Math(s)! Currently: pi is the minimum value of pi

Discussion in 'General Chatter' started by Exohedron, Mar 14, 2015.

1. ExohedronDoesn't like words

As you might be able to tell, I am slowly losing the battle against my desire to talk about algebraic topology and homological algebra. Which is bad, because algebraic topology is not one of those subjects that I'm good at.

2. ExohedronDoesn't like words

Okay, so orbitals.

If you've taken a chemistry class or cracked open a chemistry textbook, you might have seen those pictures of electron orbitals, with the p and the s and the d and the f and maybe some other ones. And they look like little lobes or spheres or whatnot, usually brightly colored, and they're supposed to be "instead" of the Bohr model where the electrons orbited the nucleus in nice circular orbits.

So what does it mean when we say that electron orbits look like those pictures?

Firstly, what can we say about electrons?

Suppose we have a single proton, a single point of positive electric charge. We get an electric field that extends outward radially, completely symmetrically if the proton isn't moving and there isn't anything else relevant around it.
So if we have a physically-feasible description of an electron orbiting that proton, we should be able to rotate that description around the proton and get another physically-feasible description.
From a quantum mechanical standpoint, a physical description consists of a function from points in 3-space to the complex numbers called the wavefunction ψ. So if we want to draw a picture of the orbit, we need to figure out what kind of wavefunctions we're looking at.

We want to talk about the angular momentum of the electron. So what is angular momentum in quantum mechanics? Well, the angular momentum operators are
Lx = -i(y(∂/∂z) - z(∂/∂y))
Ly = -i(z(∂/∂x) - x(∂/∂z))
Lz = -i(x(∂/∂y) - y(∂/∂x))
and we actually want to look at
L2 = Lx2+Ly2+Lz2
since we want a real number, not a vector, to get the squared orbital angular momentum.
So we get a big thing with some derivatives and second derivatives that will take a function and spit out a function, and the functions that correspond to electrons with well-defined orbital angular momentum, i.e. the eigenfunctions, end up being the harmonic homogeneous polynomials, with squared total angular momentum being d(d+1) where d is the degree of the polynomial.

How many homogeneous polynomials of a given degree are there? Well, lots. But how many independent harmonic homogeneous polynomials of degree d are there?
Well, let's see. The number of independent polynomials of degree d in three variables is
(d+2)(d+1)/2
(check this!)
But you'll only want the harmonic polynomials. Recalling a previous post, every polynomial is a sum of terms of the form Ck times a harmonic piece, where C has degree 2, and so stuff from degree d-2 contributes non-harmonic parts to polynomials of degree d. So we remove them:
(d+2)(d+1)/2 - (d-2+2)(d-2+1)/2 = 2d + 1
So in degree d, there are 2d+1 independent homogeneous harmonic polynomials. So for instance, in degree 2 we have
xy, xz, yz, x2-y2, 2z2 - x2 - y2
as a set of independent homogeneous harmonic polynomials.

Unfortunately, a polynomial function P can't be a wavefunction, not directly. Instead, we change coordinates so that P is written in terms of the radius r and angular coordinates φ and θ. P then decomposes into R(r)Y(φ, θ), and we can then replace R(r) with another function of r to get a genuine wavefunction; the exact replacement depends on what kind of physics we're trying to model and isn't really important. All the fun stuff is contained in Y, and in particular indicate very non-circular "orbits".
The functions Y are called spherical harmonics, because we can compute them by taking a harmonic function and pretending that it only matters on the unit sphere.

The point is that you can translate between homogeneous harmonic polynomials and wavefunctions corresponding to well-defined squared total angular momentum. So the same statement that there are 2d+1 independent homogeneous harmonic polynomials of degree d gives that there are 2d+1 independent wavefunctions with squared orbital angular momentum d(d+1).
We can further ask for wavefunctions with well-defined angular momentum in the z direction, i.e. the eigenfunctions of Lz, and we'll get particular functions; often those particular wavefunctions are the ones that get pictured in the textbooks, even though actual electrons don't care which direction is the z direction unless you ask them nicely.

3. ExohedronDoesn't like words

7 may be nonstandard, since it takes at least 7 symbols to define 7 in Peano arithmetic.

