2.1.1 Stabilizers of subspaces

For any subspace $L \subset \overline {\mathbb {Q}}^n$ , we define the following groups.

• • $\mathbf {H}_L<\mathrm {Spin}_Q$ is the identity component of the stabilizer group of L in $\mathrm {Spin}_Q$ for the action of $\mathrm {Spin}_Q$ on $\mathrm {Gr}_{n,k}$ .

• • $\mathbf {H}^{\prime }_L<\mathrm {SO}_Q$ is the identity component of the stabilizer group of L in $\mathrm {SO}_Q$ for the action of $\mathrm {SO}_Q$ on $\mathrm {Gr}_{n,k}$ .

Note that we have an isogeny $\mathbf {H}_L \to \mathbf {H}^{\prime }_L$ . Furthermore, the restriction to L (respectively, $L^\perp $ ) yields an isomorphism of $\mathbb {Q}$ -groups

(2.1) $$ \begin{align} \mathbf{H}^{\prime}_L \to \mathrm{SO}_{Q|_L} \!\times \mathrm{SO}_{Q|_{L^\perp}}. \end{align} $$

To see this, one needs to check that the image does indeed consist of special orthogonal transformations. This follows from the fact that the determinant of the restrictions is a morphism with finite image and hence its kernel must be everything by connectedness. In particular, we have the following cases.

• • If $k \geq 3$ and $n-k \geq 3$ , $\mathbf {H}^{\prime }_L$ (and hence also $\mathbf {H}_L$ ) is semisimple.

• • If $k = 2$ and $n-k \geq 3$ (or $k\geq 3$ and $n-k =2$ ), $\mathbf {H}^{\prime }_L$ is reductive.

• • If $k=2$ and $n-k=2$ (which is not a case this paper covers), $\mathbf {H}^{\prime }_L$ is abelian.

Remark 2.1. (Special Clifford groups and (2.1))

Although it might seem appealing to suspect that $\mathbf {H}_L$ is simply connected, this is actually false. The following vague and lengthy explanation is not needed in what follows. Denote by $\mathbf {M}$ the special Clifford group of Q and similarly by $\mathbf {M}_1$ (respectively $\mathbf {M}_2$ ) the special Clifford groups of $Q|_L$ (respectively, $Q|_{L^\perp }$ ) for the duration of this remark—cf. [Reference KnusKnu88]. These are reductive groups whose center is a one-dimensional $\mathbb {Q}$ -isotropic torus. We identify $\mathbf {M}_1,\mathbf {M}_2$ as subgroups of $\mathbf {M}$ and write $\mathbf {C}$ for the center of $\mathbf {M}$ which is, in fact, equal to $\mathbf {M}_1 \cap \mathbf {M}_2$ . The natural map $\phi : \mathbf {M}_1 \times \mathbf {M}_2 \to \mathbf {M}$ has kernel $\{(x,y) \in \mathbf {C}\times \mathbf {C}: xy = 1\}$ so that

Furthermore, we have the spinor norm which is a character $\chi : \mathbf {M} \to \mathbf {G}_m$ whose kernel is the spin group. Similarly, we have spinor norms $\chi _1,\chi _2$ for $\mathbf {M}_1$ (respectively, $\mathbf {M}_2$ ), which are simply the restrictions of $\chi $ . The above yields that

which is isogenous (but not isomorphic) to $\mathrm {Spin}_{Q|_L} \times \mathrm {Spin}_{Q|_{L^\perp }}$ .

The first result that we prove states that the group $\mathbf {H}_L$ totally determines the subspace L (up to orthogonal complements). This is given more precisely in the following proposition. Recall that a non-trivial subspace $W \subset V$ is non-degenerate if $\mathrm {disc}(Q|_W)\neq 0$ or, equivalently, if there is no non-zero vector $w \in W$ so that $\langle w,w'\rangle = 0$ for all $w' \in W$ . This notion is stable under extension of scalars.

The proposition follows directly from the following simple lemma.

Proof. First, we observe the following: over $\bar {\mathbb {Q}}$ , $\mathbf {H}^{\prime }_{L}$ acts transitively on the set of anisotropic lines in L and in $L^\perp $ . Indeed, by Witt’s theorem [Reference CasselsCas78, p. 20], the special orthogonal group in dimension at least two acts transitively on vectors of the same quadratic value. In any two lines, one can find vectors of the same quadratic value by taking roots.

Let $w \in W$ be anisotropic and write $w = w_1 + w_2$ for $w_1 \in L$ and $w_2 \in L^\perp $ . As w is anisotropic, one of $w_1$ or $w_2$ must also be anisotropic; we suppose that $w_1$ is anisotropic, without loss of generality. Let $h \in \mathbf {H}^{\prime }_{L}(\bar {\mathbb {Q}})$ be such that $hw_1 \neq w_1$ and $hw_2 = w_2$ . Then

$$ \begin{align*} u := hw-w = hw_1 -w_1 \in L \cap W. \end{align*} $$

We claim that we can choose h so that u is anisotropic. Indeed, as $w_1$ is anisotropic, its orthogonal complement in L is non-degenerate (as L is non-degenerate). We can thus choose h to map $w_1$ to a vector orthogonal to it by the above variant of Witt’s theorem. Then

$$ \begin{align*} Q(u) = Q(hw_1)+Q(w_1) = 2Q(w_1) \neq 0. \end{align*} $$

Now note that $L \cap W$ is $\mathbf {H}^{\prime }_L$ -invariant. By a further application of the above variant of Witt’s theorem and the fact that L is spanned by anisotropic vectors (L is non-degenerate), we obtain that $L \cap W = L$ or, equivalently, $L \subset W$ . Thus, we may write $W = L \oplus W'$ , where $W'$ is an orthogonal complement to L in W and, in particular, is contained in $L^\perp $ . The subspace $W'$ must be non-degenerate because W and L are, and hence it is trivial or contains anisotropic vectors. If $W'$ is trivial, then $W = L$ and the proof is complete. Otherwise, we apply the above variant of Witt’s theorem and obtain that $W' = L^\perp $ and $W = V$ .

An analogous statement holds for the relationship between quadratic forms and their special stabilizer groups.

For a proof, see [Reference Aka, Einsiedler and ShapiraAES16a, Lemma 3.3].

