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!



Lower bounds on the rank of a unimodular lattice, given the binlinear pairing of a subset of basis vectors
$b(n,k)\ge2k$ 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 $8$ :) 


