Posted by: matheuscmss | January 27, 2012

## SPCS 3

Today we will discuss the 3rd lecture (last Wednesday, Jan. 25, 2012) by J.-C. Yoccoz (corresponding to his 2011-2012 course on square-tiled surfaces), but before doing so, we will make a quick review of the material of this previous post.

Let $M$ be an origami associated to a finite group $G$ generated by two elements $g_r$ and $g_u$, and a subgroup $H$ of $G$ containing no nontrivial normal subgroup. Denoting by $N$ the normalizer of $H$ in $G$, we have that the automorphism group $Aut(M)$ is $Aut(M)=N/H$. For the sake of the exposition, we will denote $Aut(M)=\Gamma$.

In this language, we saw that the absolute homology group $H_1(M,\mathbb{C})$ has a natural $\Gamma$-invariant decomposition as

$H_1(M,\mathbb{C}) = H_1^{st}(M,\mathbb{C})\oplus H_1^{(0)}(M,\mathbb{C})$

where $H_1^{st}(M,\mathbb{C})$ has dimension $2$ and $H_1^{(0)}(M,\mathbb{C})$ has codimension $2$ (i.e., dimension $2g-2$).

Finally, we computed an explicit formula for the multiplicity $\ell_\alpha$ in $H_1^{(0)}(M,\mathbb{C})$ of a $\mathbb{C}$-irreducible $\Gamma$-representation of character $\chi_\alpha$, and we used this formula to prove that $\ell_\alpha\neq 1$ (for any origami $M$ and any $\alpha$).

So, this is it as far as the quick revision is concerned. Below the fold the reader will find my notes for J.-C. Yoccoz 3rd lecture: his main goal in it was to show how to use the information (derived in the previous two lectures) about $\mathbb{C}$-representations to deduce useful facts about $\mathbb{R}$-representations. As it turns out, a significant part of this classical and, in particular, the representation theory facts we’re going to use (without proof) can be found in the books of J.-P. Serre and W. Fulton and J. Harris. Nevertheless, this lecture will contain some new facts (from the forthcoming paper by J.-C. Yoccoz, D. Zmiaikou and C.M.) concerning the specific case of representations related to origamis.

Let $\mathbb{Q}\subset K\subset\mathbb{R}$ be a field. We have that

$H_1^{(0)}(M,\mathbb{C}) = H_1^{(0)}(M,K)\otimes\mathbb{C}$

Denote by $Irr_K(\Gamma)$ the set of isomorphism classes of representations defined over $K$ and $K$-irreducible.

A theorem of Brauer (see Theorem 24 in Serre’s book) says that if $K$ contains all $n$th roots of unity where $\gamma^n=1$ for all $\gamma\in\Gamma$, then $Irr_K(\Gamma)\simeq Irr_{\mathbb{C}}(\Gamma)$.

In particular, one has that $Irr_{\overline{K}}(\Gamma) \simeq Irr_{\mathbb{C}}(\Gamma)$ where $\overline{K}$ is the algebraic closure of $K$.

Recall that the Galois group $Gal(\overline{K}/K)$ acts on $Irr_{\overline{K}}(\Gamma)$, and $Irr_K(\Gamma)$ is identified to the orbit space of this action.

Let $a\subset Irr_{\overline{K}}(\Gamma)$ be such an orbit and denote by $\chi_a$ be its character (by thinking of $a$ as a $K$-representation). We can write

$\chi_a = m_a\sum\limits_{\alpha\in a}\chi_{\alpha}$

where $m_a\geq 1$ is an integer known as Schur index.

Let $V_a$ be a $K$-space where $\Gamma$ acts with character $\chi_a$. We denote by $D_a$ the commuting algebra of $\Gamma$ in $End_K(V_a)$ (i.e., the elements of $End_K(V_a)$ commuting with the action of $\Gamma$ on $V_a$). In general $D_a$ is a field (not necessarily commutative) such that

$\textrm{degree}(D_a:K_a)=m_a^2$

where $K_a$ is the center of $D_a$.

