Projective Geometry section 2: Perspecivities and Projectivities

Section 0 text: http://pastebin.com/50w9MYjD
Section 1 text: http://pastebin.com/fMdRzhTf

I mis-numbered some theorems in the Section 1 text. They have been properly numbered in the 'cheat sheets' below. I also added and cleaned up some definitions.

Definitions and notations: https://www.dropbox.com/s/ia4m7s0fy59z9jv
Axioms, Theorems, and other statements: https://www.dropbox.com/s/6ch0btu3adiz2ob

A [math]\textbf{perspectivity}[/math] from line L to a distinct line M with [math]\textbf{centre}[/math] c not incident to L or M is the function that, for any point x on L, returns the point M[ox].

A [math]\textbf{perspectivity}[/math] from point p to a distinct point q with [math]\textbf{axis}[/math] C not incident to p or q is the function that, for any point X on p, returns the point q[AX].

A [math]\textbf{projectivity}[/math] is a finite composition of perspectivities.

A [math]\textbf{half-perspectivity}[/math] from line L to a non-incident point p is the function that, for any point x on L, returns the line xp. The half-perspectivity from point p to a non-incident line L is the inverse of the function above.

A [math]\textbf{half-projectivity}[/math] is a composition of an odd number of half-perspectivities.

Given two ordered n-tuples of objects from a single incidence set each (X_1, ... , X_n) and (y_1, ... , y_n), we say these tuples are [math]\textbf{projectvely related}[/math] if there is some (half)-projectivity Pr such that Pr(X_k) = y_k for all k such that [math]1\leq k\leq n[/math].

Since we will use (half-) perspectivities and projectivities so often, we need a lot of new notation. We use two notations for (half-) perspectivities and projectivities. The two systems describe the perspectivity from L to M with centre p as follows:

L>p<M

[eqn]L\stackrel{p}{\barwedge}M[/eqn]

The perspectivity from p to q with axis L is noted as follows:

p<L>q

[eqn]p\stackrel{L}{\barwedge}q[/eqn]

Projectivities can be notated by concatenating perspectivities.

p<N>q<L>r

[eqn]A\stackrel{x}{\barwedge}B\stackrel{y}{\barwedge}C\stackrel{z}{\barwedge}D[/eqn]

The LaTeX-based notation does not deal well with half-perspectivities and half-projectivities, but the text-based notation does.

p<L

X>a<Y>b

The next post will contain some more notation, and some examples.

>>8318877
I have no idea why the math tags aren't working here.

>>8318882
It doesn't work totally like LaTeX. If you want intermediary text with spacing you need to use "\"text{}" whereas in LaTeX it would just unitalisize it.

>>8318877
>>8318882
The text appears to work properly in the TeX preview, so you can copy the text from http://pastebin.com/Lgt8FB8X and paste it into the TeX preview to see the post.

>>8318903
Yeah that looks correct I take back >>8318898
, seems like it just fucked up.

File: persp_ex.png (47KB, 1594x819px) Image search: [Google]

47KB, 1594x819px

If two tuples are (X_1, ... , X_n) and (y_1, ... , y_n) projectively related, we can note this in our two systems:

(X_1, ... , X_n) ~ (y_1, ... , y_n)

[eqn](x_{1},..., x_{n}) \doublebarwedge (Y_{1},..., Y_{n})[/eqn]

If the relating (half-)projectivity is known explicitly, we can combine the notations. Note the tuples are prefixed by the object its elements are all incident to.

p(A,B,C)<L>q<M(a,b,c)

[eqn]A(p, q, r, s)\stackrel{x}{\barwedge}B(p', q', r', s')\stackrel{y}{\barwedge}C(p”, q”, r”, s”).[/eqn]

With the notation taken care of, we can now look at a few examples of perspectivities.

Recall that the real projective plane is essentially the Euclidean plane with a line at 'infinity' whose points are uniquely associated with particular slopes. Now consider the function that takes points from the x-axis to the line y=1 such that (k,0) gets mapped to (k+1, 1). This is a perspectivity from the x-axis to the line y=1 with the centre. Note that since the two given lines are parallel in the Euclidean point, they meet at an 'ideal' point in the real projective plane.

Another example is the perspectivity from the x-axis to the y-axis with centre (1,1), as shown in the posted pic. It turns out that this perspectivity maps points (k,0) to points (0, 1 + 1/(k-1)). Note that in this case the meet of the lines is a 'real' point and unlike the first example, 'real' points get mapped to 'ideal' points and vice versa. In this case it is the points (1,0) and (0, 1) that are involved.

We can also look at the perspectivities in finite projective planes. In the projective plane generated by the PDS {0, 1, 3, 9} mod 13, let (k, 0) = p_k and (k, 1) = L_k. The point p_0 has pencil {L_0, L_1, L_3, L_9} and the point p_9 has the pencil {L_0, L_4, L_5, L_7}.

