Siegel’s Lemma Is Sharp

Abstract

Siegel’s Lemma is concerned with finding a “small” nontrivial integer solution of a large system of homogeneous linear equations with integer coefficients, where the number of variables substantially exceeds the number of equations (for example, n equations and N variables with N ≥ 2n), and “small” means small in the maximum norm. Siegel’s Lemma is a clever application of the Pigeonhole Principle, and it is a pure existence argument. The basically combinatorial Siegel’s Lemma is a key tool in transcendental number theory and diophantine approximation. David Masser (a leading expert in transcendental number theory) asked the question whether or not the Siegel’s Lemma is best possible. Here we prove that the so-called “Third Version of Siegel’s Lemma” is best possible apart from an absolute constant factor. In other words, we show that no other argument can beat the Pigeonhole Principle proof of Siegel’s Lemma (apart from an absolute constant factor). To prove this, we combine a concentration inequality (i.e., Fourier analysis) with combinatorics.

Appendix: Proof of the Third Version of Siegel’s Lemma

We combine the usual pigeonhole principle argument with probability theory; in particular, we borrow some ideas from the paper of Spencer [10]. We use the following variant of the large deviation theorem in probability theory. Bernstein’s inequality Let Z ₁, Z ₂, …, Z _N be real-valued independent random variables with zero expectation E Z _j = 0 and | Z _j | ≤ M, 1 ≤ j ≤ N. Then, for all positive τ > 0,

$$\displaystyle{\Pr \left [\left \vert \sum _{j=1}^{N}Z_{ j}\right \vert \geq \tau \right ] \leq 2\exp \left (-{ \tau ^{2}/2 \over \left (\sum _{j=1}^{N}\mathbf{E}Z_{j}^{2}\right ) + (\tau M/3)}\right ).}$$

Consider now the i-th row of the homogeneous linear system

$$\displaystyle{\sum _{j=1}^{N}d_{ i,j}x_{j} = 0,\mathrm{\ \ where\ \ }\vert d_{i,j}\vert \leq A.}$$

For a positive integer B ≥ 1 and an integer j in 1 ≤ j ≤ N, let X _j denote the random variable with $\Pr [X_{j} = b] ={ 1 \over 2B+1}$, where b runs over the integers − B, −B + 1, −B + 2, …, B. Moreover, assume that X ₁, X ₂, …, X _N are independent. We apply Bernstein’s inequality with Z _j = d _i, j X _j , $\tau =\lambda \sqrt{N}AB$ and M = AB:

$$\displaystyle{\Pr \left [\left \vert \sum _{j=1}^{N}Z_{ j}\right \vert \geq \lambda \sqrt{N}AB\right ] \leq 2\exp \left (-{ \lambda ^{2}NA^{2}B^{2}/2 \over \left (NA^{2}B^{2}\right ) + \left (\lambda \sqrt{N}A^{2}B^{2}/3\right )}\right ) =}$$

$$\displaystyle{ = 2\exp \left (-{ \lambda ^{2} \over 2 + 2N^{-1/2}\lambda /3}\right ). }$$

(109)

Let

$$\displaystyle{ \lambda _{h} = 6h,\ \ 1 \leq h \leq \log n. }$$

(110)

Then by (109), for every 1 ≤ h ≤ logn,

$$\displaystyle{\Pr \left [\left \vert \sum _{j=1}^{N}d_{ i,j}X_{j}\right \vert \geq 6h\sqrt{N}AB\right ] \leq 2\exp \left (-{ 36h^{2} \over 2 + (12N^{-1/2}h/3)}\right ) \leq }$$

$$\displaystyle{ \leq 2\exp \left (-{ 36h^{2} \over 2h + (12h/3)}\right ) = 2e^{-6h}. }$$

(111)

For every integer h in 1 ≤ h ≤ logn, define the random variable

$$\displaystyle{ Y _{h} = \left \vert \left \{i \in \{ 1,2,\ldots,n\}:\ \left \vert \sum _{j=1}^{N}d_{ i,j}X_{j}\right \vert \geq 6h\sqrt{N}AB\right \}\right \vert. }$$

(112)

By using (111), we obtain the following upper bound for the expected value of Y _h :

$$\displaystyle{ \mathbf{E}Y _{h} \leq 2e^{-6h}n,\ 1 \leq h \leq \log n. }$$

(113)

Since the random variable Y _h has non-negative values, we can use the simple Markov inequality stating that for any random variable Y with non-negative values and finite expectation

$$\displaystyle{\Pr [Y \geq a] \leq { \mathbf{E}Y \over a} \mathrm{\ \ for\ any\ }a> 0.}$$

Applying Markov inequality in (113) we have

$$\displaystyle{\Pr \left [Y _{h} \geq 2h(h + 1) \cdot 2e^{-6h}n\right ] \leq { 1 \over 2h(h + 1)},\ 1 \leq h \leq \log n.}$$

Using the telescoping sum

$$\displaystyle{\sum _{h=1}^{\infty }{ 1 \over h(h + 1)} =\sum _{ h=1}^{\infty }\left ({1 \over h} -{ 1 \over h + 1}\right ) = 1,}$$

we obtain that

$$\displaystyle{\sum _{1\leq h\leq \log n}{ 1 \over 2h(h + 1)} <{ 1 \over 2},}$$

so, with probability greater than 1∕2 we have

$$\displaystyle{ Y _{h} <2h(h + 1) \cdot 2e^{-6h}n\mathrm{\ \ for\ every\ }1 \leq h \leq \log n. }$$

(114)

(114) means that, with probability greater than 1∕2, the number of row-sums ∑ _{j = 1} ^N d _i, j X _j that have absolute value $\geq 6h\sqrt{N}AB$, is less than

$$\displaystyle{2h(h + 1) \cdot 2e^{-6h}n\mathrm{\ \ for\ every\ }1 \leq h \leq \log n.}$$

It follows that, with probability greater than 1∕2, the number of row-sums ∑ _{j = 1} ^N d _i, j X _j that have absolute value between $6h\sqrt{N}AB$ and $6(h + 1)\sqrt{N}AB$, is less than

$$\displaystyle{ 2h(h + 1) \cdot 2e^{-6h}n\mathrm{\ \ for\ every\ }1 \leq h \leq \log n, }$$

(115)

and there is no row-sum with absolute value $\geq 6\log n\sqrt{N}AB$. In the last step we used the fact that

$$\displaystyle{2h(h + 1) \cdot 2e^{-6h}n <1\mathrm{\ with\ }h = \lfloor \log n\rfloor }$$

(lower integral part).

