Posted by: matheuscmss | January 22, 2012

SPCS 2

In this previous post (about J.-C. Yoccoz 2011-2012 course at College de France), we reviewed the distinct points of view on origamis, and we started the discussion of the action of the automorphism group on the first absolute homology group of the origami. We quickly recall the main points of the previous post for the reader’s convenience.

Given an origami \pi:M\to\mathbb{T}^2, we saw how to canonically describe it (through its monodromy group) by a finite group G generated by two elements g_r and g_u, and a choice of subgroup H\subset G such that \bigcap\limits_{g\in G} gHg^{-1}=\{1\}, so that one has Sq(M)=H\backslash G. Also, g_r and g_u act on the set of squares of M by r:Hg\mapsto H g g_r and u:Hg\mapsto H g g_u. Furthermore, denoting by N the normalizer of H in G, one has Aut(M)=N/H acting as nHg=Hng, for n\in N.

Next, given a subfield K\subset \mathbb{C}, we decomposed H_1(M,K)=H_1^{st}(M,K)\oplus H_1^{(0)}(M,K) and we reduced the problem of studying the action of Aut(M) on H_1(M,K) to understanding the action of Aut(M) on H_1^{(0)}(M,K), and, to do so, we considered the set \Sigma=\Sigma_{\max}=\pi^{-1}(\{0\}), and we proved that

H_1^{(0)}(M,K)= K(M)\ominus H_0(\Sigma,K)

as Aut(M)-module. In this way, since K(M):=K^{Sq(M)}=K^{H\backslash G} and H_0(\Sigma,K)=K^{\Sigma}, we were led to the study of the action of N on \Sigma. At this stage, we noticed that \Sigma is the set of orbits of \langle c\rangle acting to the right on H\backslash G or equivalently the set of orbits of H\times\langle c\rangle acting on H\backslash G by (h,c^m)\cdot g\mapsto hgc^m.

Nevertheless, we introduced the following notation: A_g is the point of \Sigma associated to g\in G, H\subset Stab(g)\subset N is the stabilizer of A_g for the action of Aut(M) on \Sigma, and k is the order of the commutator c=g_r^{-1}g_u^{-1}g_r g_u.

Finally, if we denote by n(g)>0 the smallest integer such that g c^{n(g)}g^{-1}\in N, we showed (in the very end of the previous post) the following lemma:

Lemma. Stab(g) = H\cdot (N\cap\langle g c g^{-1}\rangle) = H\cdot\langle gc^{n(g)}g^{-1}\rangle.

After this quick revision, we will enter (below the fold) the content of Yoccoz’s 2nd lecture (on January 18, 2012).

Analogously to the definition of n(g), we denote by h(g)>0 the smallest integer such that gc^{h(g)}g^{-1}\in H. Observe that, by construction, n(g) divides h(g), and h(g) divides k. For later use, in the next proposition we collect a series of elementary facts:

Proposition. It holds:

  1. N\cap\langle g c g^{-1}\rangle is a cyclic group of order k/n(g);
  2. Stab(g)/H is a cyclic group of order h(g)/n(g);
  3. the orbit \Sigma_g of A_g in N has cardinality \Sigma_g=\#(N/H)\cdot (n(g)/h(g));
  4. \#\{g'\in G: A_{g'}=A_g\}=\#H \cdot h(g);
  5. \#\{g'\in G: A_{g'}\in\Sigma_g\}=\#N \cdot n(g).

Proof. The first two items follow from the definitions of k, n(g) and h(g). The third item follows from the second one (as the size of an orbit is the order of the group divided by the size of the stabilizer and this last quantity is given by item 2.). The fourth item is a matter of counting correctly the involved objects: A_{g'}=A_g\iff g'=hgc^m, so that the natural “reflex” is to count g' through h and c^m. However, we should be slightly careful here because the representation g'=hgc^m is not unique. To solve this (easy) problem we observe that two representations g'=hgc^m=h'gc^{m'} verify gc^{m-m'}g^{-1}=h^{-1}h'\in H, so that we have an unique representation g'=h g c^m once the condition 0\leq m <h(g), and hence the fourth item follows. Finally, the fifth item is a direct consequence of the third and fourth items. \square

Following the “general plan” mentioned in the previous post, we will dedicate today’s discussion to the study of the action of Aut(M) on H_1^{(0)}(M,K) in the case K=\mathbb{C}, i.e.,

Standing Assumption. In the sequel, K=\mathbb{C}.

