On the orthogonal group of a lattice on a quadratic space over dyadic local field

Question

Let $F$ be a local field with valuation ring $R$. $V$ is a n dimensional non-singular quadratic space over $F$ with bilinear form $B$ and quadratic map $Q$. As usual, $O(V)$ denotes the orthogonal group of $V$. Given a lattice $L$ on $V$. The orthogonal group of $L$ is defined to be $$O(L)=:\{\sigma\in O(V):\ \sigma (L)=L\}.$$

Clearly, for a maximal anisotropic vector $u\in V$, the symmetry $\tau_u\in O(L)$ if and only if $\frac{2B(u,L)}{Q(u)}\in R$. When $F$ is non-dyadic, this is equivalent to saying that the sublattice $Ru$ splits in $L$. And by induction on the dimension $n$, it is easy to show that each element of $O(L)$ is a product of at most $2n-1$ symmetries in $O(L)$.

Along this line, when $F$ is dyadic, are there any results on $O(L)$ similar to the non-dyadic case?

Answer 1

I don't think that it is true for dyadic local fields. But I think in that case $O(L)$ is generated by symmetries and Eichler-transformations. Try the paper ``Generation of integral orthogonal groups over dyadic local fields" by Fei Xu, appeared in Pacific J, 1995.