The Mansions of Science

Visual Intuition for Differential Forms in 1.5 Minutes

Look at the number 6. If you only consider the positive integers, then the prime factorization is 2 and 3. That’s it. A unique prime factorization.

Now look at all integers. Now the factorization could be 2 and 3 or -2 and -3. Now suppose you look at all numbers of the form (a + b*sqrt(-5)) where a and b are integers. You can prove that these numbers form a closed set. In fact, it forms the Z[sqrt(-5)] ring. Suddenly, (1-sqrt(-5)) and (1+sqrt(-5)) multiply to 6. Both of these number cannot be broken down any further, so they are prime. 6 no longer has a unique prime factorization. If you set b = 0, you’ll notice that 2 and 3 are also in this ring, so the original unique prime factorization is also here.

So imagine you had an elliptic curve. Any elliptic curve. Just a polynomial. Now, you have a function that takes in the elliptic curve and a complex number s. Let’s call this function L. If you plug in the elliptic curve and s = 0 or 1 or 2 or 3, you would get a number.

Now, if you plug in s = 1, you could get 0 or a nonzero number. If you plug in s = 1 and you get 0, you can take the derivative of L and then plug in s = 1 again to L prime. If you get 0 again, you could take the derivative of L prime and get L prime prime. Plug in s = 1 again and see if you still get 0. Do this until you reach a derivative that outputs a nonzero number at s = 1. The number of times you differentiated until you got a nonzero number at s = 1, this number of times is called the order of vanishing.

The Birch Swinnerton Dyer conjecture states that if you have an elliptic curve and this function L, if you plug in the elliptic curve and s = 1 to L and its derivatives, then the order of vanishing, meaning the very first time you obtain a nonzero number, that order number is the rank of the elliptic curve.

So imagine you had an elliptic curve. Literally just a polynomial of 2 variables. One variable has max degree 2 and another has max degree 3.

If you constrain the coefficients of the variables to come only from Q, meaning you only have rational coefficients, an interesting question to ask is: How many rational points exist on the curve? Remember, if you plot a 2 variable polynomial, you get a curve.

Now, are there finitely many rational points or infinitely many rational points? That’s a good question. And suppose you knew one or two or three or a few of these rational points, can you get to every other rational point? For example, if I knew there were an infinite number of rational points on a given elliptic curve, and I knew 3 of them, can I apply some kind of operation to these 3 and get to every other rational point? If yes, then we say that the elliptic curve is of rank 3. It has 3 generators. 3 independent x,y pairs that allow me to get to every other rational point.

A single pair, a single generator, and all the further rational pairs it generates, form a group. This group is isomorphic to Z. All of the pairs, all of the generators, are isomorphic to Z^r where r is the number of generators.

There are also torsion points, which are a few rational pairs that cannot be reached using the generators. These torsion points make up a finite set, and we will talk about them at another time.

Imagine you had a polynomial of two variables. Basically just x and y. They can be multiplied by each other, exponentiated to any degree, added together, so on. If you moved everything to one side, you would have f(x,y) = 0. For example, 225(x^5)(y^3) + 5y^2 + 19x + 800 = 0.

Now, when we have one equation and two variables, this forms a curve. A continuous curve. And there are infinitely many solutions since there are more variables than equations. The solutions are just the (x,y) pairs on the line, the points of the line. Some of the points happen to be rational. Meaning both x and y are rational numbers. Can be written as fractions.

An interesting question is: Are there infinitely many rational pairs?

Now, these curves don’t have to pass the vertical line test. For example, you could try plotting y^3 = x^2 + y. This curve has a part that looks like a hole. A literal hole. A hole is called a genus.

What Faltings proved was that if you have a curve of genus 2 or more, and you also constrain the coefficients of the polynomials to only rational numbers, so basically the 225, 5, 19 and 800 in our first example, it will only have finitely many rational points.

There are 2 major differences between a variety and a manifold.

1. A manifold looks like R^n if you keep zooming in. A variety can have corners, cusps, crossings (X marks), etc. You cannot zoom in and get rid of them. If a variety does not have them we call it a smooth variety. All smooth varieties are manifolds.

2. An entire variety is described by a single polynomial. This can sometimes apply to a manifold, but often not the case. Maybe only some patches can be described by polynomials. But, in any case, often an entire manifold cannot be described by a single polynomial.

Next we will discuss the Picard Variety.

Automorphic forms are functions. Functions have inputs. An automorphic form is a function that is scaled when you transform the inputs using a group G. Elements of a group G, that is. Usually a discrete group. Usually a subgroup of GL_n.

A first intro to automorphic forms:

All modular functions are modular forms. Modular functions are simply modular forms of weight 0. Modular forms have symmetry group SL(2,Z), also known as “the” Modular Group. The ambient group is SL(2,R).

Automorphic forms generalize modular forms. All modular form are automorphic forms but not all automorphic forms are modular forms. A modular form is a type of automorphic form, the same way a modular function is a type of modular form.

