Arithmetic representations of mapping class groups

In this subsection $D$ is a skew field of finite dimension over ${\mathbb{Q}}$ endowed with an anti-involution $\dagger$. We assume that the involution $\dagger$ is positive in the sense that $\lambda\in D\mapsto\operatorname{tr}_{D/{\mathbb{Q}}}(\lambda\lambda^{\dagger})$ is a positive definite form. We remark that this is so for the cases that matter here, for the given anti-involution on ${\mathbb{Q}}G$ is evidently positive: for $r\in{\mathbb{Q}}G$, the ${\mathbb{Q}}$-trace of $rr^{\dagger}$ is $|G|$ times the coefficient of $e_{1}$ in $rr^{\dagger}$ and hence is positive definite. The same is then true for its Wedderburn factors $\operatorname{End}^{D_{\chi}}(V_{\chi})$ and their associated skew fields $D_{\chi}$.

We denote the center of $D$ by $L$ (so that is a number field) and by $D_{+}$ resp. $K$ the $\dagger$-invariant part of $D$ resp. $L$. Albert’s classification of such pairs $(D,\dagger)$ (see for example [5], Ch. IV, Thm. 2) then tells us that $K$ is totally real, so that ${\mathbb{R}}\otimes_{\mathbb{Q}}D=\prod_{\sigma}{\mathbb{R}}\otimes_{\sigma}D$, where $\sigma$ runs over the distinct field embeddings $\sigma:K\hookrightarrow{\mathbb{R}}$, and that there are essentially four cases:

1. (I)

   $D=L=K$ so that ${\mathbb{R}}\otimes_{\sigma}D={\mathbb{R}}$ for each $\sigma$,

2. (II)

   $L=K$ and for each $\sigma$ there exists an isomorphism ${\mathbb{R}}\otimes_{\sigma}D\cong\operatorname{End}_{\mathbb{R}}({\mathbb{R}}^{2})$ which sends $\dagger$ to taking the transpose (so $[D:L]=4$),

3. (III)

   $L=K$ and for each $\sigma$ there exists an isomorphism ${\mathbb{R}}\otimes_{\sigma}D\cong{\mathbb{K}}$, where ${\mathbb{K}}$ denotes the Hamilton quaternions, which sends $\dagger$ to quaternion conjugation (so $[D:L]=4$),

4. (IV)

   $L$ is a purely imaginary extension of $K$ (in other words, $L$ is a CM field) and for each $\sigma$ there exists an isomorphism ${\mathbb{R}}\otimes_{\sigma}D\cong\operatorname{End}_{\mathbb{C}}({\mathbb{C}}^{d})$, which takes ${\mathbb{R}}\otimes_{\sigma}L$ to ${\mathbb{C}}$ (so $[D:L]=d^{2}$) and sends $\dagger$ to taking the conjugate transpose.

Let $M$ be a left $D$-module of finite rank. We write $M^{\dagger}$ for $M$ endowed with the structure of a right $D$-module via the rule $a\lambda:=\lambda^{\dagger}a$ ($a\in M$, $\lambda\in D$). So if $M^{\prime}$ is another left $D$-module, then $M^{\dagger}\otimes_{D}M^{\prime}$ is defined. It is a $K$-vector space with the property that $a\otimes_{D}\lambda a^{\prime}=(\lambda^{\dagger}a)\otimes_{D}a^{\prime}$ for all $\lambda\in D$, $a\in M$ and $a^{\prime}\in M^{\prime}$. In particular, we have in $M^{\dagger}\otimes_{D}M$ a $K$-linear involution defined by $(a\otimes_{D}b)^{\prime}=b\otimes_{D}a$. We denote its fixed point set $u(M)\subset M^{\dagger}\otimes_{D}M$. As a ${\mathbb{Q}}$-subspace of $M^{\dagger}\otimes_{D}M$ it is spanned by the symmetric tensors $a\otimes_{D}a$.

