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Setting Let $k$ be a real quadratic field, $\mathbb Z_k$ its ring of integers. Let $n$ be an even integer $A$ a symmetric $n$-by-$n$ matrix with coefficients in $\mathbb Z_k$. Let $L$ be the lattice $\mathbb Z_k^n$ with the quadratic form induced by $A$ and $O(L)$ its orthogonal group.

Assume that $2$ is inert in $k/\mathbb Q$ and the matrix $A$ is non-singular mod 2, and let $\bar A$ be its reduction. Then there is a map from $O(L)$ to the isometry group $G_1$ of the bilinear form $\bar f$ on $\mathbb F_4^n$ associated to $\bar A$. This is a symplectic group over $\mathbb F_4$. On the other hand there also is an orthogonal group $G_2$ associated with $A$, namely that of the quadratic form $\bar Q$ defined as follows : $$ \bar Q(x) = \sum_{1 \le i < j \le n} a_{i, j} x_ix_j. $$ Assume that all diagonal coefficients of $\bar A$ vanish, then $\bar Q$ is non-degenerate and we have the usual relation $$ (\ast) \qquad\bar f(v, w) = Q(v+w) - Q(v) - Q(w) $$ so in particular $G_2 \subset G_1$ (and this is a strict inclusion).

Question Now it is clear that $O(L)$ reduces mod 2 to a subgroup $H$ of $G_1$. On the other hand it seems not to be the case in general that it reduces to a subgroup of $G_2$. Is there a known relation between $H$ and $G_2$ in general? Has the structure of $H$ been studied?

Comments The hypothesis that all diagonal coefficients vanish mod 2 is a bit awkward but since the terms coming from them vanish in the right-hand side of $(\ast)$ it seems necessary to assume this for this relation to hold.

I used a quadratic field to be more concrete and because this is the setting i'm interested in but there's no reason not to ask the question for a general totally real number field ; likewise the hypothesis on the behaviour of 2 in the extension.

Context I have a specific quadratic lattice $L$ over the integer ring $\mathbb Z[a]$ (where $a^2-a-1=0$) associated to the matrix $$ \left(\begin{array}{rrrrrr} 2 & -a & 0 & 0 & 0 & 0 \\ -a & 2 & -1 & 0 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 & 0 \\ 0 & 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & 0 & -1 & 2 \end{array}\right) $$ and generators for $O(L)$ (a collection of seven reflections). When computing modulo 2 i found that the cardinalities of the group generated by the reductions of my generators and the cardinality of the split orthogonal group are quite close but still differ (for the record : 2036736000 versus 1974067200, the gcd is 31334400).

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  • $\begingroup$ The ratio of your orders looks like it might be $65/63$, which is equal to $|O_{12}^-(2)|/|O_{12}^+(2)|$ $\endgroup$ Commented Sep 19, 2023 at 18:32
  • $\begingroup$ That is so, in fact the cardinality of my first group is equal to $|O_6^-(4)|$, which would suggest i just made a dumb mistake and $O(L)$ reduces to this group mod 2. On the other hand the quadratic form appears to be of type +, as the basis vectors $e_1, e_3$ and $e_5$ span a totally isotropic subspace. $\endgroup$ Commented Sep 20, 2023 at 8:12
  • $\begingroup$ But do they span a totally singular subspace? $\endgroup$ Commented Sep 21, 2023 at 2:12
  • $\begingroup$ I think so, unless i'm missing some subtlety : from the matrix we have $\bar Q(e_i) = 0$ for all $i$ and $\bar f(e_i, e_j) = 0$ whenever $|i-j| > 1$. $\endgroup$ Commented Sep 21, 2023 at 7:58
  • $\begingroup$ I think the matrix gives $\bar f$ mod $2$ but not $\bar Q$ mod $2$ $\endgroup$ Commented Sep 21, 2023 at 12:35

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