Posted by: matheuscmss | February 11, 2012

## SPCS 5

As it was mentioned in the previous post of this series, today we’re going to apply our discussions so far (about origamis, their automorphisms groups and the consequences to the study of the Kontsevich-Zorich cocycle) to two concrete cases: in the first part of this post we will study a family of quasi-regular origamis with automorphism group isomorphic to the symmetric group $S_n$, and in the second part of this post we’ll study regular origamis with a rich representation theory (in the sense that real, complex and quaternionic representations appear at the same time inside the “interesting” part of homology of the origami).

-A family of quasi-regular origamis-

Fix $n\geq 2$ an integer, and let $G$ be group consisting of all permutations of the set $\{1,\dots,2n\}$ respecting its natural partition into even and odd numbers. Let $g_r:=(1,2,\dots,2n)$ and $g_u:=(24)$, so that

$c=g_r^{-1} g_u^{-1} g_r g_u = (13)(24)$

Define $H:=\{g\in G: g|_{E_{2n}} = \textrm{id}|_{E_{2n}}\}$ and $N=\{g\in G: g(E_{2n})=E_{2n}\}$, where $E_{2n}$ is the set of even number in $\{1,\dots,2n\}$.

Proposition.

1. $g_r$ and $g_u$ generate $G$;
2. $H\cap g_r H g_r^{-1} = \{\textrm{id}\}$;
3. $N$ is the normalizer of $H$ and $N/H\simeq S_n$;
4. $N$ is a normal subgroup in $G$ of index $2$.

Proof. Observe that $g_r H g_r^{-1}=\{g\in G: g(O_{2n})=O_{2n}\}$, where $O_{2n}$ is the set of odd integers in $\{1,\dots,2n\}$, so that the second item is clear. Also, it is clear that $G/N$ has order 2, i.e., $N$ is a subgroup of index 2 of $G$, and hence $N$ is normal, so that the fourth item follows. For the third item, note that $N$ normalizes $H$, and, as we just saw, $N$ is an index 2 normal subgroup of $G$, so that $N$ is the normalizer of $H$. Finally, for the first item, one notices that $g_r^i g_ug_r^{-i}$ is the transposition $(i+2,i+4)$. On the other hand, by induction on $n$, one can check that $S_n$ is generated by the transpositions $(1,2),\dots,(n-1,n)$. Thus, by letting $i$ vary amongst the even numbers, we obtain that the elements $g_r^i g_u g_r^{-i}$ generate $N$. Hence, since $g_r\notin N$ and $N$ has index 2 in $G$, we conclude that $g_r$ and $g_r^i g_u g_r^{-i}$ generate $G$, that is, $g_r$ and $g_u$ generate $G$. $\square$

This proposition says that the data $(G, H, g_r, g_u)$ generates an origami $M_n$ with automorphism group $Aut(M_n)\simeq N/H\simeq S_n$. Moreover, as we saw in this previous post here, the fact that $N$ is normal and $G/N$ is Abelian is an alternative characterization of quasi-regular origamis. In other words, $M_n$ is a quasi-regular origami with automorphism group $S_n$.

Remark. For the sake of comparison of $M_n$ with the case of regular origamis associated to $S_n$, observe that the image $\overline{c}$ of $c$ in the automorphism group $S_n\simeq N/H$ of $M_n$ is $\overline{c}=(1,2)$ (as, by definition, the image $\overline{c}$ of $c$ is computed by looking at the action of $c$ on the even numbers in $\{1,\dots,2n\}$ [in this case it is just the transposition $(2,4)$] and then renormalizing the even numbers in $\{1,\dots,2n\}$ [by multiplication by $1/2$] in order to get a permutation of $S_n\simeq N/H$), and thus $\overline{c}$ is not the commutator of a pair of elements of $S_n$ (as such commutators are necessarily even permutations).

