SINGULAR VECTORS AND $\psi$-DIRICHLET NUMBERS over function field

Following [17], we define $\psi$-Dirichlet vectors in $\mathbb{F}_{q}((T^{-1}))^{n}$ and we denote the set of those vectors as $D(\psi)$. For the definitions of norms in $\mathbb{F}_{q}((T^{-1}))^{n}$, readers are referred to §1.4.

Let $\psi_{c}(Q)=\frac{c}{Q^{n}}$. If $\mathbf{x}\in\mathbb{F}_{q}((T^{-1}))^{n}$ is $\psi_{c}$-Dirichlet for every $c>0$, then $\mathbf{x}$ is called singular vector. In recent years, $\psi$-Dirichlet vectors were studied in [17, 16, 15]. Even in the classical setting not much is known.
Diophantine approximation in function field has been a topic of interest since the work of Artin, [3], which developed the theory of continued fraction, and followed by Mahler’s work in [20], which studied geometry of numbers in function field. For recent developments, we refer readers the survey [19], and to [9, 6, 18, 12, 13, 4, 1, 2] for a necessarily incomplete set of references. There are many interesting similarities and contrasts between the theory of Diophantine approximation over the real numbers and in function field over finite fields. The main theorems in this paper show both of these features.

The above theorem shows that analogue of the main theorem in [16] over function field becomes drastically different than the real case. The main tool in proving Theorem 1.1 is the use of continued fraction expansion.

The second part of this paper deals with singular vectors in submanifolds of function fields. Note that, if $(x_{1},\cdots,x_{n})\in\mathbb{F}_{q}((T^{-1}))^{n}$ belongs to a rational affine hyperplane, then it must be singular. These are the most trivial singular vectors. In fact, the converse is also true when $n=1$ (ref.[8]). So, we have the following definition to find vectors those can not be singular in a trivial manner.

For $n>1$, in [11] Khintchine showed the existence of infinitely many totally irrational singular vectors in $\mathbb{R}^{n}$. Moreover, Kleinbock, Moshchevitin and Weiss in [14] showed that for real analytic submanifolds (of dimension greater than $2$) which are not contained inside a rational affine subspace, there are uncountably many totally irrational singular vectors. In this paper, we prove analogous result for certain submanifolds in $\mathbb{F}_{q}((T^{-1}))^{n}$.

As a special case of our Theorem 1.3, we prove the following theorem.

The main challenge comes from the lack of understanding about intersections of a surface and an affine subspace in the function field setting. Another difficulty comes due to total disconnectedness of function field. For real submanifolds, intersection of a connected analytic surface and an affine subspace is well understood due to [5], §2. Both of these facts were used in [14] in a crucial manner. The proof in [14] relies on understanding how ‘semianalytic’ sets can spilt into connected analytic sets. This becomes difficult in $\mathbb{F}_{q}((T^{-1}))^{n}$, as the notion of semianalyticity is not well defined due to the lack of order and the space $\mathbb{F}_{q}((T^{-1}))$ is totally disconnected. That is why we had to tackle case by case and we prove the theorem for a class of submanifolds which is smaller than the class of submanifolds that was taken in [14].

One can define $\hat{\omega}(\cdot)$, as follows, which quantifies singularity of a vector.

(1.2)

$$\hat{\omega}(\mathbf{y}):=\sup\left\{\omega~{}\left|~{}\text{ for all large enough }Q>0,\exists~{}(q_{0},\mathbf{q})\in\mathbb{F}_{q}[T]^{n+1}\setminus\{0\}\text{ s.t.}\begin{aligned} &\|\mathbf{q}\cdot\mathbf{y}+q_{0}\|\leq\frac{1}{Q^{\omega}},\\ &\|\mathbf{q}\|\leq Q\end{aligned}\right.\right\}$$

Dirichlet’s Theorem (ref. [7]) gives that $\hat{\omega}(\mathbf{y})\geq n$ for all $\mathbf{y}\in\mathbb{F}_{q}((T^{-1}))^{n}$. Our next theorem verifies that for a certain analytic submanifolds in $\mathbb{F}_{q}((T^{-1}))^{n}$, there are plenty of totally irrational vectors whose exponents $\hat{\omega}(\cdot)$ are infinity.

