# The number of unitary circulant matrices over a finite field $\mathbb{F}_{q^2}$

I asked this question in MSE few days ago but there was no response.

Suppose $$\mathbb{F}=\mathbb{F}_{q^2}$$, where $$q$$ is a prime power. The conjugate of elements in $$\mathbb{F}$$ is defined by $$\overline{x}=x^q$$. I need to find the number of $$n\times n$$ unitary circulant matrices over $$\mathbb{F}$$.

The number of invertible circulant matrices over a finite field can be seen elsewhere, such as when $$n,q$$ coprime and my question when $$n=\operatorname{char} q$$.

Is there any better method to calculate this number other than considering each entry?

This is equivalent to the order of the centraliser of the permutation matrix of $$(1,2,\dots,n)$$ in $$\operatorname{GU}_n(q)$$.

Added on 30 May 2020 MSE:

Let $$C$$ be the subgroup of $$\operatorname{GL}_n(q^2)$$ of all circulant matrices. Is $$C\operatorname{GU}_n(q)$$ a subgroup of $$\operatorname{GL}_n(q^2)$$? That is, is $$C\operatorname{GU}_n(q)=\operatorname{GU}_n(q)C$$? If that is correct then $$C\operatorname{GU}_n(q)=\operatorname{GL}_n(q^2)$$ and so $$|C\cap\operatorname{GU}_n(q)|$$ follows. Here we denote by $$\operatorname{GU}_n(q)$$ the general unitary group over $$\mathbb{F}_{q^2}$$.

Let $$\tau$$ denote the permutation matrix corresponding to $$(1,2,\ldots,n)$$. Consider it first as an element of $$\mathrm{M}_n(q^2)$$.
This matrix has minimal polynomial equal to $$X^n-1$$, which is equal to its characteristic polynomial. It is therefore cyclic, and its centralizer is isomorphic to $$\mathbb{F}_{q^2}$$-algebra $$\mathbb{F}_{q^2}[X]/(X^n-1)$$. For simplicity, I'll assume $$\mathbb{F}_{q^2}$$ has no $$N$$-th root of unity, so that this algebra is isomorphic to $$\mathbb{F}_{q^{2n}}$$. Mapping $$X\to \bar{X}^t$$ defines a field automorphism of order $$2$$ of $$\mathbb{F}_{q^{2n}}$$ which, by Hilbert 90 (which is an overkill, but does the job), restricts to a surjective map $$\mathbb{F}_{q^{2n}}^\times\to \mathbb{F}_{q^n}^\times$$. The centralizer you're seeking is precisely the kernel of this map, and is of carinality $$\frac{q^{2n}-1}{q^n-1}=q^n+1.$$
Now, if $$X^N-1$$ splits in $$\mathbb{F}_{q^2}$$, then $$\mathbb{F}_{q^2}[X]/(X^N-1)$$ is a product of finite fields, and the same type of argument works per coordinate, which should result in a formula of the form $$\prod_{i=1}^r q^{d_i}+1$$ for suitable $$d_i$$'s such that $$\sum d_i=n$$.
• Thank you for your answer. But what if $(n,q)\ne 1$ or in particular $n=\operatorname{char}q$? – Hongyi Huang Jun 2 '20 at 6:54