Let \(\Lambda\subset\mathbb C\) be a lattice, let \(X=\mathbb C/\Lambda\), and define \[ \operatorname{End}(\Lambda)=\{\alpha\in\mathbb C:\alpha\Lambda\subseteq\Lambda\}. \]
Show that each \(\alpha\in\operatorname{End}(\Lambda)\) induces an analytic covering map \[ f:X\to X, \qquad [z]\mapsto[\alpha z], \] of degree \(|\alpha|^2\).
For which \(\alpha\in\mathbb C\) does there exist a lattice \(\Lambda\) with \(\alpha\in\operatorname{End}(\Lambda)\)?
Conclude that \[ \operatorname{End}(\mathbb Z\oplus\mathbb Z\tau)=\mathbb Z \] for almost every \(\tau\).
Solution
1. Part (a) has one exceptional value as written: \(0\in\operatorname{End}(\Lambda)\), but the map induced by \(\alpha=0\) is constant and is not a covering map. Thus the covering assertion is correct for every nonzero \(\alpha\in\operatorname{End}(\Lambda)\).
Certainly \(0\Lambda\subseteq\Lambda\), so \(0\in\operatorname{End}(\Lambda)\). Its induced map is \([z]\mapsto[0]\), which is not a local homeomorphism. This is the only exception, since the derivative of \(z\mapsto\alpha z\) is nonzero whenever \(\alpha\neq0\).
2. If \(0\neq\alpha\in\operatorname{End}(\Lambda)\), then\[ f_\alpha([z])=[\alpha z] \]is a well-defined analytic covering of degree\[ \deg(f_\alpha)=|\alpha|^2. \]
If \(z'=z+\lambda\) with \(\lambda\in\Lambda\), then \[ \alpha z'=\alpha z+\alpha\lambda \] and \(\alpha\lambda\in\Lambda\), so \(f_\alpha\) is well-defined. Since \(\alpha\neq0\), multiplication by \(\alpha\) is locally biholomorphic on \(\mathbb C\), hence the induced map on \(X\) is locally biholomorphic. It is also surjective, since for any \([w]\in X\), \[ f_\alpha([w/\alpha])=[w]. \] Thus \(f_\alpha\) is a finite covering of the compact torus \(X\).
Its kernel is \[ \ker(f_\alpha)=\alpha^{-1}\Lambda/\Lambda. \] Multiplication by \(\alpha\) identifies this quotient with \(\Lambda/\alpha\Lambda\), so every fiber has \[ [\Lambda:\alpha\Lambda] \] points. Multiplication by \(\alpha\) scales Euclidean area by the real determinant \(|\alpha|^2\). Hence a fundamental parallelogram for \(\alpha\Lambda\) has area \(|\alpha|^2\) times that of one for \(\Lambda\). Since \(\alpha\Lambda\subseteq\Lambda\) is a full-rank sublattice, \[ [\Lambda:\alpha\Lambda] =\frac{\operatorname{covol}(\alpha\Lambda)}{\operatorname{covol}(\Lambda)} =|\alpha|^2. \] Therefore \(\deg(f_\alpha)=|\alpha|^2\).
3. If some lattice \(\Lambda\) satisfies \(\alpha\Lambda\subseteq\Lambda\), then\[ 2\operatorname{Re}(\alpha)\in\mathbb Z, \qquad |\alpha|^2\in\mathbb Z. \]Consequently either \(\alpha\in\mathbb Z\), or \(\alpha\) is a nonreal quadratic algebraic integer.