2.1.2 Maximality

We now aim to prove that, for any non-degenerate subspace L, the connected $\mathbb {Q}$ -groups $\mathbf {H}^{\prime }_L$ and $\mathbf {H}_L$ are maximal subgroups. Here, maximal means among connected and proper subgroups (as it was in Remark 1.13).

The result above is well known and due to Dynkin, who classified the maximal subgroups of the classical groups in [Reference DynkinDyn52] (see also the work of Liebeck and Seitz, for example [Reference Liebeck and SeitzLS98]). We will give an elementary proof.

Proof. Note that it suffices to prove the statement for $\mathbf {H}^{\prime }_L$ . As L is non-degenerate, we may choose an orthogonal basis of V consisting of an orthogonal basis of L and an orthogonal basis of $L^\perp $ . Let

$$ \begin{align*} M_Q = \begin{pmatrix} M_1 & 0 \\ 0 & M_4 \end{pmatrix} \quad \text{with} \ M_1,M_4 \ \text{diagonal matrices} \end{align*} $$

be the matrix representation of Q in this basis. Computing the Lie algebras of $\mathrm {SO}_Q$ and $\mathbf {H}^{\prime }_L$ we obtain

$$ \begin{align*} \mathfrak{g} := \mathrm{Lie}(\mathrm{SO}_Q) = \{ A \in \mathrm{Mat}(n) : A^TM_Q+M_QA = 0 \} \end{align*} $$

and

$$ \begin{align*} \mathfrak{h} := \mathrm{Lie}(\mathbf{H}^{\prime}_L) = \bigg\{ A \in \mathrm{Mat}(n) : A = \begin{pmatrix} A_1 & 0 \\ 0 & A_4 \end{pmatrix} \text{and } A_i^{T}M_i + M_iA_i = 0, i = 1,4 \bigg\}. \end{align*} $$

We may split $\mathfrak {g}$ in a direct sum $\mathfrak {h} \oplus \mathfrak {r}$ , where $\mathfrak {r}$ is an invariant subspace under the adjoint action of $\mathbf {H}^{\prime }_L$ on $\mathfrak {g}$ . Explicitly, we may set

$$ \begin{align*} \mathfrak{r} = \bigg\{ \begin{pmatrix} 0 & A_2 \\ A_3 & 0 \\ \end{pmatrix} : A_2^TM_1 +M_4A_3 = 0 \bigg\}. \end{align*} $$

We claim that the representation of $\mathbf {H}^{\prime }_L$ on $\mathfrak {r}$ is irreducible. Note that we may also show that the representation of $\mathrm {SO}_{Q|_L} \times \mathrm {SO}_{Q|_{L^{\perp }}}$ on $\mathrm {Mat}(k, n-k)$ given by

$$ \begin{align*} ((\sigma_1, \sigma_2) , A) \mapsto \sigma_1 A \sigma_2^{-1} \end{align*} $$

is irreducible. Over $\bar {\mathbb {Q}}$ , we may apply Lemma 2.6 below, from which this follows.

Now let $\mathbf {M}$ be a connected group containing $\mathbf {H}^{\prime }_L$ and let $\mathfrak {m}$ be its Lie algebra. Note that $\mathfrak {m} \cap \mathfrak {r}$ is an invariant subspace under the adjoint action of $\mathbf {H}^{\prime }_L$ on $\mathfrak {r}$ . Since this representation is irreducible, $\mathfrak {m} \cap \mathfrak {r} = \{0\}$ or $\mathfrak {m} \cap \mathfrak {r} =\mathfrak {r}$ . In the former case, we have that $\mathfrak {m} = \mathfrak {h}$ and in the latter $\mathfrak {m} = \mathfrak {g}$ . It follows that $\mathbf {H}^{\prime }_L$ is maximal and the proof is complete.

3.2.1 Applying Theorem 3.2

Consider the subgroup $\mathbf {J} = \mathrm {Spin}_Q \times \mathrm {SL}_n$ . Note that $\mathbf {J}$ is semisimple and simply connected so that we may apply Theorem 3.2 given a suitable sequence of subgroups.

The groups $\mathbf {H}_{L_i}$ are potentially not simply connected, so a little more care is needed in applying Theorem 3.2 to the orbit measures $\mu _i$ . We fix, for any i, some $h_i \in \boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})$ and consider the orbit measures on $g_ih_i \boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})^+\mathbf {G}(\mathbb {Q})$ . In view of the theorem, it suffices to show that these converge to the Haar probability measure on . Indeed, by disintegration, the Haar measure on $g_i\boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A}) \mathbf {G}(\mathbb {Q})$ is the integral over the Haar measures on $g_ih_i \boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})^+\mathbf {G}(\mathbb {Q})$ when $h_i$ is integrated with respect to the Haar probability measure on the compact group $\boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})/\boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})^+$ . In other words, the Haar measure on $g_i\boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A}) \mathbf {G}(\mathbb {Q})$ is a convex combination of the Haar measures on the orbits $g_ih_i\boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})^+ \mathbf {G}(\mathbb {Q})$ . To simplify notation, we replace $g_i$ by $g_ih_i$ in order to omit $h_i$ . Furthermore, we abuse notation and write $\mu _i$ for these ‘components’ of the original orbit measures.

We fix a compact open subgroup W of $\mathbf {G}(\mathbb {A}_f)$ in view of (2)(b) in Theorem 3.2 and an odd prime p as in the definition of admissibility of the sequence $(L_i)_i$ .

Let $\mu $ be any weak ${}^\ast $ -limit of the measures $\mu _i$ . Note that $\mu $ is a probability measure. Indeed, the pushforward of the measures $\mu _i$ to $\mathrm {Spin}_Q(\mathbb {A})/\mathrm {Spin}_Q(\mathbb {Q})$ has to converge to a probability measure as $\mathrm {Spin}_Q(\mathbb {A})/\mathrm {Spin}_Q(\mathbb {Q})$ is compact. We let $\mathbf {M} < \mathbf {J}$ be as in Theorem 3.2. Because $g_i \in \mathbf {G}(\mathbb {A})$ and $\boldsymbol {\Delta}{\mathbf H}_{L_i} < \mathbf {G}$ for all i, the support of the measures $\mu _i$ is contained in . Thus, $\mathbf {M}< \mathbf {G}$ .

