)= RT(0) = R(0) = -1, by construction. Interestingly, though T = az was linear, after conjugating by two non-linear Mobius Transformations, T^* is distinctly nonlinear. If you were to calculate RTR^-1 out, you would get ( z(1+a)+a-1 )/( z(a-1)+a+1).

If T looked like:

Then T* would look like:

Recall that Mobius Transformations are of the form T(z)= (az+b)/(cz+d). Thus, a fixed point under any arbitrary Mobius Transformation is any z such that z= (az+b)/(cz+d). If we do some rearrangement, we see that cz² + z(d-a) - b = 0. This is a quadratic, and we can solve for z to find the fixed points: ( (a-d) ± sqrt( (a-d)² + 4bc) ) / 2c. This means that every Mobius Transformation has either one or two fixed points, depending on the discriminant. I'll talk more about this in the next post, which will come soon.

Off the bat, we can see some interesting things. If c=0 (and d isn't, else that'd just describe the above case), for some Mobius Transformation T(z)= (az+b)/d = (a/d)z + (b/d), then T is affine (affine means a linear function translated from 0. Essentially, linear functions look like T(z) = az, and affine functions look like T(z) = az+b). If T is affine, then it only has one fixed point at infinity-- T(∞)= a(∞)+b =∞, and not at 0 (T(0) = b). This makes sense because then our quadratic formula from above would be ( (a-d) ± sqrt( (a-d)²) ) / 0, which equals ∞. 
 
Now I'd like to take a step back and talk about what it means to be a source or a sink. A source is a fixed point that is repelling, and a sink is a fixed point that is attracting. Alternatively, a fixed point can be neither, in which case it is a neutral fixed point. 

The definition of an attracting fixed point x₀ is for some L in the open interval (0,1), there exists an open interval U containing x₀, such that |f(x)-x₀| ≤ L|x-x₀|, for all x in U. Conversely, a repelling point is a fixed point x₀ such that there's a L>1 and an open interval U containing x₀ such that |f(x)-x₀| ≥ L|x-x0|, for all x in U.

Pretty much, what that means is that if you have a sink or an attracting point, then if you take some space, then undergoing iterations of the function, it will be bijectively mapped to a smaller and smaller space arbitrarily close to the point. Of course by "smaller" I don't mean "less points", but instead a smaller looking area. The reason this still works, though, is because there are uncountably many real numbers (there are as many numbers in the interval (0,1) as there are between (-∞,∞) ). The opposite is the case for repelling points.

This will come in handy to think about when next post I'll classify all types of Mobius Transformations into three categories based on how many and which kinds of fixed points they have.

Ted Mobius Transformations

Hey all, it's been a little over a week since my last post, but unlike this functor, I haven't forgotten about you. This week I'm going to introduce the titular Mobius Transformation. This bad boy, also known as a Linear Fractional Transformation, twist, turn, and stretch the sphere based on where 0, 1, and ∞ are mapped. These transformations also allow us to construct the Klienian group fractals.

As you can see the Ted Mobius Transformation adds age and a tux.

I want this post to be manageable and concise, but I also want to discuss a bunch of topics, like fixed and periodic points, and whether they're attracting or repelling, as those let us construct the Julia and Fatou set which are really awesome for building fractals. I think I'll just discuss what a Mobius Transformation is here, and then later I'll post about those points and sets.

A Mobius Transformation is a transformation from the Riemannian Sphere to the Riemmanian Sphere T(z) = (az+ b)/(cz+ d), where a,b,c,d are complex numbers such that ad-bc !=0. That last condition may remind a mathematical reader of a condition of an invertible 2x2 matrix.

Let's look at some inputs for z to see how our sphere changes. Some of the following text may be hard to parse over blogger (I'll try to format it in a way that's not too bad), but if you spend a second, you can easily convince yourself it's true. Maybe one day I'll get Latex and rewrite some of these lines.

T(∞)= (a∞ +b)/(c∞ +d) = a/c. That makes sense, as if a and c are infinitely large, then adding or subtracting a finite number won't change either of them. This also shows why we need to be on the Riemannian Sphere, because on the complex plane (without infinity) then there's no number we can plug into T(z) to get a/c.

T(0) = (a0 +b)/(c0 +d)= b/d. 

T(1) = (a1 +b)/(c1 +d)= (a+b)/(c+d).

