Non-congruence of homology Veech groups in genus two

Abstract

We study the action of the Veech group of square-tiled surfaces of genus two on homology. This action defines the homology Veech group which is a subgroup of ${\mathrm{SL}}_2(\mathcal {O}_D)$ where $\mathcal {O}_D$ is a quadratic order of square discriminant. Extending a result of Weitze-Schmithüsen we show that also the homology Veech group is a totally non-congruence subgroup with exceptions stemming only from the prime ideals lying above 2. While Weitze-Schmithüsen’s result for Veech groups is asymmetric with respect to the spin structure our use of the homology Veech group yields a completely symmetric picture.

Appendix

In the Appendix we collect some results on quadratic orders $\mathcal {O}_D$ of square discriminant. In particular, we will give proofs of the results mentioned in the main part of the paper.

Let $K ={\mathbb {Q}}\oplus {\mathbb {Q}}$. A quadratic order of square discriminant is a subring $\mathcal {O}$ of K that is also a finitely generated ${\mathbb {Z}}$-module such that $1 \in \mathcal {O}$ and $\mathcal {O} \otimes {\mathbb {Q}}= K$. Recall that any such quadratic order is of the form

$$\begin{aligned} \mathcal {O}_D= \left\{ (x,y) \in {\mathbb {Z}}\times {\mathbb {Z}}\ | \ x \equiv y \mod d \right\} \end{aligned}$$

with $D=d^2$.

Norm and trace Analogously as in the non-square case we have:

Remark A.1

An element $z \in K$ is in ${\mathbb {Q}}$ if and only if $z = z^\sigma$, since ${\mathbb {Q}}$ is diagonally embedded into K.

As in the case of non-square-discriminants, $\mathcal {O}_D$ is a Noetherian ring: this is true, because $\mathbb {1}:=(1,1)$ and $w:=(0,d)$ form a ${\mathbb {Z}}$-basis of the ${\mathbb {Z}}$-module $\mathcal {O}_D$. We call this basis the standard basis of $\mathcal {O}_D$.

Recall that if $D \in \mathbb {N}$ is square-free, then the order $\mathcal {O}_D$ is the ring of integers of $K={\mathbb {Q}}(\sqrt{D})$ and an element $z \in K$ is in $\mathcal {O}_D$ if and only if both trace and norm of z lie in ${\mathbb {Z}}$.² In accordance with this, a similar property also holds if $D=d^2$ is the square of a prime number. We can however not expect that $z \in \mathcal {O}_D$ if and only if ${\mathrm{tr}}(z)$ and ${\mathcal {N}}(z)$ are in ${\mathbb {Z}}$, as we can see by the example $d=5$ and $z=(1,2)$. Indeed, one has to additionally impose some congruence conditions on the trace and the norm.

Lemma A.2

Let $z \in K$. If $D=d^2$ and d has no prime divisor of order greater than 1, then $z \in \mathcal {O}_D$ if and only if ${\mathrm{tr}}(z), {\mathcal {N}}(z) \in {\mathbb {Z}}$ and ${\mathrm{tr}}(z) \equiv 2v \mod d$ and ${\mathcal {N}}(z) \equiv v^2 \mod d$ for some $v \in {\mathbb {Z}}$.

Proof

If $z \in \mathcal {O}_D$ then $z=(x,y)$ for some $x,y \in {\mathbb {Z}}$. Thus ${\mathrm{tr}}(z)=(x+y,x+y)$ and ${\mathcal {N}}(z)=(xy,xy)$ and hence ${\mathrm{tr}}(z), {\mathcal {N}}(z) \in {\mathbb {Z}}$. Moreover $x \equiv y \mod d$ implies that $x+y \equiv 2x \mod d$ and $xy \equiv x^2 \mod d$.

