Let \(\{e_1, \dots, e_n\}\) be a basis of a real vector space \(V\), and let \[ \Lambda = \left\{ \sum_{i=1}^n r_i e_i \;\middle|\; r_i \in \mathbb{Z} \right\} \cong \mathbb{Z}^n \] be the integer lattice generated by the basis. Let \((\cdot)\) be a non-degenerate symmetric bilinear form on \(V\) such that the Gram matrix \(M = (e_i \cdot e_j)_{1 \le i, j \le n}\) has integer entries (\(M \in M_n(\mathbb{Z})\)). Define the dual lattice by \[ \Lambda^\vee = \{v \in V \mid v \cdot x \in \mathbb{Z} \text{ for all } x \in \Lambda\}. \]
Show that \(\Lambda \subseteq \Lambda^\vee\).
Prove that \(\det M \ne 0\) and that the dual basis vectors \(e_i^* = \sum_{j=1}^n (M^{-1})_{ji} e_j\) form a \(\mathbb{Z}\)-basis for \(\Lambda^\vee\).
Prove that \(|\Lambda^\vee / \Lambda| = |\det M|\).
Solution
Goal: Prove lattice containment \(\Lambda \subseteq \Lambda^\vee\), establish invertibility of the Gram matrix \(M\), and compute the index \(|\Lambda^\vee/\Lambda|\) using the Smith normal form of \(M\).
1. Part (a): \(\Lambda \subseteq \Lambda^\vee\).
1.1. Let \(v \in \Lambda\), so \(v = \sum_{i=1}^n r_i e_i\) with \(r_i \in \mathbb{Z}\).
1.2. Let \(x \in \Lambda\), so \(x = \sum_{j=1}^n s_j e_j\) with \(s_j \in \mathbb{Z}\).
1.3. Compute the bilinear product:\[v \cdot x = \left( \sum_{i=1}^n r_i e_i \right) \cdot \left( \sum_{j=1}^n s_j e_j \right) = \sum_{i=1}^n \sum_{j=1}^n r_i s_j (e_i \cdot e_j) = \sum_{i=1}^n \sum_{j=1}^n r_i s_j M_{ij}.\]
1.4. By hypothesis, \(M_{ij} \in \mathbb{Z}\) for all \(i, j\), and \(r_i, s_j \in \mathbb{Z}\).
1.5. The sum of products of integers is an integer, so \(v \cdot x \in \mathbb{Z}\).
1.6. Since \(x \in \Lambda\) was arbitrary, \(v \in \Lambda^\vee\).
1.7. Thus \(\Lambda \subseteq \Lambda^\vee\).
2. Part (b): \(\det M \ne 0\).
2.1. Suppose \(c = (c_1, \dots, c_n)^T \in \mathbb{R}^n\) satisfies \(M c = 0\).
2.2. Define the vector \(v = \sum_{j=1}^n c_j e_j \in V\).
2.3. For each basis vector \(e_i\) (\(1 \le i \le n\)):\[e_i \cdot v = e_i \cdot \left( \sum_{j=1}^n c_j e_j \right) = \sum_{j=1}^n c_j (e_i \cdot e_j) = \sum_{j=1}^n M_{ij} c_j = (M c)_i = 0.\]
2.4. For any \(w \in V\), write \(w = \sum_{i=1}^n a_i e_i\). Then\[w \cdot v = \sum_{i=1}^n a_i (e_i \cdot v) = \sum_{i=1}^n a_i \cdot 0 = 0.\]