The line L_2 is not incident to either point, so the projectivity from p_0 to p_9 with axis L_2 exists. Working it out, it is apparent that p_0(L_0, L_1, L_3, L_9)<L_2>p_9(L_0, L_4, L_5, L_7).

I updated the notation cheat sheet with the new definitions and notations with a few corrections, so you can actually see them.

File: Thm_2-3_2.png (28KB, 1200x900px) Image search: [Google]

28KB, 1200x900px

With those examples done, we can see what the more general (half-)projectivities are like. Since in general, projectivities are compositions of perspectivities, explicitly working them out could be quite messy. Instead, what we will do is show that certain incidence set tuples are projectively related, so demonstrating what certain projectivities can do.

Theorem 2.2: Given three tuples of distinct points (a, b, c) and (a', b', c') with elements taken from I(,L) and I(,M) respectively, then (a, b, c)~(a', b', c').

Proof: If a = a' = LM, then by Axiom 1 b, c, b', and c' are all distinct, and thus [bb'][cc'] = o exists, o does not lie on L or M, and L(a, b, c)>o<M(a', b', c') holds. (see http://i.imgur.com/WedhlEe.png)

If c = b' = LM, then and a != a and c' != b, thus N = bc' exists. Also, a lies on none of M, N and a' lies on none of L, N. Thus the projectivity L>a'<N>a<M exists. It is then apparent that L(a, b, LM)>a'<N(N[aa'], b, c')>a<M(a', b', c'). (See http://i.imgur.com/EkBlhm5.png)

This second case can be generalized in steps to yield the general case by modifying the projectivity L>a'<N>a<M. If b' != LM, then replace point a in the projectivity with [bb'][aa'] while if c != LM, replace a' with [cc'][aa']. (See http://i.imgur.com/pkuBUSW.png)#

Theorem 2.3: Given four distinct yet concurrent lines A, B, C, and D, (A, B, C, D) ~ (B, A, C, D).

Proof: Choose points p*, p' on lines C, D respectively distinct from p. Choose x on line B distinct from p and not collinear to p* and p'. Then label the following lines:

E = p'p*
A* = xp*
A' = xp'
B* = p*[AA']
B' = p'[AA*]

This is shown in the posted pic and at http://i.imgur.com/lDEFXD5.png. Of the points and lines p, p', p*, A, A', and A*, only the pairs (p, A), (p', A'), and (p*, A*) are incident. Therefore, the projectivity p<A'>p*<A>p'<A*>p exists. From there one can conclude the following:

p(A, B, C, D)<A'>p*(B*, A*, C, E)<A>p'(A', B', D, E)<A*>p(B, A, D, C)

as required by the theorem.#

>>8321030
>(A, B, C, D) ~ (B, A, C, D).
This should be (A, B, C, D) ~ (B, A, D, C)

From the definitions of projectivity and perspectivity, it is apparent that perspectivities are a subcategory of projectivities. It would seem that to determine whether a projectivity is a perspectivity, one would just look at the construction of it. However, by convention (half-)projectivities are considered extensional. That means two (half-)projectivities are equal if two have the same domain and co-domain, and if for each element in the common domain, its image for each function is the same.

The upshot of that is you can't just look at a projectivity's construction and tell whether it's a perspectivity, as even if its known construction is very long, there might be a much shorter one, even up to only a single perspectivity.

Still, from looking at certain properties of a projectivity, we can often immediately rule it out as a perspectivity. First of all, it is impossible for the domain and co-domain of a perspectivity to be equal. Theorem 2.3 demonstrated that this is not the case for general projectivities.

Given that condition, the domain and co-domain have just one common element. Another condition for a perspectivity is that this common point be its own image. This is not true for projectivities in general. Consider the points a, b, c, and d which fulfil Axiom 4+. Then, by Theorem 2.3, (a, [ab][cd], b) ~ (c, d, [ab][cd]).

Now down to projectivities between distinct objects with self-corresponding meet/join, we cannot narrow our candidates any further. We know that at least some perspectivities are there, so the simplest possibility is that they are all perspectivities.

Conjecture 2.1(Perspectivity Conjecture): Every projectivity with distinct domain and co-domain such that the common element of the domain and co-domain is its own image is a perspectivity.

This conjecture will be our focus(if indirectly) for most of the next few threads. For now, the next large post will focus on some specific consequences/conditions.

For this whole post we assume that L>r<M>p<N is equal to L>q<N for some q.

Lemma 2.4: L>q<N>p<M = L>r<M

The proof of this statement is left to the reader.

Theorem 2.5: Either {L, M, N} is concurrent or {p, r, LN} collinear.