Let $\rho_{\lambda}$ be a representation of $S_n$. Since $N$ has index 2 in $G$ and $g_r\notin N$, by this previous post here, the multiplicity $\ell_{\lambda}$ of $\rho_{\lambda}$ in the interesting part $H_1^{(0)}(M_n,\mathbb{C})$ of the homology of $M_n$ is given by

$\ell_{\lambda}=\textrm{codim}Fix_{\lambda}(\overline{c})+\textrm{codim}Fix_{\lambda}(g_r\overline{c}g_r^{-1})$

On the other hand, because $g_r c g_r^{-1}=\left\{\begin{array}{cc}(2,4)(1,3) & \textrm{if }n=2\\ (2,4)(3,5) & \textrm{if } n>2\end{array}\right.$, we have that (after considering the action only on even numbers and renormalizing by multiplication by $1/2$) $g_r \overline{c} g_r^{-1}=(1,2)=\overline{c}$. Thus, the previous formula simplifies to

$\ell_{\lambda} = 2\textrm{codim}Fix_{\lambda}(\overline{c})$

In order to render this formula more useful, we’ll briefly recall some aspects of the (very classical) representation theory of $S_n$ (along the lines of the classical book “Representation theory: a first course” of W. Fulton and J. Harris).

We start with the notion of Young diagrams. Given a list $\lambda=(\lambda_1,\lambda_2,\dots)$ of integers $\lambda_1\geq\lambda_2\geq\dots\geq 0$ such that, for some $i_0=i_0(\lambda)$, $\lambda_i=0$ for $i\geq i_0$, we take $|\lambda|:=\sum \lambda_i$ boxes and we form a (Young) diagram by arranging them in a left-justified way in a manner that the first arrow consists of $\lambda_1$ boxes, the second arrow consists of $\lambda_2$ boxes, etc. For instance, we drew below the object/diagram corresponding to $\lambda=(3,2,1,1)$ :

A natural operation on $\lambda$ consists into taking its dual $\lambda^*$ by considering the list $\lambda^*_i:=\#\{j:\lambda_j\geq i\}$. Geometrically, $\lambda^*$ is the list associated to the diagram obtained from the diagram of $\lambda$ after applying the reflection with respect to the anti-diagonal. For example, the dual $\lambda^*=(4,2,1)$ of $\lambda=(3,2,1,1)$ is depicted below:

It is well-known that conjugation classes of $S_n$ are described by the list of its cycles in non-increasing order, or, in other words, by Young diagrams associated to lists $\lambda$ with $|\lambda|=n$. Furthermore, such Young diagrams $\lambda$ allow to recover all irreducible representations $V_{\lambda}$ of $S_n$ (see Fulton and Harris’ book), and, as a matter of fact, they are all real. Below we give some fairly easy examples of this correspondence between Young diagrams and irreducible representations of $S_n$

Example 1. The list $\lambda=(n)$ corresponds to the trivial representation $U$.

Example 2. The dual $(\underbrace{1,\dots,1}_{n}):=(1^n)=(n)^*$ of $(n)$ corresponds to the alternating (signature) representation $U'$.

Example 3. The list $\lambda=(n-1,1)$ gives rise to the standard representation $V^{n-1}=\{v=(a_1,\dots,a_n): \sum\limits_{j=1}^n a_j=0\}$ (such that $\mathbb{C}^n=U\oplus V^{n-1}$ is the usual permutation representation of $S_n$).

Example 4. The dual $(2,1^{n-2})=(n-1,1)^*$ of $(n-1,1)$ is the representation $V\otimes U'$ obtained by taking the tensor product of the standard representation $V$ with the alternating representation $U'$.

Remark. In general, the representation associated to the dual $\lambda^*$ of $\lambda$ can be obtained by taking the tensor product $V_{\lambda}\otimes U'$ of the representation $V_{\lambda}$ corresponding to $\lambda$ with the signature representation $U'$. In particular, Examples 2 and 4 above are concrete incarnations of this general fact.

The dimension of the irreducible representation $V_{\lambda}$ can be computed with the aid of the hook-length formula:

$\textrm{dim} V_{\lambda} = \frac{n!}{\prod\limits_{i=1}^n H_i}$

where $H_i$ is the hook length of the box of number $i$, that is, the number of boxes to the right in the same row of $i$ plus the number of boxes below in the same column of $i$ plus one (for the box itself).

In the concrete example of the list $\lambda=(3,2,1,1)$, we drew below the corresponding Young diagram and we filled each box with the number giving its hook length.