An automorphic form has as ambient group a reductive group, often reductive Lie group but DOES NOT NEED TO BE. The elements of this reductive group can always be represented as matrices in automorphic form theory. All classical reductive Lie groups used in automorphic forms are linear, so they have faithful finite-dimensional matrix representations. The entries of the matrices come from a field or ring. Now, these entries can come from Z, or Q, or R, or C. Now we know that Z is in Q and Q is in R and R is in C. The biggest ring that the matrix elements come from is called the Adele ring, which contains all real numbers and p-adic number systems. This Adele ring is called A(Q), which combines R and all p-adic completions of Q. The symmetry group is a discrete subgroup of the ambient reductive group. Often the entries come from the field or ring of integers Z. Since this symmetry group is discrete, it is not a Lie group, since Lie groups are smooth.

*The most general ring for matrix entries is the adele ring A(Q) .

*Automorphic forms can be formulated adelically, with matrices having entries in the adele ring A_Q​, which combines R and all p-adic completions of Q.

So in studying polynomials, we can increase the number of variables and/or the number of degrees.

For 1 variable, x, and any number of degrees we have ax^n + bx^n-1 + … + zx^0 = 0. The solutions are literally just a set of discrete points.

Now for two variables we have to tone it down. If y and x are both of degree 1, these are curves, but just straight curves, straight lines. If

(x,1) = One solution

(x,n) = Discrete points

(x,1,y,1) = Straight line (x,2,y,1) = Quadratic (x,1,y,2) = Quadratic (x,2,y,2) = Conic section

(x,1,y,3) = Cubic (x,2,y,3) = Elliptic curve (x,3,y,1) = Cubic (x,3,y,2) = Elliptic curve (x,3,y,3) = Fermat curve with no group law, cannot add points to find other points

(x,y,z) and beyond are surfaces.

The reason surfaces are not interesting is due to Faltings’ theorem, which I will talk about next.

For 2 vectors in the Z^2 module to be a basis, the parallelogram they make must have area 1.

Suppose I have vectors (a,b) and (c,d) in Z^2, meaning a, b, c and d are all integers. For them to be a basis of Z^2, the equation x(a,b) + y(c,d) = (g,h) must be satisfied. In this equation, (g,h) are all pairs of integers in existence, from negative infinity to infinity. Now, these are two equations with 2 unknowns, so there are solutions no matter how we vary a, b, c and d. But, as we shall see, if a, b, c and d are all integers, then a condition emerges if we demand that x and y are also integers. We demand x and y to be integers because linear combinations in Z^2 can only use scalars from the ring Z. That is why we have a module and not a vector space.

We can rewrite the equation with (a c) on top and (b d) on the bottom of a square matrix, multiplied by the vector (x y) equals a vector (g h). If we left multiply both sides by the inverse of our matrix, which is (d -c) on the top and (-b a) on the bottom divided by the determinant of our original matrix, which is ad - bc, we get that x = (dg - ch)/(ad - bc) and y = (-bg + ah)/(ad - bc). Since g and h can be any integer, we have grounds for believing that the numerators are very varied. If we fix a, b, c and d, and ad-bc, a fixed number, manages to divide a huge number of numerators and result in integers x and y, it must be because ad-bc is a divisor of everybody. The only such number we know of is, 1.

A polynomial is guaranteed to be defined on R^1. A polynomial divided by a polynomial is called a rational function. 

A holomorphic function is guaranteed to be defined on some subset of C^1. It is complex-differentiable at every point in its domain and is perfectly smooth. No spikes, tears, gaps, etc. A holomorphic function divided by a holomorphic function is called a meromorphic function. So polynomials are to rational functions as holomorphic functions are to meromorphic functions. A meromorphic function is very similar to a holomorphic function except it is not defined at points we can poles. A meromorphic function locally can be written as a fraction of holomorphic functions.

In complex analysis, we say that, locally, meromorphic functions are the fraction field of the ring of holomorphic functions. Like how Q is the field of fractions of Z.

This video builds on the fiber bundle video.

Suppose you start with a topological space B, called the base space. Could be S1, S2, R1, R2 or R3. Now attach at every point in the space a group G. This group is the fiber. Now you have a set of fibers, the same fibers, at every point in the space. This is a fiber bundle.

If you then take the group G and act on the fiber bundle, meaning you take the group G and act on every single fiber, this would mean you are acting on a group using itself. You do this for every point of B, so the number of times you do this is equal to the number of elements in B. If G happens to act freely on itself, which all groups do, then this fiber bundle is called a principal G-bundle.

So suppose G = Z20, its action is like a 20 hour clock. 20 equally spaced rotations, or 19 if you don’t consider the identity element. Ok imagine you have S1 and you attach Z20 at every point. Z20 happens to act on Z20, itself, freely. G = Z20 acting on the set X = Z20 simply rotates X = Z20. Ok, now, if you act on this fiber bundle, also known as the product space S1 x Z20, you would basically be going to every single Z20 copy and rotating it. Imagine a donut (torus) or a slinky, except instead of a circular ring at every point you have 20 points that are roughly in the shape of a ring. So you walk around the donut and at every Z20 copy, which is a 20 point circle-like, ring-like object, you rotate it. This is the action of a group G on a principal G-bundle.