Suppose given a skew-hermitian form $(a,b)\in M\times M\mapsto\langle a,b\rangle\in D$ on $M$. We denote its radical by $M_{o}$ so that the form descends to a nondegenerate one on $\overline{M}:=M/M_{o}$. We define the associated unitary group $\operatorname{U}(M)$ as the group of $D$-linear automorphisms of $M$ that preserve the form and act as the identity on $M_{o}$. It is the group of $K$-points of an algebraic group defined over $K$. If the form is nondegenerate ($M_{o}=0$), then $\operatorname{U}(M)$ is what is called in [2] (§5.2B) a classical unitary group.

If $c\in M$ is isotropic (meaning that $\langle c,c\rangle=0$), then we have the associated isotropic transvection $T_{c}=T(c\otimes_{D}c)\in\operatorname{U}(M)$ defined by $x\in M\mapsto x+\langle x,c\rangle c$ (so $c\otimes_{D}c$ is here understood as an element of $M^{\dagger}\otimes_{D}M$). It ‘generates’ an abelian unipotent subgroup of $\operatorname{U}(M)$ defined by

Isotropic transvections are particular cases of Eichler transformations. These are defined as follows. Let $c\in M$ be isotropic, $a\in M$ perpendicular to $c$ and $\lambda\in D$ such that $\lambda-\lambda^{\dagger}=\langle a,a\rangle$ (equivalently, $\lambda-\frac{1}{2}\langle a,a\rangle\in D_{+}$). Then the associated Eichler transformation is

It is a $D$-linear transformation which preserves the form. When $\lambda=\frac{1}{2}\langle a,a\rangle$, we shall write $E(c,a)$ instead. Since $T(c\otimes_{D}c)=E(\frac{1}{2}c,c)$, isotropic transvections are Eichler transformations as asserted.

One checks that each Eichler transformation lies in $\operatorname{U}(M)$ as defined above. In fact, $t\in K\mapsto E(tc,a,\lambda)=E(c,ta,t^{2}\lambda)$ is a closed one-parameter subgroup of $\operatorname{U}(M)$ whose infinitesimal generator is represented by $a\otimes_{D}c+c\otimes_{D}a\in u(M)$ (or rather by its image in $u(M)/u(M_{o})$, for if both $a$ and $c$ lie in $M_{o}$, then we get the identity). By a general property of algebraic groups ([8], Cor. 2.2.7) such subgroups then generate a closed algebraic $K$-subgroup of $\operatorname{U}(M)$. Following [2] we denote that group by $\operatorname{EU}(M)$.

We note the commutator identity

It follows that if we fix $c$, but let $a$ and $\lambda$ vary (subject to the conditions above, so with $a\in c^{\perp}$), then the $E(c,a,\lambda)$ generate a unipotent group that appears as an extension of the vector group $c^{\perp}/(M_{o}+Dc)$ by the abelian subgroup of $\operatorname{U}(M)$ defined by the $T(c\otimes_{D}\lambda c)$.

The group $\operatorname{EU}(M)$ is already generated by the isotropic transvections: When $\dagger$ is nontrivial this follows from (6.3.1) of [2]. The remaining case is the one we labelled (I): this is when $D=L=K$ and $\operatorname{U}(M)$ is a symplectic group over $K$, but then there is no issue for then every $a\in M$ is isotropic, and then $E(c,a)=T_{a+c}T_{a}^{-1}T_{c}^{-1}$.

If the form is nondegenerate (ie, $M_{o}=\{0\}$), then $\operatorname{EU}(M)$ is a $K$-form of a classical semisimple algebraic group and hence has finite center. To be precise, it is a group of symplectic type in the cases (I) and (II), of orthogonal type in case (III) and of special linear type in case (IV).