From this picture we deduce that $\textrm{dim}(V_\lambda) = \frac{7!}{6\cdot 3\cdot 1\cdot 4\cdot 1 \cdot 2 \cdot 1} = 35$.

Coming back to our concrete example of quasi-regular origami $M_n$, we had that

$\ell_{\lambda}=2\textrm{codim} Fix_\lambda(\overline{c})$

where $\overline{c}=(1,2)$ was a transposition. In particular, it follows that

$\ell_{\lambda} = \textrm{dim} V_{\lambda} - \chi_{\lambda}(\overline{c})$.

In this formula, we already know the dimension of $V_{\lambda}$ (thanks to the hook-formula), so that we need to know $\chi_{\lambda}(\overline{c})$ to determine $\ell_{\lambda}$. This is done with the aid of Frobenius formula (that we specialize to the case of a transposition $\overline{c}$):

Frobenius formula. $\chi_{\lambda}(\overline{c})=\textrm{dim} V_{\lambda}\cdot \frac{p_2(\lambda)}{|\lambda|(|\lambda|-1)}$ where $p_2(\lambda):=\sum\limits_{i\geq 1} \lambda_i(\lambda_i-2i+1)$.

Observe that $p_2(\lambda^*) = -p_2(\lambda)$. Since $\textrm{dim} V_{\lambda} = \textrm{dim} V_{\lambda^*}$ (by the hook-length formula) and $|\lambda|=|\lambda^*|$, we have the following corollary of Frobenius formula:

Corollary. $\ell_{\lambda}+\ell_{\lambda^*} = 2\textrm{dim} V_{\lambda}$.

Example 1′. For $\lambda=(n)$, $\ell_{\lambda}=0$ (this is coherent with the fact that $M_n$ is quasi-regular and hence the trivial representation has zero multiplicity in $H_1^{(0)}(M_n,\mathbb{C})$).

Example 2′. By the corollary, for $\lambda=(\underbrace{1,\dots,1}_{n}) = (n)^*$, we have $\ell_{\lambda}=2$.

Example 5. For $n\geq 4$ and $\lambda=(n-2,2)$

we have $\textrm{dim} V_{\lambda} = \frac{n(n-3)}{2}$, $p_2(\lambda)=(n-1)(n-4)$ and thus $\ell_{\lambda} = 2(n-3)$. Also, by the corollary, for $\lambda=(2,2,\underbrace{1,\dots,1}_{n-4})=(n-2,2)^*$, we have $\ell_{\lambda}=(n-2)(n-3)$.

Example 6. For $\lambda=(n-2,1,1)$

we have $\textrm{dim} V_{\lambda}=\frac{(n-1)(n-2)}{2}$, $p_2(\lambda)=n(n-5)$ and thus $\ell_{\lambda}=2(n-2)$. Also, by the corollary, for $\lambda=(3,\underbrace{1,\dots,1}_{n-3})=(n-2,1,1)^*$, we have $\ell_{\lambda}=(n-3)(n-5)$.

Remark. The examples above give the multiplicities of all irreducible representations for $n\leq 5$. For $n=6$, it remains

• $\lambda=(3,3)$  for which $\textrm{dim} V_{\lambda}=5$, $p_2(\lambda)=6$ and $\ell_{\lambda}=4$;
• $\lambda=(2,2,2)=(3,3)^*$  for which $\textrm{dim} V_{\lambda}=5$, $p_2(\lambda)=-6$ and $\ell_{\lambda}=6$;
• $\lambda=(3,2,1)=(3,2,1)^*$  for which $\textrm{dim} V_{\lambda}=8$, $p_2(\lambda)=0$ and $\ell_{\lambda}=8$.

Remark. Still concerning regular origamis associated to symmetric groups, we observe a theorem of O. Ore says that every element $c$ of $A_n$ is the commutator $c=[g_r,g_u]$ of two elements $g_r,g_u$ in $S_n$. However, it is not obvious that, for a given $c$, one can choose $g_r$ and $g_u$ with $c=[g_r,g_u]$ and $g_r, g_u$ generates $A_n$ or $S_n$.