We now focus on proving that $\mathbf {M} = \mathbf {G}$ . By Theorem 3.2, there exist $\delta _i \in \mathbf {G}(\mathbb {Q})$ such that $\delta _i^{-1}\boldsymbol {\Delta}{\mathbf H}_{L_i}\delta _i < \mathbf {M}$ for all $i \geq i_0$ . Furthermore, we fix $g \in \mathbf {G}(\mathbb {A})$ as well as $\hat {h}_i = (h_i, \Psi _{L_i}(h_i))\in \boldsymbol {\Delta}{\mathbf H}_{L_i}(\mathbb {A})^+$ , as in Theorem 3.2, such that

(3.1) $$ \begin{align} g_i \hat{h}_i \delta_i \to g. \end{align} $$

3.2.2 Individual equidistribution of subspaces and shapes

Consider the morphism

(3.2) $$ \begin{align} \mathbf{G} \to \bar{\mathbf{G}} = \mathrm{Spin}_Q \times \mathrm{SL}_k \times \mathrm{SL}_{n-k}. \end{align} $$

In the following step of the proof, we show that the image $\bar {\mathbf {M}}$ of the subgroup $\mathbf {M}$ via (3.2) projects surjectively onto each of the factors of $\bar {\mathbf {G}}$ .

Proof. We prove the proposition for each factor separately. To ease notation, $\pi $ will denote the projection of $\bar {\mathbf {G}}$ onto the factor in consideration, which we extend to $\mathbf {G}$ by precomposition.

First factor: As $\pi (\boldsymbol {\Delta}{\mathbf H}_{L_i}) = \mathbf {H}_{L_i}$ , we have, for each i,

$$ \begin{align*} \pi(\delta_i)^{-1} \mathbf{H}_{L_i} \pi(\delta_i) < \pi(\mathbf{M}). \end{align*} $$

Since $\mathbf {H}_{L_i}$ is a maximal subgroup of $\mathrm {Spin}_Q$ (see Proposition 2.5), there are two options: either $\pi _{1}(\mathbf {M}) = \mathrm {Spin}_Q$ or $\pi (\delta _i)^{-1} \mathbf {H}_{L_i} \pi (\delta _i) = \pi (\mathbf {M})$ for all $i\geq i_0$ .

Suppose the second option holds (as the proof is complete otherwise). Setting $\gamma _i = \pi (\delta _i\delta _{i_0}^{-1})$ and $L = L_{i_0}$ ,

$$ \begin{align*} \mathbf{H}_{\gamma_i.L} = \gamma_i\mathbf{H}_{L}\gamma_i^{-1} = \mathbf{H}_{L_i}. \end{align*} $$

By Proposition 2.2, we have $\gamma _i.L = L_i$ or $\gamma _i.L^\perp = L_i$ ; by changing to a subsequence and increasing $i_0$ , we may suppose that the former option holds for all $i \geq i_0$ . By (3.1) there exist $h_i \in \mathbf {H}_{L_i}(\mathbb {A})$ such that $\pi (g_i) h_i \gamma _i \to \pi (g')$ for some $g' \in \mathbf {G}(\mathbb {A})$ . Roughly speaking, this implies that $L_i = h_i\gamma _i.L \to \pi (g).L$ as $\mathbb {Q}_p$ -subspaces for any prime p contradicting the discriminant condition. More precisely, let $\varepsilon _i\to e$ be such that $\pi (g_i) h_i \gamma _i= \varepsilon _i\pi (g')$ . Then, for any prime p, the local discriminant gives

$$ \begin{align*} \mathrm{disc}_{p,Q}(L_i) = \mathrm{disc}_{p,Q}(h_{i,p}\gamma_i.L). = \mathrm{disc}_{p,Q}(\varepsilon_{i,p}\pi(g_p').L) \end{align*} $$

If i is large enough such that $\varepsilon _i \in \mathrm {Spin}_Q(\mathbb {R}\times \widehat {\mathbb {Z}})$ ,

$$ \begin{align*} \mathrm{disc}_Q(L_i) = \prod_p p^{\nu_p(\mathrm{disc}_{p,Q}(L_i))} = \prod_pp^{\nu_p(\mathrm{disc}_{p,Q}(\pi(g_p').L))}, \end{align*} $$

which is constant, contradicting that $\mathrm {disc}_Q(L_i) \to \infty $ .

Second factor: The proof is very similar to the first case, so we will be brief. By maximality of special orthogonal groups (Remark 1.13) and as $\pi (\boldsymbol {\Delta}{\mathbf H}_{L_i}) = \mathrm {SO}_{q_{{ }_{L_i\cap \Lambda _{L_i}}}}$ , we may suppose, by contradiction, that, for all $i \geq i_0$ ,

$$ \begin{align*} \pi(\delta_i)^{-1}\mathrm{SO}_{q_{{ }_{L_i \cap \Lambda_{L_i}}}}\pi(\delta_i) = \pi(\mathbf{M}). \end{align*} $$

We simplify notation and write $q_i$ for the least integer multiple of $q_{{ }_{L_i \cap \Lambda _{L_i}}}$ that has integer coefficients. Since $L_i \cap \Lambda _{L_i}$ and $L_i(\mathbb {Z})$ are commensurable with indices controlled by $\mathrm {disc}(Q)$ , we have $\mathrm {disc}(q_{i}) \asymp \mathrm {disc}(q_{L_i})$ and $\mathrm {disc}(\tilde {q}_{i}) \asymp \mathrm {disc}(\tilde {q}_{L_i})$ . In particular, by our assumption, $\mathrm {disc}(\tilde {q}_{i}) \to \infty $ as $i \to \infty $ .