If $M_{o}$ is possibly nonzero, then per convention the elements of $\operatorname{U}(M)$ act trivially on $M_{o}$. The natural map $\operatorname{U}(M)\to\operatorname{U}(\overline{M})$ is evidently onto. Its kernel consists of the transformations that act trivially on both $M_{o}$ and $M/M_{o}$ and is therefore the unipotent radical $\operatorname{R_{u}}(\operatorname{U}(M))$ of $\operatorname{U}(M)$ (recall that $\operatorname{U}(\overline{M})$ is reductive): we have an exact sequence

The elements of $\operatorname{R_{u}}(\operatorname{U}(M))$ are the Eichler transformations $E(c,a)$ with $c\in M_{o}$ and $a\in M$ arbitrary. In this case $E(c,a)$ only depends on the image of $c\otimes_{\mathbb{Z}}a$ in $M_{o}^{\dagger}\otimes_{D}\overline{M})$, so that the resulting map

is an isomorphism. So $\operatorname{R_{u}}(\operatorname{U}(M))$ is a vector group over $K$ (ie, a $K$-vector space regarded as the group of $K$-points of a $K$-algebraic group of additive type). Since $\operatorname{EU}(\overline{M})$ is a normal semisimple subgroup of $\operatorname{U}(M)$, it has the same unipotent radical: $\operatorname{R_{u}}(\operatorname{EU}(M))=\operatorname{R_{u}}(\operatorname{U}(M))$.

Since in what follows the notion of real rank of an algebraic group shows up, let us begin with reviewing this concept briefly.

Let $\mathscr{G}$ be a reductive algebraic group. Suppose first that $\mathscr{G}$ is defined over ${\mathbb{R}}$. Then the real rank $\operatorname{rk}_{\mathbb{R}}(\mathscr{G})$ of $\mathscr{G}$ is by definition the dimension of a Cartan subgroup of $\mathscr{G}$ defined over ${\mathbb{R}}$. For example, if $\mathscr{G}$ is the orthogonal group of a nondegenerate quadratic form over ${\mathbb{R}}$, then its real rank is the Witt index of this form: the dimension of a maximal isotropic subspace defined over ${\mathbb{R}}$. If $\mathscr{G}$ is defined over a number field $k$, then we restrict scalars à la Weil so that $\operatorname{Res}_{k|{\mathbb{Q}}}\mathscr{G}$ is a group defined over ${\mathbb{Q}}$. We then regard $\operatorname{Res}_{k|{\mathbb{Q}}}\mathscr{G}$ (by base change) as a group over ${\mathbb{R}}$ and define the real rank of $\mathscr{G}$ to be the real rank of the latter. Concretely, if $\sigma_{1},\dots,\sigma_{r}$ are the real embeddings of $k$ in ${\mathbb{R}}$ and $\tau_{1},\overline{\tau}_{1},\dots,\tau_{s},\overline{\tau}_{s}$ are the remaining distinct (complex) embeddings (they come in complex conjugate pairs), then the definition comes down to

(this is also the sum over all the archimedean valuations of $k$, taking as general term the real rank of the corresponding completion of $\mathscr{G}(k)$). The Dirichlet unit theorem often gives lower bounds for the rank. For example, if the skew field $D$ is as in the Albert classification, the group of units $D^{\times}$ is a reductive group defined over $K$ and its group of real points and its real rank are then as follows: putting $e:=[K:{\mathbb{Q}}]$, then

1. (I)

   $\operatorname{Res}_{K|{\mathbb{Q}}}D^{\times}({\mathbb{R}})$ is open in $({\mathbb{R}}^{\times})^{e}$; the real rank of $D^{\times}$ is $e$,

2. (II)

   $\operatorname{Res}_{K|{\mathbb{Q}}}D^{\times}({\mathbb{R}})$ is open in $\operatorname{GL}_{2}({\mathbb{R}})^{e}$; the real rank of $D^{\times}$ is $2e$,

3. (III)

   $\operatorname{Res}_{K|{\mathbb{Q}}}D^{\times}({\mathbb{R}})$ is open in $({\mathbb{K}}^{\times})^{e}$; the real rank of $D^{\times}$ is $e$,

4. (IV)

   $\operatorname{Res}_{K|{\mathbb{Q}}}D^{\times}({\mathbb{R}})$ is open in $\operatorname{GL}_{d}({\mathbb{C}})^{e}$; the real rank of $D^{\times}$ is $de$.

