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This article is cited in 1 scientific paper (total in 1 paper) On the transference principle and Nesterenko's linear independence criterion O. N. Germanab, N. G. Moshchevitinab a National Research University Higher School of Economics, Moscow b Moscow Center for Fundamental and Applied Mathematics
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     Dedicated to Yu. V. Nesterenko on the occasion of his 75th birthday § 1. IntroductionThis paper originated as a result of studying the prominent paper [1] by Nesterenko, where he proves his linear independence criterion. Given an integer $n\geqslant 2$, we fix an arbitrary $n$-tuple $\boldsymbol \theta=(\theta_1,\dots,\theta_n)\in\mathbb{R}^n$. It is well known that the problem of approximating $\theta_1,\dots,\theta_n$ simultaneously by rational numbers with equal denominators is related to the problem of approximating zero by the values of the linear form $x_0+\theta_1x_1+\dots+\theta_nx_n$ at non-zero integer points. This relation is performed by the so-called transference principle discovered by Khintchine [2]. He formulated it in terms of Diophantine exponents. Definition 1. The (regular) Diophantine exponent $\lambda=\lambda(\boldsymbol \theta)$ is defined as the supremum of real numbers $\gamma$ such that the system of inequalities admits solutions in $\mathbf x=(x_0,\dots,x_n)\in\mathbb{Z}^{n+1}$ for some arbitrarily large values of $t$. The respective uniform Diophantine exponent $\widehat\lambda=\widehat \lambda(\boldsymbol\theta)$ is defined as the supremum of real numbers $\gamma$ such that (1.1) admits solutions in $\mathbf x=(x_0,\dots,x_n)\in \mathbb{Z}^{n+1}$ for every $t$ large enough. Definition 2. The (regular) Diophantine exponent $\omega=\omega(\boldsymbol \theta)$ is defined as the supremum of real numbers $\gamma$ such that the system of inequalities admits solutions in $\mathbf x=(x_0,\dots,x_n)\in\mathbb{Z}^{n+1}$ for some arbitrarily large values of $t$. The respective uniform Diophantine exponent $\widehat\omega=\widehat \omega(\boldsymbol\theta)$ is defined as the supremum of real numbers $\gamma$ such that (1.2) admits solutions in $\mathbf x=(x_0,\dots,x_n)\in \mathbb{Z}^{n+1}$ for every $t$ large enough. It follows immediately from Dirichlet’s approximation theorem (or from Minkowski’s convex body theorem) that the Diophantine exponents satisfy the trivial relations $\lambda\geqslant \widehat\lambda\geqslant 1/n$ and $\omega\geqslant \widehat\omega\geqslant n$. It is also known (see [3]) that, unless all the $\theta_i$ are rational, we have a slightly less trivial inequality $\widehat\lambda\leqslant 1$. Let us give a brief account on the existing non-trivial relations. The aforementioned Khintchine’s transference principle was published in 1926. It can be formulated as follows:
$$
\begin{equation}
\frac{1+\omega}{1+\lambda}\geqslant n,\qquad \frac{1+\omega^{-1}}{1+\lambda^{-1}}\geqslant \frac1n.
\end{equation}
\tag{1.3}
$$
Later on, in 2007–10, Bugeaud and Laurent [4], [5] improved upon (1.3) by showing that
$$
\begin{equation}
\frac{1+\omega}{1+\lambda}\geqslant \frac{n-1}{1-\widehat\lambda},\qquad \frac{1+ \omega^{-1}}{1+\lambda^{-1}}\geqslant \frac{1-\widehat\omega^{-1}}{n-1},
\end{equation}
\tag{1.4}
$$
provided that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$. It can be easily verified that (1.4) implies (1.3), as $\widehat\omega^{-1}\leqslant 1/n\leqslant \widehat\lambda\leqslant 1$. As for the uniform exponents, Jarník [3] proved in 1938 that the following remarkable identity holds if $n= 2$ and $1,\theta_1,\theta_2$ are linearly independent over $\mathbb{Q}$: In 2012 it was shown by German [6], [7] that for arbitrary $n\geqslant 2$ we have
$$
\begin{equation}
\widehat\omega\geqslant \frac{n-1}{1-\widehat\lambda},\qquad \widehat\lambda\geqslant \frac{1-\widehat\omega^{-1}}{n-1}.
\end{equation}
\tag{1.6}
$$
Clearly, (1.6) turns into Jarník’s identity for $n= 2$. In 2013 Schmidt and Summerer [8] showed that
$$
\begin{equation}
\widehat\omega\leqslant \frac{1+\omega}{1+\lambda},\qquad \widehat\lambda\leqslant \frac{1+ \omega^{-1}}{1+\lambda^{-1}},
\end{equation}
\tag{1.7}
$$
provided that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$. An alternative proof of their result can be found in [9]. Clearly, (1.6) and (1.7) imply Bugeaud and Laurent’s inequalities (1.4). Furthermore, in 1950–1954 Jarník [10], [11] proved for $n= 2$ that
$$
\begin{equation}
\frac{\omega}{\widehat\omega}\geqslant \widehat\omega-1,\qquad \frac{\lambda}{\widehat \lambda}\geqslant \frac{\widehat\lambda}{1-\widehat\lambda}.