Set $\gamma _i = \pi (\delta _i\delta _{i_0}^{-1})\in \mathrm {SL}_k(\mathbb {Q})$ so that

(3.3) $$ \begin{align} \mathrm{SO}_{\gamma_i\tilde{q}_{i_0}} = \mathrm{SO}_{\gamma_iq_{i_0}} = \gamma_i \mathrm{SO}_{q_{{ }_{L_{i_0}}}}\gamma_{i}^{-1} = \mathrm{SO}_{q_{i}} = \mathrm{SO}_{\tilde{q}_{i}}. \end{align} $$

By Proposition 2.4, there exist coprime integers $m_i,n_i$ such that

$$ \begin{align*} m_i \gamma_i\tilde{q}_{i_0} = n_i \tilde{q}_{i}. \end{align*} $$

Using (3.1), write $\pi (g_i) h_i \gamma _i = \varepsilon _i \pi (g')$ for some $g' \in \mathbf {G}(\mathbb {A})$ and $\varepsilon _i \to e$ . By (3.3), $h_i(\gamma _i\tilde {q}_{i_0}) = \gamma _i\tilde {q}_{i_0}$ . Thus, for any prime p,

$$ \begin{align*} m_i \varepsilon_{i,p}\pi(g_p')\tilde{q}_{i_0} = m_i h_{i,p} \gamma_i\tilde{q}_{i_0} = n_i \tilde{q}_{i}. \end{align*} $$

The form $\pi (g_p')\tilde {q}_{i_0}$ is a form over $\mathbb {Q}_p$ with trivial denominators for all but finitely many p. Applying $\varepsilon _{i,p}$ for large i does not change this. Furthermore, $m_i$ needs to divide all denominators of $\tilde {q}_{i_0}$ over $\mathbb {Z}_p$ for all i as $\tilde {q}_{i}$ is primitive. Hence, $m_i$ can only assume finitely many values and, by reversing roles, one can argue the same for $n_i$ . For any prime p,

$$ \begin{align*} \mathrm{disc}_{p}(\tilde{q}_{i}) = p^{\mathrm{ord}_p({m_i}/{n_i})} \mathrm{disc}_p(\pi(g_p')\tilde{q}_{i_0}), \end{align*} $$

and hence

$$ \begin{align*} \mathrm{disc}(\tilde{q}_{i}) = \dfrac{m_i}{n_i} \prod_{p} p^{\mathrm{ord}_p(\mathrm{disc}_p(\pi(g_p')\tilde{q}_{i_0}))}, \end{align*} $$

which is in contradiction to $\mathrm {disc}(\tilde {q}_{i}) \to \infty $ .

Third factor: The proof here is the same as for the second factor. We do, however, point out that the morphism $\mathbf {G} \to \bar {\mathbf {G}}$ was constructed to satisfy that, for any $h \in \mathbf {H}_{L_i}$ , we have $\pi ((h,\Psi _{L_i}(h)) = \psi _{2,L_i}(h)$ and hence $\pi (\boldsymbol {\Delta}{\mathbf H}_{L_i}) = \mathrm {SO}_{q_{L_i^\perp \cap \Lambda _{L_i}^\#}}$ .

Remark 3.4. We recall from the beginning of §3.2 that the first three conditions in admissibility were used in this order for the three factors in the above proof. This has a consequence: if $L_i\in \mathrm {Gr}_{n,k}(\mathbb {Q})$ is any sequence of subspaces satisfying properties (1) and (4), then, for any $g_i \in \mathrm {Spin}_Q(\mathbb {R})$ , the packets

are equidistributed as $i \to \infty $ . This can be used to obtain equidistribution of $\mathcal {H}^{n,k}_Q(D) \subset \mathrm {Gr}_{n,k}(\mathbb {R})$ without any restrictions on the k-power free part of D (as opposed to our main theorems in the introduction).

3.2.3 Simultaneous equidistribution of subspaces and shapes

Proposition 3.3 shows that the image $\bar {\mathbf {M}}$ of $\mathbf {M}$ under (3.2) satisfies that the projection onto each simple factor of $\bar {\mathbf {G}}$ is surjective. We claim that this implies that $\bar {\mathbf {M}} = \bar {\mathbf {G}}$ .

We first show that the projection of $\bar {\mathbf {M}}$ to $\mathrm {SL}_k \times \mathrm {SL}_{n-k}$ is surjective. Note that any proper subgroup of $\mathrm {SL}_k \times \mathrm {SL}_{n-k}$ with surjective projections is the graph of an isomorphism $\mathrm {SL}_k \to \mathrm {SL}_{n-k}$ . In particular, the intermediate claim is finished if $k \neq n-k$ . Suppose that $ k = n-k$ and choose, for some $i \geq i_0$ , an element $h \in \mathbf {H}_{L_i}$ acting trivially on $L_i$ but not trivially on $L_i^{\perp }$ . The projection of $g_{L_i}^{-1}\rho _Q(h)g_{L_i}$ to the first (respectively, the second) $\mathrm {SL}_{k}$ is trivial (respectively, non-trivial); the projection of $\bar {\mathbf {M}}$ to $\mathrm {SL}_k \times \mathrm {SL}_{n-k}$ thus contains elements of the form $(e,g)$ with $g \neq e$ . This rules out graphs under isomorphisms and concludes the intermediate claim.

Now note that $\bar {\mathbf {M}}$ projects surjectively onto $\mathrm {Spin}_Q$ and $\mathrm {SL}_k \times \mathrm {SL}_{n-k}$ and that the latter two $\mathbb {Q}$ -groups do not have isomorphic simple factors. By an argument similar to that above, we deduce that $\bar {\mathbf {M}} = \bar {\mathbf {G}}$ .

3.2.4 Handling the unipotent radical

We now turn to proving that $\mathbf {M} = \mathbf {G}$ , which concludes the proof of the theorem. By §3.2.3, we know that $\mathbf {M}$ surjects to $\bar {\mathbf {G}}$ . In particular, by the Levi–Malcev theorem, there exists some element in the unipotent radical of $\mathbf {P}_{n,k}$

$$ \begin{align*} y_C = \begin{pmatrix} I_k & C \\ 0 & I_{n-k} \end{pmatrix} \in \mathbf{P}_{n,k}(\mathbb{Q}) \end{align*} $$

such that $\mathbf {M}$ contains $\mathrm {Spin}_Q \times y_C \mathbf {D}_{n,k} y_C^{-1}$ . By maximality of the latter group (cf. Remark 1.13), $\mathbf {M}$ is either equal to $\mathbf {G}$ or

$$ \begin{align*} \mathbf{M} = \mathrm{Spin}_Q \times y_C \mathbf{D}_{n,k} y_C^{-1}. \end{align*} $$