So $D^{\times}$ has real rank $\geq 2$, unless $D$ equals ${\mathbb{Q}}$ (I) or is a definite quaternion algebra with center ${\mathbb{Q}}$ (III) or is an imaginary quadratic extension of ${\mathbb{Q}}$.

It is clear that $\operatorname{EU}(M)$ is a normal subgroup of $\operatorname{U}(M)$. We already noticed that it is closed in $\operatorname{U}(M)$ and hence the quotient $\operatorname{U}(M)/\operatorname{EU}(M)$ is also an algebraic group. Note however that if $\overline{M}$ has no nonzero isotropic vectors, then $\operatorname{EU}(\overline{M})$ is trivial. We mention for future reference a consequence of a theorem of G.E. Wall [10]:

Wall’s result is more specific and tells us that $\operatorname{U}(M)/\operatorname{EU}(M)$ is often an anisotropic torus. But this need not be so when $\operatorname{dim}_{D}M=2$.

The excepted case is genuine, for in that case $M\cong M_{1}\otimes_{K}M_{2}$ as modules endowed with $K$-forms, where $M_{i}$ is a 2-dimensional $K$-vector space endowed with a nondegenerate symplectic form. The resulting map $\operatorname{SL}(M_{1})\times\operatorname{SL}(M_{2})\to\operatorname{GL}(M)$ has image $\operatorname{EU}(M)\cong\operatorname{O}(M)$ and its kernel has $(-1,-1)$ as its unique nonidentity element.

Our discussion of symplectic ${\mathbb{Q}}G$-modules also applies to orthogonal ${\mathbb{Q}}G$-modules. One such module is ${\mathbb{Q}}G$ itself (regarded as a left module): it comes indeed with $G$-invariant pairing ${\mathbb{Q}}G\times{\mathbb{Q}}G\to{\mathbb{Q}}$, the trace form, which assigns to the pair $(r_{1},r_{2})$ the trace of $r_{1}r_{2}^{\dagger}$ considered as an endomorphism of ${\mathbb{Q}}G$ as a ${\mathbb{Q}}$-vector space (this is simply $|G|$ times the coefficient of $e_{1}$). This pairing is symmetric and nondegenerate.

The Wedderburn decomposition ${\mathbb{Q}}G\cong\prod_{\chi}\operatorname{End}^{D_{\chi}}(V_{\chi})$ is also the isotypical decomposition, for the $K_{\chi}$-linear map

which assigns to $v_{1}\otimes_{D_{\chi}}f$ the endomorphism $v\in V_{\chi}\mapsto v_{1}f(v)$ is well-defined and is an isomorphism of left ${\mathbb{Q}}G$-modules (and also as right ${\mathbb{Q}}G$-modules). This also shows that ${\mathbb{Q}}G[\chi]\cong\operatorname{Hom}^{D_{\chi}}(V_{\chi},D_{\chi})$ as a right $D_{\chi}$-module.

We claim that $\operatorname{Hom}^{D_{\chi}}(V_{\chi},D_{\chi})\cong V_{\chi}^{\dagger}$ as left $D_{\chi}$-modules. This is based on a hermitian extension of $s_{\chi}$ to $V_{\chi}^{\dagger}$: if we follow the same recipe as in the introduction for a symplectic representation, then we find that there is a hermitian form $h_{\chi}:V_{\chi}^{\dagger}\times V_{\chi}^{\dagger}\to D_{\chi}$ characterized by the property that for all $v,v^{\prime}\in V_{\chi}$ and $f,f^{\prime}\in V_{\chi}^{\dagger}$,

This formula implies that $h_{\chi}$ is $G$-invariant (we let $G$ act on the right of $V_{\chi}^{\dagger}$). By taking $v=f=f^{\prime}$, we also see that $h_{\chi}(f,f)=s_{\chi}(f,f)$, so that $h_{\chi}$ is a hermitian extension of $s_{\chi}$. For every $v\in V_{\chi}^{\dagger}$, the expression $h_{\chi}(v,\;-)$ yields an element of $\operatorname{Hom}^{D_{\chi}}(V_{\chi},D_{\chi})$ and defines the stated isomorphism.

