 Welcome back to our lecture series Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. This lecture actually is broken up into two major sections. This first part coming from Section 6.8 in Rhodes to Geometry by Wallace and West about elliptic geometry. That I should say this portion is loosely based upon Section 6.8. Then the next part of this lecture is going to be coming from what section is it again? 7.3 about duality, which is a notion about projective geometry, but as elliptic geometry is projective, we'll make the connection there. The polar property is the way of measuring this notion of duality, which we'll explain more in the second portion right here. What we want to do is, because we only got a few more lectures here. We're at lecture 35. We only have a few more lectures in this course. As we're talking about elliptic geometry, what I want to do is try to focus on some things that are uniquely elliptic. We spent a lot of the semester developing essentially what became neutral geometry. Very briefly, did we take the road down Euclidean geometry? We spent a little bit more time in hyperbolic, but even still, we see there's a lot in common between Euclidean and hyperbolic geometries because they're both neutral geometries. But there are some important differences because of the parallel postulates. What can we say at the end of this course about elliptic geometry that gives you an appreciation of how elliptic geometry is both similar to Euclidean hyperbolic, but also very distinct from those geometries as well. The polar property is a very nice notion to show that this is a uniquely elliptic geometry type of thing. The polar property theorem states that if L is a line in an elliptic geometry, then there exists exactly one point P such that P is equidistant to all points on L. This point is often called the pole of the line. Given a elliptic line, there's one point that is equidistant from every point on the line. If we have our line L, we see right here L, we have some point P, which we claim is equidistant. If we take all these different perpendiculars dropped from P onto L, first of all, there's not a unique perpendicular drop from P onto L. There's actually multiple. In elliptic geometry, there will be uncountably many perpendicular dropped. That already is a weird thing to say. Then also, all these different lines you see here, all these different perpendicular dropped, are going to be congruent to each other. These are all the same measure. That's the idea here of this pole, this point that is equidistant from everywhere on the line. There's going to be some interesting consequences of that. The proof is pretty, it'll use some calculus notions, which I usually try to avoid, but it's somewhat unavoidable in this situation. Imagine we have two points, A and B, which are going to be indicated on the graph like this, A and B. Let's say that the distance between them is exactly one unit distance, whatever unit that means, but we have a unit distance that's between A and B. Well, there exists perpendicular lines to L that we can erect out of A and B. We get something like this, but as we're in elliptic geometry, these perpendicular lines, they are perpendicular, they have to intersect somewhere at a common point. And this point of intersection, we're going to call that point P. So this point P is the candidate for our pole that we're trying to form right here. So if this, if in fact is a pole, it has to be equidistant to all points on L. So in particular, it has to be equidistant to A and B. But the shape that we've just constructed right here is a double right triangle. There's these two right angles right here. It's kind of bizarre if you think about it because in neutral geometry, you can't have a double right triangle because of the exterior angle theorem. This is actually evidence that the exterior angle theorem does not hold in elliptic geometry. We have this double right triangle here. And because it's a double right triangle, we have two congruent angles, which actually tells us that this triangle is an Asosceles triangle, right? That is that these corresponding sides over here are congruent. This follows from the converse of the Asosceles triangle theorem, which the converse of the Asosceles triangle theorem is a consequence of the angle side angle triangle congruence condition, which is equivalent to side angle side, which is the triangle congruence we take as an axiom here. So we have an Asosceles triangle which shows that AP is congruent to BP, like so. And so that's gonna be our goal as we proceed forward. We wanna show that everything is congruent to AP. So BP is congruent to AP, all right? The next thing I wanna do is we're gonna look for a midpoint, the midpoint that lives between A and B. And so I should be cautious here because in elliptic geometry, we don't have the usual notion of between this. So when I say we have a midpoint, that midpoint is relative to some line at infinity that we're not in consideration right now. So don't worry about that. But from the picture, they're sort of like a clear candidate, but we have this midpoint right here. Well, if we take the line that goes from P, the line that goes from P to M, we'll sketch it right here. This line then cuts our triangle APB into two triangles. You have APM and BPM, like so. And what can we say about this triangle? Well, because we selected