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.

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...

  • 1
    $\begingroup$ seems to me there is a sublattice of dimension $\lceil \frac{n}{2} \rceil$ on which your nondegenerate form is definite. Does that suffice? $\endgroup$
    – Will Jagy
    Commented Aug 25, 2015 at 20:48
  • $\begingroup$ Sorry! I misstated the question: I should add an assumption that this 2-dimensional lattice is not definite! $\endgroup$ Commented Aug 25, 2015 at 21:38
  • $\begingroup$ Oh, that probably does demand some kind of diophantine approximation argument, then; I don't quite recall what you commented earlier on the proof you found not really to your liking. $\endgroup$
    – Will Jagy
    Commented Aug 25, 2015 at 22:16
  • $\begingroup$ Presumably you want your lattice not to be a multiple of any other lattice? $\endgroup$ Commented Aug 25, 2015 at 22:48
  • $\begingroup$ definitely! thanks for pointing this out, I amended the question $\endgroup$ Commented Aug 26, 2015 at 10:59

3 Answers 3


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.

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    $\begingroup$ I'm not sure I understand whether this is an actual proof or just a sketch? But it definitely seems like the sketch which comes closest to working, and I am supposed to give out the bounty, so I'll give it to you. If this is a actual proof, I'd love to understand how to fill in the gaps. (Why does this $\Lambda$ necessarily embed into $\Lambda_0$, for example?) $\endgroup$ Commented Sep 9, 2015 at 1:47
  • $\begingroup$ Thanks. There are two questions here: constructing $\Lambda_0$ and proving it embeds into $\Lambda$. (Not $\Lambda$ into $\Lambda_0$, right?) The former I gave a sketch of that can be completed with little difficulty. The latter, well it should certainly be OK for all but the smallest ranks (if $\Lambda$ has some really bad primes it might accommodate only a sublattice of $\Lambda_0$, but that's good enough. If $\Lambda$ has rank only $3$ it gets trickier but even there you can usually do anything that's not forbidden by a local obstruction, so some $M$ should work [cont'd] $\endgroup$ Commented Sep 9, 2015 at 2:13
  • $\begingroup$ ...(up to finite index and variations such as changing $M^4+1$ to $M^4-1$). $\endgroup$ Commented Sep 9, 2015 at 2:14
  • $\begingroup$ I still don't quite understand how it can be embedded to a given lattice (non-unimodular). For unimodular it's fine, except that I still cannot find a proper reference for an embedding result even in this generality (Morrison quotes Nikulin, but this result is not easy to find in Nikulin's paper). $\endgroup$ Commented Sep 9, 2015 at 6:56
  • $\begingroup$ One standard approach is to "glue" $\Lambda_0$ to an appropriate rank-1 lattice $M$, obtaining a determinant-1 lattice of rank 3 that contains $\Lambda_0 \oplus M$ with finite index, and is thus isomorphic with $\Lambda$ and contains a copy of $\Lambda_0$. This can be done for non-unimodular $\Lambda$ too as long as $\Lambda$ is the only lattice in its genus (which is common in the indefinite case). $\endgroup$ Commented Sep 10, 2015 at 0:08

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.


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.

  • $\begingroup$ In any case this approach must fail in rank $n=3$ because then there are only finitely many $N$ for which a lattice of the form $N\cdot H$ can embed primitively into $\Lambda$. $\endgroup$ Commented Sep 1, 2015 at 23:16
  • $\begingroup$ thanks! anyway, rank $\geq 6$ or $\geq 7$ is a usual assumption in these kind of applications $\endgroup$ Commented Sep 2, 2015 at 18:10
  • $\begingroup$ But then it feels like the $N \cdot H$ construction or something much like it does work even when $\Lambda$ is not unimodular. $\endgroup$ Commented Sep 2, 2015 at 18:43
  • $\begingroup$ Yes, and this seems to give an answer indeed. I will post it in a couple of days, once I am entirely sure there are no errors $\endgroup$ Commented Sep 8, 2015 at 12:40
  • $\begingroup$ strike the previous comment, my proof was based on false assumption that any lattice is commeasurable with one which is proportional to unimodular $\endgroup$ Commented Sep 9, 2015 at 6:50

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