These are some of the intuitive numbers to plug in, to see what we'd get. But now let's try some less initially intuitive but geometrically significant numbers.

T(-d/c) = (-ad/c +b)/(-cd/c +d) = (-ad +bc)/(-cd +dc) = (-ad +bc)/0) = ∞. But a tricky reader might want to test this function is well defined. "How do you know you'll never get 0/0?" Well, tricky reader, the only way we'd get 0/0 would be if -ad +bc =0, but as you'll remember by our definition of a Mobius Transformation, -ad +bc =0 can't be the case.

T(-b/a) = (-ab/a +b)/(-cb/a +d) = (-b +b)/(ad - bc) = 0/(-ad +bc) = 0. Remember, ad-bc can't equal 0, so it remains well defined.

This next one will be a bit more complicated looking, so bear with me. I'm going to color code fractions:
T((d-b)/(a-c)) = (a ((d-b)/(a-c)) +b)/(c ((d-b)/(a-c)) +d) = ((ad-ba)/(a-c) + (ba-bc)/(a-c))/((cd-cd)/(a-c) + (da-da)/(a-c)) = ((ad-bc)/(a-c))/(ad-bc)/(a-c)) = 1.

You can convince yourself now that these transformations work under composition and can be inverted.

Mobius Transformations can be thought of as a series of composition of different transformations. First take T(z) = z and scale it by c and translate it by d. Now you have T(z) = cz +d. Second, invert it so that T(z) = 1/(cz + d). Third, scale it by -1/c (almost like conjugating with the first step), and get T(z) = -1/c(cz + d). Finally, translate that by a/c: a/c - 1/c(cz + d). I'll do the algebraic manipulation below, or you can trust me that after some finagling, it comes out to T(z) = (az +b)/(cz +d). The key to the algebraic manipulation, which took me frustratingly way, way too long to see is realizing that ad-bc =1.

a/c - 1/c(cz + d) = a/c - (bc -ad)/c(cz + d)= (a(cz +d) + bc - ad)/ (c(cz+d)) = (caz + da + bc -ad)/(c(cz+d) = (caz +bc)/(c(cz+d)) = (az+b)/(cz+d).

That's really cool for several reasons. First, it makes finding the inverse very easy, because instead of composing T₄∘T₃∘T₂∘T₁(z), you can just do T^-1₁∘T^-1₂∘T^-1₃∘T^-1₄(z). Also if we take all of our Mobius Transformations to create a group under composition, with easily seen generators.

Pretty much, Mobius Transformations send three points to any other three points, and by doing so, determine the entire transformation of the sphere. This means we can determine exactly how the sphere acts based on where 0, 1, and ∞ are sent. But we can also look at other points. You can see yourself that given points a,b,c, T(z)= (z-a)/(z-b) * (c-b)/(c-a) sends a to 0, b to ∞, and c to 1.

Or going a step further, you can see that given poitns a,b,c,d,e,f, if T(z) = w is a Mobius transformation, and (z-a)/(z-b) * (c-b)/(c-a) = (w-d)/(w-e) * (f-e)/(f-d), then the transformation maps a to d, b to e, and c to f.

Take a look at this picture I drew representing a Mobius Transformation abstractly:

I'll post more later about fixing points, specific Mobius Transformations (like ones that preserve symmetries, ie, which one maps the unit circle to itself, etc), and their relation to fractals. This post should just introduce Mobius Transformations, there's still much more to go!

Abelian Grapes

Hey all! Exciting news, I've been attending a weekly differential geometry seminar at ASU, and I finally understood a lecture (about conjugacy limits in Cartan subgroups of SL(n, R), looking specifically at n=3, and then applying use of the hyperreal numbers). I don't know much about Cartan subgroups or Lie groups in general, but I could follow along the lecture and parse, understand, and reflect on examples and terms and such. Cool stuff! I guess I am learning math, who knew?