\end{equation}
\tag{1.8}
$$
It is interesting to note that the right-hand sides in (1.8) coincide for $n= 2$, by (1.5). They are both equal to $\widehat\omega\widehat\lambda$ as well as to $(1-\widehat\omega^{-1})/(1-\widehat\lambda)$. We shall mention this fact in Section 3.1. The inequalities (1.8) were recently generalized to the case of arbitrary $n\geqslant 2$ by Marnat and Moshchevitin [12]. They showed that
$$
\begin{equation}
\frac{\omega}{\widehat\omega}\geqslant G_{\mathrm{lin}}(\widehat\omega),\qquad \frac{\lambda}{\widehat\lambda}\geqslant G_{\mathrm{sim}}(\widehat\lambda),
\end{equation}
\tag{1.9}
$$
where $G_{\mathrm{lin}}(\widehat\omega)$ and $G_{\mathrm{sim}}(\widehat \lambda)$ are the largest roots of the polynomials
$$
\begin{equation}
f(x)=\widehat\omega^{-1}x^n-x+(1-\widehat\omega^{-1}), \qquad g(x)=(1-\widehat \lambda)x^n-x^{n-1}+\widehat\lambda
\end{equation}
\tag{1.10}
$$
respectively. An alternative proof of the second inequality in (1.9) can be found in [13]. Summing up, we can say that all non-trivial relations currently known follow from (1.6), (1.7), and (1.9). The main purpose of this paper is to present a geometric observation, a rather simple one, which proves (1.7) almost immediately. It appears that this observation also provides a quite simple proof of Nesterenko’s linear independence criterion. The rest of the paper is organized as follows. In Section 2 we formulate and prove our main result and apply it to prove (1.7). In Section 3 we derive some corollaries of (1.7) concerning lower bounds for the ratios $\omega/\widehat\omega$, $\lambda/\widehat\lambda$ and show that (1.7) yields the weakest of the inequalities (1.9). Moreover, we analyse how (1.6) and (1.7) split Bugeaud and Laurent’s inequalities (1.4), and compare some of our corollaries with a recent result by Schleischitz [14]. Finally, Section 4 is devoted to Nesterenko’s linear independence criterion. We present a rather simple proof of his theorem, which is based on our main geometric observation described in Section 2. We also show that in order to prove the linear independence criterion itself, it suffices to use the first step of Nesterenko’s induction. § 2. Empty cylinder lemmaLet us introduce some notation. We write $\ell$ for the one-dimensional subspace generated by $(1,\theta_1,\dots,\theta_n)$ and let $\ell^\perp$ be its orthogonal complement. Given any $\mathbf x\in\mathbb{R}^{n+1}$, we denote the Euclidean distance from $\mathbf x$ to $\ell$ (resp. from $\mathbf x$ to $\ell^\perp$) by $r(\mathbf x)$ (resp. $h(\mathbf x)$). We also write $\langle\,\cdot\,,\cdot\,\rangle$ for the inner product in $\mathbb{R}^{n+1}$. Our main geometric observation is described by the following statement. Lemma 1. Let $t$, $\alpha$, $\beta$ be positive real numbers such that $t^\beta>2t^\alpha$. Suppose that $\mathbf v\in\mathbb{Z}^{n+1}$ satisfies Consider the half-open cylinder Then $\mathcal{C}\cap\mathbb{Z}^{n+1}=\varnothing$. Proof. Consider an arbitrary $\mathbf x\in\mathcal{C}$. Let us show that $0< \langle \mathbf v,\mathbf x\rangle< 1$. To do this, we write $\mathbf y$ for the orthogonal projection of $\mathbf x$ to the two-dimensional subspace $\pi$ spanned by $\ell$ and $\mathbf v$. Then $r(\mathbf y)$ and $h(\mathbf y)$ can be interpreted as the absolute values of the coordinates of $\mathbf y$ in $\pi$ with respect to the coordinate axes $\ell$ and $\ell^\bot\cap \pi$ (see Fig. 1). Hence
$$
\begin{equation*}
\langle\mathbf v,\mathbf x\rangle= \langle\mathbf v,\mathbf y\rangle\geqslant h(\mathbf v)h(\mathbf y)-r(\mathbf v)r(\mathbf y)> t^{\alpha-\beta}-t^{\alpha-\beta}=0
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\langle\mathbf v,\mathbf x\rangle= \langle\mathbf v,\mathbf y\rangle\leqslant h(\mathbf v)h(\mathbf y)+r(\mathbf v)r(\mathbf y)< 1-t^{\alpha-\beta}+t^{\alpha-\beta}=1.