-A family of regular origamis-

Let $G=SL(2,\mathbb{F}_p)$ where $p\geq 3$ is prime. Note that $|G|=(p-1)p(p+1)$, and

$g_r=\left(\begin{array}{cc} 1 & a \\ 0 & 1\end{array}\right) \quad \textrm{ and } \quad g_u=\left(\begin{array}{cc} 1 & a \\ 0 & 1\end{array}\right)$

generate $G$ whenever $\pi:=ab\neq 0$.

Notations. $\mathbb{F} = \mathbb{F}_p$, $\mathbb{F}'=\mathbb{F}_{p^2}$, $\varepsilon$ is a generator of $\mathbb{F}^*$ (a cyclic group of order $p-1$), and $C$ is the subgroup of $(\mathbb{F}')^*$ consisting of elements with norm 1. Here, we recall that $\mathbb{F}'$ is a degree 2 extension of $\mathbb{F}$: this can be seen from the so-called Frobenius automorphism $\mathbb{F}'\to\mathbb{F}, x\mapsto x^p$. Also, using the Frobenius automorphism, we can define a norm $N(x)=x\cdot x^p=x^{p+1}$. In this language, $C=\{x\in (\mathbb{F}')^*: x^{p+1}=1\}$.  Moreover, we fix $\eta$ a generator of $C$. Finally, we think of $C$ as a subset of $SL(2,\mathbb{F}_p)$ by looking at the action of $C$ in a fixed basis $\mathbb{F}'=\mathbb{F}\oplus u\mathbb{F}$.

1. Conjugation classes in $SL(2,\mathbb{F}_p)$

The following table presents the information we will need about the conjugation classes of $SL(2,\mathbb{F}_p)$, namely, it gives representatives for each “type” of class (in the first collumn), the number of elements on each class of a given “type” (in the second collumn), and the number of classes of a given “type” (in the third collumn).

 representative number of elements in a class number of classes ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${1}$ ${1}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${1}$ ${1}$ ${\left(\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right)}$ ${(p^2-1)/2}$ ${1}$ ${\left(\begin{array}{cc}1 & \varepsilon \\ 0 & 1\end{array}\right)}$ ${(p^2-1)/2}$ ${1}$ ${\left(\begin{array}{cc}-1 & -1 \\ 0 & -1\end{array}\right)}$ ${(p^2-1)/2}$ ${1}$ ${\left(\begin{array}{cc}-1 & -\varepsilon \\ 0 & -1\end{array}\right)}$ ${(p^2-1)/2}$ ${1}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right), 0 ${p(p+1)}$ ${(p-3)/2}$ ${\eta^j, 0 ${p(p-1)}$ ${(p-1)/2}$

From this table, we see that the total number of conjugation classes is $p+4$.

2. Irreducible representations

Below we list the $p+4$ irreducible representations of $G=SL(2,\mathbb{F}_p)$.

a. the trivial representation $U$.

b. the standard representation $V$ coming from the action of $G$ on $\mathbb{P}^1(\mathbb{F}_p)$ (that is, we have a permutation representation of $G$ that we write as $U\oplus V$). The character $\chi_V$ is given in the following table

 ${\chi_V}$ ${\pm\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${p}$ ${\pm\left(\begin{array}{cc}1 & \ast \\ 0 & 1\end{array}\right)}$ ${0}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${1}$ ${\eta^j}$ ${-1}$

c. let $\tau:\mathbb{F}^*\to\mathbb{C}^*$ be a character with $\textrm{Im}\tau(\varepsilon)>0$ (there are $(p-3)/2$ possible choices of $\tau$), and let

$B=\left\{g=\left(\begin{array}{cc}a & c \\ 0 & a^{-1}\end{array}\right)\in SL(2,\mathbb{F}_p)\right\}$

be the usual Borel subgroup. Define the character $B\to\mathbb{C^*}, g\mapsto \tau(a)$ and consider the induced representation $W_{\tau}$ of $SL(2,\mathbb{F}_p)$. One has $\textrm{dim} W_\tau = p+1=[G:B]$, and the character $\chi_{W_\tau}$ is given in the following table