the midpoint, the segment AM is congruent to MB. We already know that AP is congruent to BP. And then because right angles are congruent to each other, Euclid's fourth POS would still hold in elliptic geometry. We have by side angle side that these two triangles are congruent to each other. And so as those triangles are congruent to each other, that would then tell us that this side right here, sorry, that's not what I wanna say, sorry. Because these two triangles are congruent to each other, we get that this angle M is congruent to that angle M. But as this angle is congruent to its own supplement, we get that those are right angles. And so then the triangle APM is likewise an Asosceles triangle. And because it's Asosceles, the converse of the Asosceles triangle theorem holds, and we get that the segment MP is likewise congruent to AP. So notice what here, we show that AP and BP are both congruent to each other. And now we've shown that MP is congruent to AP as well, this midpoint, all right? Another direction we wanna go is let's go outward. We went inward with the midpoint. Let's find a point C that also lives on L, right? So that the segment AB is congruent to the segment BC. So we want congruence of this segment right here. So basically B is a midpoint of A and C with respect to some line at infinity, all right? So in this situation, take the line that connects P to C, line determination holds in elliptic geometry. We have this line right here. And so what can we say about the triangles that we've constructed? Well, we do know for the following fact, what do we know? We know that angle B is a right angle because we know it's a right angle right there. And by construction, we know that BC is congruent to AB. So that side's congruent. We have right angles, those are congruent to each other. And then the segment BP is congruent to itself. So we get another side angle, side type argument which shows that this triangle is congruent to this triangle, that is APB is congruent to CPB right there. And so then corresponding parts are congruent. This will then force that angle A is congruent to angle C, which then gives that C is a right angle. And so then if you look at the triangle APC right there, it has two right angles. It's then an assosceles triangle which enforces this side to be congruent as well. So we then get that CP is congruent right there. So there's a bunch of these sides that are gonna be congruent to each other. All right, let me slide out a little bit so you can see a little bit more of the proof. So what we've shown is that if you start taking midpoints and whether the endpoints interior or the midpoints exterior, we can start constructing other segments that are congruent to AP. And so what we can do next in fact is that for any dyadic rational number. And so remember the dyadic rational numbers. These were numbers we could construct using any combination of doubling or having things. So like for example, if we took the number two times two times AB plus one half times one half of AB. Like so this right here would represent since AB is just a unit distance of one, we're looking at 4.25. This is an example of a dyadic rational number. We can construct it by multiples of two combined with multiples of one half. So we can get lots of things like that. So we can get like 3.125, that is three and eighth. These are all examples of dyadic numbers by taking linear combinations of multiples of two and multiples of one half. So we can get lots of different numbers here. And it turns out that by induction on the previous argument, if we use mathematical induction using our original segment AP right here, A and P. Well, you look at the whole number part here and so you're gonna just take copies of this. So you can copy, copy, copy to get to your point over here, which is some point BI. Well, some point we'll call it CI for a moment actually. And then so you do all the multiples of two that you have to go over there. And so basically you're writing integers in binary notation. You can get any integer in that regard. And then you might have to take a few side steps a little bit farther than that because maybe you have to take like a half step and then like a quarter step or whatever. So you take some step right here, some midpoint of some kind. So taking midpoints and double things like that, you can get any dyadic integer and this is gonna be congruent. All these intermediate steps are congruent. So by induction, you get any dyadic, any dyadic rational number that you want. And so what we're gonna do then is then pick a point D so that the distance, the measure of AD is equal to X. So it's any positive real number whatsoever. Well, if you have any real number X there as your distance A and D, the nice thing about the dyadic real numbers is this is where the real analysis part is gonna get a little bit thick. So warning, warning, the set of dyadic real numbers is dense inside of the real numbers. So our dyadic rational numbers here, they are dense inside of the real numbers. And the significance of this is that this means that every real number can be written as a, every real number is the limit of a sequence of dyadic real numbers. So if we have a specific real number in mind X, we can write that as the limit of this sequence of dyadic real numbers, Xi. And so we get that Xi will converge to X where X is a positive real number and