Projective Geometry Part 3:Projective spaces and Desargues' Theorem

So far our efforts have been focused on the general projective plane. While this has done us well so far, there are reasons we may want to change our focus, if only for a while.

Firstly, recall we based our axioms on observations of the Euclidean plane. Furthermore, the real projective plane is basically the Euclidean plane with a line at infinity. However, the Euclidean plane is just the two dimensional version of Euclidean geometry. So far, we haven't found a way to generalize projective planes in the same way, and it would be interesting to find out how, if possible.

Secondly, while we have built up a good collection of simple theorems from the plane axioms, we can't get much further with the plane axioms alone. The problem is that while the plane axioms ensure the presence of projectivities, they cannot really control their behaviour, and as such there is not much that can be proven about projectivities or projective planes from the plane axioms alone. We thus seek a natural restriction on projective planes that will allow us to 'tame' the projectivities. This control will payoff in the existence of complex structure and sophisticated tools that otherwise may be absent.

To get an idea of how to construct 'n-dimensional' analogues of projective space, we shall look at the Axioms 1-6 one by one.

Axioms 1 and 2, not only are both true in Euclidean and projective planes, but also in every Euclidean space, so that so it seems that these two axioms should remain unchanged. Note in particular this means that Theorems 1.1 and 1.2 also apply.

Axiom 3 is tricky, so we leave that for now.

As for Axioms 4-6, there is nothing wrong with Axiom 4, but for full generality, we also want projective analogs of Euclidean spaces of dimension lower than two. Axioms 5 and 6 are incompatible with this, so they must go.

That leaves us with Axioms 1, 2, and 4, but we still need to deal with Axiom 3. That will be done in the next post.

Other urls found in this thread:

pastebin.com/50w9MYjD
pastebin.com/fMdRzhTf
pastebin.com/GfdvBDML
dropbox.com/s/ia4m7s0fy59z9jv
dropbox.com/s/6ch0btu3adiz2ob
i.imgur.com/L5jAvl5.png)
i.imgur.com/yCkx45m.png).
i.imgur.com/q8WTFsA.png)
i.imgur.com/TyiJePD.png)
i.imgur.com/bWYqTU3.png)
dropbox.com/s/0sck6ris1717khn
twitter.com/SFWRedditVideos

Section 0() text: pastebin.com/50w9MYjD
Section 1 text(): pastebin.com/fMdRzhTf
Section 2 text(): pastebin.com/GfdvBDML

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

>muh axioms
into the trash

In Euclidean space dimensions 3 and higher, there are three relevant cases for two distinct lines:

-intersecting
-parallel
-skew(non-coplanar)

Axiom 3 doesn't have to deal with the skew case, so our task is to show how to correctly treat the skew case projectively. Recall that one of the arguments for the line at 'infinity' was continuity. If you 'nudge' a pair of parallel lines so they aren't skew, then they intersect at a point 'far away', which justifies the idea that they intersect 'at infinity'. On the other hand a 'nudge' to a pair of skew lines almost always keeps them skew. This suggests the correct way to handle projective 'skew' lines is to not let them meet.

This means Axiom 3 must be replaced by a statement like 'All coplanar lines meet.'. The problem with this statement as it is is that it make reference to a plane and thus the plane axioms. Since this is supposed to be a generalization of projective planes, this situation is not ideal.

To resolve this we look back at Euclidean space. Here three distinct points are always coplanar, but not so for four points. Thus, we would like our axiom to involve four distinct points. A way to ensure that the points are coplanar is to have the line through two of the points be coplanar to the line through the other two points. This then ensures all pairs of lines incident to two of the four points are coplanar.

That then suggests the following axiom in the projective case:

Axiom 3-(Veblen-Young axiom): Given 4 distinct points a, b, c, and d, if the line through a and b meets the line through c and d, then the line through a and c meets the line through b and d.

Using our point and line naming conventions, we can restate the axiom as follows:

'Given four distinct points a, b, c, and d, if [ab][cd] exists, so does [ac][bd].'

With the axioms now settled, we can define some important terms in the next post.

We now have the axioms below:

Axiom 1: For any two distinct points, at most one distinct line is incident to both.

Axiom 2: For any two distinct points, there is at least one line incident to both.

Axiom 3-: Given four distinct points a, b, c, and d, if [ab][cd] exists, so does [ac][bd].

Axiom 4: Every line is incident to at least three distinct points.