On the other hand, let $z=(x,y)$ for some $x,y \in {\mathbb {Q}}$ and let ${\mathrm{tr}}(z)$ and ${\mathcal {N}}(z)$ be in ${\mathbb {Z}}$. Since $x+y$ and xy are both in ${\mathbb {Z}}$, also x and y are in ${\mathbb {Z}}$. Then ${\mathrm{tr}}(z) \equiv 2v \mod d$ implies that $y \equiv 2v - x \mod d$. Inserting this into ${\mathcal {N}}(z) \equiv v^2 \mod d$ gives $(x-v)^2 \equiv 0 \mod d.$ Since d has no quadratic term we have $x \equiv v \mod d$ and thus $x \equiv y \mod d.$$\square$

We will from now on until the end of the Appendix exclusively restrict to the case where $D=d^2$. The following lemma describes some basic arithmetic properties in this situation.

Lemma A.3

• (i) If $b \in \mathcal {O}_D$ is not a zero divisor and $a,c \in \mathcal {O}_D$ then ab|cb if and only if a|c.

• (ii) If $p \in {\mathbb {Z}}$ is a prime number and $a,b \in {\mathbb {Z}}$ then $p|a+wb$ if and only if p|a and p|b.

• (ii) If $p,m,x \in \mathcal {O}_D$ and $p \in {\mathbb {Z}}$ is a prime number with p|d and p|mx. Then $p \not \mid {\mathcal {N}}(m)$ implies p|x.

Proof

(i) and (ii) are clear by definition.

(iii) Let $m=(m_1,m_2)$ and $x=(x_1,x_2)$. The relation p|mx implies ${\mathcal {N}}(p) | {\mathcal {N}}(mx)$ or in other words $p^2| m_1m_2x_1x_2$. Since $p \not \mid {\mathcal {N}}(m)$ hence $p|x_1x_2$ and since p|d we must have $p|x_1$ and $p|x_2$, i.e. $x_1=pk_1$ and $x_2=x_1+jd =pk_1+pk_2$ with $k_1,k_2 \in {\mathbb {Z}}$. Then $(p,p)|(pk_1m_1,(pk_1+pk_2)(m_1+ld))$ is equivalent to

$$\begin{aligned} (p,p)|(pm_1k_1,p(m_1k_1+m_1k_2+k_1ld+k_2ld)) \end{aligned}$$

which implies $d|m_1k_2+(k_1+k_2)ld$. From this it follows either that $p|m_1$ which contradicts the fact $p \not \mid {\mathcal {N}}(m)$ or $p|k_2$. In the latter case $x_2=p(k_1+d\widetilde{k_2})$ and so p|x as we claim. $\square$

Ideals Recall that a regular ideal refers to an ideal containing a non-zero divisor.

Lemma A.4

A prime ideal $\mathfrak {p}$ of $\mathcal {O}_D$ is maximal if and only if it is regular.

Proof

If $\mathfrak {p}$ is a maximal $\mathcal {O}_D$-ideal, then $\mathfrak {p} \cap {\mathbb {Z}}$ is maximal and so $\mathfrak {p} \cap {\mathbb {Z}}\ne 0$, i.e. $\mathfrak {p}$ is regular. Conversely, if $\mathfrak {p}$ is regular and $x \in \mathfrak {p}$, then $0 \ne {\mathcal {N}}(x) \in \mathfrak {p} \cap {\mathbb {Z}}$. Thus, $\mathfrak {p} \cap {\mathbb {Z}}$ is maximal and hence also $\mathfrak {p}$ is. $\square$

The definition of the norm of a regular ideal perfectly generalizes the norm of an element.

Lemma A.5

For $z \in \mathcal {O}_D$ we have ${\mathcal {N}}((z))=|{\mathcal {N}}(z)|.$

In particular, it follows that ${\mathcal {N}}((z))=z^2$ for all $z \in {\mathbb {Z}}$ and ${\mathcal {N}}_D((z))=0$, if $z \in \mathcal {O}_D$ is a zero divisor.

