2-dimensional sublattices with all vectors having very big square (in absolute value)

Question

QUESTION: Let $\Lambda\times\Lambda\rightarrow {\Bbb Z}$ be a lattice, that is, ${\Bbb Z}^n$ with a non-degenerate integer quadratic form, not definite, not necessarily unimodular, $n>2$. I want to show that for each $N>0$ there exists a 2-dimensional primitive sublattice $\Lambda_0\subset \Lambda$, also not definite, such that all nonzero vectors $x\in \Lambda_0$ satisfy $|(x,x)| > N$.

This question is motivated by considering a K3 surface (or a hyperkahler manifold) with 2-dimensional Picard lattice. It is known that the squares of minimal rational curves are bounded, and we want to find a manifold with Picard number 2 having no minimal rational curves.

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Bounty added This problem has kept me up late the last two nights, so I'm hoping some of the quadratic form experts will work on it.

I tried to find a strategy working with rational quadratic forms rather than integers, but failed for the following reason: Equip $\mathbb{Q}^3$ with the quadratic form $x^2+y^2-z^2$. Then I claim that any rank two subspace $L$ on which this form is non-degenerate contains a vector of norm $1$.

Proof: Let $L^{\perp} = \mathbb{Q} v$. Since our form is nondegenerate on $L$, we have $\langle v,v \rangle \neq 0$, say $\langle v,v \rangle = N$, and $\mathbb{Q}^3 = L \oplus \mathbb{Q} v$. Now, $x^2+y^2-z^2$ is equivalent to $N (x')^2 + (y')^2 - N (z')^2$, by the change of variables $(x,y,z) = (\tfrac{N+1}{2} x' + \tfrac{N-1}{2} z', y', \tfrac{N-1}{2} x' + \tfrac{N+1}{2} z')$. So our form is equivalent to $L' \oplus \mathbb{Q} v$, where the form on $L'$ is $(y')^2 - N (z')^2$. By Witt cancellation, the forms on $L$ and $L'$ are equivalent. Since $(y')^2 - N (z')^2$ represents $1$, so does our form on $L$. $\square$

So, when we are trying to construct rank two sub-lattices with no vectors of norm $1$, we have to do so using lattices which do represent $1$ rationally. This seems hard to me...

Answer 1

For the record: OP Misha Verbitsky writes that "[$\Lambda$ of] rank $\geq 6$ or $\geq 7$ is a usual assumption in these kind of applications", in which case Ekaterina Amerik's suggestion of using $N\cdot H$ (for some fixed indefinite rank-$2$ form $H$) surely yields the easiest construction; but there are also primitive indefinite forms $\Lambda_0$ of rank $2$ that represent no small nonzero integers of either sign, and those should work even for $\Lambda$ of rank as small as $3$ (as in the lattice that troubled the OP in the first place).

One reasonably simple construction is to take $M>N$ odd and set $$ Q(x,y) = \frac1M ((x+My)^2 - (M^4+1)x^2) = -M^3 x^2 + 2xy + M y^2, $$ which represents $\pm M$ (as for $(x,y)=(0,1)$ and $(M,M^2-1)$) but no integer of smaller absolute value other than zero. This can be checked using the fact that the associated real quadratic field ${\bf Q}(\sqrt{M^4+1})$ has fundamental unit $M^2 + \sqrt{M^4+1}$ (of norm $-1$) as small as possible. Moreover, unlike the forms $N \cdot H$, this $Q$ is primitive, so any embedding of the into $\Lambda$ of the indefinite lattice $\Lambda_0$ associated to $Q$ is automatically primitive as well.

Answer 2

This doesn't solve the question : it shows that a non-degenerate lattice always contains a primitive rank 2 sublattice with the required property regarding norms ... but the sublattice found is definite. Maybe a modification of the argument would yield the desired result, so I post it.

Without loss of generality we may assume $\Lambda$ is a $3$-dimensional maximal lattice.

Let us write $B_N=\{v\in\Lambda, (v,v)\leq N\}$.

The positive definite case seems easy : for a given $N$, the ball $B_N$ in $\Lambda$ contains only a finite number of vectors, and there certainly exists a $2$-dimensional lattice that doesn't intersect $B_N-\{0\}$.

The lorentzian case is a little subtler since then $B_N$ is not finite anymore. Here is an argument : let us see $\Lambda$ as a lattice in the standard Lorentzian space $\mathbf R^3$ with the form $q(x,y,z)=-x^2+y^2+z^2$ and let us choose the embedding so that the intersection of $\Lambda-\{0\}$ and the hyperplane $P$ defined by the equation $x=0$ is empty. The idea is that the set of vectors of $B_N$ that lie near $P$ is finite, while there are infinitely many hyperplanes of $\Lambda$ that are as near as you want from $P$.

More precisely : let $x_N$ be the minimum of $\vert x\vert $ on $B_N-\{0\}$ (the chosen embedding implies $x_N>0$), let $B_N'$ be the finite set of vectors of the form $(\pm x_N,y,z)$ in $B_N$ . Let $y_N$ (resp. $z_N$) be the maximum of $y$ (resp. $z$) on $B'_N$.

Finally let $C$ be the set of vectors $(a,b,c)$ satisfying $\vert b\vert \leq \vert \frac{x_N}{3y_N} a\vert $ and $\vert c\vert\leq \vert \frac{x_N}{3z_N} a\vert$. This is a cone with a non-empty interior in the lorentzian space. Thus its intersection with $\Lambda-\{0\}$ is non-empty. Let us pick a vector $v=(a,b,c)$ therein, with $a\neq 0$. Then for a vector $(x,y,z)$ of the ortogonal complement of $v$, the inequalities $\vert y\vert\leq y_N$ and $\vert z\vert\leq z_N$ imply the inequality $x<x_n$. Thus we have $v^\perp\cap B_N=\{0\}$, and $v^\perp\cap\Lambda$ is a solution.

Answer 3

I am grateful to Ekaterina Amerik for this remark.

This is not an answer, but in a special case (not the one I need) the solution is known. In David Morrison's paper "On K3 surfaces with large Picard number", Corollary 2.5, it is shown that any non-degenerate even lattice of appropriate signature and rank admits a primitive embedding to a given unimodular even lattice (I suppose that this is also true for odd lattices, though it's hard to find in the literature). Then we need to take a lattice $N\cdot H$ obtained from any given lattice of signature (1,1) by rescaling with a factor N, and it would be embeddable to a unimodular lattice, solving the problem for unimodular lattices.

Unfortunately, the applications that we have need lattices which are not unimodular.