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2. Lattices

2.1

Points in a lattice in dimension one generated by $\{(a),(b)\}\,\!$ have the form of $(ma+nb),m,n\in \mathbb {Z}$. Rewrite $(ma+nb)$ as:

• $\gcd(a,b)(ma'+nb'),a'={\frac {a}{\gcd(a,b)}},b'={\frac {b}{\gcd(a,b)}}$

By definition a' is coprime with b'. By Bézout's identity we know that there exist integers m and n such that $ma'+nb'=1$. It follows that for all $q\in \mathbb {Z}$, we have $q(ma'+nb')=q$.

Since $(ma'+nb')$ must be an integer, we conclude that all $(ma+nb)$ must have form $q\cdot \gcd(a,b)$. However, this means that a dimension-1 lattice generated by $\{(a),(b))\}$ can be generated equivalently by $\{(\gcd(a,b))\}$.

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 $\{(a)\},a\in \mathbb {Z}$.

2.2

Assume the clause following the "iff" is true, and that all vectors in $\mathbb {Z} ^{n}$ is expressible in the form $\sum a_{i}x_{i},x_{i}\in S,a_{i}\in \mathbb {Q}$.

Define $a_{i}={\frac {m_{i}}{n_{i}}},m_{i},n_{i}\in \mathbb {Z} ,\gcd(m_{i},n_{i})=1$, and also set $q=\operatorname {lcm} (n_{1},n_{2},n_{3}...)$. Now define $a'_{i}=a_{i}\cdot q$. We see that $a'_{i}\in \mathbb {Z}$ is true, therefore $\sum a'_{i}x_{i}\in L$ by definition. This means that for all ${\vec {a}}$, $q{\vec {a}}\in L$, since it is true that ${\vec {a}}\in \mathbb {Z} ^{n}$ and is therefore expressible as $\sum a_{i}x_{i}$.

Similarly, assume that the premise is true and that if lattice $L\,\!$ generated by $S\,\!$ is full, then by definition for all ${\vec {a}}\in \mathbb {Z} ^{n}$ there exists some $q\in \mathbb {Z} ^{+}$ such that $q{\vec {a}}\in L$. From the definition of the lattice we know that a point in $L\,\!$ is expressible as $\sum m_{i}x_{i},x_{i}\in S,m_{i}\in \mathbb {Z}$.

From the fact that $q{\vec {a}}=\sum m_{i}x_{i}$, we have ${\vec {a}}=\sum {\frac {m_{i}}{q}}x_{i}$. Define $a_{i}={\frac {m_{i}}{q}}$, and we see that $a_{i}$ must be in $\mathbb {Q}$ since $m_{i},q\in \mathbb {Z}$. Therefore, if $L\,\!$ is full, then all ${\vec {a}}$ can be expressed as $\sum a_{i}x_{i},x_{i}\in S,a_{i}\in \mathbb {Q}$.

2.3

The lattice generated by $\{(2,1,6),(5,6,8),(1,1,2)\}$ is not full; adding vectors $(2,1,6)+(5,6,8)\,\!$ gives $(7,7,14)\,\!$, which is a multiple of the third vector, $(1,1,2)\,\!$. We note that a vector of the form $(x,x,y)\,\!$ such that $y\neq 2x$ 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.

2.4

Arithmetic with vectors first give $(5,6,8)-(2,1,6)-(1,2,2)=(2,3,0)\,\!$, and then $(2,1,6)+4\cdot (1,2,2)-(2,3,0)=(0,0,14)\,\!$. We then do $3\cdot (5,6,8)+3\cdot (2,1,6)-3\cdot (0,0,14)-7\cdot (2,3,0)=(7,0,0)\,\!$, and finally $(5,6,8)+(2,1,6)-(0,0,14)-(7,0,0)=(0,7,0)\,\!$.

This means that a subset of the lattice (We denote this $L_{s}$ can be written as a combination of vectors $(7,0,0),(0,7,0),(0,0,14)\,\!$. Since $\operatorname {lcm} (7,7,14)=14$, we can see that given any vector ${\vec {a}}\in \mathbb {Z} ^{2}$, we have $14{\vec {a}}\in L_{s}\subseteq L$, which means $14{\vec {a}}\in L$, meaning the lattice given is full.