Now we consider the decomposition

$H_1^{(0)}(M,K) = \bigoplus\limits_{a\in Irr_K(\Gamma)} W_a$

into isotypical components $W_a$ (i.e., $W_a\simeq V_a^{\ell_a}$ as $K(\Gamma)$-module for some $\ell_a$). Observe that the endomorphisms of $V_a^{\ell_a}$ as $K(\Gamma)$-module have the form

$(v_1,\dots,v_{\ell_a})\mapsto (v_1',\dots,v_{\ell_a}'), \quad\quad v_i'=\sum\limits_{j} d_{ij}v_j, \quad d_{ij}\in D_a$

In the sequel, we’ll treat exclusively the case $K=\mathbb{R}$:

Standing Assumption. From now on, $K=\mathbb{R}$.

In this situation, given $a\in Irr_{\mathbb{R}}(\Gamma)$, we have three possibilities:

• $a$ is real, that is, $a=\{\alpha\}$, $m_a=1$ and $D_a=\mathbb{R}$ (and $\chi_{\alpha}$ takes only real values);
• $a$ is complex, that is, $a=\{\alpha,\overline{\alpha}\}$, $m_a=1$, $D_a\simeq \mathbb{C}$ (and $\chi_{\alpha}$ takes some complex non-real value);
• $a$ is quaternionic, that is, $a=\{\alpha\}$, $m_a=2$, $D_a\simeq\mathbb{H}$ (and $\chi_{\alpha}$ takes only real values).

(Here, $\mathbb{H}$ designs Hamilton’s quaternions)

In the specific case of origamis, recall that the absolute homology group $H_1(M,\mathbb{R})$ comes equipped with a (symplectic) intersection form

$\{.,.\}:H_1(M,\mathbb{R})\times H_1(M,\mathbb{R})\to \mathbb{R}$

Since $\Gamma$ acts on $M$ by automorphisms, we have that $\{.,.\}$ is $\Gamma$-invariant. Furthermore, since $H_1^{st}(M,\mathbb{R})$ and $H_1^{(0)}(M,\mathbb{R})$ are orthogonal (with respect to $\{.,.\}$), the restriction of $\{.,.\}$ to $H_1^{(0)}(M,\mathbb{R})$ is non-degenerate.

Proposition. The subspaces $W_a$ introduced above are mutually orthogonal with respect to $\{.,.\}$. In particular, the restriction of $\{.,.\}$ to each $W_a$ is non-degenerate.

Proof. Since $\{.,.\}$ is $\Gamma$-invariant, it defines an isomorphism (of $\mathbb{R}(\Gamma)$-modules) between $H_1^{(0)}(M,\mathbb{R})$ and its dual. On the other hand, for each $\alpha$, we have that

$\chi_{\alpha^*}(g):=\chi_{\alpha}(g^{-1})=\overline{\chi_{\alpha}(g)}$

where $\alpha^*$ is the dual of $\alpha$. Therefore, this isomorphism (induced by $\{.,.\}$) preserves each isotypical component. $\square$

This proposition says that it makes sense to define

$Sp(W_a):=\{\textrm{automorphism of } W_a \textrm{ as } \mathbb{R}(\Gamma)-\textrm{module}$

$\textrm{ preserving the symplectic form }\{.,.\}_{W_a}:=\{.,.\}|_{W_a}\}$

In the sequel, we will discuss the structure of $Sp(W_a)$ in the three possible cases ($a$ real, complex or quaternionic) in terms of adequate invariant bilinear forms. Below, the treatment of the latter objects in the quaternionic case will follow these notes here of Y. Le Cornulier.

$a$ real-

The $\mathbb{R}(\Gamma)$-module irreducible module $V_a$ of type $a$ can be equipped with a $\Gamma$-invariant scalar product $\langle.,.\rangle$, and, by irreducibility, $\langle.,.\rangle$ is unique up to multiplication by a positive scalar.