2.5. Since the form \((\cdot)\) is non-degenerate, \(w \cdot v = 0\) for all \(w \in V\) implies \(v = 0\).
2.6. Since \(\{e_1, \dots, e_n\}\) is a basis of \(V\), \(v = \sum c_j e_j = 0 \implies c_j = 0\) for all \(j\).
2.7. Thus \(\ker M = \{0\}\), so \(M\) is invertible in \(M_n(\mathbb{R})\), which proves \(\det M \ne 0\).
3. Part (b): \(\Lambda^\vee = \bigoplus_{i=1}^n \mathbb{Z} e_i^*\).
3.1. For each \(i \in \{1, \dots, n\}\), define \(e_i^* = \sum_{j=1}^n (M^{-1})_{ji} e_j \in V\).
3.2. Compute \(e_k \cdot e_i^*\):\[e_k \cdot e_i^* = \sum_{j=1}^n (M^{-1})_{ji} (e_k \cdot e_j) = \sum_{j=1}^n M_{kj} (M^{-1})_{ji} = (M M^{-1})_{ki} = \delta_{ki}.\]
3.3. Let \(v = \sum_{j=1}^n y_j e_j \in V\).
3.4. \(v \in \Lambda^\vee \iff e_i \cdot v \in \mathbb{Z}\) for all \(i \in \{1, \dots, n\}\) (since \(\{e_i\}\) spans \(\Lambda\) over \(\mathbb{Z}\)).
3.5. Compute \(e_i \cdot v = (M y)_i\).
3.6. Thus \(v \in \Lambda^\vee \iff M y = z \in \mathbb{Z}^n \iff y = M^{-1} z\) for some \(z \in \mathbb{Z}^n\).
3.7. In terms of vectors in \(V\):\[v = \sum_{j=1}^n y_j e_j = \sum_{j=1}^n \left( \sum_{i=1}^n (M^{-1})_{ji} z_i \right) e_j = \sum_{i=1}^n z_i \left( \sum_{j=1}^n (M^{-1})_{ji} e_j \right) = \sum_{i=1}^n z_i e_i^*.\]
3.8. Thus \(\Lambda^\vee = \left\{ \sum_{i=1}^n z_i e_i^* \;\middle|\; z_i \in \mathbb{Z} \right\} = \bigoplus_{i=1}^n \mathbb{Z} e_i^*\).
4. Part (c): \(|\Lambda^\vee / \Lambda| = |\det M|\).
4.1. Express the basis \(\{e_j\}\) of \(\Lambda\) in terms of the basis \(\{e_i^*\}\) of \(\Lambda^\vee\):\[e_j = \sum_{i=1}^n (e_i \cdot e_j) e_i^* = \sum_{i=1}^n M_{ij} e_i^*.\]
4.2. Thus \(\Lambda\) is the image of the \(\mathbb{Z}\)-linear map \(M: \Lambda^\vee \to \Lambda^\vee\), so\[\Lambda^\vee / \Lambda \cong \mathbb{Z}^n / M \mathbb{Z}^n.\]
4.3. By the Smith Normal Form theorem for integer matrices, there exist unimodular matrices \(P, Q \in \operatorname{GL}_n(\mathbb{Z})\) (\(\det P, \det Q \in \{\pm 1\}\)) such that\[P M Q = D = \operatorname{diag}(d_1, d_2, \dots, d_n), \quad d_i \in \mathbb{Z}_{>0}.\]
4.4. The quotient group is isomorphic to:\[\mathbb{Z}^n / M \mathbb{Z}^n \cong \mathbb{Z}^n / D \mathbb{Z}^n \cong \bigoplus_{i=1}^n \mathbb{Z} / d_i \mathbb{Z}.\]
4.5. The cardinality of the quotient group is:\[|\Lambda^\vee / \Lambda| = \prod_{i=1}^n d_i = \det D.\]
4.6. Taking determinants:\[\det D = |\det(P M Q)| = |\det P| \cdot |\det M| \cdot |\det Q| = 1 \cdot |\det M| \cdot 1 = |\det M|.\]
5. Conclusion:
$\Lambda \subseteq \Lambda^\vee$, $\det M \ne 0$, the dual basis vectors $e_i^*$ span $\Lambda^\vee$, and $|\Lambda^\vee / \Lambda| = |\det M|$.