Let \(\Lambda\subset\mathbb R^n\) be a discrete subgroup. Prove that \[ \Lambda\cong\mathbb Z^a \] for some \(a\le n\), and that \(a=n\) if and only if \(\mathbb R^n/\Lambda\) is compact.
Solution
1. Let\[ V=\operatorname{span}_{\mathbb R}(\Lambda), \qquad a=\dim_{\mathbb R}V\le n. \]Choose \(v_1,\ldots,v_a\in\Lambda\) which form a real basis of \(V\), and set\[ L=\mathbb Zv_1+\cdots+\mathbb Zv_a\subseteq\Lambda. \]Then \(L\cong\mathbb Z^a\).
Because the \(v_i\) are linearly independent over \(\mathbb R\), they are also linearly independent over \(\mathbb Z\). Hence the homomorphism \[ \mathbb Z^a\longrightarrow L, \qquad (m_1,\ldots,m_a)\longmapsto\sum_i m_iv_i \] is an isomorphism.
2. The subgroup \(L\) has finite index in \(\Lambda\).
Consider the compact parallelepiped \[ P=\left\{\sum_{i=1}^a t_iv_i:0\le t_i\le1\right\}\subset V. \] Every \(\lambda\in\Lambda\) can be written \(\lambda=\sum_i x_iv_i\). Subtracting \[ \sum_i\lfloor x_i\rfloor v_i\in L \] produces a representative of the coset \(\lambda+L\) lying in \(P\cap\Lambda\).
The set \(P\cap\Lambda\) is finite. Indeed, if it were infinite, compactness of \(P\) would give distinct \(\lambda_j\in P\cap\Lambda\) with a convergent subsequence. Then \[ \lambda_j-\lambda_k\longrightarrow0. \] Since \(\Lambda\) is discrete, some neighborhood of \(0\) contains no nonzero element of \(\Lambda\), so for sufficiently large \(j,k\) one would have \(\lambda_j-\lambda_k=0\), a contradiction. Thus only finitely many cosets of \(L\) occur in \(\Lambda\), i.e. \[ [\Lambda:L]<\infty. \]
3. One has\[ \Lambda\cong\mathbb Z^a. \]
By 2, \(\Lambda\) is finitely generated: adjoining finitely many representatives for \(\Lambda/L\) to the basis \(v_1,\ldots,v_a\) generates it. It is torsion-free because it is a subgroup of the additive group \(\mathbb R^n\). Hence the structure theorem for finitely generated abelian groups gives \[ \Lambda\cong\mathbb Z^r \] for some \(r\).
Since \(L\subseteq\Lambda\) has finite index, tensoring the inclusion with \(\mathbb Q\) gives an isomorphism \[ L\otimes_{\mathbb Z}\mathbb Q \cong \Lambda\otimes_{\mathbb Z}\mathbb Q. \] The left side has dimension \(a\) over \(\mathbb Q\), while the right side has dimension \(r\). Therefore \(r=a\).
4. If \(a=n\), then \(\mathbb R^n/\Lambda\) is compact.
When \(a=n\), the subgroup \(L\) from 1 is a full-rank lattice in \(\mathbb R^n\). Its compact parallelepiped \(P\) maps surjectively onto \(\mathbb R^n/L\), so \(\mathbb R^n/L\) is compact. Since \(L\subseteq\Lambda\), the natural map \[ \mathbb R^n/L\longrightarrow\mathbb R^n/\Lambda \] is continuous and surjective. Therefore \(\mathbb R^n/\Lambda\) is compact.
5. If \(\mathbb R^n/\Lambda\) is compact, then \(a=n\).
Because \(\Lambda\subseteq V\), there is a well-defined continuous surjection \[ \mathbb R^n/\Lambda\longrightarrow\mathbb R^n/V, \qquad x+\Lambda\longmapsto x+V. \] If \(a<n\), then \[ \mathbb R^n/V\cong\mathbb R^{n-a} \] is noncompact. It cannot be the continuous image of the compact space \(\mathbb R^n/\Lambda\). Hence \(a=n\).