Let's say you have a planar tiling. Quite often these tilings are introduced as geometrical objects with metrics; each tile having coordinates assigned to its vertices.

The tiling has an associated graph: the nodes of the graph are the vertices of the tiles, etc.

What kind of additional value is in general provided by insisting the tile vertices to have coordinates? Does not discussing the properties of the tiling graph topology give enough information?

I understand that the answer depends on the type of the tiling in question.

If one takes graphs as the starting point, then what would be some natural ways to define infinite planar graph periodicity so that there would exist periodic planar tilings corresponding to a given graph?

Can Penrose tiling be defined by its graph topology and can some of its general properties, like aperiodicity, be described purely by its tiling graph without relying on the angles and edge length of the tiles?

(If You find this post fuzzy or ignorant, that would be because the poster is not a 
professional mathematician.)

Pontus

I reject the premise of the question.  :-)

It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups.  If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$.  You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$.  A similar argument appears in Matthew Emerton's proof.  As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID.  In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field.  My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.

The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form.  Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations.  That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices.  This is one of the tricks in the proof that I give next.

Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.

Proof. By induction on $n$.  If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done.  Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation.  Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation.  (Pun intended. :-) )

The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$.  By the same argument, we can assume that $a$ does not appear in any other relation.

The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$.  By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations.  Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □

Compare the complexity of this argument to the other arguments supplied so far.

Minimizing the norm $|x|$ is a powerful step.  With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column.  Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.

Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm.  If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.

The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem.  And it only needs minor changes to apply to general PIDs.