Write (see (115))

$$\displaystyle{ k_{h} = \lfloor 2h(h + 1) \cdot 2e^{-6h}n\rfloor,\ \ 1 \leq h \leq \log n, }$$

(116)

and

$$\displaystyle{ k_{0} = n -\sum _{1\leq h\leq \log n}k_{h}. }$$

(117)

The total number of row-sum vectors (with n coordinates) satisfying (115) can be estimated from above by using the parameters k _h in (116)–(117) as follows:

$$\displaystyle{ \leq { n\choose k_{0}}\left (2 \cdot 6\sqrt{N}AB + 1\right )^{k_{0} }\prod _{1\leq h\leq \log n}\left ({n\choose k_{h}}\left (2 \cdot 6(h + 1)\sqrt{N}AB + 1\right )^{k_{h} }\right ). }$$

(118)

On the other hand, “with probability greater than 1∕2” means more than

$$\displaystyle{{ 1 \over 2}(2B + 1)^{N} }$$

(119)

possible vectors v ∈ {−B, −B + 1, −B + 2, …, B}^N.

So, if (119) is greater or equal to (118), than the Pigeonhole Principle applies, and there exist two different vectors

$$\displaystyle{\mathbf{v}_{1},\mathbf{v}_{2} \in \{-B,-B + 1,-B + 2,\ldots,B\}^{N}}$$

such that they generate the same row-vector, i.e., Dv ₁ = Dv ₂. Then x = v ₁ −v ₂ satisfies the homogeneous linear system Dx = 0 with

$$\displaystyle{ \max _{1\leq j\leq N}\vert x_{j}\vert \leq 2B. }$$

(120)

The rest is routine estimation. Clearly

$$\displaystyle{{n\choose k_{0}}\left (2 \cdot 6\sqrt{N}AB + 1\right )^{k_{0} }\prod _{1\leq h\leq \log n}\left ({n\choose k_{h}}\left (2 \cdot 6(h + 1)\sqrt{N}AB + 1\right )^{k_{h} }\right ) \leq }$$

$$\displaystyle{\leq { n\choose k_{0}}\left (13\sqrt{m}AB\right )^{k_{0} }\prod _{1\leq h\leq \log n}\left ({n\choose k_{h}}\left (13(h + 1)\sqrt{N}AB\right )^{k_{h} }\right ) =}$$

$$\displaystyle{ = \left (13\sqrt{N}AB\right )^{n}{n\choose k_{ 0}}\prod _{1\leq h\leq \log n}\left ({n\choose k_{h}}\left (h + 1\right )^{k_{h} }\right ). }$$

(121)

By using s! ≥ (s∕e)^s and (116), we have

$$\displaystyle{{n\choose k_{0}}\prod _{1\leq h\leq \log n}\left ({n\choose k_{h}}\left (h + 1\right )^{k_{h} }\right ) \leq { n\choose k_{0}}\prod _{1\leq h\leq \log n}\left (e^{6h}(h + 1)\right )^{k_{h} } \leq }$$

$$\displaystyle{\leq { n\choose k_{0}}\prod _{1\leq h\leq \log n}e^{7hk_{h} } ={ n\choose k_{0}}\exp \left (\sum _{1\leq h\leq \log n}7hk_{h}\right ) \leq }$$

$$\displaystyle{\leq { n\choose k_{0}}\exp \left (\sum _{1\leq h\leq \log n}7h \cdot 3h(h + 1)e^{-6h}n\right ) \leq }$$

$$\displaystyle{ \leq { n\choose k_{0}}\exp \left (21n\sum _{h=1}^{\infty }h^{2}(h + 1)e^{-6h}\right ) \leq 2^{n}e^{n}. }$$

(122)

Using (122) in (121), we have

$$\displaystyle{{n\choose k_{0}}\left (2 \cdot 6\sqrt{N}AB + 1\right )^{k_{0} }\prod _{1\leq h\leq \log n}\left ({n\choose k_{h}}\left (2 \cdot 6(h + 1)\sqrt{N}AB + 1\right )^{k_{h} }\right ) \leq }$$

$$\displaystyle{ \leq \left (13\sqrt{N}AB\right )^{n}2^{n}e^{n} <\left (70\sqrt{N}AB\right )^{n}. }$$

(123)

Combining (118), (119) and (123), it suffices to guarantee the inequality

$$\displaystyle{{ 1 \over 2}(2B + 1)^{N} \geq \left (70\sqrt{N}AB\right )^{n}. }$$

(124)

Inequality (124) clearly holds with

$$\displaystyle{ 2B = \left \lfloor \left (70\sqrt{N}A\right )^{n/(N-n)}\right \rfloor, }$$

(125)

and using (120) in (125), we conclude

$$\displaystyle{\max _{1\leq j\leq N}\vert x_{j}\vert \leq \left (70\sqrt{N}A\right )^{n/(N-n)},}$$

completing the proof of the Third Version of Siegel’s Lemma. □