Models of Finite Incidence Planes – Fano’s Geometry

Video Series: Incidence Geometry (Tutorials with Step-by-Step Proofs)

(D4M) — Here is the video transcript for this video.

hi everyone welcome back in this episode
models of finite incidence planes we’re going to look at uh finals geometry so
this is going to be part of the series incidence geometry tutorials with
step-by-step proofs let’s do some math so let’s first refresh our memory on
what the incident axioms are so we’ve been working with these uh axioms here
with the last several uh episodes so a1 says for every two points
there’s a unique line going through it uh a2 says for every line there’s at
least two points on it and for a3 that there’s no um
uh there exists three non-collinear points so there’s those three axioms
there we’ve worked out some models for this geometry
and in previous episodes we’ve also proven these theorems here so i
recommend going out and checking this full series out link is below in the

description now um in the uh last couple of episodes
we also talked about parallelism and we talked about these three properties here
and in the last uh two or three episodes we showed that in three point geometry
the elliptic parallel property is true this was the first statement right here
and then in another episode uh we showed that four and five point geometry
uh this one we have the euclidean parallel property holding true and in
five point geometry we have the hyperbolic parallel property holding
true so if you’re not sure about any of these three statements here and about
what these models are again i recommend that you’re going back
and checking out those episodes so uh let’s start off i mean so let’s
get on to what uh fianna’s geometry is so what we’re going to have here seven
symbols or seven points that’s way that we’re going to interpret the points it’s
going to be a b c d e f g i’m going to use those symbols and we’re going to have

these seven lines here so this will be line one made up of these three points
and line two and three and four and all the way to line 7 here and so this is a
incidence geometry in the sense that we verify that axiom a1 holds we verify
that axiom a2 holds and we verify that axiom a3 holds and so we go through each
one of these and we observe if the axiom holds or not for example a2 is probably
the easiest one here because all we got to do is check that each line has
at least two points on it so this one’s got two in fact they all got exactly
three so we can easily see that each one
has two points at least two points on it so that one’s holding true and then a3
says there exists three distinct non-collinear points uh so here’s three
distinct coding year points but if i choose a point not on this line like a b
and g and then if we check all the list here

we can see that a b and g are not all on one line so let’s double check that a b
and g so a b and g uh a b and g here’s d b and g but that’s
not a b and g right so yeah just picking the first line three points and another
point um we got a b and g are three non-collinear points so that means a
three holds so then now the next thing to check to
know that this is a model of incidence geometry
is to check that a1 holds and so here we have
pick any two points and we got to show that there’s only one line through it
so that means we need to pair up and look at a lot of cases so one case will
be a b those points a and b so i’ll put a comma b and another pair would be a c
and a d and for each pair of points we need to check that only one line goes
through it so we would check also a e a f a g

uh how far did we go to g right so these would be the first couple of cases but
we’d also we need to check bc um and then we don’t need and then we
need to check bd and so on right so we would check all of these cases
and so let’s check this case right here which line is a b on a b is on line one
and axiom one says we should only have one line here what’s the other line that
a and b are both on we check a and b yep so it’s a unique line all right and
so now let’s check this case here ac what line goes through a and c is there
only one and yep only line one goes to a and c what about a d a d is line three
so axiom a1 says there’s one and exactly one line through every just
distinct pair of points right so we check all the possible pairs

and then the last one using this ordering here that we were doing for our cases
would be what uh the point here how about g is the last so how about fg
that could be the last case there fg and so what line is that on that’s on l2
right here and can we find any other line with f and g on it
f and g anyone no so only l2 so each case that we verify
we get actually only one line you have to fill in the details here don’t
believe me go through and check out all of these
cases here and you’ll see that axiom a1 is holding here and so this that’s what
this guy did he checked them all he checked all the axioms and they all hold
so we are right in calling this an incidence geometry
and so now what we can actually say is not only is this a case when all three

axioms hold but actually in this particular geometry there might be some
additional things that actually also hold and so i’ve listed three additional
things these three things don’t hold in every um
incidence geometry but we did prove uh 11 theorems here so uh these 11 theorems
here do hold in fianna’s geometry you can go
and check each one of them but you don’t need to we we’ve already proven these
whole anytime axioms 1 2 and 3 hold all 11 of these theorems automatically hold
so for this geometry right here because we’ve shown a1 a2 and a3 hold we can
call it an instance geometry and we know
those 11 theorems must hold but actually there’s some additional statements that
also hold for this particular geometry so let’s look at these three additional
statements here so the first one is each point is incident with exactly

three lines now uh there’s you know a couple
different ways you can go about this one
way would be to try to prove this result right here by stating some simpler
axioms which you which you can verify just like we did for a1 a2 and a3 or we
can just prove this right here by looking at at all the possible cases
again so uh each point right so how many cases are there going to be right so we
have um you know seven points so we’re going to have seven cases
for each point we’re gonna check this to see if it holds so efg
so how many lines is a on one two three only three how many lines is b on one
two three only three exactly three and so on
and we check all these cases and then we
get to the last case we check g how many