These are the [math]\textbf{space axioms}[/math], and they are the most general axioms for projective geometry. Any system which follows these axioms is a [math]\textbf{projective space}[/math]. One point about the space axioms is that duality as we have defined it does not apply to projective spaces in general.

A [math]\textbf{linear set}[/math] is a set of points such that for any two distinct points in it, the range of their join is a subset of the given set.

Examples of linear sets are the empty set, a range, a single point and the entire point set itself.

A [math]\textbf{linear subspace}[/math] is a linear set coupled with the set of lines whose ranges are subsets of the linear set and an incidence relation inherited from the parent space.

Given any set of points [math]S_{points}[/math], the [math]\textbf{span}[/math] of that set is defined as follows:
[eqn]\mathrm{Span}(S_{pts}) = (| S_{pts} |) = \bigcap \left \{ S_{lin}|S_{pts} \subseteq S_{lin}, \; S_{lin}\; \mathrm{is\;linear} \right \}[/eqn]

Define the [math]\textbf{set sum}[/math] of of two linear sets [math]S_{one}[/math] and [math]S_{two}[/math] as follows:
[eqn]\bigcup \{\mathrm{In}(,ab)|a\in S_{one},b\in S_{two}\} \cup S_{one}\cup S_{two}[/eqn]
From the definitions, the following theorems are apparent.

Theorem 3.1: A linear subspace is a projective space.

Theorem 3.2: Any linear set generates a unique linear subspace.

Theorem 3.3: The span of any set of points is a linear set.

In fact, usually one doesn't make a distinction between a linear set and the linear subspace it generates.

Let the set sum be notated as follows:
[eqn]S_{one}+ S_{two}[/eqn]

Why post this image OP?
You can't draw a 4D object on a 2D plane.
It's like drawing a 3D object on a 1D line.
Even if you somehow do it, it won't make sense.
Try drawing a 4D sphere on a plane...

I didn't read your message because it's too long and in my time zone it's too late.

you should post the axioms first so we know what the heck you're talking about

That picture is irrelevant, I just needed a picture there.

The link to the axioms is right there.

What is the point of this?

Projective geometry is easy.

[math]\mathbb{P}_R^n \equiv \operatorname{Proj} R\left[ {{x_0},...,{x_n}} \right][/math]

done.

We want to know why that's true(and when it isn't).

The significance of set sums comes from the following theorem.

Theorem 3.4: The sum of two linear sets is linear.

Proof: Let the two linear sets be S_1a and S_1b

Given distinct points p and q in the sum, we need to show any point z on pq distinct from p and q is in the sum.

If p and q are in the union of S_1a and S_1b, then z is in the sum.

Now assume q is in S_1a while p is not in either set. Thus p is on ab for some a in S_1a and b in S_1b. If q = a or b then z in the sum, so assume q is not. We have p = [qz][ab] and by Axiom 3-, [zb][qa] = a' exists. The point a' is on aq so it is in S_1a, b is in S_1b, and z is on a'b, thus in the sum.(See i.imgur.com/L5jAvl5.png)

We now assume neither p nor q are in the given sets. Thus, q on a'b' and p on ab, where a, a' in S_1a and b, b' in S_1b. If we assume a = a' and b = b', then z is in the sum.

If only a = a' is true then a = [pb][qb'], so b* = [pq][bb'] exists and clearly is in S_1b. If z = b* then z is in the sum, otherwise the point q = [ab'][zb*], so [za][b'b*] = b” exists and it is also in S_1b. We now have z on ab”.(see i.imgur.com/yCkx45m.png).

If neither a = a' nor b = b' is true but [aa'][bb'] exists then so does [ab][a'b'] = e. Then [eb][zq] = p and thus f = [bz][eq] exists. If f = a' then z is in the sum, otherwise e = [fa'][ab] thus a” = [aa'][bf] exists and z lies on ba”, thus in the sum.(see i.imgur.com/q8WTFsA.png)

If [aa'][bb'] does not exist, we have p = [qz][ab], so the point c = [bz][aq] exists. If c is equal to a, a', or b', then z on cb is in the sum. Otherwise, [ac][a'b'] = q, so a” = [cb'][aa'] exists, and is on aa', thus in S_1a. The point a” cannot equal b or b' as that would force p or q on aa' thus in S_1a. If a” = z, then it is in the sum. Otherwise, c = [bz][a”b], so b” = [bb'][a”z] is on bb', thus b” in S_1b. This case forces a”b” to exist and z lies on it.(see posted pic)#

>The link to the axioms is right there.