\end{equation*}
\notag
$$
Thus, indeed, $0< \langle\mathbf v,\mathbf x\rangle< 1$, and $\mathbf x$ cannot be an integer point. Lemma 1 is proved.
Now we can prove Schmidt and Summerer’s inequalities (1.7). Proof of (1.7). We may assume that $\lambda>1/n$ and $\omega> n$. Indeed, by Khintchine’s inequalities (1.3) the equalities $\lambda=1/n$ and $\omega=n$ are equivalent, and if they hold, trivial inequalities yield that $\widehat\lambda=1/n$, $\widehat\omega=n$. With these values of the exponents both inequalities (1.7) are obviously true. So, let us suppose that $\lambda>1/n$ and $\omega> n$.
By the definition of $\lambda$ and $\omega$ we can choose a point $\mathbf v \in\mathbb{Z}^{n+1}$ and a positive real number $\gamma$ to satisfy any of the following two collections of conditions: 1) $h(\mathbf v)$ is arbitrarily large, $r(\mathbf v)=h(\mathbf v)^{-\gamma}$, $\gamma>1/n$, $\gamma$ is arbitrarily close to $\lambda$; 2) $h(\mathbf v)$ is arbitrarily small, $r(\mathbf v)=h(\mathbf v)^{-\gamma}$, $\gamma<1/n$, $\gamma$ is arbitrarily close to $1/\omega$. We shall refer to the first choice as Case 1, and to the second one as Case 2. In either case the main part of the argument is basically the same. So, let us choose $\mathbf v$ and $\gamma$ according to one of the two cases. Set $t=t(\mathbf v,\gamma)$ to be the smallest positive real number such that the cylinder
$$
\begin{equation}
\mathcal{C}_{\mathbf v}=\{\mathbf x\in\mathbb{R}^{n+1}\mid r(\mathbf x)\leqslant t,\, h(\mathbf x)\leqslant t\cdot h(\mathbf v)^{-1-\gamma}\}
\end{equation}
\tag{2.3}
$$
contains a non-zero integer point. Define also $\alpha=\alpha(\mathbf v,\gamma)$ and $\beta=\beta(\mathbf v,\gamma)$ by the relations
$$
\begin{equation}
h(\mathbf v)=t^\alpha,\qquad \alpha=\frac{1+\beta}{1+\gamma}.
\end{equation}
\tag{2.4}
$$
Then $r(\mathbf v)=h(\mathbf v)^{-\gamma}=t^{-\alpha\gamma}=t^{\alpha-1-\beta}$ and $t\cdot h(\mathbf v)^{-1-\gamma}=t^{-\beta}$. Hence $\mathbf v$ satisfies (2.1) and
$$
\begin{equation*}
\mathcal{C}_{\mathbf v}=\{\mathbf x\in\mathbb{R}^{n+1}\mid r(\mathbf x)\leqslant t,\, h(\mathbf x)\leqslant t^{-\beta}\}.
\end{equation*}
\notag
$$
In order to fulfill the hypothesis of Lemma 1, it remains to prove that $t^\beta> 2t^\alpha$. We recall that in Case 1 $h(\mathbf v)$ is arbitrarily large and $\gamma>1/n$, and in Case 2 $h(\mathbf v)$ is arbitrarily small and $\gamma<1/n$. In both cases by Minkowski’s convex body theorem the volume of $\mathcal{C}_{\mathbf v}$ is bounded (by $2^{n+1}$), and so the product $t^{n+1}h(\mathbf v)^{-1-\gamma}=h(\mathbf v)^{(n+1)/\alpha-(1+\gamma)}$ is also bounded. Hence in Case 1 we have $\alpha\gamma\geqslant (n+1)\*\gamma/(1+ \gamma)> 1$ and, therefore, $\beta=\alpha+\alpha\gamma-1>\alpha$, whereas in Case 2 we have $\alpha\gamma\leqslant (n+1)\gamma/(1+\gamma)< 1$ and, therefore, $\beta=\alpha+\alpha\gamma-1<\alpha$. Thus, in both cases we have
$$
\begin{equation*}
t^\beta= h(\mathbf v)^{\beta/\alpha}> 2h(\mathbf v)= 2t^\alpha.