Assume, by contradiction, the latter. The inclusion $\delta _i^{-1}\boldsymbol {\Delta}{\mathbf H}_{L_i}\delta _i \subset \mathbf {M}$ implies that

$$ \begin{align*} \delta_{2,i}^{-1}g_{L_i}^{-1}\rho_Q(h)g_{L_i} \delta_{2,i} \in y_C \mathbf{D}_{n,k} y_C^{-1}, \end{align*} $$

where $\delta _{2,i}$ denotes the second coordinate of the element $\delta _i \in \mathbf {G}(\mathbb {Q}) = \mathrm {Spin}_Q(\mathbb {Q})\times \mathbf {P}_{n,k}(\mathbb {Q})$ . Since $y_C \mathbf {D}_{n,k} y_C^{-1}$ stabilizes two subspaces, namely, $y_C L_0 = L_0$ and $L' = y_C(\{(0,\ldots ,0)\}\times \mathbb {Q}^{n-k})$ , the conjugated group $g_{L_i}\delta _{i,2}y_C \mathbf {D}_{n,k} y_C^{-1} \delta _{i,2}^{-1}g_{L_i}^{-1}$ fixes the subspaces

$$ \begin{align*} g_{L_i}\delta_{i,2}L_0 = g_{L_i}L_0 = L_i \quad \text{and} \quad g_{L_i}\delta_{i,2}L'. \end{align*} $$

As $\mathbf {H}_{L_i}$ fixes exactly the subspaces $L_i,L_i^\perp $ , we must have

(3.4) $$ \begin{align} L_i^\perp = g_{L_i}\delta_{i,2}L' \end{align} $$

for all i. We denote by $v_{1}^i,\ldots ,v_n^i$ the columns of $g_{L_i}$ , which is a basis of $\Lambda _{L_i}$ , and by $w_1^i,\ldots ,w_n^i$ its dual basis. Recall that $w_{k+1}^i,\ldots , w_n^i$ form a basis of $\Lambda _{L_i}^\# \cap L_i^{\perp }$ . By (3.4), there exists a rational number $\alpha _i\in \mathbb {Q}^\times $ such that

(3.5) $$ \begin{align} \alpha_i (w^{i}_{k+1}\wedge \cdots \wedge w^{i}_n) = g_{L_i}\delta_{i,2}y_C (e_{k+1} \wedge \cdots \wedge e_n). \end{align} $$

To simplify notation, we set $\eta _i = \delta _{i,2}y_C$ .

We first control the numbers $\alpha _i$ . From (3.1), we know that there are $h_i \in \mathbf {H}_{L_i}$ such that

$$ \begin{align*} g_{2,i} g_{L_i}^{-1} \rho_Q(h_i) g_{L_i} \eta_i \to g' \end{align*} $$

for some $g' \in \mathbf {P}_{n,k}(\mathbb {A})$ . For i large enough, there exist $\varepsilon _i \in \mathbf {P}_{n,k}(\mathbb {R} \times \widehat {\mathbb {Z}})$ with $g_{2,i} g_{L_i}^{-1} \rho _Q(h_i) g_{L_i} \eta _i= \varepsilon _i g'$ . We now fix a prime p so that $\rho _Q(h_{i,p}) g_{L_i} \eta _i= g_{L_i}\varepsilon _{i,p} g^{\prime }_p$ (as $g_{2,i} \in \mathbf {G}(\mathbb {R})$ ). Applying $\rho _Q(h_{i,p})$ to (3.4), we obtain

$$ \begin{align*} \alpha_i (w^{i}_{k+1}\wedge \cdots \wedge w^{i}_n) = g_{L_i}\varepsilon_{i,p} g^{\prime}_p (e_{k+1} \wedge \cdots \wedge e_n). \end{align*} $$

Considering that the vectors $w^{i}_{k+1}\wedge \cdots \wedge w^{i}_n$ and $e_{k+1} \wedge \cdots \wedge e_n$ are primitive (see, for example, [Reference CasselsCas97, Ch. 1, Lemma 2]) and that $g_{L_i}$ and $g_p'$ have bounded denominators, this shows that the denominators and numerators of the numbers $\alpha _i$ are bounded independently of i.

We now compute the discriminant of the lattice spanned by $w^{i}_{k+1}, \ldots , w^{i}_n$ in two ways. First, note that, as $w^{i}_{k+1},\ldots , w^{i}_n$ is a basis of $\Lambda _{L_i}^\# \cap L_i^\perp $ , the discriminant in question is equal to the discriminant of $\Lambda _{L_i}^\# \cap L_i^\perp $ and hence $\asymp \mathrm {disc}_Q(L_i)$ . For the second way, observe that, by (3.5), the discriminant of the lattice spanned by $w^{i}_{k+1}, \ldots , w^{i}_n$ is given by $\alpha _i^{-1}$ multiplied by the determinant of the matrix with entries

(3.6) $$ \begin{align} \langle g_{L_i} \eta_i e_j, w_{m}^i \rangle_Q \quad \text{with} \ j,m> k. \end{align} $$

(One conceptual way to see this is the following: the bilinear form $\langle \cdot ,\cdot \rangle _Q$ induces a bilinear form $\langle \cdot ,\cdot \rangle _{\bigwedge ^{n-k}Q}$ on the wedge product $\bigwedge ^{n-k}\mathbb {Q}^n$ by defining it on pure wedges through

$$ \begin{align*} \langle v_1\wedge \cdots \wedge v_{n-k},w_1\wedge \cdots \wedge w_{n-k} \rangle_{\bigwedge^{n-k}Q} = \det(\langle v_i,w_j\rangle_Q). \end{align*} $$

This definition asserts that the discriminant of a rank $n-k$ lattice is the quadratic value of the wedge product of any of its bases. Equation 3.6 is then obtained by replacing one of the wedges in $\langle w_{k+1}^i\wedge \cdots \wedge w_n^i,w_{k+1}^i\wedge \cdots \wedge w_n^i \rangle _{\bigwedge ^{n-k}Q}$ via (3.5).)

To compute this determinant, write $\eta _i e_j = \sum _{\ell } a_{\ell j}^i e_\ell $ for all $j> k$ so that

$$ \begin{align*} g_{L_i} \eta_i e_j = \sum_{\ell} a_{\ell j}^i v_{\ell}^i. \end{align*} $$

Using that $\{w^{i}_l\}$ are dual vectors to $\{v^{i}_l\}$ , we compute

$$ \begin{align*} \langle g_{L_i} \eta_i e_j, w_{m}^i \rangle_Q = \sum_{\ell} a_{\ell j}^i \langle v_{\ell}^i,w_{m}^i\rangle_Q = a_{m j}^i \end{align*} $$

for all $m,j> k$ . This implies that the determinant of the matrix with entries (3.6) is equal to the determinant of the lower right-hand block of the matrix $\eta _i$ . As the latter is equal to one, we conclude that the discriminant of the lattice spanned by $w^{i}_{k+1}\wedge \cdots \wedge w^{i}_n$ is equal to $\alpha _i^{-1}$ .