 ${\chi_{W_\tau}}$ ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${p+1}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${(p+1)\tau(-1)}$ ${\left(\begin{array}{cc}1 & \ast \\ 0 & 1\end{array}\right)}$ ${1}$ ${\left(\begin{array}{cc}-1 & \ast \\ 0 & -1\end{array}\right)}$ ${\tau(-1)}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${\tau(\varepsilon^j) + \tau(\varepsilon^{-j})}$ ${\eta^j}$ ${0}$

d. in the case of the character $\tau_*$ with $\tau_*(\varepsilon)=-1$, the induced representation $W_{\tau_*}$ is reducible: indeed, $W_{\tau_*}=W'\oplus W^{''}$. Observe that $\tau_*(\varepsilon)=-1$ implies that $\tau_*(-1)=(-1)^{(p-1)/2}$. The characters $\chi_{W'}, \chi_{W^{''}}$ are given in the following tables.

For $p\equiv 1 (\textrm{mod } 4)$,

 ${\chi_{W'}}$ ${\chi_{W^{''}}}$ ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${(p+1)/2}$ ${(p+1)/2}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${(p+1)/2}$ ${(p+1)/2}$ ${\left(\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right)}$ ${(1+\sqrt{p})/2}$ ${(1-\sqrt{p})/2}$ ${\left(\begin{array}{cc}1 & \varepsilon \\ 0 & 1\end{array}\right)}$ ${(1-\sqrt{p})/2}$ ${(1+\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -1 \\ 0 & -1\end{array}\right)}$ ${(1+\sqrt{p})/2}$ ${(1-\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -\varepsilon \\ 0 & -1\end{array}\right)}$ ${(1-\sqrt{p})/2}$ ${(1+\sqrt{p})/2}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${(-1)^j}$ ${(-1)^j}$ ${\eta^j}$ ${0}$ ${0}$

and, for $p\equiv 3 (\textrm{mod }4)$,

 ${\chi_{W'}}$ ${\chi_{W^{''}}}$ ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${(p+1)/2}$ ${(p+1)/2}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${-(p+1)/2}$ ${-(p+1)/2}$ ${\left(\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right)}$ ${(1+i\sqrt{p})/2}$ ${(1-i\sqrt{p})/2}$ ${\left(\begin{array}{cc}1 & \varepsilon \\ 0 & 1\end{array}\right)}$ ${(1-i\sqrt{p})/2}$ ${(1+i\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -1 \\ 0 & -1\end{array}\right)}$ ${(-1-i\sqrt{p})/2}$ ${(-1+i\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -\varepsilon \\ 0 & -1\end{array}\right)}$ ${(-1+i\sqrt{p})/2}$ ${(-1-i\sqrt{p})/2}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${(-1)^j}$ ${(-1)^j}$ ${\eta^j}$ ${0}$ ${0}$

e. let $\varphi: C\to\mathbb{C}^*$ be a character with $\textrm{Im}\varphi(-1)>0$ (there are $(p-1)/2$ possible choices of $\varphi$). Using $\varphi$ it is possible to construct a representation $X_{\varphi}$ whose character $\chi_{X_{\varphi}}$ is given by the following table

 ${\chi_{X_\varphi}}$ ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${p-1}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${(p-1)\varphi(-1)}$ ${\left(\begin{array}{cc}1 & \ast \\ 0 & 1\end{array}\right)}$ ${-1}$ ${\left(\begin{array}{cc}-1 & \ast \\ 0 & -1\end{array}\right)}$ ${\varphi(-1)}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${0}$ ${\eta^j}$ ${-(\varphi(\eta^j)+\varphi(\eta^{-j}))}$

f. let $\varphi_*$ be the character with $\varphi_*(\eta)=-1$. In this case, $X_{\varphi_*}$ is reducible: indeed, $X_{\varphi_*} = X'\oplus X^{''}$. Observe that $\varphi_*(\eta)=-1$ implies that $\varphi_*(-1) = (-1)^{(p+1)/2}$. The characters $\chi_{X'}, \chi_{X^{''}}$ are given in the following tables.

