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Let $S:=P\Omega_{2n}^+(q)$ with $n$ even and $q$ odd prime power be the simple orthogonal group. Then the Schur multiplier of $S$ is the Klein four-group $Z_2\times Z_2$. Therefore $S$ has three double covers.

Is there any relation between these three double covers? Are they isomorphic? When $n=4$ I know that the three involutions in $Z_2\times Z_2$ are permuted by an outer automorphism of $S$ of degree $3$ so that these three double covers are isomorphic.

Thank you in advance!

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    $\begingroup$ The Schur Multiplier is isomorphic to $Z_2 \times Z_2$ when $n$ is even. When $n$ is odd, it is isomorphic to $Z_d$ with $d=(q^n-1,4)$. $\endgroup$ – Derek Holt Aug 27 '13 at 15:08
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Sorry for editing this answer multiple times. However, as I managed to get the answer wrong I feel obliged to improve this answer and provide a few more details. I've broken this up into several parts, so you can read as much as you care about.


My original (incorrect) answer:

The three double covers of the group you consider can be described as follows (I'll assume $n>4$ as you've already described the $n=4$ case). One is the special orthogonal group $\operatorname{SO}_{2n}^+(q)$ and the remaining two are the Half-Spin groups $\operatorname{HSpin}_{2n}(q)$. The half-spin and special orthogonal groups are not isomorphic but the two half-spin groups are isomorphic. Why this is the case is described in section 5 of this paper

http://arxiv.org/abs/1211.2551

It essentially boils down to the fact that the Dynkin diagram of type $\mathrm{D}_n$ has an order 2 automorphism permuting the two nodes attached to the branch point.


Why the first answer is wrong:

The mistake comes from thinking on the level of algebraic groups. Let $\mathbf{G}$ be a simple simply connected algebraic group of type $\operatorname{D}_n$ defined over an algebraic closure $\mathbb{K} = \overline{\mathbb{F}_p}$ of the finite field of odd prime order $p$, then $\mathbf{G}$ is the spin group $\operatorname{Spin}_{2n}(\mathbb{K})$. Under the restrictions applied in the original post (namely that $p$ is odd) one can obtain the remaining simple algebraic groups of type $\operatorname{D}_n$ as $\mathbf{G}/\mathbf{K}$ where $\mathbf{K} \leqslant Z(\mathbf{G})$ is a subgroup of the centre.

In particular, taking $\mathbf{K} = Z(\mathbf{G})$ we obtain the adjoint group $\mathbf{G}_{\mathrm{ad}}$ of type $\operatorname{D}_n$. Taking $\mathbf{K}$ to be one of subgroups of index 2 we obtain three groups, which are double covers of $\mathbf{G}_{\mathrm{ad}}$. One of these groups is the special orthogonal group $\operatorname{SO}_{2n}(\mathbb{K})$ and the remaining two are isomorphic and called the half-spin groups $\operatorname{HSpin}_{2n}(\mathbb{K})$.

We now let $F$ be a split Frobenius endomorphism of $\mathbf{G}$ defining an $\mathbb{F}_q$-rational structure $G = \mathbf{G}^F$, which is isomorphic to the finite spin group $\operatorname{Spin}_{2n}^+(q)$. The Frobenius endomorphism $F$ acts trivially on $Z(\mathbf{G})$ so stabilises every subgroup of $Z(\mathbf{G})$. In particular $F$ induces a Frobenius endomorphism on the quotients $\mathbf{G/K}$, where $\mathbf{K}$ is an index 2 subgroup of $Z(\mathbf{G})$, which we again denote by $F$. The important thing to realise here is that

$$(\mathbf{G/K})^F \not\cong \mathbf{G}^F/\mathbf{K}^F.$$ This happens because $\mathbf{K}$ is not connected. In particular, the situation on the level of the algebraic groups does not directly descend to the corresponding finite groups.


The correct answer:

The finite simple group $S$ has three double covers. If $n=4$ then all three double covers are isomorphic. If $n>4$ then two of these double covers are isomorphic but they are not isomorphic to the third.

Let $\mathbf{G}$ be the spin group (as above). Let us fix a maximal torus $\mathbf{T}_0$ of $\mathbf{G}$ and a Borel subgroup $\mathbf{B}_0$ containing $\mathbf{T}_0$. We denote by $(\Phi,X,\Phi^{\vee},X^{\vee})$ the root datum of $\mathbf{G}$ with respect to $\mathbf{T}_0$. In particular, $X = \operatorname{Hom}(\mathbf{T}_0,\mathbb{K}^{\times})$ and $X^{\vee} = \operatorname{Hom}(\mathbb{K}^{\times},\mathbf{T}_0)$ are the character and cocharacter groups respectively and $\Phi \subset X$ and $\Phi^{\vee} \subset X^{\vee}$ are the roots and coroots of $\mathbf{G}$.