Didn't notice it. In that case make it more obvious that that's what you're referring to.

Sums of linear sets have other important properties.

Theorem 3.5: The sum of two linear sets is equal to the span of their union.

Proof: Let the two linear sets be S_2a and S_2b.

We first prove the sum is the subset of the span. Let x be an element of the sum. If it is in one of given linear sets, the it is obviously in the span. Thus we can assume that x is in In(,ab) for some a in S_2a and b in S_2b. The point x is in all linear sets that contain S_2a and S_2b, and therefore it is in the span.

By definition the span is a subset of any linear set containing the given sets. This includes the sum by Theorem 3.4, so the span is a subset of the sum, and thus they are equal.#

Theorem 3.6: Linear set sums are commutative.

Proof: This is apparent from the definition of the linear set sum.#

Theorem 3.7: Linear set sums are associative.

Let S1, S2, and S3 be our given linear sets. All we need to show is that (S1 + S2) + S3 is a subset of S1 + (S2 + S3), and then Theorem 3.6 and relabelling gets us to equality.

Let x be a point in (S1 + S2) + S3. We need to show that x is in S1 + (S2 + S3). If x is in the union of the given sets then that follows.

Thus we assume that x is on sr, where s is in S1 + S2 and r is in S3. If s is in S2, then x is in S2 + S3. If s is in S1, then the range of sr is in S1 + (S2 + S3) by definition, which includes x.

If s is not in either S1 or S2, then there exist points p and q in S1 and S2 respectively such that s lies on pq. We then have r = [pq][xr], thus t = [px][qr] exists(see posted pic). The point t in S2 + S3 and the point p in S1 are collinear with x, so x is in S1 + (S2 + S3).#

These theorems allow us to explicitly construct the span of any finite number of given linear sets. In particular we can construct the span of a finite number of points. Any projective space that is the span of finitely many points is a [math]\textbf{finitely generated}[/math] projective space.

You have no idea what is he talking about ,right?

We say a set of points in a projective space [math]\textbf{generates}[/math] its span. A question of interest is whether projective planes are finitely generated. To answer that, it helps to first prove a lemma.

Lemma 3.8: In a projective space generated by three non-collinear points, all pairs of lines meet.

Proof: Let p, q, and r be the given points, thus {p} + ({q} + {r}) is the projective space in question.

The lemma trivially holds for lines px with x on qr together with qr itself. Consider arbitrary distinct points a and b in {p} + ({q} + {r}) but not in {p} or {q} + {r}.

We first prove the line ab meets qr. There exist distinct points a' and b' on qr such that a and b lie on a'p and b'p respectively. We have p = [aa'][bb'] therefore c = [ab][a'b'] = [ab][qr] exists. (see i.imgur.com/TyiJePD.png)

We now prove that ab meets every line px where x lies on qr. Let c = [ab][qr], and let a' and b' be the same as before. If x in {a', c} then ab meets px, otherwise we have [pa][cx] = a', so x* = [px][ac] = [px][ab] exists. (See i.imgur.com/bWYqTU3.png)

The only case remaining is that of ab meeting another line ef where e and f are not in either {p} or {q} + {r}. Let a' and b' be the same as in the other cases. Then a” = [ef][pa'] and b” = [ef][pb'] exist. If a = a” or b = b” then we are done, otherwise we have p = [aa”][bb”], so d = [ab][a”b”] = [ab][ef] exists. (see posted pic)#

We can now prove the following:

Theorem 3.9: A projective space generated by three non-collinear points is a projective plane.

Proof: Axioms 1, 2, and 4, are inherited from the plane axioms, and Lemma 3.8 ensures Axiom 3 holds.

Call the three generating points p, q, and r. Then qr in the space verifies Axiom 6. Since the three given points are non-collinear, then any line can pass through at most two of the three given points, ensuring Axiom 5 holds.#

This theorem allows us to see projective planes as linear subspaces in parent spaces.

Holy fuck, two days ago I was thinking about this shiet hoping no body had already done it. Turns out you are doing it. Keep up with the work sir.

The converse of 3.9 also holds in general.

Theorem 3.10: A projective plane is generated by any three non-collinear points in its point set.