Proposition. The multiplicity $\ell_a$ of $V_a$ in $W_a$ is even, i.e., $\ell_a=2\ell_a'$. Also, there exists an isomorphism $\iota: V_a^{\ell_a}\to W_a$ of $\mathbb{R}(\Gamma)$-modules such that

$\{\iota(v),\iota(v')\} = \sum\limits_{m=1}^{\ell_a'}\left(\langle v_m,v_{m+\ell_a'}'\rangle - \langle v_m', v_{m+\ell_a'}\right)$

Consequently, for $A\in Sp(W_a)$, by writing

$\iota^{-1}\circ A\circ \iota(v_1,\dots,v_{\ell_a}) = \left(\sum\limits_{n=1}^{\ell_a} a_{mn}v_n\right)_{1\leq m\leq\ell_a}$

we have that the map $A\mapsto (a_{mn})$ is an isomorphism between $Sp(W_a)$ and the usual symplectic group $Sp(\ell_a,\mathbb{R})$.

Proof. Fix an isomorphism $\iota_0:V_a^{\ell_a}\to W_a$ of $\mathbb{R}(\Gamma)$-modules and define

$\{v,v'\}_{V_a^{\ell_a},\iota_0}:=\{\iota_0(v),\iota_0(v')\}_{W_a}$

Note that $\{.,.\}_{V_a^{\ell_a},\iota_0}$ defines an isomorphism of $\mathbb{R}(\Gamma)$-modules

$V_a^{\ell_a}\to (V_a^*)^{\ell_a}$

On the other hand, by using the $\Gamma$-invariant scalar product $\langle.,.\rangle$, we can identify $(V_a^*)^{\ell_a}$ with $V_a^{\ell_a}$. Hence, it follows that $\{.,.\}_{V_a^{\ell_a},\iota_0}$ induces a linear map $u:V_a^{\ell_a}\to V_a^{\ell_a}$,

$u(v_1,\dots,v_{\ell_a}) = \left(\sum\limits_{n}u_{mn}v_n\right)_m$

such that the matrix $(u_{mn})$ is antisymmetric and invertible. Thus, we have that $\ell_a$ is even, say $\ell_a=2\ell_a'$ and, furthermore, we can choose an isomorphism $\iota$ of the form $\iota=\iota_0\circ u_0$ such that $\{.,.\}_{V_a^{\ell_a},\iota}$ can be written in the standard (symplectic) form in the statement of the proposition. Finally, the last claim of the proposition is a direct verification left as an exercise to the reader. $\square$

$a$ complex ($a=\{\alpha,\overline{\alpha}\}$, $D_a\simeq\mathbb{C}$)-

We fix an isomorphism $D_a=\mathbb{C}$ and we equip $V_a$ with the corresponding structure of $\mathbb{C}$-vector space. In this way, $V_a$ is the underlying space of a $\mathbb{C}$-irreducible representation of $\Gamma$, and we can equip $V_a$ with a $\Gamma$-invariant (positive-definite) Hermitian product $\langle.,.\rangle$ (unique up to multiplication by a positive scalar).

Proposition. There exists an isomorphism $\iota: V_a^{\ell_a}\to W_a$ of $\mathbb{R}(\Gamma)$-modules and a pair of integers $p,q\geq 0$ with $p+q=\ell_a$ such that

$\{\iota(v),\iota(v')\}=\textrm{Im}\left(\sum\limits_{m=1}^{p} \langle v_m, v_m'\rangle - \sum\limits_{m=p+1}^{p+q} \langle v_m,v_m'\rangle\right)$

Consequently, for $A\in Sp(W_a)$, by writing

$\iota^{-1}\circ A\circ \iota (v_1,\dots,v_{\ell_a}) = \left(\sum\limits_{n=1}^{p+q}a_{mn}v_n\right)_{1\leq m\leq\ell_a}$

($a_{mn}\in\mathbb{C}$), we have that the map $A\mapsto (a_{mn})$ is an isomorphism between $Sp(W_a)$ and the unitary group $U_{\mathbb{C}}(p,q)$ of the Hermitian form

$\sum\limits_{m=1}^p z_m\overline{z_m'} - \sum\limits_{m=p+1}^{p+q} z_m\overline{z_m'}$