The group $X^{\vee}$ can be viewed as a $\mathbb{Z}$-module, hence we can form the tensor product $V = \mathbb{R} \otimes_{\mathbb{Z}} X^{\vee}$ which is a real vector space. Recall that we have a perfect pairing $\langle -,-\rangle : X \times X^{\vee} \to \mathbb{Z}$, which can naturally be extended to the tensor products $\mathbb{R}\otimes_{\mathbb{Z}}X$ and $\mathbb{R}\otimes_{\mathbb{Z}}X^{\vee}$, then we define the coweight lattice to be

$$\Lambda^{\vee} = \{\gamma \in V \mid \langle \alpha,\gamma\rangle \in \mathbb{Z}\text{ for all }\alpha\in\Phi\}.$$ Clearly $\mathbb{Z}\Phi^{\vee} \subset X^{\vee} \subset \Lambda^{\vee}$ (here $\mathbb{Z}\Phi^{\vee}$ denotes the $\mathbb{Z}$-span of $\Phi^{\vee}$ in $V$) but less clear is the fact that $\Lambda^{\vee}/\mathbb{Z}\Phi^{\vee}$ is isomorphic to the Klein four group. Note that, as $\mathbf{G}$ is simply connected we have $X^{\vee} = \mathbb{Z}\Phi^{\vee}$. The simple roots $\Delta$ also define a set of fundamental dominant coweights $\Omega^{\vee} = \{\omega^{\vee}_{\alpha} \mid \alpha\in \Delta\}$ by the condition that $\langle \beta,\omega^{\vee}_{\alpha} \rangle = \delta_{\alpha,\beta}$ (the Kronecker delta).

Recall that for each root $\alpha \in \Phi$ there is a minimal 1-dimensional unipotent subgroup $\mathbf{X}_{\alpha} \leqslant \mathbf{G}$ normalised by $\mathbf{T}_0$ called the root subgroup of $\alpha$. Let $\Phi^+$ be all $\alpha \in \Phi$ such that $\mathbf{X}_{\alpha} \leqslant \mathbf{B}_0$ then $\Phi^+$ forms a positive system of roots for $\Phi$, hence determines a unique system of simple roots $\Delta \subset \Phi^+$.

Each root subgroup $\mathbf{X}_{\alpha}$ is isomorphic to the additive group $\mathbb{K}^+$. We now choose an isomorphism $x_{\alpha} : \mathbb{K}^{\times} \to \mathbf{X}_{\alpha}$ for each $\alpha \in \Phi$ such that $tx_{\alpha}(c)t^{-1} = x_{\alpha}(\alpha(t)c)$ for all $t \in \mathbf{T}_0$ and $c \in \mathbb{K}^+$, then the set $\{x_{\alpha}(c) \mid \alpha \in \Delta, c \in \mathbb{K}^+\}$ forms a generating set for $\mathbf{G}$. We may now define the split Frobenius endomorphism $F$ on $\mathbf{G}$ by the condition that

$$F(x_{\alpha}(c)) = x_{\alpha}(c^q)$$ for all $\alpha \in \Delta$ and $c \in \mathbb{K}^+$. Now let $b : \Delta \to \Delta$ be a bijection induced by a graph automorphism of the root system $\Phi$. We may now define an automorphism $\tau : \mathbf{G} \to \mathbf{G}$ by setting

$$\tau(x_{\alpha}(c)) = x_{b(\alpha)}(c)$$ for all $\alpha \in \Delta$ and $c \in \mathbb{K}^{\times}$. It's important to note that $F$ and $\tau$ do not necessarily have this effect on $x_{\alpha}(c)$ for any $\alpha \in \Phi$.

Clearly $F$ and $\tau$ commute so $\tau$ restricts to an automorphism of the finite group $G = \mathbf{G}^F$. We wish to understand the action of $\tau$ on the centre of $\mathbf{G}$ and $G$. To do this we will need some more notation. Let us fix an isomorphism $(\mathbb{Q}/\mathbb{Z})_{p'} \to \mathbb{K}^{\times}$ then we obtain a surjective homomorphism $\iota : \mathbb{Q} \to \mathbb{K}^{\times}$ as the composition $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to (\mathbb{Q}/\mathbb{Z})_{p'} \to \mathbb{K}^{\times}$. We may now define a surjective homomorphism $\iota_{\mathbf{T}_0} : \mathbb{Q} \otimes_{\mathbb{Z}} X^{\vee} \to \mathbf{T}_0$ by setting

$$\iota_{\mathbf{T}_0}(r\otimes\gamma) = \gamma(\iota(r)).$$ Noting that $\Lambda^{\vee} \subseteq \mathbb{Q}\otimes_{\mathbb{Z}}X^{\vee}$ we have the following lemma.

Lemma: The restriction of the homomorphism $\iota_{\mathbf{T}_0}$ to $\Lambda^{\vee}$ induces an isomorphism $(\Lambda^{\vee}/X^{\vee})_{p'} \to Z(\mathbf{G})$.

In fact, this gives us an isomorphism $\varphi : \Lambda^{\vee}/\mathbb{Z}\Phi^{\vee} \to Z(\mathbf{G})$ because $X^{\vee} = \mathbb{Z}\Phi^{\vee}$, the quotient $\Lambda^{\vee}/\mathbb{Z}\Phi^{\vee}$ is the Klein four group and $p$ is odd. Now, any element of the quotient $\Lambda^{\vee}/\mathbb{Z}\Phi^{\vee} $ can be represented by a fundamental dominant coweight $\omega_{\alpha}^{\vee}$ for some simple root $\alpha\in\Delta$. Furthermore, it can be shown that

$$\tau(\varphi(\omega^{\vee}_{\alpha} + \mathbb{Z}\Phi^{\vee})) = \varphi(\omega^{\vee}_{b(\alpha)} + \mathbb{Z}\Phi^{\vee}).$$ Using this and the fact that $Z(G) = Z(\mathbf{G})^F$ one easily deduces the answer.

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