To summarize, we have established the identity

$$ \begin{align*} \mathrm{disc}_Q(\Lambda_{L_i}^\# \cap L_i^\perp) = \alpha_i^{-1}. \end{align*} $$

Since the left-hand side of this identity goes to infinity as $i \to \infty $ (because $\asymp \mathrm {disc}_Q(L_i)$ ) whereas the right-hand side is bounded, we have reached a contradiction. It follows that $\mathbf {M} = \mathbf {G}$ , and hence the proof of Theorem 3.1 is complete.

4.3.1 An equidistribution theorem for the pointwise stabilizers

We first establish the following proposition which shows that either orbits of the pointwise stabilizer are equidistributed or there is some arithmetic obstruction.

Proof. We prove the proposition in exactly the same way we proved the first case in Proposition 3.3; thus, we are brief. Let $\delta _i \in \mathrm {Spin}_Q(\mathbb {Q})$ and a connected $\mathbb {Q}$ -group $\mathbf {M} < \bar {\mathbf {G}}$ be as in Theorem 3.2. In particular,

$$ \begin{align*} \delta_i^{-1} \mathbf{H}^{\mathrm{pt}}_{L_i} \delta_i < \mathbf{M} \end{align*} $$

and it suffices for equidistribution to verify that $\mathbf {M} = \mathrm {Spin}_Q$ . One can see that $\mathbf {M}$ strictly contains $\delta _i^{-1} \mathbf {H}^{\mathrm {pt}}_{L_i} \delta _i$ for all i by using $\mathrm {disc}_Q(L_i) \to \infty $ and repeating the proof of the first case in Proposition 3.3.

Contrary to the case treated in Proposition 3.3, the groups $\mathbf {H}^{\mathrm {pt}}_{L_i}$ are non-maximal. The intermediate groups can, however, be understood explicitly: they are of the form $\mathbf {H}^{\mathrm {pt}}_{W}$ , where W is a rational line contained in $\delta _i^{-1}.L_i$ for all i. For a proof of this fact, we refer to [Reference Ellenberg and VenkateshEV08, Proposition 4]; see also the arXiv version of the same paper, where the authors give an elementary proof in the case $n-2 \geq 7$ . This concludes the proof of the proposition.

Corollary 4.8. Let the notation and the assumptions be as in Proposition 4.7 and suppose that the second case holds. Then

$$ \begin{align*} \min_{w \in L_i(\mathbb{Z})\setminus\{0\}} Q(w) = \min_{w \in \mathbb{Z}^2\setminus\{0\}} q_{L_i}(w) \end{align*} $$

is bounded as $i \to \infty $ .

Proof. Let $v \in \mathbb {Q}^n$ be as in Proposition 4.7 and suppose, without loss of generality, that v is integral and primitive. Suppose, also, that $g_i h_i \delta _i \to g' \in \mathrm {Spin}_Q(\mathbb {A})$ and write $g_i h_i \delta _i = \varepsilon _i g'$ , where $\varepsilon _i \to e$ . Let $i_0$ be large enough so that $\varepsilon _i \in \bar {\mathbf {G}}(\mathbb {R} \times \widehat {\mathbb {Z}})$ for all $i \geq i_0$ and let $N \in \mathbb {N}$ be the smallest integer such that $Ng^{\prime }_p$ is integral for all primes p.

We claim that $v_i:= N \delta _i.v \in L_i(\mathbb {Z})$ . To see this, first note that $v_i\in L_i(\mathbb {Q})$ . Furthermore, for any prime p, the vector $v_i$ is contained in $L(\mathbb {Z}_p)$ . Indeed, $h_{i,p} \in \mathbf {H}^{\mathrm {pt}}_{L_i}(\mathbb {Q}_p)$ necessarily fixes $v_i$ and, as $g_{i,p} = e$ ,

$$ \begin{align*} v_i = h_{i,p}.v_i = N h_{i,p}\delta_i.v = N\varepsilon_{i,p} g_{p}.v \in \mathbb{Z}_p^n. \end{align*} $$

This proves the claim and hence the corollary as $Q(N\delta _i.v) = N^2 Q(v)$ .

4.3.2 A corollary of equidistribution in the second factor

In the following, we would like to give an estimate of the measure of the set of points in $g_{i,1}\mathbf {T}_{L_i}(\mathbb {A})\mathrm {Spin}_Q(\mathbb {Q})$ whose associated point in is ‘close’ to the cusp. This will allow us to ‘wash out’ the effect of the obstructions in Proposition 4.7 on average across the full stabilizer group. To obtain said estimate, we introduce a height function that suits our needs.

Let ${\mathcal {S}}_{2}$ be the space of positive definite real binary quadratic forms up to similarity and write $[q]$ for the similarity class of a binary form q. (Two positive definite binary real quadratic forms $Q_1,Q_2$ are similar if there exist ${\unicode{x3bb} }>0$ and $g \in {\mathrm {GL}}_2({\mathbb {Z}})$ with $Q_2(\cdot ) = {\unicode{x3bb} } Q_1(g\cdot )$ . Note that the space ${\mathcal {S}}_{2}$ will be discussed in more detail in §6.2.) For $\varepsilon>0$ , we define

$$ \begin{align*} \mathcal{S}_{2}(\varepsilon) = \left\{ [q]\in \mathcal{S}_{2}: \min_{w \in \mathbb{Z}^2 \setminus\{0\}} q(w)> \varepsilon \sqrt{\mathrm{disc}(q)}\right\}\!\hspace{-0.5pt}. \end{align*} $$

Note that the condition is independent of the choice of representative of $[q]$ .

By Mahler’s compactness criterion [Reference MahlerMah46], these are compact subsets of $\mathcal {S}_{2}$ and any compact subset is contained in $\mathcal {S}_{2}(\varepsilon )$ for some $\varepsilon>0$ . Furthermore, one can show that the Haar measure of $\mathcal {S}_{2} \setminus \mathcal {S}_{2}(\varepsilon )$ is $\ll \varepsilon $ by direct integration of the hyperbolic area measure on that region.