Proof. Fix $\iota_0:V_a^{\ell_a}\to W_a$ an isomorphism of $\mathbb{R}(\Gamma)$-modules and write

$\{\iota_0(v),\iota(v')\}_{W_a}=\sum\limits_{m,n=1}^{\ell_a} b_{mn}(v_m,v_n')$

It is not hard to check that the bilinear forms $b_{mn}$ on $V_a$ are $\Gamma$-invariant. It follows that $b_{mn}$ are linear combinations of the real and imaginary parts of the (positive-definite) Hermitian form $\langle.,.\rangle$. Thus, the fact that $\langle iv,iv'\rangle=\langle v,v'\rangle$ implies that

$b_{mn}(iv,iv')=b_{mn}(v,v')$

and, a fortiori, $\{\iota_0(iv),\iota_0(iv')\}_{W_a} = \{\iota_0(v),\iota_0(v)\}_{W_a}$. Hence,

$\langle v,v'\rangle_{\iota_0}:=\{\iota_0(iv),\iota_0(iv')\} + i\{\iota_0(v),\iota_0(v')\}$

defines a $\Gamma$-invariant non-degenerate Hermitian form on $V_a^{\ell_a}$. In particular, we can write

$\langle v,v'\rangle_{\iota_0}=\sum\limits_{m,n=1}^{\ell_a} c_{mn}\langle v_m,v_n'\rangle$

where $c_{mn}\in\mathbb{C}$ and $c_{nm}=\overline{c_{mn}}$, and, since $\langle v,v'\rangle_{\iota_0}$ is non-degenerate, we can find an automorphism $u_0:V_a^{\ell_a}\to V_a^{\ell_a}$ of $\mathbb{R}(\Gamma)$-modules and a pair of integers $p,q\geq 0$ with $p+q=\ell_a$ such that $\iota=\iota_0\circ u_0$ verifies

$\langle v,v'\rangle_{\iota}:=\{\iota(iv),\iota(iv')\} + i\{\iota(v),\iota(v')\} = \sum\limits_{m=1}^p \langle v_m,v_m'\rangle - \sum\limits_{m=p+1}^{p+q} \langle v_m,v_m'\rangle$

Finally, the verification of the last claim of the proposition is straightforward (and again we leave as an exercise). $\square$

$a$ quaternionic ($a=\{\alpha\}$, $D_a\simeq\mathbb{H}$)-

We fix an isomorphism $D_a=\mathbb{H}$. Recall that given an element $u=a+bi+cj+dk\in\mathbb{H}$, its complex conjugate is $\overline{u}=a-bi-cj-dk$, and its norm $|u|$ is given by the formula $\overline{u} u := |u|^2$. Also, the complex conjugation satisfies $\overline{u\cdot v}=\overline{v}\cdot\overline{u}$.

Using the isomorphism $D_a=\mathbb{H}$ we can render $V_a$ into a right $\mathbb{H}$-vector space by imposing

$v\cdot z = \overline{z}\cdot v$

for $v\in V_a$ and $z\in\mathbb{H}$.

Definition. A Hermitian form on a right $\mathbb{H}$-vector space $V$ is a map $\langle.,.\rangle:V\times V\to \mathbb{H}$ verifying:

• $\langle v,v_1 z_1 + v_2 z_2\rangle = \langle v,v_1\rangle z_1 + \langle v,v_2\rangle z_2$ (linearity on the 2nd factor);
• $\langle v',v\rangle = \overline{\langle v,v'\rangle}$ (“usual” Hermitian condition).

Observe that the two conditions above imply $\langle v_1 z_1 + v_2 z_2,v'\rangle = \overline{z_1}\langle v_1,v'\rangle + \overline{z_2}\langle v_2,v'\rangle$ (i.e., $\langle.,.\rangle$ is antilinear in the 1st factor).

Example. The standard Hermitian form on $\mathbb{H}^n$ is $\langle z,z'\rangle = \sum\limits_{m=1}^n \overline{z_m} z_m'$.