3. Determinant and Divisor

3.1

Given that ${\vec {a}}-{\vec {b}}\in L\,\!$, then for any ${\vec {p}}\in L$, we also have ${\vec {p}}+{\vec {a}}-{\vec {b}}\in L$.

From this, we see that a lattice point ${\vec {p}}$ in $L\,\!$ corresponds to point ${\vec {p}}+{\vec {a}}$ in colattice ${\vec {a}}+L$. We also see that lattice point ${\vec {p}}+{\vec {a}}-{\vec {b}}$ in $L\,\!$ corresponds to point ${\vec {p}}+{\vec {a}}$ in colattice ${\vec {b}}+L$. Therefore, we see that every point in ${\vec {a}}+L$ corresponds to a point in ${\vec {b}}+L$ if ${\vec {a}}-{\vec {b}}\in L\,\!$, and thus they are equal.

Similarly, if ${\vec {a}}+L={\vec {b}}+L$, then the points in ${\vec {a}}+L$ and ${\vec {b}}+L$ corresponding to their original point in $L\,\!$ before the translation must differ by some sum of vectors $\sum _{x\in S}x\in L$, because if that was not the case, the two colattices would not be equal to L.

From this, we have ${\vec {a}}={\vec {b}}+\sum _{x\in S}x\in L$. Subtract ${\vec {b}}$ from both sides to get the desired result.

3.2

Given that $({\vec {a}}+L)\cap ({\vec {b}}+L)\neq \varnothing$, assume ${\vec {a}}-{\vec {b}}\notin L$. There is at least one ${\vec {q}}$ in the intersection that belongs to both ${\vec {a}}+L$ and ${\vec {b}}+L$. Without loss of generality, write ${\vec {q}}={\vec {p}}+{\vec {a}}$. Subtracting ${\vec {a}}$ and ${\vec {b}}$, respectively, yield vectors ${\vec {p}}$ and ${\vec {p}}+{\vec {a}}-{\vec {b}}$, which should both be in $L\,\!$ by definition. However, this implies that ${\vec {a}}-{\vec {b}}\in L$, producing a contradiction, therefore ${\vec {a}}-{\vec {b}}\notin L$ must be false, and thus $({\vec {a}}+L)\cap ({\vec {b}}+L)\neq \varnothing$ implies ${\vec {a}}-{\vec {b}}\in L$.

Equivalently, this implies that ${\vec {a}}-{\vec {b}}\notin L\Rightarrow ({\vec {a}}+L)\cap ({\vec {b}}+L)=\varnothing$.

3.3

This lattice $L\,\!$, when generated by $\{(1,2),(2,1)\}$, contains all $(a,b)\in \mathbb {Z} ^{2}$ such that $a+b\equiv 0\mod {3}$ since both vectors used to generate the lattice obey this. Collatices $(0,0)+L\,\!$, $(0,1)+L\,\!$, $(0,2)+L\,\!$ obey $a+b\equiv 0,1,2\mod {3}$ 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 $\mathbb {Z} ^{2}$ space, a point in a lattice must have integer coordinates. However, $x+y$ cannot possibly be anything other than $0,1,2\mod {3}$, thus we see that the three colattices cover the entire $\mathbb {Z} ^{2}$ space. Therefore, we see that there are no more colattices of $L\,\!$.

3.4

As in 4.2, a lattice in $\mathbb {Z} ^{n}$ is full iff a subset $L'\,\!$ of it can be generated by $S'=\{a_{1},a_{2},a_{3},...,a_{n}\}$, where $a_{i}=(0,0,0,...,0,p_{i},0,...,0,0)$, such that $p_{i}$ is in the $i^{th}$ position and $p_{i}\in \mathbb {Z} ^{+}$. (If it cannot be generated by one such set, then at least one vector in $\mathbb {Z} ^{n}$ does not satisfy the prerequisites of a "full" lattice. Similarly, if it can be generated by one such set, then for $N=\gcd(p_{1},p_{2},p_{3},...,p_{n})$, $N{\vec {a}}\in L$ as shown earlier in 2.

