# Concrete example of non-abelian class field theory - why Langlands program *is* a non-abelian class field theory?

Abelian class field theory generalizes quadratic reciprocity laws for general number fields with abelian Galois groups, which connects class groups and Galois groups via Artin's reciprocity map. Also, quadratic reciprocity gives us some explicit examples of simple criterions to determine whether a rational prime $$p$$ splits or inerts in a given quadratic field. (For example, $$p$$ splits in $$\mathbb{Q}(\sqrt{-1})$$ if $$p$$ is a form of $$4k +1$$.)

I also heard from a lot of people that Langlands program can be thought as a non-abelian version of class field theory, and the abelian class field theory is just 1-dimensional Langlands program (Langlands over $$\mathrm{GL}_1$$, see here for details). Also, Langlands over $$\mathrm{GL}_2$$ is about theory of modular forms (and of course Maass forms), Elliptic curves, 2-dimensional Galois representations, etc.

However, I could not find an actual example that Langlands program is the non-abelian class field theory in the way of giving a criterions for splitting primes in a number field with non-abelian Galois groups. For example, a splitting field of a random cubic polynomial over $$\mathbb{Q}$$ (let's say, $$f(x) = x^3 -x -1$$) might give a field with Galois group $$S_3$$ (if its discriminant is not a square). and it seems very hard to describe the splitting behavior of a rational prime $$p$$ in that field. What I (want to) believe is that such criterion might encoded in Fourier coefficients of a certain modular form (or automorphic forms in general). Is there any document or paper that I can find such an example: a polynomial over $$\mathbb{Q}$$, its splitting field, and corresponding modular forms? Thanks in advance.

## 2 Answers

Actually the exact cubic polynomial $$f(x) = x^3 - x - 1$$ you named is the subject of an old MO answer of Matthew Emerton's. Its splitting behavior is described by a Hecke eigenform of weight $$1$$ and level $$23$$ (the absolute value of the discriminant of $$x^3 - x - 1$$) which has a product formula

$$A(q) = q \prod_{n=1}^{\infty} (1 - q^n)(1 - q^{23n}).$$

The coefficient $$a_p$$ of $$q^p$$, for $$p$$ a prime $$\neq 23$$, is the trace of the Frobenius element at $$p$$ in the Galois group $$S_3$$ acting on the unique $$2$$-dimensional irreducible representation of $$S_3$$ (which corresponds to the Galois representation corresponding to the modular form above), which means

• $$f(x)$$ splits into linear factors $$\bmod p$$ iff the Frobenius element is the identity iff $$a_p = 2$$,
• $$f(x)$$ splits into a linear and a quadratic factor $$\bmod p$$ iff the Frobenius element is a $$2$$-cycle iff $$a_p = 0$$, and
• $$f(x)$$ is irreducible $$\bmod p$$ iff the Frobenius element is a $$3$$-cycle iff $$a_p = -1$$.

For $$p < 23$$ the coefficients are the same as the coefficients of $$q \prod_{n=1}^{\infty} (1 - q^n)$$ which is $$q$$ times the Euler function, whose coefficients are given by the pentagonal number theorem. This gives that the $$q$$-expansion of $$A$$ begins

$$A(q) = q - q^2 - q^3 + q^6 + q^8 - q^{13} - q^{16} + \dots$$

hence

• $$a_2 = -1$$, meaning $$x^3 - x - 1 \bmod 2$$ is irreducible (which is true since it has no roots),
• $$a_3 = -1$$, meaning $$x^3 - x - 1 \bmod 3$$ is irreducible (which is true since it's a nontrivial Artin-Schreier polynomial)
• $$a_5 = 0$$, meaning $$x^3 - x - 1 \bmod 5$$ splits into a linear and a quadratic factor (given by $$(x - 2)(x^2 + 2x - 2)$$)
• $$a_7 = 0$$, meaning $$x^3 - x - 1 \bmod 7$$ splits into a linear and a quadratic factor (given by $$(x + 2)(x^2 - 2x + 3)$$)

and so forth. Apparently the smallest split prime is $$p = 59$$.

This MO question might also be relevant.

• Japanese mathematician Michio Kuga was at Stony Brook when I was there. en.wikipedia.org/wiki/Michio_Kuga He made two observations to his graduate student (Susan Addington): the first was "In Japanese movies...many people die." Later he said "Many Japanese mathematicians tried to create non-Abelian class field theory. Many people die." – Will Jagy Oct 3 '20 at 1:45
• It is quite interesting that the coefficients of $A(q)$ is 2, 0, or -1. Thank you for the examples! – Seewoo Lee Oct 3 '20 at 2:19
• @Seewoo: we're only guaranteed that about the prime coefficients. I don't know about the other ones, although they can be expressed in terms of the pentagonal numbers. – Qiaochu Yuan Oct 3 '20 at 2:30
• @QiaochuYuan Oh yes you're right. I was confused. – Seewoo Lee Oct 3 '20 at 2:43
• @SeewooLee Have you looked at this article of Weinstein? ams.org/journals/bull/2016-53-01/S0273-0979-2015-01515-6/… – Alex Youcis Oct 6 '20 at 23:23

Shimura's article "A reciprocity law in non-solvable extensions" may be an example what you are looking for.

• Thank you! This is what I was looking for. – Seewoo Lee Oct 3 '20 at 2:19