For $p\equiv 1 (\textrm{mod }4)$,

 ${\chi_{X'}}$ ${\chi_{X^{''}}}$ ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${(p-1)/2}$ ${(p-1)/2}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${-(p-1)/2}$ ${-(p-1)/2}$ ${\left(\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right)}$ ${(-1+\sqrt{p})/2}$ ${(-1-\sqrt{p})/2}$ ${\left(\begin{array}{cc}1 & \varepsilon \\ 0 & 1\end{array}\right)}$ ${(-1-\sqrt{p})/2}$ ${(-1+\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -1 \\ 0 & -1\end{array}\right)}$ ${(1-\sqrt{p})/2}$ ${(1+\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -\varepsilon \\ 0 & -1\end{array}\right)}$ ${(1+\sqrt{p})/2}$ ${(1-\sqrt{p})/2}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${0}$ ${0}$ ${\eta^j}$ ${(-1)^{j+1}}$ ${(-1)^{j+1}}$

and, for $p\equiv 3 (\textrm{mod }4)$,

 ${\chi_{W'}}$ ${\chi_{W^{''}}}$ ${\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)}$ ${(p-1)/2}$ ${(p-1)/2}$ ${\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)}$ ${(p-1)/2}$ ${(p-1)/2}$ ${\left(\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right)}$ ${(-1+i\sqrt{p})/2}$ ${(-1-i\sqrt{p})/2}$ ${\left(\begin{array}{cc}1 & \varepsilon \\ 0 & 1\end{array}\right)}$ ${(-1-i\sqrt{p})/2}$ ${(-1+i\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -1 \\ 0 & -1\end{array}\right)}$ ${(1-i\sqrt{p})/2}$ ${(1+i\sqrt{p})/2}$ ${\left(\begin{array}{cc}-1 & -\varepsilon \\ 0 & -1\end{array}\right)}$ ${(1+i\sqrt{p})/2}$ ${(1-i\sqrt{p})/2}$ ${\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)}$ ${0}$ ${0}$ ${\eta^j}$ ${(-1)^{j+1}}$ ${(-1)^{j+1}}$

In resume, the complete list of irreducible representations is:

• $U$
• $V$
• $W_\tau$ (there are $(p-3)/2$ of them)
• $W',W^{''}$
• $X_{\varphi}$ (there are $(p-1)/2$ of them)
• $X',X^{''}$

From the previous tables of characters, one sees that

• $U$ and $V$ are real
• $\chi_{W_\tau}$ and $\chi_{X_\varphi}$ take real values
• $W', W^{''}$ are complex when $p\equiv 3 (\textrm{mod }4)$ and real when $p\equiv 1 (\textrm{mod 4})$ (because in this case the characters $\chi_{W'}, \chi_{W^{''}}$ take only real values, so that $W', W^{''}$ are either real or quaternionic, and the quaternionic possibility is easily ruled out as $(p+1)/2=\textrm{dim}W'=\textrm{dim} W^{''}$ is odd and quaternionic representations always have even dimension)
• $X',X^{''}$ are complex when $p\equiv 3 (\textrm{mod }4)$, and real or quaternionic when $p\equiv 1 (\textrm{mod }4)$.

In order to distinguish whether $X',X^{''}$ are real or quaternionic (when $p\equiv 1 (\textrm{mod }4)$), we apply the following general criterion (based on the Frobenius-Schur indicator):

Theorem. Let $\chi$ be a character of an irreducible representation of a finite group $G$. Then,

$\frac{1}{|G|}\sum\limits_{g\in G}\chi(g^2)=\left\{\begin{array}{cl}1, & \textrm{ for } \chi \textrm{ real } \\ 0, & \textrm{ for } \chi \textrm{ complex } \\ -1, & \textrm{ for } \chi \textrm{ quaternionic }\end{array}\right.$

Applying this criterion (with the aid of the previous character tables), we can check that

• $X',X^{''}$ are quaternionic when $p\equiv 1 (\textrm{mod }4)$
• for $\tau(\varepsilon) = \exp(2\pi i j/(p-1))$ (with $0), $W_{\tau}$ is real when $j$ is even, and $W_{\tau}$ is quaternionic when $j$ is odd
• for $\varphi(\eta)=\exp(2\pi i j/(p+1))$ (with $0), $X_{\varphi}$ is real when $j$ is even, and $X_{\varphi}$ is quaternionic when $j$ is odd

In a nutshell, our discussions so far can be resumed as follows:

• for $p\equiv 1 (\textrm{mod }4)$, there are $(p+5)/2 = 2+2+(p-3)/2$ real representations, $(p+3)/2 = 2+(p-1)/2$ quaternionic representations, and $0$ complex representations.
• for $p\equiv 3 (\textrm{mod }4)$, there are $(p+1)/2=2+(p-3)/2$ real representations, $(p-1)/2$ quaternionic representations, and $4=2+2$ complex representations.