Choose a \(\mathbb Z\)-basis \((\lambda_1,\lambda_2)\) of \(\Lambda\). Because \(\alpha\lambda_j\in\Lambda\), multiplication by \(\alpha\) is represented in this real basis by a matrix \[ A\in M_2(\mathbb Z). \] As an \(\mathbb R\)-linear map on \(\mathbb C\), multiplication by \(\alpha\) has eigenvalues \(\alpha\) and \(\overline\alpha\) after complexification. Therefore \[ \operatorname{tr}(A)=\alpha+\overline\alpha=2\operatorname{Re}(\alpha), \qquad \det(A)=\alpha\overline\alpha=|\alpha|^2. \] Both are integers, and \(\alpha\) satisfies \[ t^2-\operatorname{tr}(A)t+\det(A)=0. \] If \(\alpha\) is nonreal, this makes it a quadratic algebraic integer with negative discriminant. If \(\alpha\) is real, then the two eigenvalues of \(A\) are both \(\alpha\), so \[ 2\alpha\in\mathbb Z, \qquad \alpha^2\in\mathbb Z. \] Writing \(2\alpha=m\in\mathbb Z\) gives \(m^2=4\alpha^2\in4\mathbb Z\), hence \(m\) is even and \(\alpha\in\mathbb Z\).
4. Conversely, every integer and every nonreal quadratic algebraic integer occurs in \(\operatorname{End}(\Lambda)\) for some lattice. Hence the answer to (b) is precisely\[ \boxed{\alpha\in\mathbb Z\ \text{or}\ \alpha\text{ is a nonreal quadratic algebraic integer}.} \]Equivalently, the possible values are exactly those satisfying\[ 2\operatorname{Re}(\alpha)\in\mathbb Z, \qquad |\alpha|^2\in\mathbb Z. \]
Every integer preserves every lattice. Now let \(\alpha\notin\mathbb R\) satisfy a monic quadratic equation \[ \alpha^2-m\alpha+n=0, \qquad m,n\in\mathbb Z. \] Then \[ \Lambda=\mathbb Z\oplus\mathbb Z\alpha \] is a lattice because \(1\) and \(\alpha\) are linearly independent over \(\mathbb R\). Moreover \[ \alpha\cdot1=\alpha\in\Lambda, \qquad \alpha\cdot\alpha=m\alpha-n\in\Lambda, \] so \(\alpha\Lambda\subseteq\Lambda\).
For the equivalent numerical condition, 3 proves necessity. Conversely, if \(2\operatorname{Re}(\alpha)=m\in\mathbb Z\) and \(|\alpha|^2=n\in\mathbb Z\), then \[ \alpha^2-m\alpha+n =\alpha^2-(\alpha+\overline\alpha)\alpha+\alpha\overline\alpha =0, \] and the preceding argument applies; in the real case 3 already shows that \(\alpha\) is an integer.
5. For \(\tau\in\mathbb H\), one has\[ \operatorname{End}(\mathbb Z\oplus\mathbb Z\tau)\neq\mathbb Z \]if and only if \(\tau\) is imaginary quadratic. In particular the exceptional set is countable, so\[ \operatorname{End}(\mathbb Z\oplus\mathbb Z\tau)=\mathbb Z \]for almost every \(\tau\).
Set \(\Lambda_\tau=\mathbb Z\oplus\mathbb Z\tau\). Suppose \(\alpha\in\operatorname{End}(\Lambda_\tau)\). Since \(\alpha=\alpha\cdot1\in\Lambda_\tau\), write \[ \alpha=a+b\tau, \qquad a,b\in\mathbb Z. \] Also \(\alpha\tau\in\Lambda_\tau\), so \[ \alpha\tau=c+d\tau \qquad(c,d\in\mathbb Z). \] If \(b=0\), then \(\alpha=a\in\mathbb Z\). If \(b\neq0\), eliminating \(\alpha\) gives \[ b\tau^2+(a-d)\tau-c=0. \] Since \(\tau\notin\mathbb R\), this says that \(\tau\) is imaginary quadratic.
Conversely, if \(\tau\) is imaginary quadratic, choose integers \(A,B,C\) with \(A\neq0\) such that \[ A\tau^2+B\tau+C=0. \] Then \(\alpha=A\tau\) is nonreal and \[ \alpha\cdot1=A\tau\in\Lambda_\tau, \qquad \alpha\tau=A\tau^2=-B\tau-C\in\Lambda_\tau. \] Thus \(\alpha\in\operatorname{End}(\Lambda_\tau)\setminus\mathbb Z\).
There are only countably many integer triples \((A,B,C)\), so the imaginary-quadratic points of \(\mathbb H\) form a countable, hence Lebesgue-measure-zero, subset. This proves (c).