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  "body": "\n## Definition\n\nLet $n \\ge 1$ and consider the [[Euclidean space|0XPR1E85W0]] $\\R^n$ as a group under addition.\nA *lattice* in $\\R^n$ is a [[subgroup|0X1JXT47GA]] of the form\n$$\\Lambda = \\Z v_1 + \\Z v_2 + \\dots + \\Z v_n = \\set{k_1 v_1 + \\dots + k_n v_n : k_1, \\dots, k_n \\in \\Z},$$\nwhere $v_1, \\dots, v_n$ are [[linearly independent|0CRMGPAH83]] vectors of $\\R^n$.\nThe family $(v_1, \\dots, v_n)$ is a *basis* of $\\Lambda$.\nThe *covolume* (or *determinant*) of $\\Lambda$ is $\\det \\Lambda = \\abs{\\det(v_1, \\dots, v_n)}$.\n\n## Remarks\n\nA subgroup $\\Lambda$ of $\\R^n$ is a lattice if and only if it is discrete and spans $\\R^n$ as a real vector space.\nTwo bases of the same lattice differ by an integer matrix of [[determinant|00KDZRW4C0]] $\\pm 1$, so the covolume does not depend on the basis.\nIt equals the volume of the fundamental parallelepiped $\\set{t_1 v_1 + \\dots + t_n v_n : 0 \\le t_i \u003c 1}$.\n\nExample: $\\Z^n$ is a lattice with basis the standard vectors $e_1, \\dots, e_n$ and covolume $1$.\n\nNon-example: $\\Z + \\Z\\sqrt{2} \\subset \\R$ is a subgroup generated by two vectors, but two vectors in $\\R^1$ are never linearly independent.\nIn fact this subgroup is dense in $\\R$, so it is not discrete.\nAnother non-example: $\\Z \\times \\set{0}$ in $\\R^2$ is discrete but does not span $\\R^2$.\n",
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