Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

this is a homework question but I am pretty confused on it--just don't know where to start.

We're given a lattice basis $(a, b)$ for a lattice $L$ in $\mathbb{R}^2$, and are supposed to show that every other lattice basis $(a', b')$ can be written as $(a, b)P$ for some $2\times 2$ integer matrix $P$ with determinant $\pm 1$.

share|improve this question
    

2 Answers 2

up vote 3 down vote accepted

Since $(a,b)$ is a basis for $L$ and $a',b' \in L$, it must be the case that $a'=c_{11}a+c_{21}b$ and $b'=c_{12}a+c_{22}b$. That is

$$(a',b')=(a,b)\begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix}$$

Since $(a,b)$ is a lattice basis, everything in $L$ must be an integral linear combination of $a$ and $b$. Thus the matrix entries are integers.

Now ask yourself: Why must this matrix be invertible? Why does its inverse have integer entries? What can we say about invertible matrices with integer entries whose inverses have integer entries (i.e. elements of $\mathrm{GL}_2(\mathbb{Z})$)?

share|improve this answer
    
Thanks, this was enough to give me the insight. Guess I just needed to think it through a bit more. –  flury Nov 7 '11 at 20:20
    
Good to hear. Sometimes the first half step is the hardest by a mile. :) –  Bill Cook Nov 7 '11 at 20:58

Suppose that $(a',b')$ generates the same lattice as $(a,b)$. Since a and b are elements of this lattice we can write them as linear combination of a' and b':

$$ \begin{pmatrix} a \\ b \end{pmatrix}= \begin{pmatrix} w & x \\ y & z \end{pmatrix} \begin{pmatrix} a'\\ b' \end{pmatrix}=P\begin{pmatrix} a'\\ b'\end{pmatrix} $$

where w,x,y,z are integers. Note that this linear combination is unique since $(a',b')$ is a basis for $\mathbb{R}^2$. By the same argument we can represent a' and b' uniquely as a linear combination of a and b, where the coefficients are integers. Inverting P we get $$ \begin{pmatrix} a'\\b' \end{pmatrix}=P^{-1}\begin{pmatrix} a\\b\end{pmatrix} $$ So $P^{-1}$ must have integer coefficients. Now we have that both $\det{(P)}$ and $det{(P^{-1})}=\det{(P)}^{-1}$ must be integers, so $\det{(P)}=\pm 1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.