Recall that H_1^{(0)}(M,K)=K^{H\backslash G}\ominus K^{\Sigma}, so that it suffices to understand the action of Aut(M)=N/H on each piece.

For this reason, it is natural to introduce \chi_{\Sigma} the character of the representation of Aut(M) = N/H (or N) on K^{\Sigma}, and \chi_g the character of the representation of Aut(M) on K^{\Sigma_g}.

By the fifth item of the proposition above,

\chi_\Sigma = \frac{1}{\#N}\sum\limits_{g\in G} \chi_g n(g)^{-1}

On the other hand, since \chi_g is the induced representation of the trivial representation of Stab(g)/H, one has

\chi_g(\overline{n}) = \frac{1}{\#(Stab(g)/H)}\sum\limits_{\overline{\nu}\in N/H} 1_{Stab(g)/H} (\overline{\nu}\overline{n}\overline{\nu}^{-1})

for \overline{n}\in N/H. Here, 1_A is the characteristic function of A.

Notation. We denote by \overline{n}\in N/H the class determined by an element n\in N.

By the 2nd item of the proposition above,

\chi_{\Sigma} = \frac{1}{\#N} \sum\limits_{g\in G}\sum\limits_{\overline{\nu}\in N/H} h(g)^{-1} 1_{Stab(g)/H}(\overline{\nu}\,\overline{n}\,\overline{\nu}^{-1})

= \frac{1}{\#N \#H} \sum\limits_{g\in G}\sum\limits_{\nu\in N} h(g)^{-1} 1_{Stab(g)/H}(\nu n \nu^{-1})

Now, we consider \chi_{\alpha} a character of an irreducible representation (over K=\mathbb{C}) of Aut(M)=N/H, and we denote by m_{\alpha} the multiplicity of \chi_{\alpha} in \chi_{\Sigma}.

At this point, we observe that the action of Aut(M) on K^{\Sigma} is “determined” by the knowledge of m_{\alpha}: indeed, morally speaking, the representation theory of finite groups (such as Aut(M)) is “well-known” in the sense that irreducible characters \chi_{\alpha} are “determined”, at least when Aut(M) is a “classical” finite group (such as symmetric groups); therefore, the computation of m_{\alpha} permits (in principle) to calculate \chi_{\Sigma} from character tables of finite groups.

In any event, recall that the multiplicity m_{\alpha} is given by

m_{\alpha}=\frac{1}{\#(N/H)}\sum\limits_{\overline{n}\in N/H} \chi_{\alpha}(\overline{n}) \chi_{\Sigma}(\overline{n}) = \frac{1}{\#N}\sum\limits_{n\in N} \chi_{\alpha}(n) \chi_{\Sigma}(n)

By using the formula of \chi_{\Sigma} above, we can expand the right-hand side to get

m_{\alpha}=\frac{1}{(\#N)^2\#H}\sum\limits_{g\in G} \sum\limits_{n\in N} \sum\limits_{\nu\in N} \chi_{\alpha}(n) h(g)^{-1} 1_{Stab(g)}(\nu n \nu^{-1}) =

\frac{1}{(\#N)^2\#H}\sum\limits_{g\in G} \sum\limits_{n\in N} \sum\limits_{\nu\in N} \sum\limits_{s\in Stab(g)} \chi_{\alpha}(n) h(g)^{-1} \delta_{s,\nu n \nu^{-1}}

where \delta_{i,j} is the usual Kronecker’s delta.