Proof: Let p, q, and r be any three non-collinear points. By Theorem 3.5, all we have to show is that {p} + ({q} + {r}) is the entire point set. Let x be an arbitrary point in the plane. If x = p or is on qr, then it is clearly in {p} + ({q} + {r}). Otherwise, the line px exists and by Axiom 3, px meets qr at some point m. The point m cannot equal x or p, so x lies on the line pm, thus it is in {p} + ({q} + {r}).

Corollary 3.10.1: Given any two distinct planes that are linear subspaces of some parent space, the intersection of their point sets must be collinear in both planes.

Proof: Note that since each plane's incidence relation is inherited from the parent space, they agree on incidence and collinearity for common points and lines.

If the intersection of the point sets is non-collinear, then Axiom 3 ensures it contains three distinct non-collinear points. By Theorem 3.10 those points generate both given planes in the parent space. The span of this set of points in the parent space is unique, so the two planes are identical, in contradiction to our assumption.#

We are almost ready for the 'payoff' of this section. The next post will take care of a few small things for that payoff.

First, we mention a subtlety about Axiom 3-. These are the two statements of Axiom 3-:

'Given 4 distinct points a, b, c, and d, if the line through a and b meets the line through c and d, then the line through a and c meets the line through b and d.'

'Given four distinct points a, b, c, and d, if [ab][cd] exists, so does [ac][bd].'

Implicit in the second statement is that {a, b, c, d} is assumed non-collinear. If it is collinear, then both statements are still true, but in different ways. Practically, it makes no difference as almost any situation where Axiom 3- is used, either the points being collinear collapses the proof to triviality or the points are known to be non-collinear.

A [math]\textbf{triangle}[/math] is a configuration of three non-collinear points along with the lines that join them. These points are [math]\textbf{vertices}[/math] and the joins are the [math]\textbf{sides}[/math] of the triangle. Triangles are named after their three vertices, e.g. triangle abc. The plane generated by triangle abc is called (|abc|).

Consider the negation of Axiom 3:

Axiom ~3: There exist two non-concurrent lines.

Any space where Axiom ~3 applies contains three non-collinear points, thus it contains planes as linear subspaces, but it is not itself a plane. This lemma follows.

Lemma 3.11: For any projective plane that is a linear subspace of a parent space which respects Axiom ~3, there exists a point in the space but not in the plane.

Call a projective space where Axiom ~3 holds an [math]\textbf{extra-planar}[/math] space. An important theorem is valid for such spaces, which will be proved for the next post. There is one final lemma we need for this theorem's proof, which will be stated without proof.

Lemma 3.12: If a line's range contains one point that is outside of some linear set then the intersection of that range with the linear set has at most one distinct point.

Theorem 3.13(Desargues' Theorem): Given two triangles abc and a'b'c' with distinct vertices in an extra-planar space, if {aa', bb', cc'} is concurrent, then the points [ab][a'b'], [bc][b'c'], and [ca][c'a'] all exist and are collinear.

Proof: We with the possibility that (|abc|) = (|a'b'c'|), but other than that special cases will not be handled in detail. The theorem is usually 'easier' there than in the general case, but there are a lot of them. One special case that deserves mention is that if {[aa'][bb'], [bc][b'c'], [ca][c'a']} is collinear, then it causes no particular difficulty with the proof.

We deal with planar and non-planar cases. In both cases, the point o = [aa'][bb'] = [bb'][cc'] = [cc'][aa'] exists.

We start off with the case where the triangles are in different planes. The point o ensures f = [ab][a'b'], d = [bc][b'c'], and e = [ca][c'a'] all exist. Points d, e, and f are each both in (|abc|) and (|a'b'c'|), so by 3.10.1, they are collinear.

Now we assume that (|abc|) = (|a'b'c'|). By Lemma 3.11, there exists a point o* not in (|abc|). By Axiom 4 there exists another point o*' on the line oo*, and by 3.12, o*' is not in (|abc|). The point o = [o*o*'][aa'] = [o*o*'][bb'] = [o*o*'][cc'] exists, so a* = [o*a][o*'a'], b* = [o*b][o*'b'], and c* = [o*c][o*'c']. By 3.12, all of these points are not in (|abc|).

The set {a*, b*, c*} is non-collinear, as otherwise the span of {o*', a*, b*, c*} would be a plane distinct from (|abc|). The points a', b', and c' are in that span so by 3.10.1, they are collinear, contrary to assumption.

