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 How can we use a
simple modern puzzle
to solve Schubert's complicated
classic geometry problems?
[MUSIC PLAYING]
Here's two lines in
three-dimensional space.
Can you describe all the lines
that intersect both of these?
Or to complicate things,
here's four lines
in three-dimensional space.
How many lines can you
draw that intersect
or touch all four lines?
These are exactly the kind
of intersection problems
that fascinated Herrman
Schubert, a mathematician
in the late 1800s and the
namesake of everything
we're about to discuss.
He actually computed all sorts
of wild, enumerative geometry
problems, like the number
of twisted cubics tangent
to 12 quadrics, which is
apparently 5,819,539,783,680.
And maybe that exact number
doesn't seem particularly
important, but the fact that
Schubert was able to figure it
out is pretty amazing.
He did this calculation
with very few
modern mathematical
resources, which
is why one book
compares it to landing
a jumbo jet blindfolded.
Schubert simplified
these kind of problems
by moving lines around until
they were in a specialized
position, like this or this.
As long as we're very
careful about how we do it,
specializing lines doesn't
change certain properties,
like how many lines
intersect them,
but it does make it
easier to compute.
Schubert's justification
for doing this
was slightly sketchy.
Making his calculations
rigorous is actually
the 15th problem on David
Hilbert's famous list
of influential math problems
in the 20th century,
but here's the part
I think is wild.
Even in these specific
cases, how does one possibly
compute the number and
type of intersecting lines?
In some cases, you can make
a high school style geometry
argument, but what about in
more complicated or higher
dimensional scenarios?
This is where the puzzles help.
There are three pieces,
an equilateral triangle
with red sides, an equilateral
triangle with blue sides,
and a parallelogram
exactly twice
the size of the triangles with
two red and two blue sides.
We'll always label
the red sides with 0
and the blue sides with 1.
The goal of the puzzle is to
arrange any number of pieces
into an equilateral triangle.
Only edges of the
same color can touch.
The pieces can be rotated.
The triangles have
two orientations
and the parallelogram has
three, but not reflected.
So this is not a piece.
Here's some examples
of filled in puzzles.
What patterns do you observe?
As always, you're welcome
to pause and conjecture
or make your own puzzles.
Let's count the blue edges
along the side of a puzzle.
For example on the left
side of this big puzzle,
there are four blue edges.
The right side also
has four blue edges,
and so does the bottom side.
Each side of this puzzle
has two blue edges.
The littlest puzzle clearly
has one blue edge on each side,
and same for this one.
That seems to be a pattern.
All three sides of a puzzle have
the same number of blue edges.
We'll label that
conjecture number one.
Also, all three sides have
the same number of red edges,
but since the sides have a
fixed number of total edges,
those are equivalent
conjectures.
Now let's count the
number of blue triangles
and red triangles
within each puzzle.
Do you notice a relationship
between the number
of blue edges along a side and
the number of blue triangles,
or red edges and red triangles?
Again, there seems
to be a pattern.
The number of blue
triangles is the square
of the number of blue
edges along a side.
And the number of
red triangles is
the square of the number
of red edges along a side.
We'll call that
conjecture number two.
If you want to try and prove
the conjectures yourself,
pause here.
To help with the proof that
I'm going to give, let's
introduce some variables.
N is the total number
of edges on each side,
and K is the number of
blue edges on each side.
And N minus K is the
number of red edges.
Now start with a
filled in puzzle.
We're going to contract it
in a very particular way.
Shrink each red triangle,
maintaining its ratio
so it stays an
equilateral triangle,
but shrinking the edge lengths.
Simultaneously,
shrink the lengths
of the red edges on
the parallelograms.
Don't change the lengths
of the blue edges.
Eventually, after the edge
lengths of the red triangles
go to zero, we are only
left with blue triangles.
Notice that during and after
this shrinking process,
the entire puzzle
maintains its shape
as an equilateral triangle.
How does this prove
conjecture number one?
After the shrinking
process, the puzzle
clearly has the same number
of blue edges on each side.
Moreover, the number
of edges on each side
was preserved during
the contraction process.
So each side of
the original puzzle
must also have had the same
number of blue edges, which
we're designating
with a variable K.
Since every side has N
edges and K blue edges,
they must each have
N minus K red edges.
What about conjecture
number two?
To prove that, we're going
to use triangle area.
Normally, we think of a square
like this as having area 1,
but now let's think of a
single triangle in our puzzle
as having area 1.
The puzzle has N
edges along each side,
and so it will have N
squared triangle units.
That is, if you filled
it with only triangles
and no parallelograms,
it would contain exactly
N squared triangles.
After contracting
the puzzle, it's
made entirely of blue
triangles and there
are K edges on each side.
So it is K squared
triangle units.
It contains K squared
blue triangles.
The number of blue
triangles didn't
change during the
contraction process
so the original
puzzle also contained
K squared blue triangles.
Doing the contraction
process in reverse,
shrinking the blue
edges, shows that there
must be N minus K
squared red triangles.
Challenge problem--
using this information,
how many parallelograms
are in each puzzle?
So let's say I give you a blank
puzzle with one side colored,
like this.
You know a lot about
the filled in puzzle.
It has two blue and three
red edges on each side.
It contains four blue
triangles, nine red triangles,
and six parallelograms.
That's a lot of information
from just one boundary,
but how does this help us
with Schubert's problems?
We want a way to rigorously
compute intersections of lines.
Intuitively, there's
two basic types
of lines that intersect
both L1 and L2.
First, any line that
goes through their point
of intersection, like
this or this, works.
Now notice that any two
lines intersecting at a point
are contained within exactly
one two-dimensional plane
like this.
You can prove this with
geometry, or linear algebra,
or just observe it
in the real world.
Take two crossing sticks.
