Use the Smith Decomposition to Analyze a Lattice
Consider the lattice 
generated by integer multiples of the vectors 
and 
.
In[1]:=
b1 = {3, -3};
b2 = {2, 1};In[2]:=
ptsb = Flatten[Table[j b1 + k b2, {j, -12, 12}, {k, -12, 12}], 1];In[3]:=
graphicsb =
Graphics[{Blue, PointSize[Large], Point@ptsb}, PlotRange -> 10,
Axes -> True]Out[3]=
Let 
be the matrix whose rows are 
and 
.
In[4]:=
m = {b1, b2};The Smith decomposition gives three matrices that satisfy the identity 
.
In[5]:=
{u, r, v} = SmithDecomposition[m];In[6]:=
u.m.v == rOut[6]=
The matrices 
and 
have integer entries and determinant one.
In[7]:=
{u // MatrixForm, v // MatrixForm, Det[u], Det[v]}Out[7]=
The matrix 
is integer and diagonal. From its entries it can be seen that the structure of the group 
is 
or simply 
, as 
is the trivial group.
In[8]:=
r // MatrixFormOut[8]//MatrixForm=
Multiplying the identity 
on the right by 
gives 
. Because 
is integer and determinant 
, 
generates the same lattice as 
but is simpler.
In[9]:=
g = r.Inverse[v];
g // MatrixFormOut[9]//MatrixForm=
Visualize the lattice generated by the rows of 
.
In[10]:=
ptsg = Flatten[
Table[j First[g] + k Last[g], {j, -12, 12}, {k, -12, 12}], 1];In[11]:=
graphicsg =
Graphics[{Red, PointSize[Medium], Point@ptsg}, PlotRange -> 10,
Axes -> True]Out[11]=
Superimposing the new lattice on the original confirms that they are the same.
In[12]:=
Show[{graphicsb, graphicsg}]Out[12]=
