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For $q$ an odd prime power, one can construct $H:=O_4^-(q)<U_4(q)$ by treating the $H$-invariant bilinear form as Hermitean. On the other hand $U_4(q)$ is isomorphic to $O_6^-(q)$; I need to understand the geometric meaning of $H$ in $O_6^-(q)$.

For $q \equiv 1\pmod {4}$, it appears that $H$ can be viewed as $O_3(q^2)$; namely, the relative trace form $x\mapsto \langle x,Bx\rangle+\langle x,Bx\rangle^q$, with $x\mapsto \langle x,Bx\rangle$ being the $O_3(q^2)$-invariant form, gives a minus type form in $\mathbb{F}_q^6$, and thus the desired embedding $H<O_6^-(q)$.

However, I don't understand how to deal with the case $q \equiv 3\pmod {4}$, for the relative trace form then is of plus type. Any suggestions?

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  • $\begingroup$ Well $\Omega_n(q^2)$ ($n$ odd), or $\Omega^-_n(q^2)$ ($n$ even) is certainly a semilinear subgroup of $\Omega_{2n}^-(q)$ for all odd $q$ ($n$ odd) and all $q$ ($n$ even). I am sure you will find this all explained in Kleidman and Liebeck's book on the maximal subgroups of the finite classical groups. $\endgroup$
    – Derek Holt
    Commented Jan 3, 2016 at 12:49
  • $\begingroup$ There seems to be more than one class of these, and figuring out what $H$ corresponds to looks tricky. And the value of $q\pmod 4$ seems to matter. $\endgroup$
    – Dima Pasechnik
    Commented Jan 3, 2016 at 15:15
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    $\begingroup$ Given a $3$-dimensional quadratic form $Q$ over $\mathbb{F}_{q^2}$, any $\mathbb{F}_q$-linear function $L: \mathbb{F}_{q^2}\to \mathbb{F}_q$ will yield a $6$-dimensional quadratic form $L\circ Q$ over $\mathbb{F}_q$ (i.e. you don't need to use the trace form). What is more, you can get all possible $L$ from the trace by adding in some constant, i.e. $L(x)=L(\mu\cdot x)$ for some $\mu \in \mathbb{F}_{q^2}$. If you choose your $\mu$ carefully, the resulting linear form $L$ will give a minus type form $LQ$ for $q\equiv 3\pmod 4$. $\endgroup$
    – Nick Gill
    Commented Jan 3, 2016 at 21:07
  • $\begingroup$ (And, as Derek says, the definitive source on this is Kleidman and Liebeck, Section 4.3.) $\endgroup$
    – Nick Gill
    Commented Jan 3, 2016 at 21:09
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    $\begingroup$ $U_4(q)$ also had imprimitive sugroups isomorphic to $L_2(q^2)$, but I think they correspond to reducible subgroups of $\Omega^-_6(q)$. $\endgroup$
    – Derek Holt
    Commented Jan 3, 2016 at 22:09

1 Answer 1

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As is pointed out in the question, our job is to display an (irreducible) copy of $H=O_3(q^2)$ inside $G=O_6^-(q)$.

As mentioned in comments, this can be done by taking $H$ to be a field-extension subgroup of $G$. An explicit construction for $q\equiv 3 \pmod 4$ is as follows.

We take $H$ to be the group of isometries of $$Q: V \to \mathbb{F}_{q^2}, \,\, (x,y,z) \mapsto xy+z^2,$$ where $V$ is 3-dimensional over $\mathbb{F}_{q^2}$.

Now, viewing $V$ as $6$-dimensional over $\mathbb{F}_q$, and letting $L:\mathbb{F}_{q^2}\to \mathbb{F}_q$ be any $\mathbb{F}_q$-linear function, we observe that the elements of $H$ preserve the form $LQ:V\to \mathbb{F}_q$, and we obtain an embedding of $H$ in the isometry group of $LQ$.

Thus, to answer the question, we must exhibit $L$ so that $LQ$ is a non-degenerate form of minus type. I claim that we can take $L=L_\mu$ where

  • $\zeta$ is a primitive element of $\mathbb{F}_{q^2}$ and $\mu=\zeta^a$ for some odd number $a$.
  • $L_\mu:\mathbb{F}_{q^2}\to \mathbb{F}_q, x \mapsto \mu x + (\mu x)^q$. (In other words $L_\mu(x)= {\rm Tr}(\mu x)$.)

I won't bother showing that $L_\mu Q$ is non-degenerate as this is easy. To see that it is minus type, I must show that there exists a 2-dimensional $\mathbb{F}_q$-subspace of $V$ that is a maximal totally isotropic subspace with respect to $L_\mu Q$.

Let $W={\rm span}_{\mathbb{F}_{q^2}}((1,0,0))$. Since $Q(1,0,0)=0$, it is clear that $W$ is t.i. with respect to $Q$, and hence with respect to $L_\mu Q$.

Suppose $\langle W, (x,y,z)\rangle$ is totally isotropic. Let $\beta$ be the polarization of $Q$. Then $$ \beta((x_1,y_1,z_1), (x_2,y_2,z_2)) = x_1y_2+x_2y_1+2z_1z_2 $$ and we conclude that, for any $\lambda \in \mathbb{F}_{q^2}$, $$ L_\mu\beta((\lambda, 0,0), (x,y,z)) = L_\mu(\lambda y) = (\mu\lambda y)+(\mu\lambda y)^q = 0. $$ One concludes, immediately, that $y=0$.

Now, in addition, our supposition implies that $(x,y,z)$ is totally singular. Then $$ L_\mu Q(x,y,z)= \mu(xy+z^2)+\left(\mu(xy+z^2)\right)^q=\mu z^2+ \left(\mu z^2\right)^q = 0. $$ Our choice of $\mu$, along with the fact that $q\equiv 3\pmod 4$, now implies that $z=0$. But this means that $(x,y,z)\in W$ and so $W$ is maximal t.i., as required.

(Final note: The same argument shows that ${\rm span}_{\mathbb{F}_{q^2}}((0,1,0))$ is maximal t.i. This implies, in particular, that $L_\mu Q$ is non-degenerate.)

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