The pairs of triangles (abc, a*b*c*) and (a'b'c', a*b*c*) fall into the non-planar case, so we have f = [ab][a*b*], d = [bc][b*c*], e = [ca][c*a*], f' = [a'b'][a*b*], d' = [b'c'][b*c*], and e' = [c'a'][c*a*] such that {d, e, f} and {d', e', f'} are collinear. By 3.12, the ranges of a*b*, b*c*, and c*a* share only one point with (|abc|) so we have f = f' = [ab][a'b'], d = d' = [bc][b'c'], and e = e' = [ca][c'a'] collinear.#

Why not use a video instead of a picture?

For the OP, the picture is irrelevant. For any of the other pics, how would videos help?

Theorem 3.13 is a nice result, but it doesn't mean anything if the concept of an extra-planar projective space is inconsistent. We demonstrate a model of an extra-planar projective space here.

Let Sx be a set containing at least four distinct elements. Then define the point set P_pw, line set L_pw, and incidence relation In(p,L) as shown in the posted pic.

Lemma 3.14: If c is the symmetric difference of a and b, then a is the symmetric difference of b and c.

The proof of this lemma is left to the reader.

Theorem 3.15: The model above is an extra-planar projective space.

Proof: We confirm the relevant axioms in this model.

Axioms 1 and 2: Given distinct points a and b in P_(pw), their symmetric difference c is a non-empty subset of Sx, thus it is a point. The set {a, b, c} is thus a line that is incident to a and b. The points a and b uniquely determine a point c and thus a set {a, b, c}.

Axiom 3-: We are given four points a, b, c, d such that u = [ab][cd] exists. We assume wlog that no three of the given points are collinear. Then one can conclude that the statement below holds:
[eqn]\forall x\in \mathrm{Sx}, \neg\left ( \left (x\in \mathrm{a}\;\oplus\;x\in \mathrm{b} \right )\;\oplus\;\left (x\in \mathrm{c}\;\oplus\;x\in \mathrm{d} \right ) \right )[/eqn]Since the XOR operator is commutative and associative, one can use this statement to show that the symmetric difference of a and c is equal to the symmetric difference of b and d, thus v = [ac][bd] exists.

Axiom 4: Given a line {a, b, c}, a and b are different exactly when c is non-empty. By 3.14, the equality of any two elements is equivalent to the emptyness of the third. Since none of our points are the empty set, then the points a, b, c are all distinct.

Axiom ~3: The set Sx contains four distinct elements e1, e2, e3, e4. Then the lines {{e1}, {e2}, {e1, e2}} and {{e3}, {e4}, {e3, e4}} are non-concurrent.#

In the next thread we see how Desargues' Theorem fits with the plane axioms.

Forgot to post the pic.

Exercise:

A version of the statement of Desargues' Theorem that is appropriate for projective planes is shown below.

'Given two triangles abc and a'b'c' with distinct vertices, if {aa', bb', cc'} is concurrent, then {[ab][a'b'], [bc][b'c'], [ca][c'a']} is collinear.'

Show that (in the generic case) this statement holds in the real projective plane.

What are the aplications of this?

From the preface of Beutelspacher's book:

"So why should a person study projective geometry?
First of all, projective geometry is a jewel of mathematics, one of the outstanding
achievements of the nineteenth century, a century of remarkable mathematical
achievements such as non-Euclidean geometry, abstract algebra, and the
foundations of calculus. Projective geometry is as much a part of a general education
in mathematics as differential equations and Galois theory. Moreover, projective
geometry is a prerequisite for algebraic geometry, one of today's most vigorous
and exciting branches of mathematics.
Secondly, for more than fifty years projective geometry has been propelled in a
new direction by its combinatorial connections. The challenge of describing a
classical geometric structure by its parameters - properties that at first glance
might seem superficial - provided much of the impetus for finite geometry, another
of today's flourishing branches of mathematics.
Finally, in recent years new and important applications have been discovered.
Surprisingly, the structures of classical projective geometry are ideally suited for
modem communications. We mention, in particular, applications of projective
geometry to coding theory and to cryptography."

That said, I don't plan to cover applications at all.

I have updated the reference sheets, and have added a series index.

Definitions and notations: dropbox.com/s/ia4m7s0fy59z9jv
Axioms, Theorems, and other statements: dropbox.com/s/6ch0btu3adiz2ob
Section Index: dropbox.com/s/0sck6ris1717khn

Another pointless thread discussing "muh axioms" xD