# Symplectic group over finite field and quaternions

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 ?

• Is not standard notation Sp (rather than S)? – მამუკა ჯიბლაძე Feb 12 '18 at 9:35
• I use notation from Atlas of finite group representations: brauer.maths.qmul.ac.uk/Atlas/v3 – Marek Mitros Feb 12 '18 at 9:37
• 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 $Sp_{2n}$. – Marek Mitros Feb 12 '18 at 9:55
• 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. – Derek Holt Feb 12 '18 at 10:22
• 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 $q^2$). – Derek Holt Feb 12 '18 at 12:56