This post was originally entitled “Langlands and arithmetic”. But a friend has told me that the best easy explanation of the Langlands program is in a talk by Ed Frenkel entitled “Love and math”.

So here it is. My original post is below, under the picture of Robert Langlands. Frenkel has also written a book of the same title.

https://podcasts.ox.ac.uk/embed/5b81029c98810f045901

On 23 November I went to a lecture in Cambridge by Jack Thorne on “Langlands and arithmetic”. “Langlands” refers to the “Langlands program”, a vast web of mathematical guesses about connections between apparently distant areas of maths made by Robert Langlands (pictured above as he was in 1967, when he first formulated those guesses).

Some of the guesses have been proved; most of them remain unproved, with many mathematicians working on them from different angles.

Jack looked at the subject from the angle of number theory, i.e. the mathematics of the ordinary whole numbers … −3, −2, −1, 0, 1, 2, 3…

He started with Mordell’s equation

Y^{2} = X^{3} + a

The question is, what solutions (in whole numbers) for X and Y does this have for different values of a?

For example, for a=−2, it has only one solution. To prove even that, we have to go by way of complex numbers, or at least a particular sort of complex numbers.

Just as we can extend the real numbers ℝ to the complex numbers ℂ by adding i (square root of −1) and everything you get from adding and multiplying i and numbers from ℝ, so also we can construct the *Gaussian integers* by taking the whole numbers ℤ and adding i and everything you get from adding and multiplying i and numbers from ℤ. It is all the numbers a+bi where a and b are ordinary whole numbers, and it is called ℤ[i].

In ℤ[i] we can add and multiply pretty much as we can in ℤ. Like in ℤ, we can’t always divide. Another thing true in ℤ[i] as in ℤ is that every number has a unique prime factorisation.

Primes in ℤ[i] look different from primes in ℤ. For example, 2 is not a prime in ℤ[i].

bi, or a, are Gaussian primes (primes in ℤ[i]) when |b|, or a, are ordinary primes and have remainders of 3 when divided by 4 (like 3i, 7i, 11i… and 3, 7, 11…)

If a and bi are both non-zero, then a+bi is a Gaussian prime when (a+bi) multiplied by its conjugate (a−bi), i.e. a^{2}+b^{2}, is prime in ℤ.

However, in ℤ[i] as in ℤ, every number has a unique prime factorisation.

The same can be proved for ℤ[√−2], all the numbers you get by by taking the whole numbers ℤ and adding i and everything you get from adding and multiplying i and numbers from ℤ.

This is not true for ℤ[√−3] or ℤ[√−a] for any bigger a: see http://www.cut-the-knot.org/arithmetic/int_domain4.shtml for all this.

Now let’s go back to Mordell’s equation for a=−2. It is

Y^{2} = X^{3} − 2

or

Y^{2} + 2 = X^{3}

or, if we jump into ℤ[√−2]

(Y+√−2)(Y−√−2)=X^{3}

Every number in ℤ[√−2] can be factorised uniquely (in only one way) into primes. Also, in ℤ[√−2] if a number is prime then its *conjugate* is also prime. Let p be a Gaussian prime dividing (Y+√+2). Say it divides into it n times. Then the conjugate of p divides (Y+√−2) n times. pp* is a prime in ℤ, and (pp*)^{n} divides X^{3}. So n must be a multiple of 3, and so (Y+√−2)(Y−√−2) must both be cubes.

So for some whole numbers m and n

(m+n√−2)^{3} = (Y+√−2)

m^{3} −6mn^{2} + 3m^{2}n√−2 − 2n^{3}√−2 = Y+√−2

so, equating the √−2 bits

3m^{2}n − 2n^{3} = 1

so n=1 and 3m^{2} − 2n^{2} = 3m^{2} − 2 = 1, and so m=±1

Y = m^{3} −6mn^{2} = ±5

So the *only* solution of this equation for a=−2 are (±5)^{2} = 3^{3} − 2.

Since unique factorisation is not true for ℤ[√−3] or ℤ[√−a] for any bigger a, to find out about solutions for bigger a we need different methods. In 1967, before Langlands, Alan Baker proved that there is a finite number of solutions of Mordell’s equation for any a, but then there was no systematic way of finding the actual solutions.

This is where Langlands comes in.

Langlands’ guesses show connections between:

**Elliptic curves**, which are curves (ordinary continuous curves, drawn on graphs with real-number coordinates) of form **y ^{2}=x^{3}+ax+b**. (They are

*not*ellipses; they are called elliptic curves because their mathematics grew out of the study of lengths of arcs on ellipses).

**Modular forms**, which are functions f(z) defined for complex numbers z in the upper half-plane (with imaginary part > 0) which are “very” smooth (differentiable, that is you can differentiate them “in every direction” at every point) and grow as z gets bigger in a particular way defined in terms of matrix theory.

**Algebraic number theory**, the theory of the collection of all the numbers which you can get from adding to ℤ all the numbers which can be got as real or complex roots of polynomial equations in which the coefficient of the highest power is 1 and all the other coefficients are in ℤ.

So they show connections between “continuous” maths and “discrete” maths.

The proved connections enable mathematicians to draw on knowledge about elliptic curves to compile lists of solutions to Mordell’s equation for various values of a. So far 2.3 million solutions, for hundreds of thousands of values of a, have been tabulated.

The proved connections are also strong enough that Jack was able to summarise the proof of Fermat’s last theorem in one PowerPoint slide:

That there is no whole-number solution to X^{n} + Y^{n} = Z^{n} for n>2

This is a proof which Andrew Wiles famously laboured non-stop for seven years to complete, and took 150 pages to write out in 1995, after previous generations of mathematicians had failed to find it despite 358 years of effort.

*Below: Robert Langlands’ covering note to his letter to Andre Weil in 1967 in which he first set out the “Langlands program”*