In this section and in the following sections, we will use $|\cdot|$ (resp. $\|\cdot\|$ ) to denote norm in $\mathbb{F}_{q}((T^{-1}))$ (resp. $\mathbb{F}_{q}((T^{-1}))^{n}$), unless otherwise mentioned. Let $p$ be a prime and $q:=p^{r}$, where $r\in\mathbb{N}$ and consider the finite field $\mathbb{F}_{q}$. We consider the integral domain $\mathbb{F}_{q}[T]$, the set of polynomials with coefficients in $\mathbb{F}_{q}$. Then we consider the function field $\mathbb{F}_{q}(T)$. We define a norm $|\cdot|$ on $\mathbb{F}_{q}(T)$ as follow:

$$|0|:=0;\left|\frac{P}{Q}\right|:=e^{\mathrm{degP}-\mathrm{degQ}}$$

for all nonzero $P,Q\in\mathbb{F}_{q}[T]$ . Clearly $|\cdot|$ is a nontrivial, non-archimedian and discrete absolute value in $\mathbb{F}_{q}(T)$. The completion field of $\mathbb{F}_{q}(T)$ with respect to this absolute value is $\mathbb{F}_{q}((T^{-1}))$, i.e. the field of Laurent series over $\mathbb{F}_{q}$. We will denote the absolute value of $\mathbb{F}_{q}((T^{-1}))$ by the same notation $|\cdot|$, is given as follows. Let $a\in\mathbb{F}_{q}((T^{-1}))$,

$$|a|:=\left\{\begin{aligned} &0\text{ if }a=0,\\ &e^{k_{0}}\text{ if }a=\sum_{k\leq k_{0}}a_{k}T^{k},k_{0}\in\mathbb{Z},a_{k}\in\mathbb{F}_{q}\text{ and }a_{k_{0}}\neq 0.\end{aligned}\right.$$

This clearly extends the absolute value $|\cdot|$ of $\mathbb{F}_{q}(T)\to\mathbb{F}_{q}((T^{-1}))$ and moreover, the extension remains non-archimedian and discrete. In the above, we call $k_{0}$ to be the degree of $a$, $\mathrm{deg}~{}a$. It is obvious that $\mathbb{F}_{q}[T]$ is discrete in $\mathbb{F}_{q}((T^{-1}))$. For any $n\in\mathbb{N}$, throughout $\mathbb{F}_{q}((T^{-1}))^{n}$ is assumed to be equipped with the supremum norm which is defined as $\|\mathbf{x}\|:=\max_{1\leq i\leq n}|x_{i}|$ for all $\mathbf{x}=(x_{1},x_{2},...,x_{n})\in\mathbb{F}_{q}((T^{-1}))^{n}$ , and with the topology induced by this norm. Clearly $\mathbb{F}_{q}[T]^{n}$ is discrete in $\mathbb{F}_{q}((T^{-1}))^{n}$. Since the topology on $\mathbb{F}_{q}((T^{-1}))^{n}$ considered here is the usual product topology on $\mathbb{F}_{q}((T^{-1}))^{n}$, it follows that $\mathbb{F}_{q}((T^{-1}))^{n}$ is locally compact as $\mathbb{F}_{q}((T^{-1}))$ is locally compact. Note this construction $\mathbb{F}_{q}[T]\subset\mathbb{F}_{q}(T)\subset\mathbb{F}_{q}((T^{-1}))$ is similar to $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$. Let $\lambda$ be the Haar measure on $\mathbb{F}_{q}((T^{-1}))^{n}$ which takes the value 1 on the closed unit ball $\|\mathbf{x}\|=1$.