We define

to be the preimage of $\mathcal {S}_{2}(\varepsilon )$ under the composition

By the previous discussion, this is a compact set whose complement has Haar measure $\ll \varepsilon $ . For

, we call the supremum over all $\varepsilon>0$ with $x \in K_\varepsilon $ the minimal quadratic value for x.

The following is a direct corollary of equidistribution in the second factor.

4.3.3 Using the shape in the subspace

In the following, we identify the minimal quadratic value for the points on the orbits in the context of proving Theorem 4.1. As $\mathrm {Spin}_Q(\mathbb {R} \times \widehat {\mathbb {Z}})$ is a compact open subgroup, it has finitely many orbits on $\mathrm {Spin}_Q(\mathbb {A})/\mathrm {Spin}_Q(\mathbb {Q})$ (these correspond to the spin genus of the quadratic form Q). We choose a finite set of representatives $\mathcal {R} \subset \mathrm {Spin}_Q(\mathbb {A}_f)$ such that

(4.4) $$ \begin{align} \mathrm{Spin}_Q(\mathbb{A})/\mathrm{Spin}_Q(\mathbb{Q}) = \bigsqcup_{\mathsf{r} \in \mathcal{R}} \mathrm{Spin}_Q(\mathbb{R} \times \widehat{\mathbb{Z}}) \mathsf{r}\, \mathrm{Spin}_Q(\mathbb{Q}). \end{align} $$

Note that, in $\mathrm {SL}_2$ (or $\mathrm {SL}_{n-2}$ ), any $g \in \mathrm {SL}_2(\mathbb {A})$ can be written as $g = b \gamma $ , where $b \in \mathrm {SL}_2 (\mathbb {R} \times \widehat {\mathbb {Z}})$ and $\gamma \in \mathrm {SL}_2(\mathbb {Q})$ .

Lemma 4.10. Let $h \in \boldsymbol{\Delta \bar {\mathbf H}}_{L_i}(\mathbb {A})$ and write $h \gamma = b \mathsf {r}$ for some $\gamma \in \bar {\mathbf {G}}(\mathbb {Q})$ , $b \in \bar {\mathbf {G}}(\mathbb {R} \times \widehat {\mathbb {Z}})$ and $\mathsf {r} \in \mathcal {R}$ . Then $\gamma _1^{-1}.L_i$ is a rational subspace of discriminant $\asymp D$ . Furthermore, the minimum

$$ \begin{align*} \min_{w \in \mathbb{Z}^2 \setminus\{0\}} \frac{q_{\gamma_1^{-1}.L_i}(w)}{\sqrt{\mathrm{disc}_Q(\gamma_1^{-1}.L_i)}} \end{align*} $$

is comparable to the minimal quadratic value for $g_{i,2}\psi _{1,L}(h)\mathrm {SL}_2(\mathbb {Q})$ .

Note that a lemma in this spirit will later be used to deduce the main theorems from their dynamical counterparts (cf. Proposition 7.1). The statement here is more technical in nature (as it needs to treat different genera) and the reader is encouraged to return to the proof after reading Proposition 7.1. We note that such a treatment has appeared in different contexts in the literature [Reference Aka, Luethi, Michel and WieserALMW22, Reference Ellenberg and VenkateshEV08].

Proof. The ingredients for this proof are all contained in the proof of Proposition 7.1, so we are brief. Write $L = L_i$ for simplicity. Note that $h_{1,p} \gamma _1 = b_{1,p}\mathsf {r}_p$ and hence

$$ \begin{align*} \mathrm{disc}_{p,Q}(\gamma_1^{-1}.L) &= \mathrm{disc}_{p,Q}(\gamma_1^{-1}h_{1,p}^{-1}.L) = \mathrm{disc}_{p,Q}(\mathsf{r}_p^{-1}b_{1,p}.L)\\ &\asymp_{\mathsf{r}} \mathrm{disc}_{p,Q}(b_{1,p}.L) = \mathrm{disc}_{p,Q}(L). \end{align*} $$

As the discriminant is a product of the local discriminants (1.5), this proves the first claim.

For the second claim, we let $L' = \gamma _1^{-1}.L$ and consider $m = g_{L'}^{-1}\rho _Q(\gamma _1^{-1})g_L \gamma _2\in \mathrm {GL}_n(\mathbb {Q})$ . Observe that

$$ \begin{align*} m L_0 = g_{L'}^{-1}\rho_Q(\gamma_1^{-1})g_L L_0 = g_{L'}^{-1}\rho_Q(\gamma_1^{-1})L = g_{L'}^{-1}L' = L_0. \end{align*} $$

As we will now see, m is ‘almost integral’ and invertible. For this, compute

$$ \begin{align*} \rho_Q(\gamma_1^{-1})g_L \gamma_2 = \rho_Q(\gamma_1^{-1} h_{1,p}^{-1})g_L h_{2,p} \gamma_2 = \rho_Q(\mathsf{r}_p^{-1} b_{1,p}^{-1})g_L b_{2,p}. \end{align*} $$

This implies that there exists some $N \in \mathbb {N}$ independent of L such that $N\rho _Q( \gamma _1^{-1})g_L \gamma _2$ and $N\rho _Q(\gamma _1^{-1})g_L^{-1} \gamma _2$ are integral. Recall that $\mathrm {disc}(Q)g_{L'},\mathrm {disc}(Q)g_{L'}^{-1}$ are integral so that $N\mathrm {disc}(Q) m$ and $N \mathrm {disc}(Q)m^{-1}$ are integral. This discussion implies that, for any two positive definite real binary quadratic forms $q,q'$ with the property that $\pi _1(m)q$ and $q'$ are similar,

$$ \begin{align*} \min_{w \in \mathbb{Z}^2\setminus\{0\}} \frac{q(w)}{\sqrt{\mathrm{disc}(q)}} \asymp \min_{w \in \mathbb{Z}^2\setminus\{0\}} \frac{q'(w)}{\sqrt{\mathrm{disc}(q')}}. \end{align*} $$

Here, recall that $\mathrm {GL}_2(\mathbb {R})$ acts on binary forms via $gq(x) = q(g^t x)$ .

