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Symplectic group over finite field is defined as group preserving non-degenerate antisymmetric bilinear form on $\mathbb F_q^{2n}$. How could we define this group using quaternions ? This should be group of $n\times n$ matrices with quaternion entries preserving hermitian form.

I know that there might be troubles with definition of quaternions over finite field. One approach is to use Cayley-Dickson formula. Second is to use some 4-dimensional subalgebra of octonions.

$q=2$

Let us define quaternions $\mathbb H_2$ as algebra $M_2\mathbb F_2$. It is generated by two invertible $t,u$ with conjugation and relations $\bar t=1+t, \bar u =u, ut=\bar tu$. This algebra contains $6$ invertible elements (forming group $S_3$) and $9$ zero divisors. Conjugation for matrix $x$ having single $1$ on diagonal is expressed as $\bar x=1+x$ otherwise it fixes element.

In this case I claim that group of $n\times n$ quaternion matrices $\{A A^*=I\}$ form $S_{2n}(2)$ finite simple group. $A^*$ denote transposed conjugated matrix to $A$. I have verified this in GAP for $n=2,3$.

Complex case

Let $\mathbb C_1$ be subalgebra of $\mathbb H_2$ generated by zero divisor $a$ such that $aa=a$. In this case $\bar a=1+a$ is second zero divisor in algebra. Let us call this algebra split complex numbers over field with two elements. Subgroup of $\{A A^*=I\}$ with elements from $\mathbb C_1$ form $L_n(2)$ finite simple group.

Let $\mathbb C_2=\mathbb F_4$ be subalgebra of $\mathbb H_2$ generated by $t$. Let us call this algebra complex numbers over field with two elements. Subgroup of $\{A A^*=I\}$ with elements from $\mathbb C_2$ form $U_n(2)$ finite simple group.

$q=3$

For field with three elements define quaternions as $M_2\mathbb F_3$. Conjugation exchanging elements on main diagonal and change sign for off-diagonal elements. In this case $n\times n$ quaternion matrices $\{A A^*=I\}$ form group $Sp_{2n}(3)$. It is tested in GAP for dimensions 2 and 3.

Geometry

Let us define projective space over quaternions in following way:

$\mathbb HP_q^n=\frac{Sp_{2n+2}(q)}{Sp_{2n}(q)\times Sp_2(q)}$

The size is $20, 336, 5440$ for $n=1,2,3$ and $q=2$. General formula in case of $q=2$ is $2^{2n}\frac{2^{2n+2}-1}{3}$.

Question

Have anyone thought of defining symplectic or unitary group over finite field using algebra of complex or quaternion numbers as I tried to sketch above ?

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    $\begingroup$ Is not standard notation Sp (rather than S)? $\endgroup$ – მამუკა ჯიბლაძე Feb 12 '18 at 9:35
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    $\begingroup$ I use notation from Atlas of finite group representations: brauer.maths.qmul.ac.uk/Atlas/v3 $\endgroup$ – Marek Mitros Feb 12 '18 at 9:37
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    $\begingroup$ Yes, in context of my question I prefer Sp as well in order to have connection to compact Sp(n). Another drawback of Atlas notation is doubled index for symplectic groups. In n-dimensional quaternion space we obtain $S_{2n}$ or $Sp_{2n}$. $\endgroup$ – Marek Mitros Feb 12 '18 at 9:55
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    $\begingroup$ I think you are wrong about the Atlas notation. In the Atlas, $S_4(3)$ denotes the simple group ${\rm PSp}_4(3) = {\rm Sp}_4(3)/Z({\rm Sp}_4(3))$. The Atlas uses ${\rm Sp}_4(3)$ or sometimes $2.S_4(3)$ to denote the symplectic group itself. $\endgroup$ – Derek Holt Feb 12 '18 at 10:22
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    $\begingroup$ I don't know whether this is relevant, but there is an embedding ${\rm Sp}_n(q) \to {\rm SU}_n(q)$ for even $n$ (where ${\rm SU}_n(q)$ is defined over the field of order $q^2$). $\endgroup$ – Derek Holt Feb 12 '18 at 12:56

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