Here and in the rest of this paper, we write $R$ for the integral group ring ${\mathbb{Z}}G$. Let $M$ be a finitely generated (left) $R$-module, free over ${\mathbb{Z}}$ and let $(a,b)\in M\times M\mapsto a\cdot b\in{\mathbb{Z}}$ be a nondegenerate (but not necessarily unimodular) $G$-equivariant symplectic form. We extend this in the standard manner to a form

This form is skew-hermitian: it is $R$-linear in the first variable and $\langle a,b\rangle=-\langle b,a\rangle^{\dagger}$. A ${\mathbb{Z}}$-linear automorphism of $M$ is $G$-equivariant and preserves the symplectic form if and only if it is an $R$-module automorphism which preserves this skew-hermitian form. We denote the group of such automorphisms by $\operatorname{U}(M)$.

Let $R_{+}$ stand for the fixed point set of $\dagger$ in $R$; this is an additive subgroup of $R$. If $a\in M$ is $R$-isotropic in the sense that $\langle a,a\rangle=0$, then for every $r\in R_{+}$ the isotropic transvection

lies in $U(M)$ and $r\in R_{+}\mapsto T_{a}(r)$ is a homomorphism from (the additively written) $R_{+}$ to (the multiplicatively written) $U(M)$. Since $T_{a}(r)$ only depends on $a\otimes ra\in M^{\dagger}\otimes_{R}M$, we also denote this transformation by $T(a\otimes ra)$.

Let $A$ be a (not necessarily commutative) unital ring with unit endowed with an anti-involution $\dagger$. The basic hyperbolic $A$-module $\mathcal{H}^{2}(A)$ is the free left $A$-module of rank $2$ (whose generators we denote $\mathbf{e}$ and $\mathbf{f}$) endowed with the skew-hermitian form defined by $\langle\mathbf{e},\mathbf{e}\rangle=\langle\mathbf{f},\mathbf{f}\rangle=0$ and $\langle\mathbf{e},\mathbf{f}\rangle=1$. It can be regarded as the $A$-form of the standard symplectic module ${\mathbb{Z}}^{2}$. In vector notation:

It is unimodular in the sense that $a\in\mathcal{H}^{2}(A)\mapsto\langle-,a\rangle\in\operatorname{Hom}_{A}(\mathcal{H}^{2}(A),A)$ is an antilinear isomorphism. We will write $\operatorname{U}_{2}(A)$ resp.  $\operatorname{EU}_{2}(A)$ for $\operatorname{U}(\mathcal{H}^{2}(A))$ resp. $\operatorname{EU}(\mathcal{H}^{2}(A))$. The latter contains $\operatorname{SL}_{2}({\mathbb{Z}})$ in an obvious manner. Let $A_{+}\subset A$ be the set of $\dagger$-invariant elements. One verifies that $T_{\mathbf{e}}:x\in\mathcal{H}^{2}(A)\mapsto x+\langle x,\mathbf{e}\rangle\mathbf{e}$ and the similarly defined $T_{\mathbf{f}}$ have the matrix form

where $\rho_{a}$ stands for right multiplication with $a$.

We take $A=R(={\mathbb{Z}}G)$. The elements of the form $r+r^{\dagger}$ with $r\in R$, make up a subgroup $R_{++}$ of $R_{+}$ such that $R_{+}/R_{++}$ is a finite dimensional ${\mathbb{F}}_{2}$-vector space. Let $\Gamma_{+}(G)$ resp.  $\Gamma_{-}(G)$ denote the subgroup of $\operatorname{EU}_{2}(R)$ generated by $T_{\mathbf{e}}(R_{++})$ resp. $T_{\mathbf{f}}(R_{++})$ and let $\Gamma_{o}(G)\subset\operatorname{EU}_{2}(R)$ stand for the subgroup generated by $\Gamma_{+}(G)$ and $\operatorname{SL}_{2}({\mathbb{Z}})$. Since $\Gamma_{-}(G)$ is a $\operatorname{SL}_{2}({\mathbb{Z}})$-conjugate of $\Gamma_{+}(G)$, the group $\Gamma_{o}(G)$ contains $T_{\mathbf{f}}(R_{++})$. The right (inverse) action of $G$ on $\mathcal{H}^{2}(R)$ defines an embedding of $G$ in $\operatorname{U}_{2}(R)$. One checks that

and so $G$ normalizes $\Gamma_{o}(G)$. We put $\Gamma(G):=\Gamma_{o}(G).G$.