There's one way to lay a
piece of paper flat over them.
The second type of
line that intersects
both L1 and L2 are the lines
contained within this plane,
like this or this.
But to solve this
more rigorously,
we'll describe different sets
of lines using a template.
The collection of all lines
that touch a fixed blank
and are contained
within a blank.
And fill each blank in with the
smallest one of the following--
0D 1D line, 2D
plane, or 3D space.
We'll encode the template in a
string of two 0's and two 1's,
where the 1's indicate how
you fill in the blanks.
That's a lot to follow, but
let's see some examples.
This is the collection of all
lines that touch a fixed line
and are contained
within 3D space--
so it's 0 1 0 1.
If all the lines were
contained in a plane,
like this, we'd move
the second one forward--
so it's 0 1 1 0.
This is the collection of all
lines that touch a fixed point
and are contained
within a plane--
so it's 1 0 1 0.
What is 1 1 0 0 indicate?
It's all the lines that
touch a fixed point
and are contained within a
line, but there's only one line
contained within a line.
So the collection has size 1.
We want to compute lines that
are in multiple collections.
For example, what's the
overlap between all the lines
through this point and all
the lines through this point?
Well, remember
from geometry class
that there's only one
line between two points.
So we can write that as 1001
times 1001 is equal to 1100.
The collection of lines through
this point and this point
are just one line, but these
problems can get much trickier.
This is where the
puzzles come in handy.
We'll use the same
example from before.
We read each side of the
puzzle from left to right.
Write the code for the first
condition, intersecting L1,
on the left side, and the
code for the second condition,
intersecting L2,
on the right side.
Then start to fill
out the puzzle.
There's two ways to do
it, like this or this.
The bottom of the puzzle
is either 1 0 0 1,
all the lines through a
particular point, or 0 1 1 0,
the lines contained in a plane.
This corresponds precisely
to our geometric picture.
This puzzle, invented by
mathematicians Alan Knutson
and Terence Tao, is
computing the result
of imposing multiple conditions.
We can write this algebraically
0101 times 0101 is equal
to 1001 plus 0110, where
essentially multiplication
means "and" and
addition means "or."
That is, the lines which
intersect a fixed line
and intersect another
fixed line are either
the lines that intersect
a point or the lines
contained in a plane.
How many lines intersect
all four of these lines?
That means we want to compute
0101 times 0101 times 0101
times 0101, which computes the
overlap between all the lines
that intersect each
of these four lines.
Using the last computation,
we can simplify it to this.
Just like normal addition
and multiplication,
we can distribute.
And any time we need to solve
a multiplication problem,
we just use the
puzzle like this.
Then we repeat that
process until finally,
we get to 2 times 1100--
two lines.
We can actually see
this geometrically.
One line goes between
the two crossing points,
and the other line
is the intersection
of the two planes containing
the pairs of lines,
which brings me to a quick
caveat, a technical detail.
Most of today's
episode is actually
about complex projective
space, but to simplify matters,
we fudged it by
working in real space.
That means I was
actually using line
in three dimensions
as a stand-in
for two-dimensional subspace and
four-dimensional complex space.
Speaking of higher
dimensions, a puzzle
with N edges on each
side, K of which are ones,
gives us information about
the K-dimensional subspaces
within N-dimensional space.
That's part of what's amazing--
these puzzles work even
when we can't literally
envision the geometry in
higher dimensions and far more
abstract situations.
See you next time
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during the signup process.
You all had a ton to say
about the worrisome lack
of a foundation for mathematics.
Alan asked, "Can pure logicists
prove almost everything
using all but the two
axioms they don't like?
And then maybe they should
be unsure of the validity
of whatever is remaining if
that's the logicist position."
That's a good question.
Unfortunately, some of
the non-logical axioms
are really necessary to build
all of modern mathematics.
For example, the
axiom of infinity
gives the existence of
the natural numbers.
Certainly, one could study
the remaining axioms,
but the logicists really wanted
to find a logical foundation
for all of modern mathematics.
Goldfish asked a wonderfully
detailed and curious question
about Dedekind cuts.
It's all fun to read, but
here's my favorite part.
"Things get even more
confusing in my head
when I bring to the table the
fact that rational numbers are
countable.
If rational numbers
are countable
and every irrational
number can be defined
as a gap between two
rational numbers,
then it seems to me
that there must be a way
to map every irrational
number to a natural number
via its corresponding
rational number, right?"
I find this
countable/uncountable thing
sort of amazing too.
I'm with you, Goldfish.
The real number line is weird,
but there are a few responses
to your question.
Here's one-- a Dedekind cut is
a gap in the rational numbers,
but not a gap between two
rational numbers, which
is a weird distinction.
A Dedekind cut is
simply a division
of the rationals
into two sets, where
all the elements in one
set are bigger than all
the elements in another set.
Using the square root
of 2 as an example,
there is no rational
number immediately above
or below the square root of 2.
For any rational
number you select
that's bigger than
the square root of 2,
there will always be some
rational number smaller
than that, but bigger
than the square root of 2.
But there's another
answer to your question
that might be related
to rngwrld's question.
"How do you get all
transcendental numbers
from Dedekind cuts?
It would seem it would
only allow you to construct
algebraic numbers?"
Algebraic numbers are
the roots of polynomials
with rational coefficients,
and transcendental numbers
are everything else.
Dedekind cuts are
all possible ways
to split the rational
numbers into two sets.
And many-- actually,
most of them--
aren't describable in the
way the square root of 2 was.
Those are the
transcendental ones.
Finally, Steven recommended
this wonderful graphic novel
about the history of
logic and the search
for math's foundation
called "Logicomix."
I can't believe I
forgot to mention it.
The book is so good,
and I'm glad you did.
Thanks, Steven.