Proof: We assume that {L, M, N} is non-concurrent. The image of LN under the component perspectivity L>r<M is not itself. Call the image a. Then we have {r, a, LN} collinear. Since L>r<M>p<N is a perspectivity, then it must be the case that LN is the image of a under M>p<N, so we have {p, a, LN} collinear. Then our conclusion follows from Col. L.#

Theorem 2.6: If {L, M, N} is concurrent, then {p, q, r} is collinear.

Proof: We first assume that {r, p, LN} is not collinear, so L[pr], M[pr], N[pr],are all distinct and the set {p, r, L[pr], M[pr], N[pr]} is collinear. The image of L[pr] under L>r<M>p<N is N[pr]. Therefore, the image of L[pr] under L>q<N is N[pr], so {q, L[pr], N[pr]} is collinear, so {p, q, r} is collinear by Col. L.

If {r, p, LN} is collinear, you can try L>q<N>p<M and L>r<M instead by Lemma 2.4. If {p, q, LM} is collinear too, since LM = LN, the conclusion again follows Col. L.#

Lemma 2.7: If {L, M, N} is non-concurrent, then none of the pairs (p, L), (q, M), or (r, N) is incident.

Proof: By Theorem 2.5, {p, r, LM} is collinear, so if p lies on L, so does r, which is not possible since L>r<M exists. Similar reasoning carries for the pair (r, N). Looking at L>q<N>p<M = L>r<M with this reasoning, then (q, M) non-incident also follows.#

Theorem 2.8: If {L, M, N} is non-concurrent, then q = [p[LM]][r[MN]].

Proof: From Lemma 2.4 and Theorem 2.5, {p, q, LM} is collinear. The inverses N>q<L and N>p<M>r<L are equal as well, and applying 2.4 and 2.5 to that we get {r, q, MN} collinear. By Lemma 2.7, the lines p[LM] and r[MN] exist. If they were equal, then {p, r, q, MN, LM} is collinear and thus a subset of In(,M), contradicting Lemma 2.7, so p[LM] and r[MN] are distinct. The point q lies on both, so q = [p[LM]][r[MN]].#

The Theorem and Definition cheat sheets have been fully updated with minor corrections to the newest entries.

The next thread will be a bit of an interlude, as we introduce a generalization to projective planes and a new axiom.

>>8324574
P.S. The next thread will start in about a week, but it might be delayed a bit. After that I plan to swap to a two-week thread schedule.

Hey man, I am taking a projective geometry undergrad course, and the professor is mostly following this book: http://www.maths.ed.ac.uk/~aar/papers/beutel.pdf.
While your texts here facinate me, I must say that my professor is taking a much more abstract aproach to projective geometry.
Do you think the Beutelspacher is a good book to learn it?
My previous geometry experience was a 1-year course on Modern Geometry, an introductory course to linear algebra and I am taking actual linear algebra this semester. this Projective Geometry course would be 6-months.

bretty gud OP
always love intuition stuff

What's projective geometry? Is it useful for chemistry? If not then idc about it

Can this help me learn how to use python?

this sucks

Coming back to this, all I can see are the typos.

>>8322481
>Then, by Theorem 2.3, (a, [ab][cd], b) ~ (c, d, [ab][cd])
That should be Theorem 2.2

>>8323735
>By Theorem 2.5, {p, r, LM} is collinear,
That should be {p, r, LN}

>>8325851
That book happens to be one of my sources, and in fact the next thread will partly based on it. If you could only use one book it's pretty good, but personally I'm not fond of the style. Also, I'm not exactly "Modern Geometry" entails.

>>8325993
>>8326005
Sorry, probably not.

Posted is a diagram of the projective plane generated by the PDS {0, 1, 3, 9} mod 13. You can use this to explore projectivities and perspectivities. If you don't like the colour, try the diagram at http://i.imgur.com/gD02sgz.png.

An interesting fact is that in this particular projective plane, the Perspectivity Conjecture is true.

File: o3_plane.png (69KB, 853x836px) Image search: [Google]

69KB, 853x836px

>>8326855
Forgot to post the diagram.

>>8326472
Modern Geometry is a course that goes through most of the geometry developed from the euclidian axioms.
I think that the most projective-geometry relevant thing would be Desarge's Theorem seen with euclidean geometry, and the ratio between line segments (we saw harmony from pencils too), but other than that, I saw Menelaus, Ceva, The 9 point circunference and Simpson lines, Circunference theorems, and a bit of an intro into hyperbolic geometry based on Poison's disk.

>>8327347
Desargues' Theorem is next, and I plan to eventually get to harmonic conjugates and the cross-ratio eventually. Because of the way I'm building up the series, the cross-ratio and the explicit use of algebra will be very much delayed.

>>8328505
I'll wait warmly then.