Why are there vectors that cannot span the Bravais lattice?

Question

I'm taking a course in condensed matter physics and have just been introduced to the Bravais lattice. From a mathematical point of view this could be thought of as a vector space where the scalar multipliers of the spanning set must belong to the integers. However, to me this does not seem to be quite true. Unlike a regular $n$ dimensional vector space where all sets of $n$ linearly independent vectors belonging to that vector space can span it, there are sets of $n$ linearly independent vectors that cannot span the Bravais lattice.

For example, the image below shows a Bravais lattice and a set of two linearly independent vectors $\{a_1,a_2\}$ that cannot span the lattice.

My question to you is, why is this? Is it because regular vector spaces require a field, but because the integers don't meant the definition of a field (no multiplicative identity), the Bravais lattice does not behave like a vector space?

Answer 1

The vectors cannot span the space due to the discrete values of the scalars which are allowed. If the scalars could take all real values, the vectors could span the whole space. But as you can take only integral multiple of the vectors, they can only span some discrete points.

Integral scalars restricting the spanning of the full space

Answer 2

It's not a vector space because the integers don't form a field, because nonzero integers aren't always invertible. The Bravais lattice is only an abelian group, or equivalently a module over $\mathbb{Z}$, and these don't behave quite like vector spaces. As you've seen, it can happen that a maximal linearly independent subset isn't a basis; instead it spans a submodule (in this context "sublattice" might be a more appropriate term) of finite index.

Answer 3

Suppose $A$ is a linearly independent set of vectors with the same finite number of elements as a set that spans a given lattice or space. Let $v$ denote a point not in $A$'s span, assuming one exists. In a vector space, we can prove it doesn't with a simple contradiction: $A\cup\{ v\}$ is linearly independent, yet larger than a basis. For the Bravais lattice, this argument fails: while $v=\sum_i c_ia_i$ for an enumeration $a_i$ of $A$ fails for integers $c_i$, $\sum_ic_ia_i+Cv=0$ need not. This is actually where the fact that a vector is over a field of scalars is crucial: with field elements, if $C\ne0$ we can cancel it, which in general is impossible in integers.