In general, given an Hermitian form $H$, we can write:

$H =: H_0 + H_i i + H_j j + H_k k$

Observe that $H_0$ is symmetric while $H_i, H_j, H_k$ are antisymmetric. Moreover, they verify the relations

$H_0(v,v')=H_i(v,v'i)=H_j(v,v'j)=H_k(v,v'k)$

allowing to express $H_a, H_b, H_c$ in terms of $H_d$ for any $\{a,b,c,d\}=\{0,i,j,k\}$. For later use, we will focus on antisymmetric forms, e.g., $H_i$ for sake of concreteness, because in the setting of origamis we will be interested in producing Hermitian forms from the symplectic intersection form.That being said, observe that $H_i$ satisfies

$(1)\left\{\begin{array}{c} H_i(vi,vi)=H_i(v,v') \\ H_i(vj,v'j)=-H_i(v,v') \\ H_i(vk,v'k) = -H_k(v,v')\end{array}\right.$

because $H(v\varepsilon, v'\varepsilon) = \overline{\varepsilon} H(v,v')\varepsilon$, and hence $H(v\varepsilon,v'\varepsilon)$ is derived from $H(v,v')$ by conjugation by $\varepsilon$ whenever $\varepsilon^2=-1$ (i.e., $\varepsilon$ is a purely imaginary quaternion with unit norm.

Conversely, given a bilinear antisymmetric $H_i:V\times V\to\mathbb{R}$ verifying (1) above, then

$(2) H(v,v'):= H_i(v,v'i) + H_i(v,v')i + H_i(v,v'k)j - H_i(v,v'j)k$

is a Hermitian form.

Next, we observe that, by irreducibility, $V_a$ has an unique (up to multiplication by a positive real number) $\Gamma$-invariant positive-definite Hermitian form $\langle.,.\rangle$ whose components $\langle.,.\rangle_0$, $\langle .,. \rangle_i$, $\langle.,.\rangle_j$, $\langle.,.\rangle_k$ form a basis of the space of $\Gamma$-invariant $\mathbb{R}$-bilinear forms (and, in particular, the space of symmetric $\Gamma$-invariant $\mathbb{R}$-bilinear forms has dimension 1, while the space of antisymmetric $\Gamma$-invariant $\mathbb{R}$-bilinear forms has dimension 3).

At this point, we are ready to state the proposition whose proof will occupy complete today’s discussion.

Proposition. There exists an isomorphism $\iota: V_a^{\ell_a}\to W_a$ of $\mathbb{R}(\Gamma)$-modules and a pair of integers $p,q\geq 0$ with $p+q=\ell_a$ such that the Hermitian form $H(v,v')$ corresponding to $H_i(v,v'):=\{\iota(v),\iota(v')\}$ by the formula (2) above satisfies

$H(v,v')=\sum\limits_{m=1}^p\langle v_m,v_m'\rangle - \sum\limits_{m=p+1}^{p+q}\langle v_m,v_m'\rangle$

Consequently, for $A\in Sp(W_a)$, by writing

$\iota^{-1}\circ A\circ\iota(v_1,\dots,v_{\ell_a}) = \left(\sum\limits_{n} v_n a_{nm}\right)$,

$a_{nm}\in\mathbb{H}$, we have that $A\mapsto (a_{mn})$ is an isomorphism between $Sp(W_a)$ and the unitary group $U_{\mathbb{H}}(p,q)$ of the Hermitian form

$\sum\limits_{m=1}^p \overline{z_m} z_m - \sum\limits_{m=p+1}^{p+q} \overline{z_m} z_m$

We will show this proposition with the aid of the following three lemmas:

Lemma 1. Let $W$ be a right $\mathbb{R}$-vector space and assume that $W$ is an isotypical $\mathbb{R}(\Gamma)$-module of quaternionic type. Let $B$ be a $\Gamma$-invariant antisymmetric non-degenerate $\mathbb{R}$-bilinear form. Then, there exists $x\in W$ and $\gamma\in\Gamma$ such that

