2. Lattices

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Points in a lattice in dimension one generated by   have the form of  . Rewrite   as:

  •  

By definition a' is coprime with b'. By Bézout's identity we know that there exist integers m and n such that  . It follows that for all  , we have  .

Since   must be an integer, we conclude that all   must have form  . However, this means that a dimension-1 lattice generated by   can be generated equivalently by  .

Therefore, we see that all dimension-1 lattices generated by a set of vectors can be equivalently generated by a set of one vector. One can achieve this by repeatedly replacing one pair of vectors within the set with an "equivalent" vector, reducing the total number of vectors in the set by one for each step, until only one vector is remaining. Thus, all the lattices in dimension one can be generated by  .

Assume the clause following the "iff" is true, and that all vectors in   is expressible in the form  .

Define  , and also set  . Now define  . We see that   is true, therefore   by definition. This means that for all  ,  , since it is true that   and is therefore expressible as  .


Similarly, assume that the premise is true and that if lattice   generated by   is full, then by definition for all   there exists some   such that  . From the definition of the lattice we know that a point in   is expressible as  .


From the fact that  , we have  . Define  , and we see that   must be in   since  . Therefore, if   is full, then all   can be expressed as  .

The lattice generated by   is not full; adding vectors   gives  , which is a multiple of the third vector,  . We note that a vector of the form   such that   is not in the lattice since it is not possible to combine vectors such that the first two coordinates remain equal while the third varies. Because every integer multiple of such a vector will remain in the same form, the lattice is therefore not full.

Arithmetic with vectors first give  , and then  . We then do  , and finally  .

This means that a subset of the lattice (We denote this   can be written as a combination of vectors  . Since  , we can see that given any vector  , we have  , which means  , meaning the lattice given is full.


3. Determinant and Divisor

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Given that  , then for any  , we also have  .

From this, we see that a lattice point   in   corresponds to point   in colattice  . We also see that lattice point   in   corresponds to point   in colattice  . Therefore, we see that every point in   corresponds to a point in   if  , and thus they are equal.


Similarly, if  , then the points in   and   corresponding to their original point in   before the translation must differ by some sum of vectors  , because if that was not the case, the two colattices would not be equal to L.

From this, we have  . Subtract   from both sides to get the desired result.

Given that  , assume  . There is at least one   in the intersection that belongs to both   and  . Without loss of generality, write  . Subtracting   and  , respectively, yield vectors   and  , which should both be in   by definition. However, this implies that  , producing a contradiction, therefore   must be false, and thus   implies  .

Equivalently, this implies that  .

This lattice  , when generated by  , contains all   such that   since both vectors used to generate the lattice obey this. Collatices  ,  ,   obey   respectively due to the additional y-value contributed by the shifting vector, and thus it is impossible for points in each lattice to overlap and they are therefore distinct. Because a lattice must be within the   space, a point in a lattice must have integer coordinates. However,   cannot possibly be anything other than  , thus we see that the three colattices cover the entire   space. Therefore, we see that there are no more colattices of  .

As in 4.2, a lattice in   is full iff a subset   of it can be generated by  , where  , such that   is in the   position and  . (If it cannot be generated by one such set, then at least one vector in   does not satisfy the prerequisites of a "full" lattice. Similarly, if it can be generated by one such set, then for  ,   as shown earlier in 2.

We find that   due to the fact that such a lattice represents a regular grid in  , which must be finite. If it is infinite,   cannot possibly be a full lattice as the product used to compute   is not an infinite product. Therefore, a lattice is full iff its determinant is finite.

Given lattice   generated by   and  , the lattice   generated by   is a homothety of   about the origin by a factor of  . The determinant of this lattice will clearly be an integer. Now apply a dilation about the origin by a factor of  . Every point in the   space of   now corresponds to   points in the  space of  . Therefore, every colattice of   corresponds to   colattices of  . Therefore,  , and we see that it must, therefore, be divisible by   since  .

4. Finite Generation

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Given that  ,   will contain more points than   for some arbitrary region of  . In such a region, the number of colattices that   can take on is equal to the number that   can take on, minus the number of points in   but not in  . Take a smallest region containing all vectors   such that   describe all distinct colattices of L_2. Here, every point not in   represents a distinct colattice, but for  , its number of distinct colattices is less than that of   as   contains more points (and therefore less empty spaces). Therefore, the determinant of   must be lower than that of  .

Oh the other hand, if the determinant of   is infinite, then the determinant of   is either also infinite, or takes on a finite value. The lattice containing all lattice points in   space would be a superset of all  , and yet have a determinant of 1.

A lattice in   is full iff a subset   of it can be generated by  , where  , such that   is in the   position and  . (If it cannot be generated by one such set, then at least one vector in   does not satisfy the prerequisites of a "full" lattice. Similarly, if it can be generated by one such set, then for  ,  .

Given a finitely generated subset of the full lattice, the remaining points may be accounted for by the addition of vectors to the finite set. As full lattices are ordered, only a finite number of vectors is needed or allowed to account for all points in the lattice, thus only a finite number of vectors is needed to completely describe the lattice.

5. Isomorphism Types of Lattices

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Given 5.6, we conclude that the two lattices are isomorphic as their determinants are both 15 and divisors both 1.

Assume   is generated by   and   is generated by  . Since  , by 5.1 we see that the lattices are not isomorphic, since the divisors would be equal if they were.

6. Canonical Form

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As the isomorphism effectively relies on a projection of  , the signature is (1,1,0) as the GCD of 2 and 3 is 1.

The signature can be found by breaking down the lattice to the combination of  , giving the signature