Since \chi_{\alpha} is the character of a N-representation, one has \chi_{\alpha}(n)=\chi_{\alpha}(\nu n\nu^{-1}), and, hence,

\chi_{\alpha}(n) h(g)^{-1}\delta_{s,\nu n \nu^{-1}} = \chi_{\alpha}(\nu n\nu^{-1}) h(g)^{-1} \delta_{s,\nu n \nu^{-1}} = \chi_{\alpha}(s) h(g)^{-1} \delta_{s,\nu n \nu^{-1}}

It follows that

m_{\alpha} = \frac{1}{(\#N)^2\#H}\sum\limits_{g\in G} \sum\limits_{n\in N} \sum\limits_{\nu\in N} \sum\limits_{s\in Stab(g)} \chi_{\alpha}(s) h(g)^{-1} \delta_{s,\nu n \nu^{-1}}=

\frac{1}{(\#N)^2\#H} \sum\limits_{g\in G}\sum\limits_{s\in Stab(g)}\sum\limits_{\nu\in N} \chi_{\alpha}(s) h(g)^{-1} = \frac{1}{\#N\#H} \sum\limits_{g\in G} \sum\limits_{s\in Stab(g)} \chi_{\alpha}(s) h(g)^{-1}

On the other hand, \sum\limits_{s\in Stab(g)} \chi_{\alpha}(s) = \#H \sum\limits_{\overline{s}\in Stab(g)/H} \chi_{\alpha}(\overline{s}). By the second item of the proposition above, Stab(g)/H = \langle c_g\rangle where c_g= gc^{n(g)}g^{-1}, so that

\sum\limits_{s\in Stab(g)} \chi_{\alpha}(s) = \#H \sum\limits_{\overline{s}\in Stab(g)/H} \chi_{\alpha}(\overline{s}) = \#H\sum\limits_{0\leq j<h(g)/n(g)}\chi_{\alpha}(c_g^j)

Now, denoting by d_{\alpha} the degree of \chi_{\alpha}, we note that if \lambda_1,\dots,\lambda_{d_{\alpha}} are the eigenvalues of \chi_{\alpha}(c_g), then \lambda_1^j,\dots,\lambda_{d_{\alpha}}^j are the eigenvalues of \chi_{\alpha}(c_g^j). On the other hand, c_g^{h(g)/n(g)}=1, so that \lambda_i^{h(g)/n(g)}=1 and, hence,

\sum\limits_{0\leq j<h(g)/n(g)}\lambda_i^j = \left\{\begin{array}{cc}h(g)/n(g) & \textrm{ if } \lambda_i=1 \\ 0 & \textrm{ otherwise }\end{array}\right.

In particular

\sum\limits_{s\in Stab(g)} \chi_{\alpha}(s) = \#H\sum\limits_{0\leq j<h(g)/n(g)}\chi_{\alpha}(c_g^j) = \#H \cdot \frac{h(g)}{n(g)} \cdot \textrm{dim}(\textrm{Fix}_{\alpha}(c_g))

where \textrm{Fix}_{\alpha}(c_g) is the subspace of fixed vectors under the action of c_g (a subspace of dimension \#\{1\leq i\leq d_{\alpha}: \lambda_i=1\}, by definition).

By putting these identities together, we obtain the following statement:

Proposition. The multiplicity m_{\alpha} of \chi_{\alpha} in \chi_{\Sigma} is

m_{\alpha} = \frac{1}{\#N}\sum\limits_{g\in G}\frac{1}{n(g)} \textrm{dim}(\textrm{Fix}_{\alpha}(gc^{n(g)}g^{-1}))

Next, we observe that K^{H\backslash G} is the sum of [G:N] copies of the regular representation of N/H. Since the regular representation contains (all) irreducible representations with a multiplicity equal to its degree, we have that the multiplicity of \chi_{\alpha} in K^{H\backslash G} is [G:N]\textrm{dim}\chi_{\alpha}.

