Posted by: matheuscmss | January 27, 2012


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 nth 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


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


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


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


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


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



(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


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)

so that its adjunction verifies

(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


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.



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


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})


\{\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


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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