 Welcome back to our lecture series Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. In lecture three, we started talking about finite geometries, for which we introduced the three-point geometry and the four-point affine geometry as examples of finite geometries. Because after all, a geometry is essentially just some type of axiomatic system that gives some meaning to lines and points, in which case there's no requirement that the number of lines, the number of points has to be infinite or anything like that. Now in this lecture four, we're going to talk about the so-called phano geometry. Now more precisely what we're going to talk about is called the phano plane. But in this lecture series, we're only going to be worried with planar geometry. We really won't worry about things of higher dimensions. So for our purposes, we can get away calling this the phano geometry. But this can be generalized to higher dimensions. And I should mention that the axiomatic approach we're going to take to phano geometry, this is not the approach that's typically taken to describe phano geometry. Again, this is meant for us as a toy that we're playing around the axioms. That's actually what our interest is here. How can we make geometric inferences about these geometries from the axioms? The axiomatic method is then the key that we're going for in these videos, at least for right now. So phano geometry is often discussed very differently in purely geometric settings. That's okay. Also part of it is really just to emphasize what do these axioms do, but there are perhaps more efficient axioms that can describe the phano plane, but they're far less intuitive for a beginning geometry student. So we're going to accept these five axioms as we see on the screen, which we call the five phano axioms. I should mention that phano geometry, of course, is named after Genofano, an Italian mathematician who is essentially the founder of finite geometry, as the name seems to suggest, sort of invented what we now call the phano geometry. And so the axioms of phano geometry are five-folder than the following. The first axiom says that there exists at least one line. I should mention that with the geometries you've studied so far, there's often a statement of existence. Like with three-point geometry, there was a statement that says three points exist. With four-point geometry, there was a statement that says four points exist. Those existence statements are very, very important. If we don't have some statement of existence, how do we know we don't have just some empty geometry? Phano geometry is going to take things a little bit different. This time we're going to take the existence of lines instead of the existence of points. We actually could have done existence of points, but just to see a different flavor, our existence statement will be that lines exist. At least one line exists. Then axiom two says that there exists exactly three points on every line. The three-point and four-point geometry studied in lecture three were examples where every line had exactly two points on it. So phano geometry is a little bit graduated in that regard that it has three points on each line. But we're still going to have a finite geometry. Every line has three points. We're also going to see that there's a finite number of lines. In particular, there's going to be seven lines when we're done with this thing. Axiom three says that not all points lie on the same line. We've seen this axiom before. Three-point geometry had it as an example. There's no one line that is incident to all the points. Axiom four says that for each two distinct points, there exists a unique line containing both of them. Now, this is an axiom we've seen a couple times already. Both three-point and four-point geometry had it as an axiom, and they were both axiom two in their respective situations. This seems like a pretty important property. There's something inherently important about this axiom. We'll come back to it, of course, in the future. We'll just accept it for right now. And then axiom five says that all lines intersect. In particular, another way of saying that is there are no parallel lines in phano geometry. The three-point geometry also had this as an axiom that all lines intersect. So in some regard, phano geometry is one step above three-point geometry. In terms of its complexity, the axioms are very, very similar, but there are some very important differences like, hey, all lines will have three points instead of two points. And that's the main difference between three-point geometry and this phano geometry. It's basically our lines are now bigger, and everything else is basically the same to a point. But that's a good comparison to put here. Now, this is a diagram that shows that phano geometry is consistent. This model satisfies all five axioms. When you look at it at the surface, it kind of looks like the insignia of the deathly hollows from the Harry Potter series, for which, of course, you have your cloak of invisibility, you have your resurrection stone, and you have the elder wand here. The only difference, of course, is we have now three elder wands. And that makes sense, because in Harry Potter, the elder wand starts off with double door, then it goes to Voldemort, then it ends up with Harry Potter at the end. So yeah, you think of it that way. It's just three elder wands with our deathly hollows insignia. As usual, our little circles will indicate points on the graph. So there are seven points in this model. And then likewise, there are seven lines here. Here's one line, two, three. So the outer cloak of invisibility makes three lines. The three elder wands make lines as well. So there's three lines here. So that's six total. And then this circle, the resurrection stone is considered one single