each Xi is a dyadic rational number. And so using the process we did before, we take X1. X1 will correspond to some, there's some BI right here so that the distance, sorry, B1, the distance from A to B1 is a dyadic real number X1. And so if you think about the lines that connect you to this potential pole right there, those are congruent right there. X1 is what that was supposed to say, not Xi. And so then the next one, okay, then there's like a B2 so that the distance there is X2. And if we think of the distance from there, since the distance between A and B2 is a dyadic real number, we can say that it's congruent to AP, that is B2P is congruent to AP. And then we take the next dyadic number in this sequence, right? So you get some number over here, B3. Think of the line connecting B3 to P. Since the distance between A and B is a dyadic real number X3, we can then include by induction that B3P is congruent to AP. And this process just continues. We continue going in our sequence, getting closer and closer and closer to D. There's these numbers B4, B5, B6, et cetera, continuing. Our sequence of BIs is going to converge continuously towards the point D. And so as this happens, the sequence ABI, which is equal to Xi, this will converge to the distance AD. And this is some continuous sequence going on right here. And most importantly, throughout each step along the way, the segment, the segment, let's see, fix that, the segment BIP, which in every single step is equal to AP, this will converge towards the segment DP. And so since we have a sequence of measures, so we have a sequence of lines, segments, whose measurements will converge to the measure of DP, as that's a constant sequence, it can converge to only one thing. And that thing it converges to has to be the distance AP. So there's a lot of calculus notions going on there, and I was kind of loose about it. I mean, we could pull out some type of epsilon delta proof right now, but not gonna do it. So because it's constant, this distance is constant for all these dyadic numbers, we can actually converge to all real numbers and get that it's actually the same for everything. So we've now found a point P, which is equidistant to all points on L. Well, what if there's a second one? What if there's a second point P prime that's equidistant to all these points? And in that situation, I have to scroll this up a little bit more. If that's our situation, we have L, and we have these two points, P and P prime. We'll consider our two points A and B from before. We have A and we have B. So we have that the distance from P to A and the P to B is right. We know the angle's right, but we also know the distance is the same. And then we have this other point, I was gonna switch the color there. We have these other points. This is right. And this is right right here. And so those are all the same distance. Well, the distance being measured is that these are gonna be right angles, right? We measure distance by looking at the length of perpendicular lines. And in this construction we did before, we were focusing on the measure, but all of these lines were constructed in such a way that they were right angles. So by uniqueness of segment translation, sorry, the uniqueness of angle translation, the only way we can have two right angles on the same half plane like this is that they are equal to each other. Well, one potential escape to this is, well, who's to say these two points aren't in different half planes, right? What if you have P up here and you have P down here, right? You have your picture A and B. So you can have these equidistant points. If they lived in two half planes, there's no contradiction going on here whatsoever. But of course the issue is an elliptic geometry in the real projective plane. We take the single point elliptic model. There is no two half planes. There's only one half plane. There's only one side of the line. And as such, by uniqueness of angle translation, these two angles have to equal the same thing. These angles are the exact same, that is the rays associated them are the same thing. So the picture looks more like we're coming up, coming up, you would have to do something like this. You have this point P and this point P prime, right? You have A and B right there. So the rays have to coincide because there's only one half plane. But then that forces the line to intersect at two different places if there was, if there was two different points. So I don't know what the heck I'm talking about with uniqueness of segment translation. This is coming from uniqueness of intersection, which is a consequence of line determination, which we have in the single point elliptic model. That is the real projective plane. Now I wanna mention over here that the situation, what if you have two different half planes? That's not a horrible idea. If you have the two point model, that is the case if you take spherical geometry to be your elliptic geometry, that actually is perfectly acceptable. You actually get two poles in that situation because the way you wanna think about it is with the spherical model, if your line is the equator, your pole is the North Pole and the South Pole. You get these two poles, which are equidistant from that equator right there. And the one point elliptic geometry, we could think of this with the hemisphere model, where of course we glue together the sides of