My favorite part about math is the terminology--for example, take the hyperreal numbers, *R, (equivalence classes of sequences of real numbers constructed by taking all real series and denoting their convergent point a hyperreal number--and apparently if it doesn't converge like (0,1,0,1,0...) then you just assign it to either 0 or 1? I don't know all the ins and outs of them). The hyperreal numbers *R are cool because they're non-Archimedian. Numbers like the reals are Archimedian because if there's a number a < 1/n, where n is any natural number, then a as to be 0. But the hyperreals have infinitesimals like ϵ= (1, 1/2, 1/4, ...), which is smaller than everything, but bigger than 0. This also means the hyperreals have infinitely large numbers, like 1/ϵ--which should remind you of how the Riemannian sphere acts with infinity and 0. And then *R has appreciable numbers, which are just real numbers plus infinitesimals α = a+ ϵ, a in R. Pretty cool stuff. Turns out the hyperreals are a much easier way to conjugate by a series of matrices, which is needed to do to find conjugacy limits of things. But the reason I bring this up is because some of the terms are pretty cool. If you have an appreciable number, α = a+ ϵ, and you want to go back to the reals, you just "take the shadow of α," which, though it just means to take the real part, a, of α, makes it sound really cool, like it's from some children's TV show about children's card games. And another cool term, galaxy, is represented Galϵ(x) = {y in *R, | |x- y| <= kϵ}, k in R. During the lecture, when she introduced "galaxies," I turned to the guy to my right and said "I love math," and he nodded in agreement.

Anyway, this post should be fairly short and easy to grasp, especially compared to the other posts, and the posts about to come. This will just introduce the concept of a group.

A group is a set G with a binary operation, * : G x G -> G, such that the following three properties

1. For g,h,i in G, (g * h) * i = g * (h * i), ie, * is associative
2. There exists an identity element, e, in G such that, g in G, g * e = e * g = g
3. For each g in G, there exists an inverse g-1 in G such that g * g-1  = g-1 * g = e

But remember that because * is a binary operation, then for G = {a,b,c} to be a group, a*b has to also be in G as well. This property is called closure and is essential to bear in mind.

Furthermore, a group G is abelian if its binary operation is commutative, that is, gh=hg, g,h in G.        

Common groups include the integers under addition represented as (Z, +) or referred to Z mod n, consisting of the elements {0,1,2,...,n-1}. To form the multiplicative group (Z, •), one must remove the 0 element, as to satisfy the inverse requirement of a group--g in Z, 0•g = 0 =/= 1.

Symmetry involves and necessitates group theory. Given a finite set A, its permutations (bijections from A->A) result in interesting properties, with application to geometry. If A = {1,2,...,n}, the group of possible permutations of A is called Sn, the symmetry group of n.

If n=3, then we may model geometrically the elements of S3 with the vertices of a triangle. 

from Troy University

Here, various permutations of Sn yield different transpositions. For example, mapping 1->2, 2->3, and 3->1 results in a rotation by 120°, but the mapping 1->1, 2->3, 3->2 results in a reflection across the angle bisector of vertex 1.

Also, groups are key for complex analysis. Not only are there many groups involving C, but the big one, which this project is titled for, is called the Linear Fractional Group, ie all the Mobius Transformations together. Matrices are a great way to represent groups, and in the next post about Mobius Transformations, I will probably also introduce the analogous matrices.

EDIT== By the way, the title is a joke based on

Wish You Were Sphere

First of all, I just found this link, and thought it'd really nicely explain some of the purpose questions behind this whole project for anyone out there skeptical of the point of math, especially when it's so abstract.

This is the continuation of the last post, How Complex Could Geometry Be? Here, I'll talk about mappings in the complex plane, the Riemannian sphere, and stereographic projection. I'll take this time now to apologize preemptively to my blogging group, for the quite frankly ridiculous amount of posts you have to read. All I ask is that you remember a lot of it is pictures, and I like to think this is pretty interesting stuff.

Of course, huge shout out to the wonderflorious Mrs. Bailey, who is largely responsible for my knowing what I do. To ease the non-math readers, this post will be more heuristic, but if anyone wants more mathematical rigor, make a comment and I will either answer there or make a new post if need be.

I don't want to take about complex mappings too much, but I want to point out some significant, overarching geometric implications.

Remember back to complex multiplication. When two complex vectors, w= a*e^ix and v= b*e^iy, are multiplied, the resulting product wv= ab*e^i(x+y). This would geometrically look as if it were spiraling out, as the modulus keeps getting larger and the argument does too, but loops back to 0 every time it reaches 2π.

Now, mapping may get kind of, heh, complex, sometimes--especially with lots of numbers to consider, but it becomes easier if we use colors to heuristically describe it (foreshadowing, this color mapping may be relevant later when it comes to determining fractals...). The following pictures of color mappings are from Levente Locsi's seminar Colorful visualization of complex functions.

