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Just need to understand basics of Projective Plane: any two lines intersect at exactly one point. R^2 parallel lines intersect at “ideal” point at infinity. There is one line “ideal line”, going through all the ideal points. For any two points, there exists exactly one line going through them. Any two lines intersect at one point. Various theorems hold in P. 

Spherical geometry: 

	S^2 is vectors in R^3 with length 1. 

	Points are on this surface. 

	Lines are S^2 intersections with lines passing through the origin. These are called “great circles”. Essentially, they’re full circumferences. 

	For mostly any two points, x, y, there is one unique line that connects them. We can see this sort of by knowing that three points define a unique plane. Because you take the two points and 	the origin, and there’s only one plane that goes through them all, and that’s the unique line that connects them. 

	But when the two points are anti-polar (that is, the origin, x, y are collinear), there are infinitely many planes that go through them, and therefore infinitely many lines. 

	Any two lines intersect at two (antipolar) points. 

	The poles of a great circle, l, are vectors in S^2 that are perpendicular to the plane that defines l (obviously, there are only two). 

	Polar triangle of triangle x,y,z is defined as x’y’z’ where x’ = pole of great circle yz, on the same side as x. y’, z’ defined similarly. 

	Every triangle has a polar triangle, and the polar of that triangle is the original triangle. 

	T_xS^2 is the tangent space to S^2 at vector x. 

	Say you have spherical segment xy, v is the unit tangent vector at x if |v|=1, v \in T_x S^2, y= alpha*x + beta*v where beta > 0. So essentially you can draw the path to y using only x and v (linear combination). 

	The length of spherical segment xy is arccos(x \dot y). Because we’re dealing with S^2, this means that length is equal to angle. 

	Angle of two segments is defined as angle of their unit tangent vectors. 

	You can parameterize as you’d like. So S(t) t -> cos t * x + sin t * v, where v is unit tangent at x, will always be on S^2. 

	Law of Cosines: you know two sides (a, b) and the angle between them (gamma), you try to find opposite side (c). cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(gamma). 

	***missed some class*** 

	Side lengths of triangle abc are pi - angles of a’b’c’, and vice versa. 

	                                 x \dot y
	Angle(x,y) = arccos(	————) 
				 ||x|| ||y||

	s(t) = xcos(t) + vsin(t) has speed 1 for all values t. 
	
	Alternate Law of Cosines: -cos(gamma) = cos(alpha)cos(beta) - sin(alpha)sin(beta)cos(c). 

	Angles in spherical don’t add up to 180, so with two angles fixed, the third depends on the size of the opposite side.  

	Law of sines: sin(a)/sin(alpha) = sin(b)/sin(beta) = … 


Hyperbolic Geometry: 
		
	Define inner product Phi((a_1, a_2, a_3), (b_1, b_2, b_3)) = a_1b_1 + a_2b_2 - a_3b_3. 
	
	Phi(a+b, c) = Phi(a,c) + Phi(b,c). 

	The sign of vectors Phi’d with themselves make a cone. The surface of this cone are “light-like vectors”, outside this cone is “space-like”, and inside is “time-like”. Time-like vectors Phi’d 	with themselves are <0, which makes sense because they have a large third component. 

	H^2 = {x \in R^3 : Phi(x,x)=-1 and x_3 > 0} So this is a small hyperboloid that sits inside the cone. Its vertex is at +1. 

	A line is the intersection of a plane (passing through the origin) and this hyperboloid. 

	 Define K (Klein-model) to be a disk sitting tangent to the vertex of H^2, which is defined by {(x_1, x_2, x_3) \in R^3: x_3=1 and x_1^2 + x_2^2 <1}. So an open disk (no boundary) with about radius 1. Then there’s a 1-to-1 mapping between H^2 and K. The correspondence is you take the point in H^2 and connect it with the origin and that line’s intersection with K is the corresponding K point. 
	
	For any two points, there’s a unique line connecting. 

	For any two lines, there are 0 or 1 points in their intersection. 

	For any line and a point (not on line), there are infinity many lines going through that point that are parallel to the line. 

	Metric in H^2 (can’t use Klein model): we need angle and distance in H^2. 

	Recall sinh(x) = (e^x - e^-x)/2 
	Recall cosh(x) = (e^x + e^-x)/2
	Recall cosh(x)^2 - sinh(x)^2 = 1 
	They are each other’s derivative. 
	