\end{equation*}
\notag
$$
Having the hypothesis of Lemma 1 fulfilled, we conclude that there are no integer points in the cylinder $\mathcal{C}=\mathcal{C}(t,\alpha,\beta)$ defined by (2.2). At the same time there are some non-zero integer points in $\mathcal{C}_{\mathbf v}$, but all of them lie on the boundary of $\mathcal{C}_{\mathbf v}$. Moreover, every such point satisfies the condition $r(\mathbf x)=t$ since
$$
\begin{equation*}
\{\mathbf x\in\mathcal{C}_{\mathbf v}\mid r(\mathbf x)<t,\, h(\mathbf x)=t^{-\beta}\}= \mathcal{C}_{\mathbf v}\cap(\mathcal{C}\cup(-\mathcal{C}))
\end{equation*}
\notag
$$
and $\mathcal{C}\cup(-\mathcal{C})$ is empty. Thus, there is an integer point satisfying
$$
\begin{equation*}
r(\mathbf x)=t,\qquad h(\mathbf x)\leqslant t^{-\beta},
\end{equation*}
\notag
$$
and there are no non-zero integer points satisfying
$$
\begin{equation*}
r(\mathbf x)<t,\qquad h(\mathbf x)\leqslant t^{-\alpha}-t^{-\beta}.
\end{equation*}
\notag
$$
Suppose now that Case 1 holds. Then $\beta>\alpha$ and $t$ can be arbitrarily large. Hence $\omega\geqslant \beta$, $\widehat\omega\leqslant \alpha$, and
$$
\begin{equation*}
\widehat\omega\leqslant \alpha= \frac{1+\beta}{1+\gamma}\leqslant \frac{1+\omega}{1+ \lambda-\varepsilon}
\end{equation*}
\notag
$$
with $\varepsilon$ positive and arbitrarily small. This gives us the first inequality in (1.7).
Suppose that Case 2 holds. Then $\beta<\alpha$ and $t$ can be arbitrarily small. Hence $\lambda\geqslant 1/\beta$, $\widehat\lambda\leqslant 1/\alpha$, and
$$
\begin{equation*}
\widehat\lambda\leqslant \alpha^{-1}= \frac{1+\gamma}{1+\beta}\leqslant \frac{1+\omega^{-1}+ \varepsilon}{1+\lambda^{-1}}
\end{equation*}
\notag
$$
with $\varepsilon$ positive and arbitrarily small. This gives us the second inequality in (1.7).
§ 3. Corollaries of Schmidt and Summerer’s inequalities3.1. Lower bounds for the ratios $\omega/\widehat\omega$ and $\lambda/\widehat\lambda$The inequalities (1.7) can be rewritten as
$$
\begin{equation}
\frac{\omega}{\widehat\omega}\geqslant \frac{1+\lambda}{1+\omega^{-1}},\qquad \frac{\lambda}{\widehat\lambda}\geqslant \frac{1+\lambda}{1+\omega^{-1}},
\end{equation}
\tag{3.1}
$$
giving thus new lower bounds in the problem of estimating the ratio between the regular and uniform exponents. Note that, as in Jarník’s inequalities (1.8) for $n= 2$, the right-hand sides in (3.1) coincide. However, the lower estimate for $\omega/\widehat\omega$ and $\lambda/\widehat\lambda$ provided by (3.1) is a function of $\omega$ and $\lambda$, whereas it has been more traditional to think in this context of functions of $\widehat\omega$ and $\widehat \lambda$. The right-hand side $(1+\lambda)/(1+\omega^{-1})$ can be weakened to $(1-\widehat\omega^{-1})/(1-\widehat\lambda)$, thus generalizing Jarník’s inequalities (1.8) to arbitrary dimension. Proposition 1. Suppose that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$. Then Proof. Set $B=(1+\lambda)/(1+\omega^{-1})$. Then (3.1) reads as $\omega\geqslant \widehat\omega B$ and $\lambda\geqslant \widehat\lambda B$. Inserting these inequalities into the definition of $B$, we get or, equivalently, The proposition is proved. Corollary 1. Suppose that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$. Then The inequalities (3.3) coincide with Jarník’s inequalities (1.8) for $n= 2$. Another generalization of (1.8) to arbitrary dimension is provided by (1.9). An important difference between (1.9) and (3.3) is that the lower bounds $G_{\mathrm{lin}}(\widehat \omega)$ and $G_{\mathrm{sim}}(\widehat\lambda)$ in (1.9) are functions of only one exponent, whereas (3.3) requires both. If $G_{\mathrm{lin}}(\widehat\omega)=G_{\mathrm{sim}}(\widehat\lambda)$, then (1.9) is equivalent to (3.3), but if $G_{\mathrm{lin}}(\widehat \omega)\neq G_{\mathrm{sim}}(\widehat\lambda)$, then (3.3) is stronger than the weakest of (1.9) and weaker than the strongest. Proposition 2. Suppose that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$. Then either or Moreover, if any two of the four quantities under comparison coincide, then so do all of them. Proof. Let $f(x)$ and $g(x)$ be the polynomials defined by (1.10). Then, given $t\geqslant 1$, we have
$$
\begin{equation*}
t>G_{\mathrm{lin}}(\widehat\omega)\ \Longleftrightarrow \ f(t)>0,\qquad t>G_{\mathrm{sim}}(\widehat\lambda)\ \Longleftrightarrow \ g(t)>0.