The notion of a basic hyperbolic module generalizes in a straightforward manner to $\mathcal{H}^{2}(V_{\chi}^{\dagger})$, the skew-hermitian form being given by

So we have defined $\operatorname{U}_{2}(V_{\chi}^{\dagger})$; it is the group of $K_{\chi}$-points of a reductive algebraic group defined over $K_{\chi}$. If we write an element of $\operatorname{U}_{2}(V_{\chi}^{\dagger})$ in block form $(\begin{smallmatrix}A&B\\ C&D\\ \end{smallmatrix})$, with $A,B,C,D$ in $\operatorname{End}_{D_{\chi}}(V_{\chi}^{\dagger})$, then the subgroup defined by $C=0$ is a parabolic subgroup. Its unipotent radical is given by requiring that in addition $A$ and $D$ are the identity. The corresponding subgroup is then the vector group $(\begin{smallmatrix}1&B\\ 0&1\\ \end{smallmatrix})$ for which $B$ is hermitian relative to $h_{\chi}$. In other words, $h_{\chi}$ identifies $B$ with an element of $u(V_{\chi}^{\dagger})$ (that is, a symmetric element of $V_{\chi}\otimes_{D_{\chi}}V_{\chi}^{\dagger}$) and hence defines an isotropic transvection. In particular, this is also the unipotent radical of the corresponding subgroup of $\operatorname{EU}_{2}(V_{\chi}^{\dagger})$. An opposite parabolic subgroup is defined by $B=0$ and has a similar description of its unipotent radical. Let us denote these unipotent radicals ${\mathscr{R}}_{+}(\operatorname{U}_{2}(V_{\chi}^{\dagger}))$ resp.  ${\mathscr{R}}_{-}(\operatorname{U}_{2}(V_{\chi}^{\dagger}))$.

We run into this when we consider the isotypical decomposition of $\mathcal{H}^{2}({\mathbb{Q}}G)$. The isogeny space $\mathcal{H}^{2}({\mathbb{Q}}G)[\chi]=\operatorname{Hom}_{{\mathbb{Q}}G}(V_{\chi},\mathcal{H}^{2}({\mathbb{Q}}G))$ is then a left $D_{\chi}$-module that is naturally identified with $\mathcal{H}^{2}(V_{\chi}^{\dagger})$. This gives rise to a decomposition

The image of $\Gamma_{\pm}(G)$ in $\operatorname{U}_{2}(V_{\chi}^{\dagger})$ clearly lands in ${\mathscr{R}}_{\pm}(\operatorname{U}_{2}(V_{\chi}^{\dagger}))$. Since $\operatorname{End}_{D_{\chi}}(V_{\chi}^{\dagger})=\operatorname{End}^{D_{\chi}}(V_{\chi})$ is a Wedderburn factor of ${\mathbb{Q}}G$, it follows that the image of $R$ in $\operatorname{End}_{D_{\chi}}(V_{\chi}^{\dagger})$ is an order (a lattice that is also a unital algebra). This is compatible with the anti-involutions and hence the image of $R_{++}$ in $\operatorname{End}_{D_{\chi}}(V_{\chi}^{\dagger})$ is a lattice in the subspace of hermitian matrices. So the image of $\Gamma_{\pm}(G)$ in $\operatorname{U}_{2}(V_{\chi}^{\dagger})$ is a lattice in ${\mathscr{R}}_{\pm}(\operatorname{U}_{2}(V_{\chi}^{\dagger}))$.

It is clear that $\Gamma_{o}(G)$ maps to $\operatorname{EU}_{2}(V_{\chi}^{\dagger})$.