Consider $\Lambda$ a lattice in $\mathbb{R}^2$. Let $S \in O(\Lambda)$ be a reflection, i.e. $\det S = -1$. Set $S_{1}= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$ and $S_{2}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$.

I now want to prove that there is a basis of $\Lambda$ such that the matrix of $S$ is $S_1$ or $S_2$.

Until now I was only able to prove that since $S, S_1, S_2$ all have the same spectrum and thus I can deduce that there is a Basis of $\mathbb{R}^2$ such that the condition is fulfilled.

Question: How can I now prove that this basis is also a basis of $\Lambda$?

Any help is very appreciated!

Definition of a lattice and its basis:

$\lambda_1 \mathbb{Z} \oplus \dots \oplus \lambda_n \mathbb{Z} = \{m_1 \lambda_1 + \dots m_n \lambda_n | m_i \in \mathbb{Z} \} \subset V$ is the lattice generated by the linearly independent vectors $\lambda_1, \dots, \lambda_n \subset V$. Those are the basis.

Let $ \Lambda = \lambda_1 \mathbb{Z} \oplus \dots \oplus \lambda_n \mathbb{Z}$ and $ \Lambda' = \lambda_1' \mathbb{Z} \oplus \dots \oplus \lambda_n' \mathbb{Z}$

$\Lambda = \Lambda'$ if and only if there is a $C \in GL(n, \mathbb{Z}$ such that $\lambda_i' = \sum\limits_{j=1}^n C_{ij} \lambda_j$ for all $n$.


Here is one way to do it (I will write $L$ instead of $\Lambda$):

We already know that $S$ has Eigenvalues $1,-1$ and (due to those being integral) there is a basis $\{l_1,l_2\}\subset L$ of $\mathbb{R}^2$ such that $Sl_1=l_1$ and $Sl_2=-l_2$. We will take $l_1$ and $l_2$ reduced, i.e. $\frac{m}{n} l_i \notin L$ for $n \nmid m $.

Now there are two possibilities:

$(l_1,l_2)$ is already a basis of $L$. Then $S$ takes diagonal form with respect to this basis and we are done.

Otherwise there is some $l \in L - \langle l_1,l_2 \rangle_\mathbb{Z}$. But certainly $l \in \langle l_1,l_2 \rangle_\mathbb{Q}$ so write $l=\frac{a_1}{b_1}l_1+\frac{a_2}{b_2}l_2$ (both fractions completly reduced).

Then $b_1l-a_1l_1=\frac{a_2b_1}{b_2} l_2$ is an element of $L$, hence $b_2 \mid b_1$ and by an analogous computation $b_1 \mid b_2$. So $b_1=b_2=:b$.

Now $l+Sl=\frac{2a_1}{b}l_1$ is in $L$ hence $b=2$.

So every element of $L$ which is not in the sublattice generated by $l_1$ and $l_2$ is of the form $\frac{1}{2}(a_1l_1+a_2l_2)$ with $a_1,a_2$ odd.

But in that case $\frac{1}{2}(l_1+l_2), \frac{1}{2}(l_1-l_2)$ is a basis of $L$ and $S$ takes the second form with respect to this basis.

Edit: I just realized one can completly skip the step where I concluded $b_1=b_2$ and just have a look at $l+Sl$ and $l-Sl$ to get to $b_1=b_2=2$.

  • $\begingroup$ Thank you for your answer. Is it completely obvious to state that the eigenvectors of S have to be in the Lattice? Because I'm not sure I can see that. $\endgroup$ – stebu92 Jul 28 '14 at 8:30
  • $\begingroup$ Could you please clarify what $L- \langle l_1,l_2 \rangle_{\mathbb{Z}}$ means? Is this the set of points in $L$ not reached by $l_1$ and $l_2$? I suppose that $\langle l_1,l_2 \rangle_{\mathbb{Z}} = \{ml_1+nl_2 | m,n \in \mathbb{Z}\}$, right? $\endgroup$ – stebu92 Jul 28 '14 at 8:43
  • $\begingroup$ On your second comment: You are correct. by "$-$" I denote the "set difference" and $\langle l_1,l_2 \rangle_\mathbb{Z}$ is as you stated the lattice generated by $l_1$ and $l_2$. $\endgroup$ – Sebastian Schoennenbeck Jul 28 '14 at 8:53
  • $\begingroup$ On your first comment: Write $S$ with respect to some basis of $L$ then the entries of that matrix are integral. Hence the this matrix plus (or minus) the identity matrix are also integral. So the kernels of these matrices (which are exactly the Eigenspaces of $S$) have generators with rational entries and after multiplication with a suitable integer there is in fact a generator with integral entries. But then the corresponding linear combination of your basis lies in the lattice. $\endgroup$ – Sebastian Schoennenbeck Jul 28 '14 at 8:55
  • $\begingroup$ @stebu92: Do you have any more concerns? $\endgroup$ – Sebastian Schoennenbeck Aug 1 '14 at 12:16

The definition of a reflection is not $\det(S) = -1$. For example, $\pmatrix{2&0\\0&-1/2}$ is not a reflection.

What, then, is the definition of a reflection?

Once you have a proper definition, you'll be able to state that since the matrices $S,S_1,S_2$ have the same spectrum (which consists of $n$ distinct eigenvectors), they are similar, which means that the desired statement holds.

  • $\begingroup$ thanks for your comments. From the conditions given one can deduce that the eigenvalues of $S$ are indeed 1 and -1. So it indeed is a reflection. As written in my question I already found that $S,S_1,S_2$ are similar and that I can express them in a basis, such that the statement holds. However I want to prove that this basis is also a basis of $\Lambda$. $\endgroup$ – stebu92 Jul 23 '14 at 17:47
  • $\begingroup$ @stebu92 why would a basis of $\Bbb R^2$ not be a basis of $\Lambda$? $\endgroup$ – Omnomnomnom Jul 23 '14 at 17:49
  • $\begingroup$ @stebu92 I'm not actually sure what it means for a set to be a basis of $\Lambda$ as opposed to one of the general space, so if you could clarify this in either a comment or the question, it would help. $\endgroup$ – Omnomnomnom Jul 23 '14 at 17:53
  • $\begingroup$ The way I understand the exercise we have given a fixed lattice. On this lattice there are orthogonal maps from the lattice to itself. So with those we could construct another basis of the same lattice. I will add a definition to the question in a minute. $\endgroup$ – stebu92 Jul 23 '14 at 18:11
  • $\begingroup$ please see my edit, i hope it clarifies things $\endgroup$ – stebu92 Jul 23 '14 at 18:22

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