This concludes our quick review of the representation theory of $G=SL(2,\mathbb{F}_p)$. Now, we pass to the study of the regular origami.

Recall that we have chosen

$g_r=\left(\begin{array}{cc}1 & a \\ 0 & 1\end{array}\right) \quad \textrm{and} \quad g_u=\left(\begin{array}{cc}1 & 0 \\ b & 1\end{array}\right)$

with $\pi:=ab\neq 0$. Their commutator $c=g_r^{-1}g_u^{-1}g_r g_u$ is

$c=\left(\begin{array}{cc}1+ab+a^2b^2 & a^2b \\ -a^2b & 1-ab\end{array}\right)$

and hence its trace is $\textrm{tr}(c)=2+\pi^2$. The nature of the eigenvalues of $c$ are described by the quantity

$(\textrm{tr}(c))^2-4=\pi^2(\pi^2+4)$

It follows that

• if $\pi^2=-4$ (this can only happen when $p\equiv 1 (\textrm{mod }4)$ as $-1$ must be a square), then $c$ is conjugated to $c\sim\left(\begin{array}{cc}-1 & \ast \\ 0 & -1\end{array}\right)$ (and the order of $c$ is $2p$)
• if $\pi^2+4\neq 0$ is a square, $c$ is conjugated to $c\sim\left(\begin{array}{cc}\varepsilon^j & 0 \\ 0 & \varepsilon^{-j}\end{array}\right)$, and the order $\textrm{ord}(c)$ of $c$ satisfies $\textrm{ord}(c)>2$ and $\textrm{ord}(c)|(p-1)$
• if $\pi^2+4$ is not a square, $c$ is conjugated to $c\sim \eta^j$, and the order$\textrm{ord}(c)$ of $c$ satisfies $\textrm{ord}(c)>2$ and $\textrm{ord}(c)|(p+1)$

As we saw in this post here, the genus $g$ of the regular origami $(G,g_r,g_u)$ is

$g=1+\frac{1}{2}|G|\left(1-\frac{1}{\textrm{ord}(c)}\right)$

Also, note that

$\#\{\pi^2: \pi^2+4\neq 0 \textrm{ is a square} \}=\left\{\begin{array}{cc}(p-5)/4, & \textrm{ if } p\equiv 1 (\textrm{mod }4) \\ (p-3)/4, & \textrm{ if } p\equiv 3 (\textrm{mod }4)\end{array}\right.$

and

$\#\{\pi^2: \pi^2+4\neq 0 \textrm{ is not a square} \}=\left\{\begin{array}{cc}(p-1)/4, & \textrm{ if } p\equiv 1 (\textrm{mod }4) \\ (p+1)/4, & \textrm{ if } p\equiv 3 (\textrm{mod }4)\end{array}\right.$

Finally, note that the order of $c$ can be computed by the following recurrence equation

$t_{n+1} = t_1 t_n - t_{n-1}$

on $t_n:=\textrm{tr}(c^n)$ (with $t_0=2$, $t_1=2+\pi^2$). (This recurrence formula follows from the fact that $\textrm{tr}(A^2B)=\textrm{tr}(A)\cdot\textrm{tr}(AB)-\textrm{tr}(B)$ for $A,B\in SL(2)$).

For the first values of $p$, one gets the following tables (where below $\left(\begin{array}{c} a \\ p\end{array}\right)$ is the Legendre symbol).

