Equivalent definitions of a lattice in a real vector space of finite dimension

Question

I'm currently trying to work my way through chapter seven of Serre's book "A Course in Arithmetic" with a view to learning about modular forms. During the course of this chapter the book begins to talk about lattices in a real vector space of finite dimension and defines them as follows

A lattice in a real vector space $ V $ of finite dimension is a subgroup $ \Gamma $ which satisfies one of the following equivalent conditions:

1. $ \Gamma $ is discrete and $ V/\Gamma $ is compact;
2. $ \Gamma $ is discrete and generates $ V $;
3. There exists an $ \mathbb{R} $-basis of $ V $ which is a $ \mathbb{Z} $-basis of $ \Gamma $.

I'm not really comfortable with taking things such as the equivalence of these conditions on blind faith. However I have not been able to see how these conditions are equivalent. If anyone could provide some insight it would be much appreciated.

Answer 1

$(1)\implies (2)$

All we need to show is that $\Gamma$ generates $V$, i.e. $\text{span}_{\Bbb R}\Gamma = V$. If not, then there is a dimension $v\in V$ such that $v\ne \sum_n c_n \gamma_n$ for $\gamma_i\in \Gamma$. But then $V/\Gamma$ contains a vector subspace, hence is not compact.

Proof of this fact

If $\text{span}_{\Bbb R}\Gamma = W\subset V$ is proper, then since we're in finite dimensions we can find a complement to $W$ of positive dimension, and call it $W^\perp$ and we know that

$$V=W\oplus W^\perp$$

So looking just at the topological projection, we have $V/\Gamma\supseteq V/W\oplus W^\perp$ since $\Gamma\subseteq W$, and $W^\perp$ is a non-trivial closed vector space, hence $V/\Gamma$ contains a closed subset which is not compact, and so must not be compact itself.

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$(2)\implies (3)$

$\Gamma$ generates $V$, so select any $\Bbb Z$-basis (one exists by discreteness, since all discrete subgroups of $\Bbb R^n$ are isomorphic to $\Bbb Z^m$ for some $m$), $\beta$, for $\Gamma$ and consider $\text{span}_{\Bbb R}\beta$. It's some closed subspace of $V$, if not everything then $\Gamma$ does not generate $V$, since any element of $\Gamma$ is some combination of elements from $\beta$.

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$(3)\implies (1)$

This is the easiest, write the basis $\beta=\{v_1,\ldots, v_n\}$ then it's immediately observed that $V/\Gamma \cong \Bbb R^n/\Bbb Z^n$ which is a product of compact sets.

Answer 2

Here's an argument for $2$ implies $3$, presumably filling in some details from Adam's answer above.

Definition: Let $b_1, \ldots, b_m \in \Gamma$. Then $P(b_1, \ldots, b_m) = \{\sum r_i b_i : 0 \leq r_i < 1\}$.

$P$ the going to be the device that we use to measure the difference between a $\mathbb{R}$ basis an a $\mathbb{Z}$ basis. To quantify that, we have the following lemma, which amounts to saying that if $(b_1,\ldots, b_n)$ is a basis, then $P(b_1, \ldots, b_n)$ is a fundamental domain for the lattice:

Lemma: If $B = \{b_1 , \ldots b_n\} \subset \Gamma $ is a $\mathbb{R}$ basis for $V$, then $B$ is a $\mathbb{Z}$ basis for $\Gamma$ iff $P(B) \cap \Gamma = \{0\}$. Proof:

$\leftarrow$: Suppose that $P(B) \cap \Gamma = \{0\}$. $B$ is clearly linearly independent over $\mathbb{Z}$, since this was true over $\mathbb{R}$. It suffices to show that it $\mathbb{Z}$ spans $\Gamma$. Let $\gamma \in \Gamma$, then for some $t_i \in \mathbb{R}$, $\gamma = \sum t_i b_i$, and $\gamma' = \sum \text{floor}(t_i) b_i \in \mathbb{Z} B$, and $\gamma - \gamma' \in P(B) \cap \Gamma = \{0\}$. Hence $\gamma = \gamma'$ and $\mathbb{Z} B = \mathbb{R} B$.