4. ExohedronDoesn't like words

The Permut0hedra and the Associahedra

Suppose you have an object A. There is one way to order the list {A}, i.e. as A. Draw a single point.
Suppose you have two objects, A and B. There are two ways to order the list {A, B}, i.e. as AB and BA. Draw a point for each ordering, and connect them. You end up with a line segment.
Suppose you have three objects, A, B and C. There are six ways to order the list {A, B, C}, i.e. as ABC, ACB, etc. Draw a point for each ordering, and connect a pair of points if the corresponding orderings differ by swapping two adjacent elements of the ordering, i.e. ABC connects to ACB and to BAC, but not to anything else,. You end up with a hexagon.
Suppose you have four objects, A, B, C and D. There are twenty-four ways to order the list {A, B, C, D}. Draw a point for each ordering, and connect a pair of points if they differ by swapping two adjacent elements of the ordering, and fill in a. You end up with a shape that's usually called a truncated octahedron. Note that some of the faces of this shape are square and some are hexagonal. The squares represent swaps that are disjoint, i.e. a swap of the first two objects and a swap of the last two objects. We can go from ABCD to BADC either by stopping first at BACD or ABDC, without having to go through any other steps.
Etc.
Suppose you have n objects. There are n! ways to order the list of objects. Draw a point for each ordering, and connect the pairs whose orderings differ by a swap of two adjacent elements. If we fill in the surfaces and volumes and so on, we get a shape called the permutohedron on n objects. It tracks how one gets from one ordering of n objects to another ordering.
If a k-dimensional face can be written as the Cartesian product of an m-dimensional face and a (k-m)-dimensional face, then it indicates that the relevant set of swaps for the edges of the k-dimensional face can be decomposed as swaps on a set of m+1 adjacent objects and a set of k-m+1 adjacent objects that's disjoint from the first set. So the shape of the faces tells us about disjoint suborderings.
You can build a permutohedron on n objects in n-dimensional space by looking at all of the points whose coordinates are orderings of the first n positive integers and then connecting the appropriate edges.
Honestly, though, I find this guy a little boring.

Consider an ordered list of objects that belong to a set with a multiplication operation. We don't ask for the multiplication to be associative, so we need parentheses.
If we have one object A, no big deal, our parenthesized list is just A. Draw a single point.
Similarly, if we have two objects, our parenthesized list is just AB. Draw a single point.
If we have three objects, we have two ways to put in parentheses: (AB)C and A(BC). We draw a point for each, and connect them by a line.
If we have four objects, we have five ways to put in parentheses: ((AB)C)D, (A(BC))D, (AB)(CD), A((BC)D), and A(B(CD)). Draw a point for each, and connect two points if you can get from one parenthesization to the other by a move that looks like (XY)Z -> X(YZ), i.e. the associativity rule, where X, Y and Z can themselves be parenthesizations of lists.
So we would connect ((AB)C)D to (A(BC))D and to (AB)(CD), but not to anything else.
We end up with a pentagon; in particular, there are two ways to get from ((AB)C)D to A(B(CD)):
((AB)C)D -> (AB)(CD) -> A(B(CD))
((AB)C)D -> (A(BC))D -> A((BC)D) -> A(B(CD))
If we have five objects, we have fourteen ways to put in parentheses. I'll let you write them out. We end up with a shape with six five-sided faces and three four-sided ones; there isn't a good way to embed this guy into 3-dimensional Euclidean space that has all the faces regular.
If we have n objects in our list, we end up with what's called the associahedron on n objects. This is also called a Stasheff polytope, after the guy who (re)discovered them.
I like the associahedra because they're a little weirder than the permutohedra. For instance, the number of vertices in the associahedron on n objects is given by what is called the (n-1)-st Catalan number, given by
Cn-1 = (2n-2 choose n-1)/n
The combinatorics of other parts of the associahedra is similarly more complicated than the corresponding parts of the permutohedra.
As noted above, you can't usually embed the associahedra in Euclidean space with the faces being regular-ish.

I mostly approach these guys from a (higher)-algebraic standpoint, where we say "this product is commutative up to equivalence" or "this product is associative up to equivalence" and then the associated -hedra tell us how loose that notion of "up to equivalence" is.
Suppose that AB = BA, i.e. we can shrink our 2-object permutohedron to a single point. Then for all higher permutohedra, we shrink their edge to points and we get a single point, i.e. the ordering doesn't matter, our product is commutative.
Now suppose that while AB is not equal to BA, there is an map fA,B: AB -> BA, and then the permutohedron on three objects measures the difference between
fB,CIA◦IBfA,C◦fABIC
and
ICfA, B◦fA, CIB◦IAfB,C
where IX means the identity on X; these are the two ways of going around the hexagon from ABC to CBA. Does your algebraic structure care about the difference between those two paths?
So for instance, in our usual notion of a noncommutative multiplication, the permutohedron on 3 objects always yields the identity map, i.e. that the difference between ABC and CBA doesn't depend on how we get from ABC to CBA. This corresponds to flattening the hexagon to a single line going from ABC to CBA. Taking the permutohedron on 4 objects and flattening the A,B,C hexagons to single lines going from ABCD to CBAD and from DABC to DCBA, and flattening the A,B,D hexagons to single lines going from ABDC to DBAC and from CABD to CDBA, and so on. You get a funny-looking shape that's basically just a single line from ABCD to DCBA, with some lobes attached that you can ignore. So you get that there's only one way to get from ABCD to DCBA, i.e. no path dependence for four object products. And similarly no path dependence on for more objects.
But what if there was a path dependence for three objects, but no path dependence for four objects? Then again we get no path dependence for five objects and so on.
These are called "coherence" theorems, that if your -hedra on n objects all collapse to single lines, then so do all of your -hedra on >n objects. These kind of path-dependent algebraic properties shows up in topology and higher category theory and string theory and basically a lot of places that are trying to do algebra in a "higher-dimensional" setting. For me, personally, I focus on the analogues of Lie algebras, where the Jacobi identity doesn't hold exactly, and the discrepancy is controlled by various extra structures that can be integrated to get analogues of Lie groups; these are the actual automorphism objects of string theories, not Lie groups.