$B(x,\gamma x)\neq 0$

Proof of Lemma 1. Otherwise, $B(x,\gamma x)=B(y,\gamma y)=B(x+y,\gamma(x+y))=0$ for all $x,y\in W, \gamma\in\Gamma$, so that, by bilinearity,

$B(x,\gamma y)+B(y,\gamma x)=B(x+y,\gamma(x+y))=0$

i.e., $B(x,\gamma y) = - B(y,\gamma x)$. Since $B$ is antisymmetric, we deduce that

$B(x,\gamma y) = - B(y,\gamma x) = B(\gamma x,y)$

Since $B$ is $\Gamma$-invariant, it follows that

$B(x,y)=B(x,\gamma^2 y)$

for all $x,y\in W, \gamma\in \Gamma$. Since $B$ is non-degenerate, we get that $\gamma^2 y=y$ for all $y\in W, \gamma\in\Gamma$, that is, the action of $\Gamma$ factors through the action of a subgroup $\overline{\Gamma}$ whose elements have order two, i.e., $\Gamma\to\overline{\Gamma}\to Aut(W)$, and $c^2=1$ for all $c\in\overline{\Gamma}$. Since $abab=aabb=1$ for all $a,b\in\overline{\Gamma}$, we have that $\overline{\Gamma}$ is Abelian and, in particular, it doesn’t have representations of quaternionic type. Of course, since the action of $\Gamma$ on $W$ factors through $\overline{\Gamma}$, we conclude that $W$ can’t be a representation of quaternionic type, a contradiction with our hypothesis. $\square$

Lemma 2. Under the assumptions of Lemma 1, we can write

$W=W_1\oplus\dots\oplus W_{\ell}$

where the $W_i$‘s are $\Gamma$-invariant, irreducible, and mutually orthogonal with respect to $B$.

Proof of Lemma 2. We proceed by induction on $\ell$. For $\ell=1$ there is nothing to prove. For $\ell>1$, we choose $x$ satisfying the conclusion of Lemma 1, and we define $W_1 = \mathbb{R}(\Gamma)\cdot x$. We have that $W_1$ is $\Gamma$-invariant and isotypical, and $B|_{W_1}\neq 0$ is $\Gamma$-invariant. In particular, since $\mathbb{R}(\Gamma)\cdot x$ is irreducible, we also have that $B|_{W_1}$ is non-degenerate (otherwise its annihilator would be a non-trivial $\Gamma$-invariant subspace of $\mathbb{R}(\Gamma)\cdot x$). In particular, we can write

$W=W_1\oplus W_1^{\perp}$

where $W_1^{\perp}$ is $B$-orthogonal to $W_1$, and hence $\Gamma$-invariant of multiplicity $\ell-1$. $\square$

Lemma 3. Let $b$ be a $\Gamma$-invariant antisymmetric $\mathbb{R}$-bilinear form on $V_a$. Then, there exists $u\in\mathbb{H}$, $\overline{u}u=1$ such that

$b_u(v,v'):=b(vu,v'u)$

verifies

$(3) \left\{\begin{array}{c}b_u(vi,v'i)=b_u(v,v') \\ b_u(vj,v'j)=-b_u(v,v') \\ b_u(vk,v'k)=-b_u(v,v') \end{array}\right.$

Proof of Lemma 3. For a $\Gamma$-invariant antisymmetric $\mathbb{R}$-bilinear form $B\neq 0$, we define its adjunction $\sigma_B$ by the formula

$B(v,v'a) = B(v\sigma_B(a),v')$

The adjunction $\sigma_B$ has the following properties:

• $\sigma_B(a)\in\mathbb{H}$ because it commutes with the action of $\Gamma$;
• $\sigma_B^2=1$ because $B$ is antisymmetric;
• $\sigma_B(aa')=\sigma_B(a')\sigma_B(a)$;
• $\sigma_B(a)=a$ for all $a\in\mathbb{R}$;
• if $\overline{u}=-u$, then $\overline{\sigma_B(u)}=-\sigma_B(u)$ because $\overline{u}=-u$ and $\overline{u}u=1$, i.e., $u$ is purely imaginary with unit norm, implies $u^2=-1$ and hence $\sigma_B(u)^2=-1$, so that $\sigma_B(u)$ is purely imaginary with unit norm as well.