\end{equation*}
\notag
$$
Since $\widehat\omega\geqslant n$ and $\widehat\lambda\geqslant 1/n$, both $(\widehat \omega\widehat\lambda)^{1/(n-1)}\geqslant 1$ and $(1-\widehat\omega^{-1})/(1-\widehat \lambda)\geqslant 1$. By simple substitution and regrouping it is easily verified that
$$
\begin{equation*}
g\biggl(\frac{1-\widehat\omega^{-1}}{1-\widehat\lambda}\biggr)\geqslant 0\ \Longleftrightarrow \ \frac{1-\widehat\omega^{-1}}{1-\widehat\lambda}\leqslant (\widehat\omega\widehat\lambda)^{1/(n-1)}\ \Longleftrightarrow \ f\bigl((\widehat \omega\widehat\lambda)^{1/(n-1)}\bigr)\leqslant 0
\end{equation*}
\notag
$$
and that the respective equalities are equivalent. This proves the statement of the proposition. Corollary 2. Denote Then, assuming that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$, we have Moreover, $B(\omega,\lambda)\geqslant A(\widehat\omega,\widehat\lambda)\geqslant \min \bigl(G_{\mathrm{lin}}(\widehat\omega),G_{\mathrm{sim}}(\widehat \lambda)\bigr)$. Remark 1. In case $n= 2$ all the four quantities in Proposition 2 coincide giving rise to the relation 3.2. The inequalities by Bugeaud and LaurentThe inequalities (1.4) obviously follow from (1.7) and (1.6):
$$
\begin{equation}
\frac{1+\omega}{1+\lambda}\geqslant \widehat\omega\geqslant \frac{n-1}{1-\widehat \lambda},\qquad \frac{1+\omega^{-1}}{1+\lambda^{-1}}\geqslant \widehat\lambda\geqslant \frac{1-\widehat\omega^{-1}}{n-1}.
\end{equation}
\tag{3.7}
$$
Bugeaud and Laurent proved (1.4) with the help of the so-called intermediate Diophantine exponents by splitting Khintchine’s inequalities (1.3) into “going up” and “going down” chains of inequalities between consecutive intermediate exponents and improving on the first inequality in each chain with the respective uniform exponent. An analogous splitting of German’s inequalities (1.6) can be found in [6]. It would be interesting to establish a suitable splitting of Schmidt and Summerer’s inequalities (1.7). Such a splitting, combined with the existing splitting of (1.6), would provide a splitting of (1.4), alternative to the one of Bugeaud and Laurent. As for (3.7) itself, the way it splits (1.4) with the help of the respective fourth exponent gives much additional information in the case when one of the relations (1.4) is an equality. In this case the respective pair of the inequalities (3.7) becomes a pair of equalities and we get a whole range of equalities for the intermediate exponents mentioned above. It is worth mentioning in this context that there is an explicit description (obtained recently by Schleischitz [15]) of triples of exponents for which either the first pair of the inequalities (3.7), or the second one, is a pair of equalities. 3.3. The inequalities by SchleischitzAnalysis of the quantity $(\widehat\omega\widehat\lambda)^{1/(n{-}1)}$ appearing in Proposition 2 provides the following observation. Proposition 3. Suppose that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$. Then Proof. Applying (3.6) and (3.2) in both cases (3.4) and (3.5), we get
$$
\begin{equation*}
\frac{\omega}{\widehat\omega}\geqslant \max\biggl(\frac{1+\lambda}{1+ \omega^{-1}},\,G_{\mathrm{lin}}(\widehat\omega)\biggr)\geqslant \max\biggl(\frac{1-\widehat \omega^{-1}}{1-\widehat\lambda},\,G_{\mathrm{lin}}(\widehat\omega) \biggr)\geqslant (\widehat\omega\widehat\lambda)^{1/(n-1)}.
\end{equation*}
\notag
$$
Hence the first inequality in (3.8) follows immediately.