Suppose $a=\sum_{k\leq k_{0}}a_{k}T^{k}\in\mathbb{F}_{q}((T^{-1}))$ where $a_{k_{0}}\neq 0$, we call $[a]:=\sum_{k\leq k_{0}}^{0}a_{k}T^{k}$ as the integer part of $a$ and $\langle a\rangle=\sum_{k<0}a_{k}T^{k}$ as the fractional part of $a$. Note that $|[a]|=e^{k_{0}}\geq 1$ if $a_{k_{0}}\neq 0$, and otherwise we have $[a]=0$. Also, note that $|\langle a\rangle|\leq 1$. This observation leads us to construct continued fraction expansion of $a$. An expression of the form $a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\dots}}$ where $a_{0},a_{1},a_{2}\in\mathbb{F}_{q}[T]$ is called a simple continued fraction; see §$1$ in [22]. An expression of the form $\frac{p_{n}}{q_{n}}=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+{\dots}_{+\frac{1}{a_{n}}}}}$, where $p_{n},q_{n},a_{0},a_{1},\dots,a_{n}\in\mathbb{F}_{q}[T]$ is called a finite continued fraction. An element of $\mathbb{F}_{q}(T)$, can be represented as an unique finite continued fraction. An $\alpha\in\mathbb{F}_{q}((T^{-1}))\setminus\mathbb{F}_{q}[T]$ can be represented as a simple continued fraction in the form of $[a_{0},a_{1},a_{2},\dots]$ and we call the numbers $\frac{p_{n}}{q_{n}}=[a_{0},a_{1},\dots,a_{n}]$ the convergents of $\alpha$. Note that $|q_{n}|$ is increasing as $n\to\infty$. The relation between two consequitive covergents is given by the following equation:

$$p_{i}q_{i+1}-p_{i+1}q_{i}=(-1)^{i+1}\text{ for }i\in\mathbb{Z},i\geq-2.$$

Hence we have,

$$\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_{n}}{q_{n}}\right|=\left|\frac{p_{n+1}q_{n}-p_{n}q_{n+1}}{q_{n}q_{n+1}}\right|=\left|\frac{\pm 1}{q_{n}q_{n+1}}\right|=\frac{1}{|q_{n}|\cdot|q_{n+1}|}\leq\frac{1}{|q_{n}|^{2}}.$$

In fact, by Equation $1.12$ in [22] we have

(2.1)

$$\left|\alpha-\frac{p_{n}}{q_{n}}\right|=\left|\frac{1}{q_{n}q_{n+1}}\right|,$$

where $\frac{p_{n}}{q_{n}}$ is a convergent of $\alpha\in\mathbb{F}_{q}((T^{-1}))\setminus\mathbb{F}_{q}(T)$. We recall the following definition and theorem of best approximation [21], §1.2.

We want to recall the following Lemma $2.1$ from [16] which was stated for real numbers. The verbatim proof will give the following lemma for function field. The proof uses the fact that convergents are best approximations, which we have by Theorem 2.1. In what follows $\frac{p_{n}}{q_{n}}$ are convergents of $x$.

It is easy to see that $\mathbb{F}_{q}(T)\subset D(\psi)$. We want to show that $D(\psi)\subset\mathbb{F}_{q}(T)$. By Equation (2.1) for $x\in\mathbb{F}_{q}((T^{-1}))\setminus\mathbb{F}_{q}(T)$ we have $|\langle q_{n-1}x\rangle|=\frac{1}{|q_{n}|},$ $\forall~{}n$. Since $\psi(t)<\frac{1}{t}$ for all large enough $t$, by Lemma 2.1 we conclude that there is no $x\in\mathbb{F}_{q}((T^{-1}))\setminus\mathbb{F}_{q}(T)$ such that $x$ is $\psi$-Dirichlet.