line. All three points are here. We just drew the line as a circle instead of a usual straight line, just for the sake of convenience. And I should also point out that these points where the circle intersects the elder wands, those are not intersections. There's no points there. Those lines, we should think of, no intersections there. The intersection might be somewhere else. Of course, it's somewhere else, because no lines intersect on this geometry. Just so you're aware. I also want you to know by the end of this video, we'll essentially have proven that this is the only model of phanogeometry, that this axiomatic system is complete. All models are isomorphic. Now, we're going to do this in basically one big theorem, which is going to be a construction theorem, a constructive theorem. Basically, we're going to build everything from scratch. I'm not quite there yet. I want to prove a helper theorem first. So this is actually be like a lima for phanogeometry. We'll just call it a theorem here. So in phanogeometry, two distinct lines intersect in exactly one point. I should mention that, okay, let's take two different lines. Their intersection is exactly one. By axiom five, which is still on the screen, all lines intersect. So there's at least one point of intersection. We've seen that before. But I also want to point out that by axiom four, if our lines intersected at two different points, this would be a violation of axiom four. Because if the intersection was two, then that means given this pair, the two points here, there will be two lines. At least two lines that contain both of them, that violates axiom four. So this idea that lines intersect at exactly two points is by combining these two axioms together. Axiom four with axiom five. When we did three point geometry, three point geometry had equivalence to axiom four and axiom five. We called them something different, but it had those exact same two axioms. This was a theorem of three points geometry as well, because it had those same axioms. At four point geometry, we didn't quite have that. We don't have axiom four, but we had axiom five. So we've got was that if two lines intersect, it would be exactly one point. So there was a conditional there because we had no guarantee that lines intersect. So I want you to see the similarities between these theorems and their proofs because they come from the axioms. Three point geometry and final geometry have these same two axioms, so that they have the exact same proof. Therefore, they have this exact same theorem. Four point geometry has a similar theorem because you could produce a similar proof from axiom five. So this idea, excuse me, exact same four was the same thing about four point geometry. So these results are similar because their axiomatic structures are similar. I want you to think of these axioms as sort of like this stack of Jenga blocks, this tower of blocks for which we're studying what happens if we pull out one block. Does the whole thing fall apart? Right? Not so much, but we're also considering putting one block back inside of them. These are interchangeable parts. As we start changing the axioms, how does that change the geometry? That's one of our principal interests in studying these things right now. So okay, in phano geometry, all lines intersect at exactly one point. Now with that lima now out of the way, we're ready to prove our major result. In the statement, we're going to prove that phano geometry always has seven points and seven lines, but I want you to be aware that by the end of this proof, we'll prove actually a lot more than that. We'll prove in that the way that the lines in the seven points, the seven lines interact with each other, the incidence relation has to be in agreement with the deadly hollows insignia we had drawn on the screen previously, and that's going to get this completeness that every model is isomorphic to the one we started off with. All right, but how are we going to do this? So axiom one, and again, you're going to start to see why I phrase the axioms the way I did when we go through this proof here. By axiom one, there exists at least one line. So we can think of it right here. We have one line and let's call it L1, like so. By axiom two, there exists exactly three points on every line. So in particular, L1 has three points. So we'll call these, we have these three points, and then we'll give them labels A, B, and C, like so. Axiom three told us that not all points lie on the same line. So there has to be a point that's not on L1, call that point D. So now we have four points, one line, just keeping an Italian where we are so far. So we've used axiom one so far, axiom one to give us a line, axiom two gave us three points, and then axiom three gave us a fourth point off of the line. So we're working through our list pretty well. Then we get to axiom four. By axiom four, there exists a unique line incidence to D and A. So there's some line that's between D and A. But then if we use axiom four again, there's a line that's incident to D and B, and there's a line incidence to D and C. Let's call these lines L2, L3, and L4 respectively. What we can say is that lines two, three, and four are distinct from L1. So we can say for a fact that L1 does not equal L2, it does not equal L3, it does not equal L4. Because L1 does not contain the point D by construction D is not on L1, but D is on lines two, three, and four. So far, that's what we know. All right, what else can we say? Well, L2 as a line contains three points, right? That was axiom two. By axiom two, it contains a third point. What is the third point on line two? It would be nice to say that that third point is some new point, but what if it's not? We have A and D on the line. What if B were on the line as well? So what if this happens to be the