In particular, the multiplicity \ell_{\alpha} of \chi_{\alpha} in H_1^{(0)}(M,K)=K^{H\backslash G} \ominus K^{\Sigma} is \ell_{\alpha}=[G:N]\textrm{dim}(\chi_{\alpha})-m_{\alpha}. By the previous proposition, we can write

\ell_{\alpha} = \frac{1}{\#N}\sum\limits_{g\in G}(\textrm{dim}(\chi_{\alpha}) - \frac{1}{n(g)}\textrm{dim}(\textrm{Fix}(gc^{n(g)}g^{-1})))

Observe that the right-hand side of this identity depends only on the class Ng: indeed, g'=ng, n\in N, implies n(g')=n(g) and, a fortiori, g'c^{n(g')}g'^{-1}=ngc^{n(g)}g^{-1}n^{-1}, i.e., the elements c_g and c_{g'} are conjugated by an element n\in N. Therefore, we can rewrite

\ell_{\alpha} = \sum\limits_{N\backslash G}(\textrm{dim}(\chi_{\alpha}) - \frac{1}{n(g)}\textrm{dim}(\textrm{Fix}(gc^{n(g)}g^{-1})))

Now we proceed to rewrite this formula by using induced representations. More precisely, we think of the irreducible representation \chi_{\alpha} of N/H as a representation of \chi_{\alpha} of N on a vector space E_{\alpha}, and we denote by \psi_{\alpha} the induced representation of G on F_{\alpha}=\bigoplus\limits_{i=1}^r g_i^{-1}\cdot E_{\alpha}, where g_1,\dots, g_r are representants of the classes Ng and r=\#G/N. Denote by \rho the permutation of \{1,\dots,r\} determined by the action of c, i.e., Ng_i c:=Ng_{\rho(i)}. In this notation, we have, by definition, \psi_{\alpha}(c)(g_i^{-1}E_{\alpha}) = g_{\rho(i)}^{-1}E_{\alpha}.

Observe that n(g_i) is precisely the length of the cycle C of \rho containing g_i, and the restriction of c^{n(g_i)} to g_i^{-1}E_{\alpha} is conjugated to the action of g_i c^{n(g_i)}g_i^{-1} on E_{\alpha}. Thus, we get the following statement:

Theorem. The multiplicity \ell_{\alpha} is given by the formula

\ell_{\alpha}=\textrm{dim}\psi_{\alpha} - \sum\limits_{C \textrm{ cycle of } \rho} f_\alpha(C)

where f_{\alpha}(C) is the dimension of the subspace of g_i^{-1}E_{\alpha} fixed by c^{n(C)}.

Even though this statement is a simple consequence of our discussion so far, we decided to call it a “theorem” because it has some interesting consequences. In fact, the rest of today’s post will be dedicated to the proof of the following “corollary” to this theorem:

Corollary 1. For any origami M, and for any \alpha, one has \ell_{\alpha}\neq 1.

We begin the discussion of the proof of the previous statement by showing the following result:

Corollary 2. The multiplicity \ell_{\alpha} is always greater than or equal to the multiplicity \ell_0 of the trivial representation.

Proof. By the theorem, \ell_{\alpha} = \sum\limits_{C \textrm{ cycle}}(n(C)\textrm{dim}\chi_{\alpha} - f_{\alpha}(C)). Since f_{\alpha}(C)\leq \textrm{dim}(\chi_{\alpha}) and \textrm{dim}(chi_{\alpha})\geq 1 for every \alpha, we conclude

\ell_{\alpha}\geq \sum\limits_{C \textrm{ cycle}}(n(C)-1)\textrm{dim}\chi_{\alpha}\geq \sum\limits_{C \textrm{ cycle}}(n(C)-1) = \ell_0

so that the argument is complete. \square

Let us consider now the case of regular origamis (in the sense of D. Zmiaikou), i.e., H=\{1\}, G=N=Aut(M). In this context, the formulas for the multiplicities simplify to m_{\alpha}=\textrm{dim}(\textrm{Fix}_{\alpha}(c)) and \ell_{\alpha} = \textrm{codim}(\textrm{Fix}_{\alpha}(c)).

Proposition. In the case of regular origamis:

  • If \textrm{dim}\chi_{\alpha}=1 then \ell_\alpha=0;
  • If \textrm{dim}\chi_{\alpha}>1 then \ell_\alpha>1.