Let us consider

$$S=\{(x,y,p_{3}(x,y),\cdots,p_{n}(x,y))\ \mid\ x,y\in U\},$$

where $U$ is an open subset of $\mathbb{F}_{q}((T^{-1}))$, and

$$p_{i}(x,y)=b_{i,1}x^{2}+b_{i,2}xy+b_{i,3}y^{2}+b_{i,4}x+b_{i,5}y+b_{i,6}$$

with $b_{1,i},b_{i,2},\dots,b_{i,6}\in\mathbb{F}_{q}((T^{-1}))$ and $b_{i,1},b_{i,2},b_{i,3}$ not being zero simutaneously for $i=3,\cdots,n$. Let us take $A$ to be a rational affine hyperplane in $\mathbb{F}_{q}((T^{-1}))^{n}$ and we assume that $S$ is not contained inside $A$. We can define $A$ by the linear equation $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}=a_{n+1}$, where $(a_{1},a_{2},\cdots,a_{n+1})\in\mathbb{F}_{q}[T]^{n+1}$ is primitive. Note that $S\cap A$ is given by the solutions to the equation;

(3.1)

$$f(x,y):=0,$$

where

$$f(x,y)=\sum_{i=3}^{n}a_{i}p_{i}(x,y)+a_{1}x+a_{2}y-a_{n+1}.$$

We see that $f$ is a polynomial of degree less than or equal to $2$. Now note that

(3.2)

$$\frac{\partial f}{\partial x}=\sum_{i=3}^{n}(2a_{i}b_{i,1}x+a_{i}b_{i,2}y)+(a_{1}+\sum_{i=3}^{n}a_{i}b_{i,4})$$

and

(3.3)

$$\frac{\partial f}{\partial y}=\sum_{i=3}^{n}(a_{i}b_{i,2}x+2a_{i}b_{i,3}y)+(a_{2}+\sum_{i=3}^{n}a_{i}b_{i,5}).$$

If $\frac{\partial f}{\partial x}(x_{0},y_{0})\neq 0$, then by Theorem 3.3 we get a neighborhood of $(x_{0},y_{0})$, where $y$ is a $\mathbb{F}_{q}((T^{-1}))$-analytic function of $x$. If $\frac{\partial f}{\partial y}(x_{0},y_{0})\neq 0$ then locally we can write $x$ as a $\mathbb{F}_{q}((T^{-1}))$-analytic function of $y$. Hence in order to find out all possible $(x_{0},y_{0})$ such that there is no neighborhood of $(x_{0},y_{0},p_{3}(x_{0},y_{0}),\cdots,p_{n}(x_{0},y_{0}))\in S\cap A$ that is analytic curve in $S\cap A$, we consider the linear system

(3.4)

$$\begin{cases}\frac{\partial f}{\partial x}=0;\\ \frac{\partial f}{\partial y}=0.\end{cases}$$

The corresponding coefficient matrix $M\in\text{Mat}_{2\times 2}(\mathbb{F}_{q}((T^{-1})))$ of the system is

$$\begin{bmatrix}\sum_{i=3}^{n}2a_{i}b_{i,1}&\sum_{i=3}^{n}a_{i}b_{i,2}\\ \sum_{i=3}^{n}a_{i}b_{i,2}&\sum_{i=3}^{n}2a_{i}b_{i,3}\end{bmatrix}.$$

First note that if $a_{3},\cdots,a_{n}$ are zero, then $S\cap A$ is an analytic curve as the equation of $A$ would be $a_{1}x+a_{2}y=a_{n+1}$. Therefore one of $a_{3},\cdots,a_{n}$ must be nonzero, and without loss of generality we assume that $a_{3}\neq 0$. Next let us denote,

$$b_{k}=\sum_{i=3}^{n}a_{i}b_{i,k}$$

for $k=1,\cdots,6$. With this setting, we have the following lemma.

From the proof above we know that the key is to solve the following equations;

(3.5)		$\displaystyle f(x,y)$	$\displaystyle=\sum_{i=3}^{n}a_{i}p_{i}(x,y)+a_{1}x+a_{2}y-a_{n+1}$
			$\displaystyle=b_{1}x^{2}+b_{2}xy+b_{3}y^{2}+(a_{1}+b_{4})x+(a_{2}+b_{5})y+(b_{6}-a_{n+1})=0,$

(3.6)

$$\frac{\partial f}{\partial x}(x,y)=2{b_{1}}x+{b_{2}}y+(a_{1}+{b_{4}})=0,$$

(3.7)

$$\frac{\partial f}{\partial y}(x,y)=b_{2}x+2b_{3}y+(a_{2}+b_{5})=0.$$