# What finite simple groups we can obtain using octonions?

Rearranged on 2017-05-31

What I am missing is a uniform definition of finite simple groups. Especially sporadic groups are difficult to define.

Many of the finite groups are defined using machinery from Lie groups. For example compact group $$F_4$$ is defined using Albert algebra over octonions. Finite group $$F_4(q)$$ is defined similarly by using octonions over field $$\mathbb F_q$$. See for example Wilson - Albert algebras and construction of the finite simple groups $$F_4(q)$$, $$E_6(q)$$ and $$^2E_6(q)$$ and their generic covers.

Lie group $$G_2$$ is defined as group of automorphisms of octonions. Finite group $$G_2(q)$$ is group of automorphisms of finite octonions $$\mathbb O(q)$$. I observed that this group is generated by elements $$(L_xR_y)^2$$ where $$x$$, $$y$$ are imaginary perpendicular octonions. Such element fix subalgebra $$\langle x,y\rangle$$. $$L_x$$ is left multiplication and $$R_x$$ is right multiplication by octonion $$x$$. Octonion $$x$$ is imaginary when $$x\bar x=1$$ and $$\bar x=-x$$ and $$x \neq 1$$. Two imaginary octonions are perpendicular when they commute or anticommute (equivalently product of them is imaginary). The number of imaginary octonions is $$q^6-1$$ for $$q$$ even; $$q^6+(-1)^{(q - 1)/2}q^3$$ for $$q$$ odd. By experiment in GAP I conjecture that number of perpendicular imaginary elements to fixed imaginary $$u$$ is $$q^5-2$$ for $$q$$ even; $$q^5+(-1)^{(q + 1)/2}q^2$$ for $$q$$ odd. As next step we can count number of involutions equal to number of quaternion subalgebras. We need to be careful in characteristic two, what the quaternion subalgebra is. There are two involution conjugacy classes in this case. The stabilizer of involution is $$D_2(q)$$ in case of $$q$$ odd. Another step would be to find $$SU_3$$ equivalent there. Surprisingly there are two of them $$U_3(q)$$ and $$SL_3(q)$$.

In compact Lie group $$G_2$$ each element fixes some complex subalgebra of octonions. The automorphisms fixing a given subalgebra form $$SU_3$$. In the finite case each element of $$G_2(q)$$ fixes some 2-dimensional subalgebra generated by one element. The automorphisms fixing a given 2-dimensional subalgebra generated by one element form $$U_3(q)$$ or $$SL_3(q)$$ (the details to be clarified).

I found generators of $$S_8(2)$$ using octonions $$\mathbb O(2)$$. The conjugacy class of size 255 consists of elements $$\{L_xR_xS\}$$ for invertible $$x$$, $$I+L_uR_u$$ for $$uu=0$$ and $$I+(S+L_\bar v)R_v$$ for other zero divisor $$v$$ (satisfying $$vv=v$$) ($$S$$ is matrix of conjugation). Group $$S_8(2)$$ is equal to $$O_9(2)$$, so it corresponds to $$Spin_9$$ subgroup in $$F_4$$. Look at $$F_4$$ as automorphisms of projective plane $$\mathbb OP^2$$ and $$Spin_9$$ as automorphisms of $$\mathbb OP^1$$. How could we define $$O_9(q) \subset F_4(q)$$ using finite octonions $$\mathbb O(q)$$?

Similarly how to define $$O_{10}^+(q) \subset E_6(q)$$ and $$O_{10}^-(q) \subset {^2E_6(q)}$$ using finite octonions? In compact case we may look at $$E_6$$ as automorphisms of "projective plane $$\mathbb C \otimes \mathbb OP^2$$". See Baez. I put quotes, because not all mathematicians believe in this "projective plane". In compact case we have $$Spin_{10}$$ and $$E_6$$ as automorphisms of $$1$$- and $$2$$-dimensional planes over $$\mathbb C \otimes \mathbb O$$.

Has anybody tried to define $$E_7(q)$$ and $$E_8(q)$$ using octonions? What are the $$O_{12}$$ and $$O_{16}$$ embedded?

Leech lattice can be seen as result of 819 points on $$\mathbb OP^2$$. See the definition in Wilson - Octonions and the Leech lattice of the Leech lattice. It is defined as union of 819 distinct $$E_8$$ lattices.

I don't know how to ask about application of finite octonions to define sporadic groups. I encountered big criticism here.