Imagine that the complex plane is this picture:

We are going to see, visually, how that color-plane changes based on applying different transformations.

If T(z) = z, then obviously the result is:

That's just the identity function. If we look at the conjugate function (where every point becomes it's conjugate), the color-map looks like:

Remember, the conjugate to a+ ib is a- ib, which is just the reflection over the real axis, which is what happens in the picture. Pink swaps with red, and blue swaps with black. Now let's look at the square function, T(z) = z².

Notice now how essentially we've duplicated out map! If we were to cube it, we'd have three iterations of the pattern pink-blue-black-red, if we were to raise z to the fourth, we'd have four, and you get the picture. The reason this is super interesting is that it connects complex analysis further to geometry and also to group theory! As I briefly mentioned last post, if we have the equation x⁴- 1= 0, there are 4 solutions: 1, -1, i, and -i. If we were to plot these solutions on the complex plane, we'd get a square!

And if we were to have the equation, say, x⁵- 1= 0, then there'd be 5 solutions, which means instead of a square, we'd get a pentagon!

We could keep going, or we could generalize this to say for some real number a, xⁿ- a= 0, in the complex plane, it corresponds to an n-gon! How cool is that?!

Now this next picture is totally irrelevant to anything else in this post, and only will be included because it's an example of color mapping on the complex plane AND it's a really cool picture, and I promised you really cool pictures.

If we start out with the following as our color representation of the complex plane:

Then applying the transformation T(z) = 1/cos(z) gives us:

Again, I don't really have analysis to go along with this picture, I just think it looks cool. :)

Anyway! Though it's really hard to imagine mappings on the complex plane numerically, fortunately, it becomes much easier and solves the whole "infinity" problem with the Riemmanian sphere!

What we do is essentially wrap the complex plane around the unit sphere, and then glue it together at the north pole, and tada! we have the Riemannian Sphere, where everything is better. This is done by a process called stereographic projection.

Let's go through this again, slower. Imagine the unit sphere, such that the unit circle lies on the complex plane. 

Just focus on the picture of the sphere for right now, the rest will be relevant in a bit.

Now, imagine creating a line that goes through the north pole, (0,0,1) and towards some point (x,y, 0) on the complex plane. If we ignore the north pole, then the line from the north pole to the point (x,y, 0) on the complex plane only intersects one point on the sphere, (a, b, c). Stereographic projection essentially is the process is moving all the points on the complex plane to their corresponding point on the sphere.

Now a colorful picture:

This shows you which parts on the plane map to which parts on the sphere. The light blue ring seems to be
the unit circle--or everything with modulus 1. Notice how all the points with a modulus less than 1 are
mapped to the lower half of the sphere, and everything with modulus greater than one to the top half

But what about infinity? In order for us to account for infinity, then we have to pretend like whenever on our complex plane a line would go through infinity--in any direction--then we just map it to the north pole, (0,0,1). Think about it; it makes sense. If you look at the colorful picture above, you could imagine that the black line would "intersect" infinity when it's tangent to north pole. Also, this let's us think of 0 and ∞ as opposites, after all, they are the opposite poles on the Riemannian Sphere.

Remember how your math teacher would always yell at you for trying to divide by zero? Well, in the Riemannian Sphere, the following properties hold:

Let a be a finite, run-of-the-mill number, a+ ∞ = a- ∞ = a*∞ = ∞/a = ∞. (Obviously. If you're confused about a- ∞ = ∞, remember, as we just defined ∞, it's infinity in all directions on the plane, which includes what you might think of as -∞).

But now a/ ∞ = 0 and a/ 0 = ∞! This makes sense, as if you took, say, 1 and parceled it up infinite times, you'd have nothing left. As for a/ 0 = ∞, think about how 1/.5 = 2, because if, for two numbers a,b, b is less than 1, then a/b will be larger than a. Logically, if you keep making b smaller, a/b will get larger. So, it reasons, when you get to 0, and have an appropriately defined ∞ term, you'll get ∞.

But, alas, you still can't divide 0/0 or add, subtract or divide ∞ from itself. Sorry, but us math geeks haven't figured that one out just yet.

It's worth quickly noting how cool circles are on the Riemmanian sphere. Rather than me just tell you everything about it, try looking at the below picture I drew to think about when which circles and lines map to circles from the complex plane to the Riemannian sphere! If you want a tip, the answer, to discuss the proof, or anything related--head on down to the comments!