Furthermore,1 by the first inequality in (1.7) we have
$$
\begin{equation*}
\lambda\leqslant \frac{\omega-\widehat\omega+1}{\widehat\omega}= \frac{\omega^n}{\widehat \omega^{n+1}}-f\biggl(\frac{\omega}{\widehat\omega}\biggr),
\end{equation*}
\notag
$$
where $f(x)$ is as in (1.10). Since $f\bigl(\omega/\widehat\omega \bigr)\geqslant 0$ by (1.9), we get the second inequality in (3.8). The proposition is proved. In [14], Schleischitz proves two inequalities that very much resemble (3.8). They are of the form
$$
\begin{equation}
\widehat\omega\leqslant \frac{\lambda^{n-1}}{\widehat\lambda^n}+\psi(\lambda,\widehat \lambda),\qquad \omega\leqslant \frac{\lambda^n}{\widehat\lambda^{n+1}}+ \chi(\lambda,\widehat\lambda)
\end{equation}
\tag{3.9}
$$
with some non-negative $\psi(\lambda,\widehat\lambda)$ and $\chi(\lambda,\widehat \lambda)$ turning into zero only if $\lambda/\widehat\lambda=G_{\mathrm{sim}}(\widehat \lambda)$. One might wonder whether $\psi(\lambda,\widehat\lambda)$ or $\chi(\lambda,\widehat\lambda)$ can be removed from (3.9), so that it would resemble (3.8) closer. However, the answer is negative since it follows from the results of Kleinbock, Moshchevitin, and Weiss [16] that there exist many $\boldsymbol\theta$ with small $\lambda$ and infinite $\widehat\omega$. Their argument can be modified to show that, preserving $\lambda$ small, we can make $\widehat\omega$ finite, but however large. Thus, it is impossible to substitute the extra summands with zero in the general case.
§ 4. Nesterenko’s linear independence criterionIn 1985 Nesterenko published his famous linear independence criterion. His original proof was rather involved. A simpler argument can be found in [17] and [18]; see also [19]. Our Lemma 1 provides an even simpler proof. As before, let us fix $\boldsymbol\theta=(\theta_1,\dots,\theta_n)\in \mathbb{R}^n$. Let $\ell$, $\ell^\perp$, $r(\,\cdot\,)$, $h(\,\cdot\,)$ be defined as at the beginning of Section 2. Given any subspace $\mathcal{L}$ of $\mathbb{R}^{n+1}$, we write $\varphi(\mathcal{L})$ for the tangent of the angle between $\ell$ and $\mathcal{L}$. If $\mathcal{L}$ is defined over $\mathbb{Q}$, we write $H(\mathcal{L})$ for its height, that is, the covolume of the lattice $\mathcal{L}\cap \mathbb{Z}^{n+1}$. Theorem 1 (Nesterenko, 1985). Let $\alpha$, $\beta$, $c_1$, $c_2$, $\varepsilon$ be positive real numbers, $\beta\geqslant \alpha$. Let $(t_k)_{k\in\mathbb{N}}$ be an increasing sequence of positive real numbers such that 4.1. A simple proof of Nesterenko’s theoremLet us derive Theorem 1 from Lemma 1. We split our argument into three steps. Suppose that the hypothesis of Theorem 1 holds. Step 1: Adjusting the hypothesis. Let $\varepsilon'$ be a positive real number sufficiently small with respect to $\varepsilon$. Set $\alpha'=(1-\varepsilon')\alpha$, $\beta'=(1+\varepsilon')\beta$. Let us show that for every real $t$ large enough there exists $\mathbf x\in \mathbb{Z}^{n+1}$ such that
$$
\begin{equation}
r(\mathbf x)<t,\qquad t^{-\beta'}<h(\mathbf x)<t^{-\alpha'}-t^{-\beta'}.
\end{equation}
\tag{4.4}
$$
For every large $t$ choose $k$ so that $t_k<t\leqslant t_{k+1}$. It follows from (4.1) that $t_{k+1}<t_k^{1+\varepsilon'/2}$ if $k$ is large enough. Hence for $\mathbf x\in\mathbb{Z}^{n+1}$ satisfying (4.2) we have $r(\mathbf x)\leqslant t_k< t$ and
$$
\begin{equation*}
t^{-\beta'}< c_1t_k^{-\beta}\leqslant h(\mathbf x)\leqslant c_2t_k^{-\alpha}< c_2t^{-\alpha(1-\varepsilon'/2)}< t^{-\alpha'}-t^{-\beta'}
\end{equation*}
\notag
$$
if $k$ and $t$ are large enough. Thus (4.4) holds. Step 2: Proof for $d= 1$. Let $\mathcal{L}$ be the one-dimensional subspace generated by a primitive vector $\mathbf v\in\mathbb{Z}^{n+1}$ (primitive means that the coordinates of $\mathbf v$ are coprime). Then $\varphi(\mathcal{L})=r(\mathbf v)/h(\mathbf v)$ and $H(\mathcal{L})=\sqrt{r(\mathbf v)^2+h(\mathbf v)^2}$. Let us suppose that $h(\mathbf v)$ is large and that
$$
\begin{equation*}
r(\mathbf v)<h(\mathbf v)^{1-\delta'},\qquad \delta'=\frac{1+\beta'}{\alpha'}.
\end{equation*}
\notag
$$
Set $t=h(\mathbf v)^{1/\alpha'}$. Then $h(\mathbf v)=t^{\alpha'}$ and $r(\mathbf v)=t^{\alpha'-1-\beta''}$ with some $\beta''>\beta'$. By Lemma 1 there are no integer points $\mathbf x$ such that
$$
\begin{equation*}
r(\mathbf x)<t,\qquad t^{-\beta''}\leqslant h(\mathbf x)\leqslant t^{-\alpha'}-t^{-\beta''}.