Proof. If \textrm{dim}\chi_{\alpha}=1, we have that \chi_\alpha:G\to\mathbb{C}^* is a homomorphism and hence \chi_\alpha(c)=1 as c is a commutator. Therefore, \ell_{\alpha}=0 in this case.

If \textrm{dim}\chi_{\alpha}>1, we have that \det\pi_\alpha(c)=1, where \pi_\alpha denotes the representation with character \chi_\alpha. We claim that \ell_\alpha>0. Otherwise, \pi_\alpha(c)=1, and hence the action of G factorizes into the action of a commutative group (as c=[g_r,g_u] and g_r,g_u generate G). However, this would force \textrm{dim}\chi_{\alpha}=1 (since commutative groups only have 1-dimensional irreducible representations), a contradiction proving our claim. Now, we observe that the claim and the fact that \det\pi_\alpha(c)=1 imply that \pi_{\alpha} has at least two eigenvalues \neq 1, so that \ell_{\alpha}\geq 2. \square

Partly motivated by this discussion, we introduce the following definition:

Definition. An origami M is quasi-regular if the multiplicity \ell_0 of the trivial representation is zero.

Remark. By the previous proposition, any regular origami is quasi-regular (so that the nomenclature is coherent).

The following proposition allows to characterize quasi-regular origamis in terms of the commutator c:

Proposition 1. M is quasi-regular \iff c\in\bigcap\limits_{g\in G}gNg^{-1} \iff the normal subgroup generated by c is contained in N.

Proof. We know that \ell_0=\sum\limits_{C} (n(C)-1). Thus, \ell_0=0 \iff n(C)=1 for all C cycle \iff c acts trivially on N\backslash G. \square

Proposition 2. If M is not quasi-regular, then \ell_{\alpha}\geq 2 for all \alpha.

Proof. By Corollary 2 above, it suffices to check that \ell_0\geq 2. Since \ell_0=\sum\limits_C (n(C)-1) and M is not quasi-regular (i.e., \ell_0>0), we have that the permutation \rho (associated to the action of c) is not the identity. On the other hand, since c is a commutator, the permutation \rho is even. In particular, \rho is not a transposition, and, consequently, \ell_0\geq 2. \square

Before proceeding further, let us give an example of a quasi-regular but not regular origami.

Example. Let G=\left\{\left(\begin{array}{ccc}1&a&d \\ 0 &1&b \\ 0 & 0 & 1\end{array}\right):a,b,d\in\mathbb{Z}/p\mathbb{Z}\right\} be a finite Heisenberg group. Observe that G is generated by the two elements

g_r:=\left(\begin{array}{ccc}1&1&0\\ 0 & 1& 0 \\ 0 & 0&1\end{array}\right), g_u:=\left(\begin{array}{ccc}1&0&0\\ 0 & 1& 1\\ 0 & 0&1\end{array}\right)

We choose H=\langle g_u\rangle. Note that \bigcap gHg^{-1}=\{1\} because

g_rg_ug_r^{-1} = \left(\begin{array}{ccc}1&0&1\\ 0 & 1& 1 \\ 0 & 0&1\end{array}\right)

Hence, this data defines an origami M. Since H\neq\{1\}, M is not a regular origami. On the other hand, the normalizer N of H is

N= \left\{\left(\begin{array}{ccc}1&a&d \\ 0 &1&b \\ 0 & 0 & 1\end{array}\right): a=0\right\}

is a normal subgroup (coinciding with the centralizer of H). Note also that

c=g_u^{-1}g_r^{-1}g_u g_r=\left(\begin{array}{ccc}1&0&1 \\ 0 &1&0 \\ 0 & 0 & 1\end{array}\right)\in N

By Proposition 1, M is a quasi-regular origami. However, this is not the most interesting example of quasi-regular origami (from the representation theory point of view) because Aut(M)=N/H\simeq\langle c\rangle\simeq\mathbb{Z}/p\mathbb{Z} is an Abelian group. In any case, this quasi-regular example shows two interesting features: N is normal and G/N\simeq \mathbb{Z}/p\mathbb{Z}. As we are going to see below, this is a “general phenomenon” for quasi-regular origamis.