	Remember, arccosh inverse. f(x)=y, that is. 

	Define distance between x,y to be arccosh(-\Phi(x,y)). This means two points are distance 0 if they’re scalar multiples of each other, but no two points in H^2 are scalar multiples of each other. 

	x^perp is the set of {y \in R^3: \Phi(x,y)=0}. This is a plane tangent to H^2 at x, and all these vectors are space-like (you have to translate the plane to the origin). 

	Angle is arccosh(\Phi(u,v)) where u,v \in T_xH^2. And ||u|| and ||v|| = 1, that is \Phi(u,u)=1 and \Phi(v,v)=1. So this is measuring the angle at point x on H^2, where u and v are pointing different directions. 

	Parameterize line in H^2, is given \Phi(u,u)=1 so u is a unit tangent vector to x. t -> xcosh(t) + usinh(t). 

	Law of cosines: cosh(c) = cosh(a)cosh(b) - sinh(a)sinh(b)cos(gamma). 

	Dual law of cosines: cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cosh(c). 

	Law of sines: sin(alpha)/sinh(a) = sin(beta)/sinh(b) = sin(gamma)/sinh(c). 

	Locally, for small triangles,  H^2 looks like E^2. This makes sense because sh and ch are close to sin and cos at small values. 

	***** missed 15 minutes of class **** 

	(Imagine this in K): for a line, there are infinitely many parallel lines. Define the limiting parallel lines to be the lines closest to intersecting with that line (that is, they are the limits of the parallel lines). 

	Given some line, and looking at all of the lines parallel to it. You can define a function \Pi(d), where d is the distance between the two lines (draw an open triangle), to be the angle between the two lines. \Pi(d) = arcsin(1/ch(d)). We see some interesting properties, like this function is < pi/2 for all d>0. Should look at this more.  

	Sum of angles is not constant. 

	Now, what is the sum of the angles? 

	You can have three types of triangles. You can have 1x ideal, 2x, ideal, and 3x ideal. The 1x ideal is where you have 1 point on the boundary of K, essentially. And similarly for the others. Imagining them on the actual space, you sort of see the lines diverge where they would meet at a vertex. 
	
	The area of 3x ideal triangles is finite, and its area is pi. 

	Let’s define f(alpha) to give the area of these non-triangles, in a sense. We know f(pi)=0, because this doesn’t create a triangle with any area. We know f(alpha+beta) = f(alpha)+f(beta) - pi. 

	f(x)=pi-x, for any angle x. This defines the area of a triangle in H^2. 

	Small triangles are around area pi

	Large triangles are around area 0
	
		
	Proofs (not comprehensive): 

		Polar of polar triangle is the original: use quadrants (xy’ is a quadrant, by definition of y’, and same for z’, so x is pole of y’z’. Then you have to do a bit more). 

		Correspondence between H^2 and K. 

		Proof that curve of xcosh(t)+vsinh(t) stays in H^2, and similarly xcos(t)+vsin(t) stays in S^2, where v are the unit tangent vectors to H^2/S^2 at x. 

		Proof xcosh(t) + usinh(t) belongs to a line in H^2, where u is a unit tangent vector to H^2 at x. 

		

	












REVISED: 

	P^2:   Desargues’ and Pappus’ hold in P^2. 
	

	S^2: 
		Any two points x, y, define a unique line unless x,y are antipodal. Imagine the fact that three points uniquely define a plane in R^3 (x,y, and origin) unless they are collinear. 

		Polar triangle of triangle x,y,z is defined as x’y’z’ where x’ = pole of great circle yz, on the same side as x. y’, z’ defined similarly. 

		Polar(Polar(abc))=abc. (proof using quadrants and same side conclusions). 
		
		T_xS^2 is tangent space to S^2 at vector x. v is unit tangent vector to segment xy if ||v|| = 1, and y=alpha*x + beta*v. 
	
		Length of Spherical Segment xy is arccos(x \dot y). Length = angle. 

		s: t —> cos(t)x+ sin(t)v is a curve such that any |t1-t2| = s(t1)s(t2). 

		Side lengths of triangle abc are pi - angles of a’b’c’. 