the equator. And that situation you have only one pole, which is this point at the top. And that's actually why we call it a pole. It's from this sort of geographic analogy. We're describing the North or South Poles of our spheres, our half sphere right here. And so since there's a unique pole given every elliptic line and the distance between P to the line is just you pick any point. It's this value AP we kept on seeing over and over and over again. We call this the polar distance between the line and P. And every elliptic line has a polar distance because it has a pole. What I want to mention next is that the polar distance is the same for every line that's in elliptic geometry. So if you have any elliptic line, the polar distance is a constant. It doesn't matter which one you have. So imagine you have two elliptic lines. So you have some line L and you have some line L prime, okay? Something like this. And let's take two points A and B, which are on the line L. So you have A right here and B right here. And let A prime and B prime be two points on L prime so that the distance between the distance between A prime and B prime is the same as the distance between A and B. So those are congruent segments right there. Alrighty. Well, let's consider the poles of these things. So L has a pole. You have something like this. So here's your pole P. These will be right angles because we're talking about the pole. You'll have some other pole for P prime. We'll call it P prime the pole of L prime right here. So you get this right here. And so then you can see that these two triangles, it's a double right triangle again. They're congruent by side angle side. It's kind of nice having double right triangles because you can pull out, not side angle side, whoops a daisy. Angle side angle, whoo, no one noticed that one. So by angle side angle, these two triangles are gonna be congruent to each other. And particularly the corresponding sides will be congruent with the corresponding sides are the polar distances of the lines. And so this polar distance will be constant for any elliptic line. And so since polar distances are congruent and they're equal when you have a specific measure applied here, we can speak instead of the polar distance in the elliptic geometry. So not just the polar distance of a line but the distance for the whole geometry. And so if you have a specific measure in mind when you're talking about distance, I mean, the measure function is an artifact of the geometry. It can change the same set of points could have a different measure functions applied to it. So don't worry about that too much. But kind of think of it this way. If you're thinking of like elliptic geometry with this hemisphere model that we were just drawing on the other slide just a moment ago. If you have this hemispherical model, well, you could think of elliptic geometry this way or you can think of elliptic geometry this way, right? You could have a smaller hemisphere, okay? Again, this is the hemisphere model of the projected plane. And so you could see then that the distance, the polar distance is the distance between the North Pole and the equator, right? It's this distance here. But what if you're on a smaller planet, so to speak? The polar distance would be smaller. And so the model will have a smaller one potentially but this is just an artifact of the model itself. Choosing different distance functions is like you're just rescaling your hemisphere or if you think of two point elliptic geometry with a sphere, it's just talking about like, okay, the polar distance on Jupiter is much different than the polar distance on Earth, right? The polar distance has to do with essentially the diameter of the sphere but if you're like on Jupiter, this is not drawn to scale. You know, your polar distance could be much bigger because your sphere has a much larger diameter. So yeah, but you do have a unique polar distance for elliptic geometry but it depends on the measure function. Choosing different measure functions for different elliptic geometries is basically like taking spheres of different sizes which is perfectly acceptable for us. So we've seen that every line has a pole but it's also true that every point has, every point in elliptic geometry is a pole of a line, right? So that there's, this process is reversed. So starting with a point, we can construct two segments. We can construct two segments of polar length that emanate from P. So if we know what the polar length is we could take our point P right here and we just draw two segments, right? That are that polar distance, right? We call it D. This, there's a point down here, there's a point down here, call that point A, call that point B. Well by line determination you connect the dots and there's some line L that harbors A and B then P will be the pole of that line. And so for every line there is a pole and for every point there is a line. I guess what I'm saying is if you give me an equator I'll give you a pole. If you give me a pole, I'll give you an equator. And so these things are related to each other in a principle which we call duality, which is the other part of this lecture here. And so I'm gonna pause this video so you can go potty or something like that. And we'll talk some more about that in the next, in the next video. So take a look for that in the link that hopefully you should see right now.