\end{equation*}
\notag
$$
This contradicts the existence of an integer point $\mathbf x$ satisfying (4.4) if $t$ is large enough. Thus, if $h(\mathbf v)$ is large enough, we have This yields (4.3) for $d= 1$ since $\varepsilon'$ can be arbitrarily small. Step 3: Proof for $d\geqslant 2$. Suppose that $2\leqslant d<(1+\beta')/(1+\beta'-\alpha')$. Then, in particular,
$$
\begin{equation}
\beta'>\alpha'>1,\qquad \smash[t]{1<\delta'<\frac{d}{d-1}}.
\end{equation}
\tag{4.6}
$$
Given a $d$-dimensional subspace $\mathcal{L}$ defined over $\mathbb{Q}$, let $\ell'$ be the one-dimensional subspace generated by the point in $\mathcal{L}$ closest to the point $(1,\theta_1,\dots,\theta_n)$. Then the angle between $\ell$ and $\mathcal{L}$ equals the angle between $\ell$ and $\ell'$. Furthermore, given any $\mathbf x\in\mathcal{L}$, we write $r'(\mathbf x)$ and $h'(\mathbf x)$ for the Euclidean distances from $\mathbf x$ to $\ell'$ and $\mathcal{L}\cap\ell^\perp$ respectively. By Minkowski’s convex body theorem, for each positive $t$, there is a non-zero integer point $\mathbf x$ in $\mathcal{L}$ such that
$$
\begin{equation}
h'(\mathbf x)\leqslant t,\qquad \smash[t]{r'(\mathbf x)\leqslant c_d \biggl(\frac{H(\mathcal{L})}{t}\biggr)^{1/(d-1)}}
\end{equation}
\tag{4.7}
$$
with some positive number $c_d$ depending only on $d$. Set
$$
\begin{equation}
t_0=\bigl((4c_d)^{d-1}H(\mathcal{L})\bigr)^{1/(d-(d-1)\delta')}.
\end{equation}
\tag{4.8}
$$
Choose a non-zero integer point $\mathbf v$ in $\mathcal{L}$ (see Fig. 2), so that, in accordance with (4.7),
$$
\begin{equation}
h'(\mathbf v)\leqslant t_0,\qquad r'(\mathbf v)\leqslant c_d \biggl(\frac{H(\mathcal{L})}{t_0}\biggr)^{1/(d-1)}.
\end{equation}
\tag{4.9}
$$
It follows from (4.6) that $t_0/H(\mathcal{L})\to \infty$ and $r'(\mathbf v)\to 0$ as $H(\mathcal{L})\to \infty$. We may also assume that $\varphi(\mathcal{L})\to 0$ as $H(\mathcal{L})\to \infty$, neglecting most of subspaces. Then $h'(\mathbf v)\to \infty$ as $H(\mathcal{L})\to \infty$ and
$$
\begin{equation}
h(\mathbf v)\leqslant h'(\mathbf v)< 2h(\mathbf v).
\end{equation}
\tag{4.10}
$$
Having thus chosen $\mathbf v$ and assuming $H(\mathcal{L})$ to be large enough, we apply (4.5), (4.6), (4.8)–(4.10) and estimate $\varphi(\mathcal{L})$ as follows:
$$
\begin{equation*}
\begin{aligned} \, \varphi(\mathcal{L}) &\geqslant \frac{r(\mathbf v)}{h(\mathbf v)}-\frac{r'(\mathbf v)}{h(\mathbf v)}> h(\mathbf v)^{-\delta'}-2\frac{r'(\mathbf v)}{h'(\mathbf v)}\geqslant h'(\mathbf v)^{-\delta'}-\frac{2c_d(H(\mathcal{L}))^{1/(d-1)}}{t_0^{1/(d-1)}h'(\mathbf v)} \\ &\geqslant h'(\mathbf v)^{-\delta'}\biggl(1-\frac{2c_d(H(\mathcal{L}))^{1/(d-1)}}{t_0^{1/(d-1)}h'(\mathbf v)^{1-\delta'}}\biggr) \geqslant t_0^{-\delta'}\biggl(1-\frac{2c_d(H(\mathcal{L}))^{1/(d-1)}}{t_0^{-\delta'+d/(d-1)}}\biggr) \\ &=\frac{t_0^{-\delta'}}2= \frac12 \bigl((4c_d)^{d-1}H(\mathcal{L})\bigr)^{-\delta'/(d-(d-1)\delta')}. \end{aligned}
\end{equation*}
\notag
$$
Taking into account that $\delta'/(d-(d-1)\delta')=(1+\beta')/(1+\beta'-d(1+ \beta'-\alpha'))$, we get (4.3), as $\varepsilon'$ can be arbitrarily small. 4.2. Concerning the linear independence criterion itselfTheorem 1 yields the estimate
$$
\begin{equation}
\smash[t]{\dim_{\mathbb{Q}}(\mathbb{Q}+\mathbb{Q}\theta_1+\dots+\mathbb{Q}\theta_n)\geqslant \frac{1+\beta}{1+\beta-\alpha}}.