So now you know enough complex analysis for me to introduce the eponymous Mobius Transformations! Without stealing thunder from a post which will be able to devote more time and words to the subject, Mobius Transformations are just transformations which act on the Riemannian Sphere to itself in the form T(z) = (az +b)/(cz +d)!

These have many cool properties I look forward to sharing at a later time!

Connor

How Complex Could Geometry Be?

Okay I spent way too long trying to think of an interesting title. I initially wanted to make a pun based on psychological complexes--"Argand had a real zero complex" (see, I've got absolutely nothing... if you've got something, feel free to use the comments...)--but then I said the word "complex" too many times that I completely forgot what it meant, and I had to google "complex plane" to make sure that was a thing.

See, math, it's dangerous (maybe more than I thought). Oh well, this title should be sufficient. If anything, it thematically connects this post to my last post's brief mention of how complex analysis is awesome for geometry, symmetry, and pretty much everything.

Last post I gave a brief introduction to symmetry. I figured I could take the time and make another post to detail the basics of complex analysis. Again, this will be long and will simply give an introduction to the use of complex numbers, so feel free to skip through it if you know it already.. Soon, I'll make another post for group theory. 

Complex numbers are awesome. They are very similar to the real numbers, but by the way we construct them, allow for greater geometric significance. First let's begin with i.

i is the square root of -1, defined by the equation i² = -1. This allows us to define complex numbers, where we can more fully account for algebra. For example, when we are restricted to just the reals, the equation x⁴ - 1 = 0 has just two solutions--1 and -1--but in the complex numbers it has 4 (the same amount as the degree of the polynomial)--1, -1, i, and -i.

Now let's define complex numbers

Take R2

No, not you

Excuse me, the real plane, R²

It has an x-axis and a y-axis, and every point in R² can be represented by a coordinate pair, (a, b).

The complex plane is very similar, instead of an x- and a y-axis, it has a "real" and an "imaginary" axis:

Here, still every point can be represented (a, b), if we understand (a, b) --> a + ib. This is the form of a complex number--a is the real part, and ib is the imaginary part.

Every complex number, as I alluded to in my last post, has an argument and a modulus. The argument is the angle, and the modulus is the distance between (a, b) and the origin, (0, 0). Thus, the modulus of z = a + ib is |z|² = a² + b².

Complex numbers can also be represented by z = r cis(θ)= r(cos(θ)+ isin(θ)) = r e^iθ, where r is a real number and represents the modulus, and 0≤ θ≤ 2π where θ represents the argument.

The usual algebraic operations work with complex numbers, and it is easiest to view i as a variable with the special case that i² = -1.


Let w = a + ib and v = c + id be complex numbers. Addition and subtraction are clear: w+ v = (a+b) + i(b+ d). Multiplication, less so, but still intuitive: wv = (a + ib)(c + id) = ac + iad+ ibc - bd = (ac - bd) + i(ad + bc).

Now that we see multiplication, let me introduce you to our friend the complex conjugate! Each complex number z = a + ib has a complex conjugate = a - ib. Notice how this is the reflection of z across the real axis!

From wikipedia

Why this is so special is that if we ever want to get a real number from a complex one , all we have to do is multiply z by is conjugate (hence the line in the sonnet in my first post "Conjugating in C, you made me realize)! (a + ib)(a - ib) = a² + b².

Now, this gives us motivation behind division. v != 0, w/v = (a+ ib)/(c+ id), and multiplying by the conjugate of v over the conjugate of v, gives us ((ac+ db)+ i(bc - ad)) / c² + d².

In polar coordinates, these operations becomes much smoother. Multiplication of w= r₁ e^iθ₁ and v = r₂ e^iθ₂, wv= (r₁r₂) e^{i(θ₁+ θ₂)}. Notice that the multiplication of the conjugate would result in some z = (r r ) e^{i(θ- θ} = r² e⁰ = r², which is exactly what we'd think it'd be.

Inversion is easy to see. We are looking for a z^-1 such that zz^-1 = 1. If z = r e^iθ then by what we just saw before, we are looking for some z^-1= p e^iΦ such that rp = 1 and θ+ Φ= 0. Thus, p = 1/r and Φ= -θ.

This post got pretty long, so I'll give you a break and will post about mapping in the complex plane and the Riemmanian sphere in a separate post later.

Connor