We find that $\operatorname {det} (L')=\prod p_{i}$ due to the fact that such a lattice represents a regular grid in $\mathbb {Z} ^{n}$, which must be finite. If it is infinite, $L\,\!$ cannot possibly be a full lattice as the product used to compute $\operatorname {det} (L')$ is not an infinite product. Therefore, a lattice is full iff its determinant is finite.

3.5

Given lattice $L\,\!$ generated by $S\,\!$ and $d=\operatorname {div} (L)$, the lattice $L'\,\!$ generated by ${\frac {S}{d}}$ is a homothety of $L\,\!$ about the origin by a factor of $1/d\,\!$. The determinant of this lattice will clearly be an integer. Now apply a dilation about the origin by a factor of $d\,\!$. Every point in the $\mathbb {Z} ^{n}$ space of $L'\,\!$ now corresponds to $d^{n}$ points in the $\mathbb {Z} ^{n}$space of $L\,\!$. Therefore, every colattice of $L'\,\!$ corresponds to $d^{n}$ colattices of $L\,\!$. Therefore, $\operatorname {det} (L)=\operatorname {det} (L')\cdot d^{n}$, and we see that it must, therefore, be divisible by $d^{n}$ since $\operatorname {det} (L')\in \mathbb {Z}$.

4. Finite Generation

4.1

Given that $L_{1}\supsetneq L_{2}$, $L_{1}$ will contain more points than $L_{2}$ for some arbitrary region of $\mathbb {Z} ^{n}$. In such a region, the number of colattices that $L_{1}$ can take on is equal to the number that $L_{2}$ can take on, minus the number of points in $L_{1}$ but not in $L_{2}$. Take a smallest region containing all vectors ${\vec {l}}$ such that ${\vec {l}}+L_{2}$ describe all distinct colattices of L_2. Here, every point not in $L_{2}$ represents a distinct colattice, but for $L_{1}$, its number of distinct colattices is less than that of $L_{2}$ as $L_{1}$ contains more points (and therefore less empty spaces). Therefore, the determinant of $L_{1}$ must be lower than that of $L_{2}$.

Oh the other hand, if the determinant of $L_{2}$ is infinite, then the determinant of $L_{1}$ is either also infinite, or takes on a finite value. The lattice containing all lattice points in $\mathbb {Z} ^{n}$ space would be a superset of all $L_{2}$, and yet have a determinant of 1.

4.2

A lattice in $\mathbb {Z} ^{n}$ is full iff a subset $L'\,\!$ of it can be generated by $S'=\{a_{1},a_{2},a_{3},...,a_{n}\}$, where $a_{i}=(0,0,0,...,0,p_{i},0,...,0,0)$, such that $p_{i}$ is in the $i^{th}$ position and $p_{i}\in \mathbb {Z} ^{+}$. (If it cannot be generated by one such set, then at least one vector in $\mathbb {Z} ^{n}$ does not satisfy the prerequisites of a "full" lattice. Similarly, if it can be generated by one such set, then for $N=\gcd(p_{1},p_{2},p_{3},...,p_{n})$, $N{\vec {a}}\in L$.

4.3

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

5.3

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

5.4

Assume $L_{1}$ is generated by $\{(2,0)(0,4)\}$ and $L_{2}$ is generated by $\{(1,0)(0,8)\}$. Since $\operatorname {div} (L_{1})=2\neq \operatorname {div} (L_{2})=1$, by 5.1 we see that the lattices are not isomorphic, since the divisors would be equal if they were.

6. Canonical Form

6.1

As the isomorphism effectively relies on a projection of $\mathbb {Z} ^{2}to\mathbb {Z} ^{3}$, the signature is (1,1,0) as the GCD of 2 and 3 is 1.

6.2

The signature can be found by breaking down the lattice to the combination of $\{(450,0,0,0),(0,3,0,0),(0,0,30,0),(0,0,0,90)\}$, giving the signature $(3,30,90,450)\,\!$