In particular, we have that $\sigma_B$ is an anti-involution (i.e., $\sigma_B^2=1$ and $\sigma_B(aa')=\sigma_B(a')\sigma_B(a)$) preserving the space of purely imaginary quaternions. Since an anti-involution can’t act by the identity on the space of purely imaginary quaternions (as $\mathbb{H}$ is not commutative), it follows that there exists a purely imaginary $u\in\mathbb{H}$ such that

$\sigma_B(u)=-u$

Also, recall that $\langle .,.\rangle_i$ denotes the $i$-component of the (unique up to multiplication by a positive real number) $\Gamma$-invariant positive-definite Hermitian product $\langle .,. \rangle$ on $V_a$. We saw that $B_i:=\langle .,.\rangle_i$ satisfies (1) above, that is,

$B_i(vi,v'i)=B_i(v,v'), B_i(vj,v'j)=-B_i(v,v')=B_i(vk,v'k)$

$(4) \sigma_{B_i}(i)=-i, \sigma_{B_i}(j)=j, \sigma_{B_i}(k)=k$

(Also, similar formulas hold for $j$ and $k$ in the place of $i$)

Now we come back to the bilinear form $b$ of the statement of the lemma. As we saw, there exists a purely imaginary $u_0$ with unit norm such that

$\sigma_b(u_0)=-u_0$

On the other hand, the reader can check that the conclusion (3) of the lemma is equivalent to $\sigma_{b_u}=\sigma_{B_i}$.

The adjunction $\sigma_{b_u}$ of $b_u$ (when $|u|^2=u\overline{u}=1$) can be computed as follows.

$b_u(v,v'a)=b(vu,v'au)=b(vu,v'u\overline{u}au)=b(vu\sigma_{b}(\overline{u}au),v'u)=$

$b(vu\sigma_{b}(\overline{u}au)\overline{u}u,v'u)=b_u(vu\sigma_{b}(\overline{u}au)\overline{u},v')$

so that $\sigma_{b_u}(a)=u\sigma_{b}(\overline{u}au)\overline{u}$. In particular, by choosing $u\in\mathbb{H}$, $u\overline{u}=1$ such that $\overline{u} i u=u_0$, we obtain that

$(5) \sigma_{b_u}(i)=-i$.

It follows that $\sigma_{b_u}$ preserves the subspace $\{h\in \mathbb{H}: hi+ih=0\}$ generated by $j$ and $k$. This leaves us with three possibilities:

• the restriction of $\sigma_{b_u}$ to $\mathbb{R}j\oplus\mathbb{R}k$ is a reflection;
• the restriction of $\sigma_{b_u}$ to $\mathbb{R}j\oplus\mathbb{R}k$ is $-\textrm{id}$;
• the restriction of $\sigma_{b_u}$ to $\mathbb{R}j\oplus\mathbb{R}k$ is $\textrm{id}$.

The case of reflection is easily excluded as follows: up to conjugation, we may assume that $\sigma_{b_u}$ has the form

$a+bi+cj+dk\mapsto a-bi+cj-dk$

and this last map is not an anti-involution (as it is the conjugation by $j$ and thus a “true” involution).

The case of $-\textrm{id}$ can also be excluded because it forces $\sigma_{b_u}$ to the the complex conjugation, and this can’t be the adjunction of a non-trivial antisymmetric form (but only of a symmetric one). For instance, given an antisymmetric form $B$, we can write $B=\beta\langle.,.\rangle_i+ \gamma\langle.,.\rangle_j+\delta\langle.,.\rangle_k:=\beta B_i+\gamma B_j +\delta B_k$. If the adjunction $\sigma_B$ of $B$ is the complex conjugation, we would deduce

$-\beta B_i(vi,v')-\gamma B_j(vi,v') - \delta B_k(vi,v') = -B(vi,v') = B(v\sigma_B(i),v') =$