\end{equation}
\tag{4.11}
$$
However, this estimate easily follows from (4.3) with $d= 1$, that is, we may actually avoid Step 3 if (4.11) is our aim. Indeed, if $\ell$ is contained in a $d$-dimensional subspace $\mathcal{L}$ defined over $\mathbb{Q}$, then Dirichlet’s approximation theorem guarantees that there are infinitely many integer points $\mathbf x$ in $\mathcal{L}$ such that
$$
\begin{equation*}
r(\mathbf x)\leqslant c(d,H(\mathcal{L}))h(\mathbf x)^{-1/(d-1)}
\end{equation*}
\notag
$$
with some positive $c(d,H(\mathcal{L}))$. But if $d<(1+\beta')/(1+\beta'-\alpha')$, we have $\delta'<d/(d-1)$, where $\alpha'$, $\beta'$, $\delta'$ are as in Section 4.1, and Step 2 tells us that the opposite holds for every integer point $\mathbf x$ with $h(\mathbf x)$ large enough:
$$
\begin{equation*}
r(\mathbf x)\geqslant h(\mathbf x)^{1-\delta'}> c(d,H(\mathcal{L}))h(\mathbf x)^{-1/(d-1)}.
\end{equation*}
\notag
$$
Thus, there is no such $\mathcal{L}$, and we get (4.11). 4.3. A slight refinement in the case of linear independenceTheorem 1 is mainly applied to prove linear independence of the numbers $1,\theta_1,\dots,\theta_n$ over $\mathbb{Q}$. But if it is already known that they are linearly independent, the estimate (4.3) in Theorem 1 can be improved for $d=n$. In this case, the restriction on $\alpha$ and $\beta$ in Theorem 1 implies that $\alpha>n-1$. For $\alpha> n$ the following statement provides a stronger inequality than (4.3). It also provides a lower bound for $\omega=\omega(\boldsymbol\theta)$. Proposition 4. Under the hypothesis of Theorem 1, suppose additionally that $1,\theta_1,\dots,\theta_n$ are linearly independent over $\mathbb{Q}$ and that Proof. It follows from (4.4) and (4.5) of Steps 1 and 2 that
$$
\begin{equation*}
\widehat\omega\geqslant \alpha,\qquad \lambda\leqslant \frac{1+\beta}{\alpha}-1.
\end{equation*}
\notag
$$
By (1.4) we also have
$$
\begin{equation*}
\frac{1+\omega^{-1}}{1+\lambda^{-1}}\geqslant \frac{1-\widehat\omega^{-1}}{n-1}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\omega\leqslant \frac{n-1}{(1-\widehat\omega^{-1})(1+\lambda^{-1})-(n-1)}\leqslant \frac{(n-1)\alpha(1+\beta-\alpha)}{(\alpha-1)(1+\beta)-(n-1)\alpha(1+ \beta-\alpha)}.
\end{equation*}
\notag
$$
Here we have used the fact that (4.12) implies positiveness of the denominators in both fractions above. This proves (4.13).
Furthermore, let $\mathbf w$ be the integer normal of $\mathcal{L}$, that is, the unique (up to sign) non-zero primitive integer vector orthogonal to $\mathcal{L}$. Bounds for $\varphi(\mathcal{L})$ and $\omega$ are obviously related, as $H(\mathcal{L})=\sqrt{h(\mathbf w)^2+r(\mathbf w)^2}$ and the angle between $\mathcal{L}$ and $\ell$ is equal to the angle between $\mathbf w$ and $\ell^\perp$. Hence
$$
\begin{equation*}
\varphi(\mathcal{L})= \frac{h(\mathbf w)}{r(\mathbf w)}\geqslant cH(\mathcal{L})^{-\omega-1-\varepsilon}\geqslant cH(\mathcal{L})^{-\delta-\varepsilon}
\end{equation*}
\notag
$$
with some positive $c$ depending only on $n$ and $\varepsilon$. Thus, (4.14) is also proved. The proposition is proved. AcknowledgementsThe authors are grateful to Johannes Schleischitz for a series of useful comments, especially, for drawing the authors’ attention to Schmidt and Summerer’s paper [8] once again. The first author is a winner of the “Junior Leader” contest conducted by the Foundation for the Advancement of Theoretical Physics and Mathematics “BASIS” and would like to thank its sponsors and jury.
Citation in format AMSBIB
\Bibitem{GerMos23} |
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