Now, note that

$$ \begin{align*} [q_{L'}] = [Q(g_L' \cdot)] = [Q(\rho_Q(\gamma_1) g_{L'}\cdot)] \end{align*} $$

whereas the similarity class belonging to $g_{2}\psi _{1,L}(t)\mathrm {SL}_2(\mathbb {Q})$ is

$$ \begin{align*} [\gamma_2^{-1}q_{L}]= [Q(g_L \gamma_2 \cdot)] = [Q(\rho_Q(\gamma_1) g_{L'}m \cdot)] \end{align*} $$

The claim follows.

4.3.4 Proof of Theorem 4.1

As explained, it now suffices to prove that the packets for  $L_i$ ,

equidistribute as $\mathrm {disc}_Q(L_i) \to \infty $ . Similarly to the situation in the proof of Theorem 3.1, we need to circumvent the problem that $\mathbf {H}_L$ for $L \in \mathrm {Gr}_{n,2}(\mathbb {Q})$ is not exactly isomorphic to $\mathbf {H}^{\mathrm {pt}}_L \times \mathbf {T}_L$ (see Remark 2.1 for a more careful discussion). Denote by $\mathbf {H}_L(\mathbb {A})^\star $ the image of $\mathbf {H}^{\mathrm {pt}}_L(\mathbb {A}) \times \mathbf {T}_L(\mathbb {A}) \to \mathbf {H}_L(\mathbb {A})$ ; this is a normal subgroup of $\mathbf {H}_L(\mathbb {A})$ with the property that

is compact and abelian. By an argument as at the beginning of the proof of Theorem 3.1, it suffices to show that, for any $k_i \in K_{L_i}$ , the orbits

are equidistributed as $i \to \infty $ . We let $\mu _i$ be the Haar measure on the ith such orbit and let

$$ \begin{align*} \mu_{i,1} = m_{\mathbf{H}^{\mathrm{pt}}_{L_i}(\mathbb{A})\mathrm{Spin}_Q(\mathbb{Q})},\quad \mu_{i,2} = m_{\mathbf{T}_{L_i}(\mathbb{A})\mathrm{Spin}_Q(\mathbb{Q})} \end{align*} $$

be the Haar measure on the closed orbits of $\mathbf {H}^{\mathrm {pt}}_{L_i}(\mathbb {A})$ (respectively, $\mathbf {T}_{L_i}(\mathbb {A})$ ). Then, for any function

,

(4.5) $$ \begin{align} \int f \, \mathrm{d} \mu_i = \int \int f(g_{i,1} k_i h t) \, \mathrm{d} \mu_{1,i}(h) \, \mathrm{d} \mu_{i,2}(t). \end{align} $$

In the following, we identify $k_i$ with a representative in a fixed bounded region of $\mathbf {H}_{L_i}(\mathbb {A})$ .

For a fixed $t_i\in \mathbf {T}_{L_i}(\mathbb {A})$ , the inner integral is the integral over the orbit

$$ \begin{align*} g_{i,1}k_i \mathbf{H}^{\mathrm{pt}}_{L_i}(\mathbb{A})t_i \mathrm{Spin}_Q(\mathbb{Q}) = g_{i,1}k_i t_i\mathbf{H}^{\mathrm{pt}}_{L_i}(\mathbb{A}) \mathrm{Spin}_Q(\mathbb{Q}). \end{align*} $$

Writing $t_i\gamma _i = b_i\mathsf {r}$ as in (4.4), we see that

$$ \begin{align*} g_{i,1}k_i t_i\mathbf{H}^{\mathrm{pt}}_{L_i}(\mathbb{A}) \mathrm{Spin}_Q(\mathbb{Q}) &= g_{i,1}k_ib_i\mathsf{r} \gamma_i^{-1} \mathbf{H}^{\mathrm{pt}}_{L_i}(\mathbb{A}) \mathrm{Spin}_Q(\mathbb{Q})\\ &= g_{i,1}k_ib_i\mathsf{r} \mathbf{H}^{\mathrm{pt}}_{\gamma_i^{-1}.L_i}(\mathbb{A}) \mathrm{Spin}_Q(\mathbb{Q}), \end{align*} $$

which is equidistributed if and only if $\mathbf {H}^{\mathrm {pt}}_{\gamma _i^{-1}.L_i}(\mathbb {A}) \mathrm {Spin}_Q(\mathbb {Q})$ is equidistributed (as $g_{i,1}k_ib_i$ is bounded). By Proposition 4.7 and its corollary, it suffices to show that the minimal non-zero value of $q_{\gamma _i^{-1}.L_i}$ goes to infinity. This minimum is comparable to the minimal quadratic value for $g_{i,2} \psi _{1,L}(t_i) \mathrm {SL}_2(\mathbb {Q})$ by Lemma 4.10.

Motivated by this observation, we define, for $\varepsilon>0$ ,

so that the complement of $\mathcal {B}_i(\varepsilon )$ has $\mu _{i,2}$ -measure $\ll \varepsilon $ for all i large enough (depending on $\varepsilon $ ), by Corollary 4.9. In view of (4.5), this implies that

$$ \begin{align*} \int f \, \mathrm{d} \mu_i = \dfrac{1}{\mu_{i,2}(\mathcal{B}_i(\varepsilon))} \int_{\mathcal{B}_i(\varepsilon)} \int f(g_{i,1}k_i h t) \, \mathrm{d} \mu_{1,i}(h) \, \mathrm{d} \mu_{i,2}(t) + \mathcal{O}(\varepsilon). \end{align*} $$

By the previous paragraph, the orbits $g_{i,1}k_i \mathbf {H}^{\mathrm {pt}}_{L_i}(\mathbb {A})t_i \mathrm {Spin}_Q(\mathbb {Q})$ are equidistributed for any sequence $t_i \in \mathcal {B}_i(\varepsilon )$ . The integral on the right-hand side is a convex combination of such orbital integrals and hence must converge to the integral of f over the Haar measure. Letting $\mu $ be any weak ${}^*$ -limit of the measures $\mu _i$ , we obtain

As $\varepsilon $ is arbitrary, this implies the claim.

5.1.1 Definitions

For any $\mathbb {Z}$ -lattice $\Gamma \subset \mathbb {Q}^n$ , we define the dual lattice

$$ \begin{align*} \Gamma^\# = \{x \in \Gamma \otimes \mathbb{Q}: \langle x,y\rangle_Q \in \mathbb{Z} \text{ for all } y \in \Gamma\}. \end{align*} $$