$\rightarrow$: Suppose that $B$ is a $\mathbb{Z}$ basis. Suppose that $\gamma \in P(B) \cap \Gamma$. Then $\gamma = \sum t_i b_i$ for $0 \leq t_i < 1$. Since the linear combination to express $\gamma$ this way is unique, and since there must be an integral combination, each $t_i = 0$. Hence $\gamma = 0$. QED

Thus, in order to show that a discrete, spanning subgroup $\Gamma$ contains a $\mathbb{Z}$ basis for $\Gamma$, we will inductively build up a basis for $V$, and use the previous lemma to ensure that it is also a $\mathbb{Z}$ basis for $\Gamma$.

In particular, we are going to use induction in order to build a linearly independent set $B_k$ with $P(B_k) \cap \Gamma = \{0\}$ and $|B_k| = k$.

The main topological fact we will need is the following:

Topological Lemma: Let $X$ be a discrete subgroup of $\mathbb{R}^n$, and let $K$ be any compact region. Then $|K \cap X| < \infty$. Proof: We will show that $X$ is closed. Suppose otherwise, that $a_n \in K$ is a sequence with $a_n \to a$ and all $a_i$ distinct. Then, since $X$ is a subgroup, $x_n := a_{n+1} - a_{n} \in X$ for all $n$. Moreover, $x_n \to 0 \in X$, which, as no $x_n = 0$, contradicts the assumption that $X$ is discrete. Thus, $K \cap X$ is a discrete and compact, and therefore finite.

Base step: We are going to pick a $b_1 \in \Gamma$ so that $P(\{ b_1 \}) \cap \Gamma = \{0\}$. We do this by taking $b_1$ to be a shortest nonzero vector in $\Gamma$: simply take a ball of radius the size of some nonzero vector in $\Gamma$, and solve the finite optimization problem.

We set $B_1 = \{\gamma\}$.

Inductive step: Assume that $B_{k - 1}$ is built and $k \leq n = dim(V)$. Let $V_{k - 1} = \mathbb{R}\text{Span}B_{k-1} < V$. Since $\Gamma$ spanned $V$, there are vectors in $\Gamma \cap V_{k-1}^c$. Any such vector $b_k$ will satisfy the linear independence condition, but we need to pick $b_k$ so that $P(B_k) \cap \Gamma = \{0\}$.

We choose $b_k$ minimizing the Euclidean distance to the plane $V_{k-1}$. We remark on why we can do this: because of the topological lemma, every ball around $0$ contains only finitely many points of $\Gamma$. Thus, we take a ball of radius larger than the minimum distance of $\Gamma \setminus V_{k-1}$ to $V_{k-1}$, and solve the corresponding finite optimization problem.

A point $z \in P(B_k)$ can be written $z = z' + a b_k$, for $0 \leq a < 1$, and $z' \in P(B_{k-1})$. If, assuming there is an errant lattice point, $z = a b_k + \sum_{i = 1}^{k -1 } r_i b_i \in \Gamma \setminus \{0\}$, then $a > 0$, since by inductive hypothesis, $P(B_{k-1}) \cap \Gamma = \{0\}$. In particular, $z$ is a lattice point $z \not \in V_{k-1}$. The distance of $z$ to $V_{k-1}$ is the distance of $a b_k$, which is less than the distance of $b_k$, a contradiction.

End induction step.

Thus, running this procedure, we end up with a $B_n$ satisfying the requirements of the induction process:

0. $B_n \subset \Gamma$
1. $B_n$ is a linearly indepednent set of vectors.
2. $|B_n| = n$.
3. $P(B_n) \cap \Gamma = \{0\}$.

1 and 2 imply that $B_n$ is an $\mathbb{R}$ basis for $V$, and the lemma and $3.$ imply that $B_n$ is a $\mathbb{Z}$ basis for the lattice $\Gamma$.

We have proven:

Theorem: A discrete subgroup of $\mathbb{R}^n$ that spans has a $\mathbb{Z}$ basis given by a basis of $\mathbb{R}^n$.

Corollary: A discrete subgroup has $\mathbb{Z}$ rank bounded by the ambient vector space dimension.