same? This third point is actually a point we've already encountered. Well, then that would give us that L2. L2 goes through A and B, but then L1 goes through A and B, but as L1 and L2 are not the same line, this would then give us two lines that both contain A and B, which is in violation of axiom four. So this tells us that the third point on L2 cannot be B. Well, could it be C? Well, you get the same problem. C, excuse me, line two would contain A and C, but line one would also contain A and C, which would be a violation of axiom four again. So it turns out that the third point on L2 cannot be B or C. Clearly, it's a third point distinct from A and D. So it has to be a third point, which continuing on in the alphabet here, I guess we call this point E, right? Like so. So then we can ask the same questions as before. Let's look at now L2. It has a third point as well. Who is that third point? Well, if it was A, we would then have two lines that contain A and B. That's a problem. If it was C, then we'd have two lines that contain B and C, L1 and L3. That would be a problem as well. So it can't be C. Could it be E now? That's a new point, right? Well, L3 contains both D and E, but L2 also contains D and E. Are those the same line? Are they different? We haven't yet considered that. Well, L2 contains A, and so notice L2 contains A, E, and D. L3 contains B, E, and D. It's the bed line there for which those are not the same three points. So those would have to be two different lines. So the fact is now we can say that L3 is not L2 because it has three points, and it doesn't have the same three points as L2 is. So they have to be different lines. So we can actually now say that. Let me write this down. We've now proven that L2 at this moment doesn't equal L3 because, like I said, L2 contains this new point E that was not on, well, maybe it's on the next line, so we're trying to figure that out. But we'll start considering if E was here, we get these two lines between D, E, but they don't agree on A and B. So they're different lines at that moment. So that would be a contradiction to Axiom 4 as well. Same thing going on here. What if this other point, well, we've done all the other points, so it has E, can't be this new point, so it has to be some new point we'll call it F. So but we also know that L2 and L3 are different. Now let's come over to L4. It has a third point. What is that point? Well, it can't be A for the same issues we saw before. There'll be two lines containing AC. It can't be B because then we have two lines containing BC. It can't be E because then we have two lines that contain D and E, right? But why do L4 and L2 disagree with each other, right? If E was the third point, then we know that third point's not A. A is on L2, but it's not on L4, so that can't be the case. There'll be two different lines. And so then that other point, I guess another simpler way of saying this, we want to conclude that A is not on L4. It's on L2, so it can't be L2. So we know that L2 is not equal to L4 anymore, right? But could L3 and L4 be the same line somehow, right? Maybe by accident. Could this point be F? Well, the same issue here is that if this points F, which is not B, that means B is not on L4, so they're different. And then you have two lines that contain D and F that would contradict Axiom 4. So basically, I'm going really quickly through these proofs by contradiction. It can't be F. So it's going to have to be some new point here. Let's call it now G, for which then this proves, like we said before, L2 is not 3 and 4 because they have different points, but also 3 and 4 are not the same line. So we've now at this point have constructed seven points. Notice them, ABC, DEFG. We have seven points. We also have four lines, L1, 2, 3, and 4. That's what we've constructed so far. We're going to continue on, of course. We still need to find three more lines. We'll get there in a moment. But what I want us to do, at least for the moment B, is analyze what it is we've constructed so far. So if I keep track of just the lines we have here, L1, L1 contains A, B, and C. L2 contained A, D, and E. L3 contains B, D, and F. And then L4 contains C, D, and G. So we have to come up with these three remaining lines, but this is what we've constructed so far. I'm going to zoom out a little bit so we can try to see the next paragraph in our diagram. It's going to be a little bit hard to do, so I have to zoom it out. I know it's going to be very difficult to read the next one, but really if you're just listening to what I'm saying, you're going to get the proof right here. This is one of our lengthier proofs in this lecture series. So what have we done so far? So we've now accommodated for G and all these seven points. So we know these are all different points now. So what I want us to consider next is this paragraph of the proof. Sorry, this gets a little bit lengthy. This is a very long one here. And so what we have to do next is consider the line, because by Axiom 4 there's a line that connects together A and F, right? There has to be some line that connects them. But what's the third point on this line that's given by Axiom 2? I have to point out that we haven't yet used Axiom 5. This is the point where we started using Axiom 5, that all lines intersect to each other. For which this line, let's call it a L5 here, L5 has to intersect L4. Clearly it intersects L2, L1 at A, and it intersects L3 at F. But where does it intersect L4? That's the kicker here. Well, it's got to be one of these three points here. Does it intersect at D? Well, if L5 contained D, since it contains A, then we'd have two lines that contain AD, which this is a different line from L2, since L2 doesn't have F. So it doesn't work. Does it contain C? Well, no. Same thing, another problem with Axiom 2 