Proof

The linear map $T_z: \mathcal {O}_D\rightarrow \mathcal {O}_D$ given by $T_z(\alpha ) = z\alpha$ has determinant ${\mathcal {N}}(z)$. Thus, ${\mathcal {N}}((z))= [\mathcal {O}_D: z\mathcal {O}_D] = |\det (T_z)| = | {\mathcal {N}}(z)|$. $\square$

Ideals as modules Let us now describe all ideals in $\mathcal {O}_D$. It is well-known that every ideal of $\mathcal {O}_D$ is also a ${\mathbb {Z}}$-module. The point of view that ideals are modules is very useful for giving a list of prime ideals. Moreover it allows us to calculate $\text {Spec} \ \mathcal {O}_D$. As in the case of non-square discriminants, it is essential that every ${\mathbb {Z}}$-module in $\mathcal {O}_D$ is generated by at most two elements.

Proposition A.6

Let $M \subset \mathcal {O}_D$ be a ${\mathbb {Z}}$-module in $\mathcal {O}_D$. Then there exist integers $m,n \in {\mathbb {Z}}_{\ge 0}$ and $a \in {\mathbb {Z}}$ such that

$$\begin{aligned}M = [n\mathbb {1}; a\mathbb {1}+mw]:= n\mathbb {1}{\mathbb {Z}}\oplus (a\mathbb {1}+mw){\mathbb {Z}}.\end{aligned}$$

Proof

Consider the subgroup $H:= \left\{ s \in {\mathbb {Z}}: \exists r \in {\mathbb {Z}}\ \text {with} \ r\mathbb {1}+sw \in M \right\}$ of ${\mathbb {Z}}$. As H is a subgroup of ${\mathbb {Z}}$, it is of the form $m{\mathbb {Z}}$ for some $m \ge 0$. By construction, there exists an $a \in {\mathbb {Z}}$ with $a\mathbb {1} + mw \in M$. Furthermore we know that $M \cap {\mathbb {Z}}\mathbb {1}$ can be regarded a subgroup of ${\mathbb {Z}}$ and so $M \cap \mathbb {1}{\mathbb {Z}}= n \mathbb {1} {\mathbb {Z}}$ for some $n \ge 0.$ We claim that $M = n\mathbb {1}{\mathbb {Z}}\oplus (a\mathbb {1} + mw) {\mathbb {Z}}.$ The inclusion $\supseteq$ is evident. Hence let us assume that $r\mathbb {1} + sw \in M$. Since $s \in H$ we have $s = um$ for some $u \in {\mathbb {Z}}$, and thus

$$\begin{aligned} r\mathbb {1} -ua\mathbb {1} = r\mathbb {1} +sw - u(a \mathbb {1} +mw) \in M \cap \mathbb {1}{\mathbb {Z}}. \end{aligned}$$

Hence $r-ua = nv$. But then

$$\begin{aligned} r\mathbb {1} + sw = (r-ua)\mathbb {1} + u(a\mathbb {1}+mw) = nv\mathbb {1} + u(a\mathbb {1}+mw) \in n\mathbb {1}{\mathbb {Z}}\oplus (a\mathbb {1} + mw) {\mathbb {Z}}. \end{aligned}$$

$\square$

As it simplifies notation and cannot cause any confusion the symbol $\mathbb {1}$ is usually omitted when embedding ${\mathbb {Z}}$ into $\mathcal {O}_D$. In other words, we write every ${\mathbb {Z}}$-module in $\mathcal {O}_D$ as $[n;a+mw]$ for some $a,n,m \in {\mathbb {Z}}.$

Since every ideal of $\mathcal {O}_D$ is also ${\mathbb {Z}}$-module, it is generated by at most two elements. The converse is not true since e.g. $M=[1;0] = {\mathbb {Z}}$ is a ${\mathbb {Z}}$-submodule of $\mathcal {O}_D$, but not an ideal. We therefore now describe under which conditions on a, m, n the ${\mathbb {Z}}$-module M is also an ideal. These conditions are just the same as in the case of non-square discriminants.

