number of zeros of a holomorphic function

One basic problem in mathematics is to solve equations. Giving a polynomial non-trivial of complex coefficients on one variable, we know that it has as many zeros as its degree. The simplest holomorphic function which has infinite zeros is perhaps the sinus and cosinus functions.

Of course, from complex analysis we know that a holomorphic function on the whole plane can not have infinite zeros on a bounded region. So we can talk about the number of zeros of a holomorphic function inside a circle of certain radius.

Jacobi sums and Beta functions

(0)In this post we will talk a little about the similarity between these two terms appearing in the title. It is, in fact, what I learned in a course on Galois theory in Ecole Polytechnique.

(1)In number theory one common problem is to determine the number of solutions to some equation.

For example, suppose that $p$ is a prime number, and thus $\mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p$ is a finite field.The simplest case of these problems is a linear equation in indeterminant $x$:$ax=b(a,b\in \mathbb{F}_p)$. Of course to determine the number of its solutions is a very simple problem.

Then what about this quadratic one: $x^2=b(b\in \mathbb{F}_p)$(we suppose that $p\neq 2$)? This, again, is not difficult. We just have to utilize the method of quadratic residue: if $b\neq 0$ is a quadratic residue modulo $p$, then this equation has two solutions, while otherwise it has no solution(for $b=0$, there is always only one solution). Using the Legendre symbol, we can say that the number of solutions to this quadratic equation is just $1+(\frac{b}{p})$.

(2)There are three directions to generalize what we have said, generalize the polynomial in the equation either  to higher degrees, either to more variables, or generalize the equation to a larger field.

Let’s do the first kind of generalization. So the problem becomes now: determine the number of solutions to the equations $x^n=b(n>1)$ in $\mathbb{F}_p$.There is a little problem here: whether $n|(p-1)$? If $n\not|(p-1)$, then consider the map $f:\mathbb{F}_p\longrightarrow \mathbb{F}_p, a\mapsto a^n$ and its restriction $f'$ to $(\mathbb{F}_p)^*$. It’s easy to see that $f'$ is a group homomorphism, and observing that $n$ doesn’t divide the cardinal of the group $(\mathbb{F}_p)^*,i.e. p-1$, we get that $f'$ is just a bijection, and thus so is $f$. So the problem in this case($n\not |(p-1)$) is not very interesting. Therefore in the rest we want to consider the more interesting case, $n|(p-1)$.(In case that $n>p-1$, we can apply Fermat’s little theorem to the case $n\leq p-1$) . In order to get a formula similar to the above one, we need to introduce the concept of characters of an abelian group. Basically, a character $\xi$ of an abelian group $G$ is just a group homomorphism $\xi:G\longrightarrow \mathbb{C}^*$. In our case, we define a character $\xi$on the finite field $\mathbb{F}_p$ to be a character on the group $(\mathbb{F}_p)^*$ and extend to the whole field by setting $\xi(0)=0$. Furthermore we can define some kind of composition between two characters $\xi,\phi$ over a same group: the point-wise multiplication, $(\xi\circ\phi)(a)=(\xi(a))(\phi(a))$. It’s easy to verify that the resulting mapping is again a character. In fact, in noting that $(\mathbb{F}_p)^*$ is a cyclic group, we verify easily that the set of all the characters over $\mathbb{F}_p$ forms a cyclic group of order $p-1$(We denote this set as $C_p$,and the neutral element as $e$, we can regard $C_p$ as the dual group of $(\mathbb{F}_p)^*=\mathbb{Z}/(p-1)\mathbb{Z}$). Now we can state the corresponding formula for the number $NS(x^n=b)$ of solutions to the equation $x^n=b$:

$NS(x^n=b)=\sum_{\xi\in C_p,\xi^n=e}\xi(b)$

We can prove this assertion by simply writing down all the $\xi$ explicitly.

In all these formulas, we see the same form: a sum of some characters satisfying some conditions. What does this remind us? In fact, in some way, this reminds us of generating functions: if we have a set of numbers $\{a_n\}$ indexed by positive integers, then the most natural way is to give this set a generating function $f(x)=\sum_n a_n x^n$, sometimes under some conditions for convenience we write $x=e^{2i\pi\theta}$. What we will do next is similar: we want to give $C_p$ a generating function. Noting that there are just $p-1$ elements in this set, so this generating function is a finite sum. We can try this one $g_{\xi}(a)=\sum_{b\in\mathbb{F}_p}\xi(b)e^{2i\pi ab/p}(a\in\mathbb{F}_p)$(it is indeed well defined). It is easy to find the correspondence between this generating function and the generic one.

This generating function has a special name, it is called the Gauss sum of $\xi$.

It is worth pointing out that the Gauss sum has some similarity with the $\Gamma$ function: $\Gamma(s)=\int_0^{\infty}t^se^{-t}\frac{dt}{t}$. Just comparing term by term: integral v.s sum, $t^s$ v.s $\xi(b)$, $e^{-t}$ v.s $e^{2i\pi ab/p}$.

(3)Now we want to deal with the second kind of generalization: adding more variables in the equation.

The simplest case is this one: find the number of solutions to the equation $x^2+y^2=b(b\neq 0)$. Now we make use of a little trick to find out the number $NS(x^2+y^2=b)$. In fact we can write $NS(x^2+y^2=b)=\sum_{a\in\mathbb{F}_p}NS(x^2=a)NS(y^2=b-a)$$=\sum_a(1+(\frac{a}{p}))(1+(\frac{b-a}{p}))$. Expanding out, and noting that $\sum_{a\in\mathbb{F}_p}(\frac{a}{p})=0$(the number of non zero quadratic residues equals that of non-quadratic residues), we obtain $NS(x^2+y^2=b)=p+\sum_{a,a\neq 0}(\frac{a(b-a)}{p})$. To evaluate the second term, we write $\sum_{a\neq 0}(\frac{a(b-a)}{p})=\sum_{a\neq 0}(\frac{b/a-1}{p})=\sum_{a\neq 0}(\frac{a-1}{p})=-(\frac{-1}{p})=-(-1)^{(p-1)/2}$.

So here we see another form of sum, that is $\sum_a \xi(a)\psi(b-a)$. In other words, if we define $J(\xi, \psi)=\sum_{a\in\mathbb{F}_p}\xi(a)\psi(1-a)$, then after some similar calculations, we have that $NS(x^n+y^n=1)=\sum_{\xi^n=e,\psi^n=e}\sum_a\xi(a)\psi(1-a)=\sum_{\xi^n=e,\psi^n=e}J(\xi,\psi)$. This definition gives the Jacobi sum $J(\xi,\psi)$.

(4)So after a lengthy introduction, we define the Jacobi sum, and understand that Jacobi sum can be used to calculate the number of solutions to some kind of equations.

Now since the title of this post includes another name, ‘Bessel functions’, next we want to formally talk a little bit about the similarity between Jacobi sum and Bessel functions.

The Beta function is defined to be $B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt$. Compare this function with the Jacobi sum, their similarity is very clear.

Furthermore, we know that there is some relation between the Beta function and the $\Gamma$ function, that is

$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.

In fact, we also have such relation between the Jacobi sum and the Gauss sum, that is

$J(\xi, \psi)=\frac{g_{\xi}(1)g_{\psi}(1)}{g_{\xi\circ\psi}(1)}$.

the principals that I write blogs

I write blogs about what I have learned (perhaps mathematics, perhaps physics, etc.) and my reflexions on these things.