Proposition. M is quasi-regular if and only if N is normal and G/N is Abelian generated by two elements \overline{g_r}, \overline{g_u} (i.e., G/N is either cyclic group or a product of two cyclic groups).

Proof. If N is normal and G/N is Abelian, then the commutator c is mapped into the identity under the natural map G\to G/N, and, by Proposition 1, M is quasi-regular.

Conversely, if M is quasi-regular, then c\in N_0:=\bigcap\limits_{g\in G}gNg^{-1}. Thus, G/N_0 is generated by two elements \overline{g}_r, \overline{g}_u satisfying \overline{g}_u \overline{g}_r = \overline{g}_r \overline{g}_u inside G/N_0, i.e., G/N_0 is an Abelian group (generated by two elements). Since G/N is a subgroup of the Abelian group G/N_0, it follows that N is normal and G/N is Abelian. \square

At this point, we are ready to complete today’s discussion by proving Corollary 1. We start by noticing that, by Proposition 2, it suffices to consider the case of quasi-regular origamis. The idea of the proof in this case is similar to the one in the case of regular origamis (see the proposition before Proposition 1), despite the fact that we need the following little trick.

Let M be a quasi-regular origami and denote by k_r and k_u the orders of \overline{g}_r and \overline{g}_u in G/N. Define n_r:=g_r^{k_r}\in N and n_u:=g_u^{k_u}\in N.

Note that \{g_r^{-i}g_u^{-j}: 0\leq i<k_r, 0\leq j<k_u\} is a complete system of representatives of G/N. Define c(i,j):=g_r^{-i}g_u^{-j}g_r^i g_u^j, for 1\leq i\leq k_r, 1\leq j\leq k_u, so that c=c(1,1).

Observe that

c(1,j)=c(1,1)(g_u^{-1}c(1,1)g_u)\dots(g_u^{1-j}c(1,1)g_u^{j-1})

and

c(i,j)=(g_r^{1-i}c(1,j)g_r^{i-1})\dots(g_r^{-1}c(1,j)g_r)c(1,j)

so that c(k_r,k_u)=n_r^{-1} n_u^{-1} n_r n_u is the product of c_{i,j}:=g_r^{-i} g_u^{-j} c(1,1) g_r g_u, 0\leq i<k_r, 0\leq j<k_u.

In this language, the Theorem implies that

Proposition. If M is a quasi-regular origami, then \ell_\alpha = \sum\limits_{i,j}\textrm{codim}(\textrm{Fix}_\alpha(c_{i,j})).

On the other hand, since c(k_r,k_u) is a commutator of elements of N, we obtain

\prod\limits_{i,j}\det\pi_{\alpha}(c_{i,j})=1

We have the following possibilities:

  • \det\pi_{\alpha}(c_{i,j})=1 for all i,j: here either \pi_{\alpha}(c_{i,j})=1 for all i,j and, a fortiori, \ell_{\alpha}=0, or \pi_{\alpha}(c_{i,j})\neq 1 for some i_0,j_0, so that \textrm{codim}\textrm{Fix}_{\alpha}(c_{i_0,j_0})\geq 2 (by the condition on the determinant);
  • \det\pi_{\alpha}(c_{i,j})\neq 1 for at least two pairs of indices (i_1,j_1), (i_2,j_2): in this situation, \textrm{codim}\textrm{Fix}_{\alpha}(c_{i,j})\geq 1 for (i,j)\in\{(i_1,j_1),(i_2,j_2)\} and, a fortiori, \ell_{\alpha}\geq \textrm{codim}\textrm{Fix}_{\alpha}(c_{i_1,j_1})+\textrm{codim}\textrm{Fix}_{\alpha}(c_{i_2,j_2})\geq 2.

In any event, we find that \ell_{\alpha}\neq 1. This completes the proof of Corollary 1.


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