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 , 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

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

In particular

$\sum\limits_{s\in Stab(g)} \chi_{\alpha}(s) = \#H\sum\limits_{0\leq j

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 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, $0\leq j.

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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