# Connection between Fourier analysis and Galois theory

Let $$x\ \%\ m$$ be the residue of $$x$$ modulo $$m$$, i.e.

$$x \equiv x\ \%\ m\pmod{m}$$

Let $$\mu^n_m(x)$$ denote multiplication by $$n$$ modulo $$m$$, i.e.

$$\mu^n_m(x) = nx\ \%\ m$$

Consider the Fourier transform of $$\mu^n_m(x)$$, i.e.

$$\tilde{\mu}^n_m( k) = \frac{1}{m}\sum_{x=0}^{m-1}e^{i2\pi k x/m}\mu^n_m(x)$$

From user GH from MO's answer to another MO-question I've learned that

$$\tilde{\mu}^n_m(k) = -\frac{1}{2}-\frac{i}{2}\cot\frac{\pi \bar nk}{m}$$

with $$\bar n$$ the multiplicative inverse of $$n$$ modulo $$m$$.

A suggested analogy between the Fourier transform $$\tilde{\mu}(k)$$ and Riemann's hypothesis seemed too far-fetched, and there's only little evidence that there is a deeper connection:

• "The one has to do with coprimes, the other with primes."

• "Both exhibit some irregularity and quasi-randomness."

• "In both $$\operatorname{Re}(x) = \pm 1/2$$ plays a somehow similar role."

But I wonder if a deeper connection between the Fourier transform $$\tilde{\mu}(k)$$ and the roots $$\zeta(k)$$ of polynomials – especially their Galois groups – may be motivated by the following analogies.

Let $$\zeta_m(k)$$ be the complex roots (indexed by $$k$$) of a monic polynomial $$P_m(x) = x^m + \sum_{k=0}^{m-1}a_kx^k$$ with coefficients from $$[m] = \{0,1,\dots,m-1\}$$.

$$\begin{array}{r|l|l|} & \tilde{\mu}^n_m & \zeta_m \\ \hline \text{as functions} & [m] \rightarrow \mathbb{C} & [m] \rightarrow \mathbb{C} \\ \hline \text{sum} & \sum_{k=0}^{m-1} \tilde{\mu}^n_m( k) = 0 & \sum_{k=0}^{m-1} \zeta_m(k) = -a_{m-1} \\ \hline \text{conjugate pairs} & z \in \tilde{\mu}^n_m \equiv \overline{z} \in \tilde{\mu}^n_m & z \in \zeta_m \equiv \overline{z} \in \zeta_m \\ \hline \text{permutation symmetry among} & \operatorname{Re}(\tilde{\mu}^{n_1}_m) = \operatorname{Re}(\tilde{\mu}^{n_2}_m) & \text{?} \\ \hline \text{possible multiplicities for} & \tilde{\mu}^n_m(k) = 0\ & \text{??} \\ \hline \text{closed form} & \text{depends on } \overline{n} & \text{depends on Galois group} \\ \hline \end{array}$$

There's one major difference: The Fourier coefficients $$\tilde{\mu}(k)$$ are genuinely ordered – and $$\tilde{\mu}(0)$$ plays a special role – while the roots $$\zeta(k)$$ are not – and $$\zeta(0)$$ is not even well defined.

It might be revealing to especially consider monic polynomials

• $$P^n_m(x)$$ with coefficients $$a_k = \mu^n_m(k)$$ for some $$n,m$$ – with roots $$\zeta^n_m(k) \in \mathbb{C}$$

• $$P^n_m(x) \equiv 0 \pmod{m}$$ instead of $$P^n_m(x) = 0$$ – with roots $$\zeta^n_m(k) \in [m]$$

My questions are:

How can the gaps $$?$$ and $$??$$ in the table above be closed in a concise way?

What might knowing $$\tilde{\mu}^n_m(k)$$ tell us about $$\zeta^n_m(k)$$ and vice versa?

• From a very general point of view, the Fourier transform is an automorphism of the Schwartz space, so sends a function therein to some "conjugate" thereof, like an element of the Galois group of a polynomial, i.e. a field automorphism, sends a root thereof to a conjugate thereof (yes, I like those therestuff words). The difference is that if the FT is linear, it exchanges multiplication and convolution. – Sylvain JULIEN Jan 28 '19 at 20:59