Let $M$ be a $n$-dimensional smooth manifold and let $\pi:E\to M$ be a $r$-dimensional smooth vector bundle over $M$. Now assume that we have the following situation:
$\Lambda$ is a submanifold of $E$ such that each fibre of the map $\pi_\Lambda:\Lambda\to M$, denoted by $\Lambda_p\subset \mathbb R^r$, is a lattice of maximal rank.
This situation is quite common and happens for instance in the case of fibrations of abelian varities.
I don't know how to prove the following statement:
There exist $r$ smooth section $s_i\colon U\to E$ ( with $U\subset M$, $i=1,\ldots r$) such that $s_1(p),\ldots, s_r(p)\in\Lambda_p$ is a $\mathbb Z$-basis for the lattice $\Lambda_p$ for any $p\in U$.
In other words it should be possible to find a basis for the lattices in a smooth way. Do you have any suggestion?
As is, the claim is false. One needs a sort of transversality condition, i.e. that the restriction $\pi_{\Lambda}:\Lambda\to M$ has nowhere vanishing derivative, by the following argument.
Consider the trivial line bundle $\mathbb{R}\times (-1,1)\to(-1,1)$. The set $\Lambda=\{(a(x^{1/3}+1),x)\in \mathbb{R}| x\in (-1,1), a\in\mathbb{Z}\}$ defines a submanifold of $\mathbb{R}\times (-1,1)$, parametrized by $t\mapsto(a(t+1), t^3)$. As $\Lambda$ is generated by $(x^{1/3}, x)$, it always forms a maximal lattice in the trivial bundle. The only sections of a trivial bundle are given by $(f(x),x)$, but in order for such a section to lie in $\Lambda$ it necessarily is of the form $(a(x^{1/3}+1),x)$ which is not smooth.
If we assume that $\pi_{\Lambda}$ has nowhere vanishing deriviative, i.e. it is a local diffeomorphism, then given $p\in M$ and $q\in \pi_{\Lambda}^{-1}(p)$, there exists $U_q\subset \Lambda$ open such that $\pi_{\Lambda}\vert_{U_q}: U_{q}\to \pi(U_q)$ is a diffeomorphism. Take $B$ an ordered basis for $\Lambda_p$ Define $U=\cap_{q\in B}\pi_{\Lambda}(U_q)$. We see that $\pi_{\Lambda}\vert_{U_{q}\cap \pi^{-1}(U)}:U_{q}\cap \pi^{-1}(U)\to U $ is still a diffeomorphism for $q\in B$. These maps have smooth sections $\sigma_q:U\to \pi_{\Lambda}^{-1}(U)$ given by taking the inverse of $\pi_{\Lambda}\vert_{U_q\cap\pi^{-1}(U)}$. Now since these $\sigma_q$ are sections of $\pi_{\Lambda}$, if they are $\mathbb{R}$ linearly independent over $U$, they form a $\mathbb{Z}$ basis for $\Lambda$ at each point. Now with respect to a local trivialization $V\subset U$ of $E$, each $\sigma_q$ is represented by a map $f_q:V\to\mathbb{R}^r$ and so the whole package of the basis gives a map $f_B: V\to \mathrm{Mat}_{n\times n}(\mathbb{R})$ such that $(\sigma_{q_1}(p'),\cdots, \sigma_{q_r}(p'))$ are linearly independent in $E_p$ if and only if $f_B(p')$ is invertible. Taking $g=\det\circ f_B$, we see that the points at which $f_B$ is linearly independent are given by $W=g^{-1}(\mathbb{R}^*)$ which is an open set of $V$ containing $p$. Over $W$, the $\sigma_q$'s give the desired sections.