For $p=5$:

 ${\pi^2}$ ${\pi^2+4}$ ${\left(\begin{array}{c} \pi^2+4 \\ p\end{array}\right)}$ ${\textrm{ord}(c)}$ ${1}$ ${0}$ ${0}$ ${10}$ ${0}$ ${3}$ ${-1}$ ${6}$

For $p=7$:

 ${\pi^2}$ ${\pi^2+4}$ ${\left(\begin{array}{c} \pi^2+4 \\ p\end{array}\right)}$ ${\textrm{ord}(c)}$ ${1}$ ${5}$ ${-1}$ ${8}$ ${4}$ ${1}$ ${1}$ ${3}$ ${2}$ ${6}$ ${-1}$ ${8}$

For $p=11$:

 ${\pi^2}$ ${\pi^2+4}$ ${\left(\begin{array}{c} \pi^2+4 \\ p\end{array}\right)}$ ${\textrm{ord}(c)}$ ${1}$ ${5}$ ${1}$ ${5}$ ${4}$ ${8}$ ${-1}$ ${12}$ ${9}$ ${2}$ ${-1}$ ${4}$ ${5}$ ${9}$ ${1}$ ${5}$ ${3}$ ${7}$ ${-1}$ ${6}$

For $p=13$:

 ${\pi^2}$ ${\pi^2+4}$ ${\left(\begin{array}{c} \pi^2+4 \\ p\end{array}\right)}$ ${\textrm{ord}(c)}$ ${1}$ ${5}$ ${-1}$ ${14}$ ${4}$ ${8}$ ${-1}$ ${14}$ ${9}$ ${0}$ ${0}$ ${26}$ ${3}$ ${7}$ ${-1}$ ${14}$ ${12}$ ${3}$ ${1}$ ${6}$ ${10}$ ${1}$ ${1}$ ${3}$

In the case of $p\equiv 1 (\textrm{mod }4)$ and $\textrm{tr}(c)=-2$ (parabolic), one can use the character tables above (and the fact that, as $c$ has order $2p$, $\sum\limits_{j=0}^{2p-1}\chi_{\lambda}(c)=2p\cdot\textrm{dim}\textrm{Fix}_{\lambda}(c^j)$) to deduce that

$\textrm{dim}\textrm{Fix}_{\lambda}(c)=\left\{\begin{array}{cl}1 & \textrm{ for } \lambda = U, V, W', W^{''} \\ 0 & \textrm{ for } \lambda=X_{\varphi}, X', X^{''} \\ 1+\tau(-1)\in\{0,2\} & \textrm{ for } \lambda = W_{\tau}\end{array}\right.$

and hence

$\ell_{\lambda}=\textrm{codim}\textrm{Fix}_{\lambda}(c)=\left\{\begin{array}{cl}0 & \textrm{ for } \lambda = U \\ p-1 & \textrm{ for } \lambda=V (\textrm{of dim.}=p) \\ (p-1)/2 & \textrm{ for } W', W^{''} (\textrm{of dim.}= (p+1)/2) \\ p-1 & \textrm{ for } \lambda=X_{\varphi} (\textrm{of dim.}=p-1) \\ (p-1)/2 & \textrm{ for }\lambda = X',X^{''} (\textrm{of dim.}=(p-1)/2) \\ p-\tau(-1) & \textrm{ for } \lambda=W_{\tau} (\textrm{of dim.}=p+1)\end{array}\right.$

Of course, while this give the multiplicities $\ell_{\lambda}$ of all irreducible representations in the interesting part of the homology of the regular origami associated to $(SL(2,\mathbb{F}_p), g_r,g_u)$ when $p\equiv 1 (\textrm{mod }4)$ and $\textrm{tr}(c)=-2$), there are several interesting questions left open here: for instance, what are the signatures of the Hermitian forms associated to the quaternionic representations $X', X''$, and $X_{\varphi}, W_{\tau}$ (when $\varphi(\eta)=\exp(2\pi i j/(p+1))$ and $\tau(\varepsilon)=\exp(2\pi ij/(p-1))$ for $j$ odd)?

It is likely that some of these questions will be dealt with in the forthcoming paper (by C.M., J.-C. Yoccoz and D. Zmiaikou), but J.-C. Yoccoz decided to stop here the comments on this project, so that the first “part” of the course ends now.

Next time, we will discuss simplicity of Lyapunov exponents of the Kontsevich-Zorich cocycle (along the lines of the works of A. Avila and M. Viana) in “general”, and then later we will restrict this discussion to the setting of square-tiled surfaces (by following the lines of a work in progress by C.M., M. Moeller and J.-C. Yoccoz).

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