0
$\begingroup$

I have some questions on lower bounds on the rank of unimodular lattices given the bilinear pairing of a subset of its basis is known.

  1. Let $\Lambda$ be an odd, unimodular matrix of signature $(1,T)$. In particular, there exists a basis in which the bilinear form $B$ of the lattice is given by \begin{equation} B = \rm diag(1,-1,\cdots,-1) \,. \end{equation} Now let us assume that there exists $k$ mutually orthogonal lattice vectors $x_1,\cdots,x_k$ with norm $-n < -1$, i.e., \begin{equation} x_i \cdot x_j = -n \delta_{ij} \,, \end{equation} that are further assumed to be elements of a basis of the lattice $\Lambda$. Given the existence of these vectors, how big does $T$ have to be to accommodate them? In other words, what is the lower bound $b(n,k)$ on $T$?

  2. There are some trivial bounds we can get. For example, \begin{equation} b(n,k) \geq k \,. \end{equation} Also, it is clear that \begin{equation} b(n,k) \leq nk \,. \end{equation} Can we get a tighter range or better lower-bounds on $b(n,k)$? For example, it can be shown that \begin{equation} b(n,k) \geq 2k \,. \end{equation} Can one do better? (One might mistakenly think that $b(m^2,k)=k < 2k$, but the assumption is that $\{ x_i : i =1,\cdots, k \}$ is a subset of the basis of the lattice. Hence, for example, $b(4,1) = 3$, rather than $1$.)

  3. Can one find better linear lower-bounds? In other words, does there exist a number $\alpha_n >2$ such that \begin{equation} b(n,k) \geq \alpha_n k +\beta \end{equation} for some constant $\beta$?

  4. Now let us assume there exist $k$ mutually orthogonal basis vectors $x_1, \cdots, x_k$ of $\Lambda$, all of which have negative norms, i.e., \begin{equation} (x_i \cdot x_j) = {\rm diag} (-n_1, \cdots , -n_k) \,, \end{equation} with $n_i > 1$. What is the smallest allowed value $b(n_1,\cdots,n_k)$ for $T$ in this case?

  5. Can one imagine there being good linear bound(s) \begin{equation} b(n_1,\cdots,n_k) \geq \gamma_{n_1} + \cdots + \gamma_{n_k} +\beta \end{equation} for some $\beta$ in this more generalized case as well? In other words, does there exist $\gamma_n >1$ for each $n$ that gives such linear bounds?

  6. I have only been able to get some very loose bounds using only very elementary tools so far. Where should I look to learn methods that can be used to effectively tackle with problems of this sort?

Thank you!

$\endgroup$
1
$\begingroup$

$b(n,k)\ge2k$ is a pretty good bound. You definitely have $b(n,k)\le2k+\mathrm{const}$ with a constant not very large (something like at most $8$, but you can make it better depending on $kn\bmod8$). Here is the argument.

The situation with even form is somewhat simpler and everything follows from

MR0525944 (80j:10031) Nikulin, V. V. Integer symmetric bilinear forms and some of their geometric applications. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 43 (1979), no. 1, 111–177, 238. 10C05 (14G30 14J17 14J25 57M99 57R45 58C27)

(There is an English translation.)

Edit: Here is a proof of the fact that $2k+8$ is enough. (All refs are to the above paper.) It suffices to construct an odd lattice $T$ (the prospective orthogonal complement) with inertia indices $(1,\sigma_-)$ and discriminant $k\langle\frac1n\rangle$, cf. Proposition 1.6.1 (where discriminant are considered with bilinear forms only, disregarding the quadratic form). Let $r=1+\sigma_-$ be the rank. We start with an even lattice $T$, see Theorem 1.10.1. (If $n$ is even, we can play with recovering the quadratic discriminant form; if $n$ is odd, it is unique.) Since signature $\bmod 8$ is determined by the discriminant form (Brown invariant), we may have to add up to $7$ to $\sigma_-$ to make it right. Next, since the length of the discriminant form is $k$, Nikulin's theorem will guarantee the existence (provided that the signature is right) whenever $r>k$. Thus, we can always find an even lattice $T$ of rank at most $k+7$. Now, just add $[-1]$ to make it odd (of rank at most $k+8$). By Proposition 1.6.1, there is a primitive extension of your lattice $S$ to a unimodular lattice such that $S^\perp\cong T\oplus[-1]$.

If you are fighting for every single unit, one can compute more precisely the Brown invariant (which depends on $k$ and the prime factors of $n$), there is no need in $[-1]$ if $n$ is odd, and, if $n$ is even, one can probably try to avoid that $[-1]$ by a "wrong" gluing of the two lattice (choosing an anti-isometry of the bilinear forms which is not an anti-isometry of quadratic ones). However, constructing an odd form $T$ in the first place would probably give a better result; unfortunately, I do not know any existence theorems for this case. And, as you've correctly noticed yourself, there is no way to improve that more than by $8$ :)

$\endgroup$
  • $\begingroup$ Thanks for the answer. I'm surprised at your claim that $b(n,k) \leq 2k + {\rm const}$---can you explain this in more detail? I expected $\alpha_n$ to be larger than $2$ for, say, $n=12$. Note that I've fixed the signature of the lattice to be $(1,T)$ and assumed that the lattice is odd. $\endgroup$ – D. S. Park Dec 23 '14 at 15:35
  • $\begingroup$ I will try to put a short proof to the answer. $\endgroup$ – Alex Degtyarev Dec 23 '14 at 17:59
  • $\begingroup$ Accidentally, the answer to your Question 5 is also $\le2k+8$, with the same proof :) $\endgroup$ – Alex Degtyarev Dec 23 '14 at 18:19
  • $\begingroup$ Thank you very much! Your explanation is crystal clear now. :) $\endgroup$ – D. S. Park Dec 23 '14 at 21:12
  • $\begingroup$ @D.S.Park: Just one last addition: sometimes, you have to add $8$ to get the signature right and satisfy the hypothesis of Nikulin's theorem (without computing the determinant for $\mathrm{length}=\mathrm{rank}$). On the other hand, there's no need to add $[-1]$: if $n$ is odd, the result is odd automatically, if $n$ is even, you take the quadratic extension $\langle(n+1)/n\rangle$ rather than $\langle1/n\rangle$, so the quadratic forms are not identified via an anti-isomorphism and the ambient lattice is still odd. Now, everything seems taken care of, and it's still $2k+8$ :) $\endgroup$ – Alex Degtyarev Dec 23 '14 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.