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!

  • 1
    $\begingroup$ 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

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.