• What precisely do you denote by $L_x$ and $R_y$, and what do you mean by imaginary perpendicular octonions? – Stefan Kohl May 26 '17 at 19:04
• I am afraid it has been closed, but I voted to reopen. It's a pity that it's not possible to vote against closing. – Derek Holt May 30 '17 at 21:45
• @Derek: Sure you cannot literally 'vote' against closing -- but I think your comment actually counts more than a vote! – Stefan Kohl May 31 '17 at 9:28
• Hi Marek. The incorporation of discoveries, additions, etc. into the question begins to make the question hard to read, or even to locate in the text. An alternative might be to collect the information you have subsequently obtained in an answer below the question. You can make the answer CW if you like. – Todd Trimble Jan 1 at 14:43
• @Todd : you mean no CW complex is needed ? : -) – Sylvain JULIEN Jan 1 at 16:51

Discovery 2019-01-01

Let $$u$$ be order 3 octonion of norm $$1$$ and trace $$-1$$ (in case of characteristic 2 we have $$-1=1$$; norm is $$x\bar x$$, trace is $$x+\bar x$$). Then element $$L_uR_{u^{-1}}$$ is automorphism. This way we obtain order 3 generators of $$G_2(q)$$. I tested it in GAP for $$q=2,3,4,5,7,8,9$$. I don't know how to prove it yet but it seems to be rather easy. I define finite octonions as $$\mathbb H_q+\mathbb H_q\iota$$ using Cayley-Dickson formula. We define quaternions $$\mathbb H_q=M_2\mathbb F_q$$ as matrices $$2\times 2$$ over finite field. Conjugation is defined as adjugate matrix - exchange elements on the diagonal and reverse sign off-diagonal. The complex numbers algebra is finite field $$\mathbb F_{q^2}$$ represented by subalgebra generated by matrix $$\pmatrix{n&1\\ 1&0}$$ of order $$q+1$$ when $$n$$ is order $$q-1$$ generator of the field multiplicative group. This matrix satisfy equation $$x^2=nx+1$$.

Having $$\mathbb C_q=\mathbb F_{q^2}$$ we can also define quaternions using Cayley-Dickson formula $$\mathbb H_q=\mathbb C_q+\mathbb C_qj$$. Conjugation is then defined as $$\overline{x+yj}=\bar{x}-yj$$. It coincides with Frobenius automorphism $$x\to x^q$$ of the field $$\mathbb F_{q^2}$$.

Similar law is valid for compact octonions. Take $$u=e^{2πi/3}$$ for some imaginary octonion $$i$$. Then $$L_uR_{u^{-1}}$$ is automorphism fixing complex plane $$\langle1,i\rangle$$ and rotating by angle $$\frac{2π}{3}$$ in perpendicular space.

End of discovery

Let $$T_u$$ denote $$L_uR_{\bar u}$$. Then we have following formula $$T_u^{T_v}=T_{u^v}$$. In this formula $$g^h=h^{-1}gh$$ valid in the group or in octonion algebra (elements of norm 1 are invertible).
Recipe for $$J_2$$
Let $$u,v,w$$ be three octonions of norm 1 and trace 1 over field $$F_4$$ such that order of product of each pair is 5 (for fixed $$u$$ there are 2016 such $$v$$; for fixed $$u,v$$ there are 960 such $$w$$). It is equivalent to the fact that pair generate group $$A_5$$ or 4-dimensional subalgebra $$M_2F_4$$. In such case elements $$T_u, T_v, T_w$$ generate group $$J_2$$ of size 604800. From the formula above we have that 3A conjugacy class in the group corresponds to constellation of 560 order 3 octonions closed by conjugation $$x\to \bar yxy$$. What is the meaning of such constellation ? I am looking for geometrical object preserved by group $$J_2\subset G_24$$.
• Does this work regardless of whether the field contains square roots of $-1$ or not? – მამუკა ჯიბლაძე Jan 2 at 12:50
• Yes, in field GF(5) we have $2*2=-1$ and it works as well. – Marek Mitros Jan 2 at 12:59
• So there is no analog of the fact that over $\mathbb R$ there are split and non-split octonions which only become isomorphic over $\mathbb C$? – მამუკა ჯიბლაძე Jan 2 at 14:23
• I am afraid of editing my answer - it seems editing is persecuted on this forum - so I add new information as a comment. The number of norm 1 and trace -1 octonions is $q^6-q^3, q^6+q^3, q^6-1$ and stabilizer is $3.L_3(q), 3.U_3(q)$ or other for $q=1,2,0(mod 3)$. It would be interesting to connect maximal subgroups of $G_2(q)$ to subgroups fixing some subalgebras of octonions. What objects are fixed by sporadics $J_2$ and $J_1$ embedded in $G_2(4)$ and $G_2(11)$ ? – Marek Mitros Jan 3 at 9:38