Proposition A.7

A regular ${\mathbb {Z}}$-module $M=[n;a+mw]$ is an ideal if and only if m|n, m|a, i.e. $a=mb$ for some $b \in {\mathbb {Z}}$, and $n|m{\mathcal {N}}(b+w)$.

Proof

Suppose that M is an ideal and consider the group H from the Proof of Proposition A.6. Then $c \in M \cap {\mathbb {Z}}$ implies $cw \in M$ and hence $c \in H$. This shows that $n{\mathbb {Z}}= M \cap {\mathbb {Z}}\subset H = m{\mathbb {Z}}$ or in other words that m|n. Observe that $w^2 = dw$. Since M is an ideal, $a+mw \in {\mathrm{M}}$ implies that $(a+mw)w=(a+md)w \in M$. By definition of H we therefore have that $a \in H$ and hence m|a. Finally, we set $\beta :=a+mw=m(b+w)$. Then $\beta \in M$ yields $\beta (b+w^\sigma ) \in M$. Hence $n| m{\mathcal {N}}(b+w).$

Now suppose that all the divisibility relations are fulfilled by M. It suffices to check that nw and $(a+mw)w$ both lie in M. We have

$$\begin{aligned} nw=\frac{n}{m}mw = \frac{n}{m} (a+mw) - \frac{n}{m}a = \frac{n}{m}(a+mw)-bn \end{aligned}$$

and so $nw \in M$ since m|n. And

$$\begin{aligned} (a+mw)w&= aw+mw^2 = m(b+d)w\\&= (b+d)(a+mw) - mb(b+d)\\&= (b+d)(a+mw)-m{\mathcal {N}}(b+w) \end{aligned}$$

implies $(a+mw)w \in M$, because $n|m{\mathcal {N}}(b+w).$$\square$

For an arbitrary ideal $\mathfrak {a} = [n;a+mw]$, it is straightforward to check (by giving an explicit list of representatives) that ${\mathcal {N}}(\mathfrak {a})=|mn|$. Note that if $M=[n;a+mw]$ is an ideal then its conjugated module is given by $[n;a+mw]^\sigma =[n;a+md-mw].$ If $\mathfrak {a}=((x,y))$ is a principal ideal then also $\mathfrak {a}^\sigma =((y,x))$ is a principal ideal.

Corollary A.8

Every ideal of prime norm p is of the form $[p;a+w]$ for some $a \in {\mathbb {Z}}$ with $p|{\mathcal {N}}(a+w).$ These ideals are indeed prime ideals.

Proof

The first assertion is clear from Proposition A.7. The second assertion follows from the fact that if $[p;a+w]$ is an $\mathcal {O}_D$-ideal, then it has index p in $\mathcal {O}_D$ and so it is a maximal ideal. $\square$

The corollary shows that there does not exist any inert prime number if D is a square because it is always possible to find an $a \in {\mathbb {Z}}$ such that $p|{\mathcal {N}}(a+w)$, e.g. $a=0$. Furthermore, it puts us into the position to count the number of different prime ideals of norm p if p is a prime number. This paves the way towards a ramification theory of prime numbers over $\mathcal {O}_D$.

Theorem A.9

Let $p \in {\mathbb {Z}}$ be a prime number. If p|D then there exists exactly one prime ideal $\mathfrak {a}$ of norm p and $\mathfrak {a}^\sigma =\mathfrak {a}$. Otherwise there exist exactly two different prime ideals $\mathfrak {a}, \mathfrak {b}$ of norm p and $\mathfrak {a}^\sigma =\mathfrak {b}$.