$B(v,v'i) =\beta B_i(v,v'i)+\gamma B_j(v,v'i)+\delta B_k(v,v'i) =$

$\beta B_i(v\sigma_{B_i}(i),v')+\gamma B_j(v\sigma_{B_j}(i),v')+\delta B_k(v\sigma_{B_k}(i),v')=$

$-\beta B_i(vi,v')+\gamma B_j(vi,v')+\delta B_k(vi,v')$

Here, in the last equality, we used our explicit knowledge of the adjunctions $\sigma_{B_i}, \sigma_{B_j}, \sigma_{B_k}$ (see (4) above). Since $B_i, B_j, B_k$ is a basis, one has that $B=\beta B_i$, a contradiction (as $B=\beta B_i$ implies that $\sigma_B=\sigma_{B_i}$ but $\sigma_{B_i}$ is not the complex conjugation as we can see from (4) above).

It follows that $\sigma_{b_u}(j)=j$ and $\sigma_{b_u}(k)=k$. By combining this with (5) above, we conclude that $\sigma_{b_u}=\sigma_{B_i}$, and, as we already observed, this is equivalent to (3), so that the lemma is proved. $\square$

At this stage, we can complete the proof of the proposition (and today’s post). By Lemma 2, we can write

$W_a = W_1\oplus\dots\oplus W_{\ell_a}$

where $W_i\simeq V_a$ are irreducible and mutually orthogonal with respect to the symplectic intersection form $\{.,.\}$. Using this we can build an isomorphism $\iota_0:V_a^{\ell_a}\to W_a$ such that

$\{\iota_0(v),\iota_0(v')\}=\sum\limits_{m=1}^{\ell_a}b^{(m)}(v_m,v_m')$

where $b^{(m)}$ are $\Gamma$-invariant antisymmetric non-zero forms on $V_a$.

By Lemma 3, we can choose $u_1,\dots,u_{\ell_a}$ such that the isomorphism

$\iota_1(v_1,\dots,v_{\ell_a})=(v_1 u_1,\dots,v_{\ell_a} u_{\ell_a})$

satisfies

$\{\iota_1(v),\iota_1(v')\}=\sum\limits_{m=1}^{\ell_a} \hat{b}^{(m)}(v_m,v_m')$

where each $\hat{b}^{(m)}$ verifies (3) above. In particular, we have that $H_i(v,v'):=\{\iota_1(v),\iota_1(v')\}$ verifies (1) above and thus

$H(v,v'):=H_i(v,v'i)+H_i(v,v')i+H_i(v,v'k)j-H_i(v,v'j)k$

is a $\Gamma$-invariant Hermitian form. By uniqueness (up to real scalars), we have that

$H(v,v') = \sum\limits_{m=1}^{\ell_a} c_m \langle v_m,v_m'\rangle$

where $c_m\in\mathbb{R}-\{0\}$.

Finally, by permuting coordinates and performing appropriate scaling, we can replace $\iota_1$ by an isomorphism $\iota$ such that the numbers $c_m\neq 0$ become $\pm1$ and, therefore,

$H(v,v')= \sum\limits_{m=1}^{p} \langle v_m,v_m'\rangle - \sum\limits_{m=p+1}^{p+q} \langle v_m,v_m'\rangle$

for a pair of integers $p,q\geq 0$ with $p+q=\ell_a$. Finally, the verification that the map $Sp(W_a)\ni A\mapsto (a_{mn})\in M_{\ell_a\times\ell_a}(\mathbb{H})$ where

$\iota^{-1}\circ A\circ \iota(v_1,\dots,v_{\ell_a}) = (v_1',\dots,v_{\ell_a}'), v_m'=\sum\limits_{n}v_n a_{nm}$

is an isomorphism between $Sp(W_a)$ and the unitary group $U_{\mathbb{H}}(p,q)$ of the standard Hermitian form $\sum\limits_{m=1}^p \overline{z_m}z_m-\sum\limits_{m=p+1}^{p+q}\overline{z_m}z_m$ is left as an exercise to the reader.

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