Since it might come up: no, as far as I know, "exohedron" doesn't actually refer to anything other than me.

Last edited: Jun 30, 2019
5. ExohedronDoesn't like words

A formal power series is like a polynomial with infinite degree. In calculus class you often care about where and when a power series converges, but for formal power series you don't worry about convergence.

Consider formal power series of the following type:

C(x) = x + c1x2 + c2x3 + ...

You can consider what happens when you plug one such formal power series into another. If C(x) and D(x) both have the above form, then their composition C(D(x)) also has the above form.

You can also ask about the inverse with respect to composition: given C(x), find D(x) such that

C(D(x)) = x

How do you relate the coefficients of C and D?

d1 = -c1
d2 = -c2 + 2c12
d3 = -c3 + 5c2c1 - 5c13
d4 = -c4 + 6c3c1 + 3c22 - 21c2c12 + 14c14

Where do these numbers come from? Well, for a given coefficient dk, the terms on the right side all have total index k, where we consider c1m to give total index m. Also the sign of a term is given by how many factors there are in the term. But what about the number itself?

Consider A1 to be a point, A2 to be a line, A3 to be a pentagon, and in general, Ak to be the associahedron on k+1 objects.
Then A1 is a point and has one point, i.e. one A1.
A2 is a line segment, and thus has one line, i.e. A2, and two points; we view the points as A12 to get the correct total index.
A3 is a pentagon, and thus has one pentagon A3, five lines which we view as A2A1, and five points which we view as A13, again multiplying by points to get the correct total index.
A4 has one A3, six pentagonal faces A3A1, three square faces A22, twenty one lines A2A12, and fourteen points A14.

So you can see a match between the coefficients in D and the pieces of the associahedra.

dk = Σk1 + ... + kl = k (#of pieces of Ak of the form Ak1Ak2...Akl)(-ck1)...(-ckl)

Last edited: Jun 29, 2019
6. ExohedronDoesn't like words

Not exactly math but I figure that a bunch of people here would appreciate twocubes trying to make manga in LaTeX.

Also twocubes has some interesting discussions about how to cut an infinite deck of cards, but that's somewhat less hilarious.

8. ExohedronDoesn't like words

Oh, sure, digressions are super allowed! I mean, the only reason it looks like I'm actually running this thread is that I get to change the name and I make the majority of the posts, but that's because I like...whatever is the text equivalent of the sound of my own voice.

9. ExohedronDoesn't like words

Today a bunch of coworkers and I were joking about using your fingers to count in binary and why we didn't teach that to kids, and I mentioned that kids naturally count in base 1 rather than base 2. One of the group expressed shock at the notion of base 1.
I'm going to have fun presenting next week.

10. evilasCustom title cause all the cool kids are doing it

How does... Can you represent 0 in base 1?
Can you represent decimals?
I mean, yes I know you're talking about kids but I imagine it's a properly defined base, right? Basically tallies?
Like. I can see negative numbers and fractions but. Is it ever used for stuff that isn't natural numbers?

Last edited: Sep 8, 2019
11. ExohedronDoesn't like words

No, it''s not really a well-developed base all. It's just tallies.
The issues are probably related to the fact that the field with 1 element doesn't exist.

Last edited: Sep 8, 2019
• Informative x 1
12. TheSeer37 Bright Visionary Crushes The Doubtful

Yeah, having a place value system at all is a fairly sophisticated idea. A young beginner's number system is just a one-to-one correspondence between the group to be counted and an ordered list of words. "Ten" has no special significance, and is just the number after "nine."

I suppose if you think of the names of place values in base ten as the same sort of ordered list, you could roughly describe the basic beginner's number system as "base 1". But the only real utility of the term is to troll other nerds.

13. ExohedronDoesn't like words

Speaking of fun stuff with fields, I was reading about the q-analog of the Moebius inversion formula when I was struck with the thought of "why don't they just call it inqlusion-exqlusion?"

14. ExohedronDoesn't like words

Spent some time trying to prove from Euclid's axioms that a line that intersects a circle at one point without being tangent there must intersect the circle again at a different point. Using algebra is cheating.

I've managed to prove that a line segment that intersects a triangle transversely at some point that isn't a vertex must intersect the triangle again. Circles are harder.

Last edited: Sep 11, 2019 at 10:22 PM