lines is g on one two three and so we we
verify i i didn’t do it here you have to go do this verify that each point is on
exactly three lines and then you can claim this right here holds
and uh the way this is just brute force just by inspection if you have a
concrete model then this is one way to get something done assuming you have a
finite model right uh in the upcoming series we’re going to look at some
geometries which have infinitely many points on them and so that’ll be uh real
fun and exciting but for this one right here
um you know we just we can just check how about this one right here number two
for every two distinct points there’s exactly two lines
that are not incident with these points so what does that all mean
so uh for every two distinct points let’s pick on a and b
this will be one of our cases um and in fact let’s just write out
let’s just write them out uh rook real quickly or at least some of them so two

distinct points right so we have the point here let’s say a b
b c would be another one or let’s stick with a first a c a d a e
so two distinct points so for all of these cases here and then we have bc
and then bd and so on until we get to the last two distinct points f and g
so for each of these for every two distinct points there’s exactly two
lines that don’t go through these two points here so let’s see if i can find
them for a and b so a and b all right so what are the two
lines that don’t go through a and b this one goes through a that that one goes
through a that one goes through b here’s
one line that doesn’t go through a and b so i’ll put it over here l5 and
uh let’s see here where we were here where where l5 all right so l6 doesn’t
go through a and b either so there’s another one l6

and l7 goes through b all right so we we got for a and b there’s exactly two
lines that don’t go through a and b so that’s what number two says for every
two distinct points no matter which pair you pick there’s exactly two lines that
don’t go through these points here so let’s just do one more
case right here what about f and g so here’s one line that doesn’t go
through f and g l1 this goes to f this this doesn’t go through either l3
and this goes through [Music] g and g and f and f all right so in this
case it worked again exactly two lines that don’t go through f and g
so here’s statement number two the way that you can prove and this is just one
way i mean you can take different approaches to this geometry but this is
a very concrete model i just have everything laid out and so you can prove
things right here just by going through all the cases

again we checked a axiom a1 a2 and a3 holds and these are three additional
statements here that we can verify now by going through all those cases you can
verify statement one here and statement to hold
now here i have a question for you the uh viewer of this episode here if you
can show if if you know that axiom a1 a2 and a3 hold
and statement i holds and statement two holds
if all of these statements hold one two three four five if all five of these
statements hold can you then show that this one right here has to hold also so
that’s an interesting question i look forward to your answer in the comments
but if you cannot do that it’s okay we can show this third statement here and
here’s how let’s remember what the eclip elliptic parallel property says

it says that if you draw any line and you take any point not on the line
then there’s no line that goes through p that’s parallel to l no line through p
and you have to be able to do that for every line and every point not on the
line so in order to verify uh number three here how many cases would we have
well one case would be to pick line one and then we would have to pick a point
not on line one how many points are not on line one
well all together we have seven points line one’s got three of them
so we’re going to be able to choose four points not on the line
so the four points are d e and f and g so here would be four cases we
would have to go verify and then we can do this for each of the

lines not only for line one there will be four points off of it but also for
line two so here would be the next four cases line two
and then what’s the point not on it right so b line 2 and c right
and line 2 and d and line two and e and so
there’s eight cases but actually we’re gonna have four cases for each line and
there’s seven lines so we’re gonna have 28 cases to verify 28 cases
and what do we do for each case so remember you uh elliptic parallel
property says for each line in each point not on the line there are no lines
going through d that are parallel to l so i need to check let’s just check the

first case here we won’t check all 28 cases in this episode here let’s just
check this one case here and see how it works so here’s a line
and here’s a point not on the line now every line that goes through d
must have an a b or c on it so here’s a point that goes through d
and it’s got a on it here’s a point d that has a b on it
and that doesn’t go through d and this has a d on it and it goes
through c so every point so uh sorry every line that goes through d is not
parallel to l so uh that right there holds um i’ll just put a check there
and let’s just check one more you know let’s check uh l2 and b just to make sure
that uh you’re understanding here so now we have l2 is our line and point b is
off of it and the elliptic parallel property says that there’s no line through b

that is parallel to l2 in other words every line through b
has to not be parallel to l or i said differently every line through b
has to have a point in common with l2 so let’s check that here’s l2 right here
and now let’s check lines that pass through b every line that passed through
b will have an a g or f on it this goes through b it has an a on it this goes
through b it has a g on it this goes through right here this goes through b
and this has an f on it so they’re not parallel either
so every line that went through b was uh non-parallel to l2
so uh we again we so we check that case right there and again you have to check
28 cases in order to get to number three here so that’s why i was asking the
question is if you if once you know that a1 a2 and a3 hold and these other two

statements hold can we then go and prove the third um
statement right there that’s a question for you to answer in the comments below
but you know you can check it out and do all 28 cases actually 28 cases sounds
like a lot but it’s actually not that many you could probably do it within say i
don’t know 30 minutes for sure but in any case uh here’s fianna’s geometry and
now we’re going to look at young’s geometry click right here and we’ll
see you in that episode

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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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