Proof

Let $p \in {\mathbb {Z}}$ be a prime number with p|D. Then every ideal of norm p is of the form $\mathfrak {a}=[p;a+w]$ with $p|a(a+d)$. As $p \in \mathfrak {a}$ we may without loss of generality assume that $0 \le a \le p-1$. Since p|d we get p|a and therefore $a=0$. So there exists exactly one ideal of norm p if p|d. It is then clear that $\mathfrak {a}^\sigma =\mathfrak {a}$.

If $p \in {\mathbb {Z}}$ is a prime number with $p\not \mid d$ then we may again assume that $0 \le a \le p-1$. So there remain the two possibilities p|a and $p|(a+d)$. These ideals $\mathfrak {a}$ and $\mathfrak {b}$ are indeed different since $p \not \mid d$ and $\mathfrak {a}^\sigma =\mathfrak {b}$. $\square$

Ramification From this the ramification theory for prime numbers over $\mathcal {O}_D$ can be deduced. In order to do this, we have to analyze the multiplication of two prime ideals. Let us first assume that p is a prime number with p|d and let $\mathfrak {a}=[p,w]$ be the unique prime ideal of norm p. Then $\mathfrak {a}^2=[p,w][p,w] = (p)[p, w, w, d/pw]$. Hence $\mathfrak {a}^2 = (p)[p,w] \ne (p)$. In particular, (p) cannot be further decomposed and the norm is not multiplicative.

If $p \not \mid d$ let $\mathfrak {a}=[p,w]$ and $\mathfrak {b}=[p,d+w]$ be the two different ideals of norm p. Hence

$$\begin{aligned} {[}p,w][p,d+w]=[p^2,pw,pd+pw,dw+w^2]=p[p,w,d] = (p). \end{aligned}$$

Recall that an ideal is called irreducible if it cannot be written as the intersection of two larger ideals. Hence we have established:

Theorem A.10

Let $p \in {\mathbb {Z}}$ be a prime number.

• (i) If $p \not \mid d$ then $(p)=\mathfrak {p}\mathfrak {p}^\sigma$ for a prime ideal $\mathfrak {p}$ of norm p with $\mathfrak {p} \ne \mathfrak {p}^\sigma$, i.e. p splits.

• (ii) If p|d then (p) is an irreducible ideal which is not prime.

Corollary A.11

Every ideal $\mathfrak {a} \subset \mathcal {O}_D$ with $\gcd ({\mathcal {N}}(\mathfrak {a}),d) = 1$ can be uniquely written as product of prime ideals.

Proof

We have $\mathfrak {a}\mathfrak {a}^\sigma = {\mathcal {N}}(\mathfrak {a})$. From Theorem A.10 it follows that ${\mathcal {N}}(\mathfrak {a})$ can be uniquely written as product of prime ideals. Thus, this is also true for $\mathfrak {a}$. $\square$

The special linear group In this paragraph, we only prove the claim of Proposition 2.5 with the help of the following two lemmas.

Lemma A.12

Let R, S be two commutative rings such that there exists a surjective homomorphism of rings $f: S \rightarrow R$. If ${\mathrm{SL}}_2(R)$ is generated by elementary matrices the induced map ${\mathrm{SL}}_2(S) \rightarrow {\mathrm{SL}}_2(S)$ is also surjective.

Proof

Any elementary matrix over R lifts to an elementary matrix of S. $\square$

Lemma A.13

If R is a finite commutative ring, then ${\mathrm{SL}}_2(R)$ is generated by elementary matrices.

Proof

Every finite commutative ring is a direct product of local rings. Since the claim is true for local rings (see e.g. [21, Chapter 2.2]) this finishes the proof. $\square$

Proof of Proposition 2.5

By definition $\Gamma ^D(\mathfrak {a})$ is the kernel of the projection ${\mathrm{SL}}_2(\mathcal {O}_D) \rightarrow {\mathrm{SL}}_2(\mathcal {O}_D/\mathfrak {a})$. The projection map is surjective by the preceding two lemmas. $\square$