		Law of Cosines: you know two sides (a, b) and the angle between them (gamma), you try to find opposite side (c). cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(gamma). 

		Triangle inequality: in any triangle, c < a+b	
		
		                                  x \dot y
		Angle(x,y) = arccos(	————) 
				 	 ||x|| ||y||

		Dual law of Cosines: -cos(gamma) = cos(alpha)cos(beta) - sin(alpha)sin(beta)cos(c). Note that the angle, gamma, depends on the opposite side, not just the other angles!
		
		Law of sines: sin(a)/sin(alpha) = sin(b)/sin(beta) = sin(c)=sin(gamma)

		Theta-lune: two intersecting great circles at angle theta, that area (for both sides) = 4*theta. Could think about 2pi case.

		(alpha + beta + gamma) =  pi + area
	
		Sum of angles go between pi< and <3pi

         H^2: 	
		Inner product: Phi((a_1, a_2, a_3), (b_1, b_2, b_3)) = a_1b_1 + a_2b_2 - a_3b_3.
		
		H^2 is all vectors u such that Phi(u,u)=-1. 

		Klein model: sits at z=1, open disk with radius 1. 

		Recall sinh(x) = (e^x - e^-x)/2 
		Recall cosh(x) = (e^x + e^-x)/2
		Recall cosh(x)^2 - sinh(x)^2 = 1 
		They are each other’s derivative. 

		Length of curve d(x,y) = arccosh(-\Phi(x,y))
		
		For x \in R^2, y is perp to x if \Phi(x,y)=0. This is space-like plane in R^3. 

		phi: t—> xcosh(t) + usinh(t) where Phi(u,u)=1, phi stays in H^2 for all t. Also, d(x,s(t))=t. 

		Angle is arccosh(\Phi(u,v)) where u,v \in T_xH^2. And ||u|| and ||v|| = 1, that is \Phi(u,u)=1 and \Phi(v,v)=1		
		Law of cosines: cosh(c) = cosh(a)cosh(b) - sinh(a)sinh(b)cos(gamma). 

		Dual law of cosines: cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cosh(c). Same thing as in S^2: gamma depends on opposite side, not just the other two angles. 
		
		Triangle inequality: in any triangle, c < a+b

		Law of sines: sin(alpha)/sinh(a) = sin(beta)/sinh(b) = sin(gamma)/sinh(c). 

		You can have three types of triangles. You can have 1x ideal, 2x, ideal, and 3x ideal. 

		Pi(d) = arcsin(1/ch(d)), where d is the vertical distance between a line and its limiting parallel. Pi defines the angle opposite the line. 
	
		The area of 3x ideal triangles is finite, and its area is pi. 		

		(alpha + beta + gamma) = pi - area. Likewise, area = pi - (alpha + beta + gamma). 		
		





Summary of things to remember: 

                          	            x \dot y
	Angle(x,y) = arccos(	————) 
			 	 ||x|| ||y||

	S^2: 
		
		Spherical Segment length between vectors x,y: arccos(x \dot y). 

		Unit vector u \in T_xS^2, can parameterize curve cos(t)x + sin(t)u, to create line in H^2. Moving t some amount moves this curve that same amount. 

		Triangle abc. Sides of a’b’c’ are pi-angles of abc. Angles of a’b’c’ are pi-sides. 

		cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(gamma).
	
		-cos(gamma) = cos(alpha)cos(beta) - sin(alpha)sin(beta)cos(c). 

		sin(a)/sin(alpha) = sin(b)/sin(beta) = sin(c)/sin(gamma). 

		Lune with degree theta, area = 4*theta. 

		(alpha + beta + gamma) = pi + area. 

	H^2: 

		Length of curve between vectors x,y: arccosh(—Phi(x,y))

		cosh(t)^2 - sinh(t)^2 = 1. 

		cosh(t)x + sinh(t)u is standard parameterization. Change t a bit, and get same change for this function. 

		Angle between two unit vectors is arccosh(\Phi(u,v)). 

		cosh(c) = cosh(a)cosh(b) - sinh(a)sinh(b)cos(gamma). 

		cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cosh(c). 
		
		sinh(a)/sin(alpha) = sinh(b)/sin(beta) = sinh(c)/sin(gamma). 

		Pi(d) = arcsin(1/cosh(d))

		(alpha + beta + gamma) = pi - area.