there. We'd get a point that, sorry, we have two points, which are contained on two different lines. That's not a possibility. So then we, from there, have to infer that the third point on L5 must in fact be this point G. Let's play this same argument here with the points B and E. There's got to be a line that contains B and E together. Let's call that line L6 now. But L6, same type of game as before, L6 cannot be parallel to L4. Because L6 contains B, it intersects L1 and L3, and since it contains E, it'll intersect L2. So we can't let L6 be parallel to L4. We also have the problems. We can't have L6 be parallel to L5 right here, but we'll actually be able to get that automatically when we consider the next part here. Same thing as before. How does L6 intersect L4? Is it by C? Well, then we get two lines that contain both B and C. Is it D? Well, then we get two lines that contain D and E, which doesn't work. So then the third point has got to be G, that's the only point available to us. So we fill in L6 in this following ray right here. So let's note what we have. L5 is containing A, F, and G, and then we also had that L6, bring this down. It contains, lower the points, B, E, and G. So now we have one line left to go with, and you can probably see how we're going to play this game here. So we're going to take the line that goes between C and F, like so. C and F. Well, C and F intersects L4 and L1. That's great. It'll also intersect L3, but what about L2? How does it intersect there? Well, it can't be D, because we have two lines that contain D and F. It can't be A, because then we have two lines that contain A and F, but also A and C problems there. So it's got to be E. It's the only other way that they could not be parallel. And so this is going to be our line L7 for which let's add that onto the screen here. L7 is going to have to contain the points C. Oh boy, what are the other points? I can't see them on the screen anymore. C, F, and E. And so we've now constructed our lines, our seven lines, and the seven points. So you'll now notice that this is the picture we've constructed. This looks like the picture we had before with our seven points and seven lines. Now when we look at the last paragraph, why can't there be, and I'm going to zoom in on this one now, why can't there be another point or another line? Those are things that have to legitimately be considered in this situation. So I'm going to kind of redraw our diagram really quick. So we have our points, lines, something like so. So we have A, B, C, D, oh boy, D, E, F, and G. There's our picture there. Why couldn't there be an eighth point? What if there's some other point H that's external to the seven that we've already considered? So the fact that we've constructed seven lines and seven points does not mean that's the end of the story. We have to provide an argument on why there can't be another one. Well, if there's some new point H, then there has to be a line between A and H. This comes from axiom two, but this line can't be parallel to any other line, right? It has to intersect every line. Now I'll color these ones out here like this does intersect this line. It does intersect this line. It does intersect this line because those lines all contain A. So what I actually want to focus on or what are the lines that it's so far parallel to? So this line doesn't contain A. So that's a possibility. It would be parallel to this, parallel to this, and parallel to these lines over here. It kind of depends on, I mean, I say they're parallel at the moment, but that's because there's a third point we haven't yet accounted for. So notice that if this point were E, that would take care of this line and this line, but then the line BFD and CGD, that is the two that I didn't call over, would be parallel. So that doesn't work. It can't be E. What if this point was F, right? Which I should mention that if this was some other point I, it would be parallel to too many things. So it can't be a new point. So it has to be a point we already used. What about F? Well, if this was F, you take care of this one and this one, but you still have two lines parallel to this yellow line, which that can't be the case. What if it was like G? Well, if it was G, you would take care of this line and this line, but then we'd still have two lines parallel to that one. That's not a possibility. What if it was B? Well, if it was B, you get this line, this one, there's still two parallel lines. Yikes. If it was C, you get this one and this one, but there'd still be two parallel lines. And if you did D, you get these two, but there'd still be two parallel lines. So in every situation, we've exhausted all the possibilities. If there was a new point H, then the line between H and A would be parallel to at least two other lines in our geometry, which would violate Axiom 5. So that's not a possibility. So there's not any new points. This gives us that there's exactly seven points, but could there be an eighth line? Why not? Could there be an eighth line over here? Is there, let me clean this thing up a little bit. Could there be some other line, right? Some L8? Well, it has to contain three points and these can't be new points because we have the same problem as before. So it would have to be some of the points we already have here, but as there's already a unique line that contains any pair of points, by Axiom 4, any new lines would have to either have new points, or which we can't do, or it would contradict Axiom 4. So we have exactly seven lines with the seven points. And so that brings us to the end of this mega theorem here, that essentially this is the only picture we can draw, at least up to Isomorphism. And so we get the 7.7 lines. And so this is a complete Axiomatic system. This is up to Isomorphism, the only phano geometry.