Return to Question

Discovery 2019-01-01

Let $u$ be order 3 octonion of norm 1 and trace 1 for $q$ even, trace -1 for $q$ odd. 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.

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

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 $\mathbb O(q)$?

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 $\mathbb O(q)$?

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 this paper.

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; $q^6-q^3$ for $q=4k+3$; $q^6+q^3$ for $q=4k+1$. By experiment in GAP I conjecture that number of perpendicular imaginary elements to fixed imaginary $u$ is $q^5-2$ for $q$ even; $q^5+q^2$ for $q=4k+3$; $q^5-q^2$ for $q=4k+1$. 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 fix some complex subalgebra of octonions. All automorphisms fixing given subalgebra form $SU_3$. In finite case each element of $G_2(q)$ fix some 2-dimensional subalgebra generated by one element. All automorphisms fixing given 2-dimensional subalgebra generated by one element form $U_3(q)$ or $SL_3(q)$ (the details to be clarified).

Let $u$ be order 3 octonion of norm 1 and trace 1 for $q$ even, trace -1 for q odd. 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.

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 $<1,i>$ and rotating by angle $\frac{2π}{3}$ in perpendicular space.

I found generators of $S_8(2)$ using octonions $\mathbb O(2)$. The conjugacy class of size 255 consist 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 $2-\text{dimensional}$ planes over $\mathbb C \otimes \mathbb O$.

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

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).

Let $u$ be order 3 octonion of norm 1 and trace 1 for $q$ even, trace -1 for $q$ odd. 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.

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.

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 $\mathbb C \otimes \mathbb O$.

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.

Let $u$ be order 3 octonion of norm 1 and trace 1 for $q$ even, trace -1 for q odd. 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. Similar law is valid for compact octonions. Take $u=e^{2πi/3}$ for some imaginary octonion $i$. Then $L_uR_{u^{-1}}$ is automorphism.

I found generators of $S_8(2)$ using octonions $\mathbb O(2)$. The conjugacy class of size 255 consist 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 $vv=v$). 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 $\mathbb O(q)$ ?

Let $u$ be order 3 octonion of norm 1 and trace 1 for $q$ even, trace -1 for q odd. 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.

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 $<1,i>$ and rotating by angle $\frac{2π}{3}$ in perpendicular space.

I found generators of $S_8(2)$ using octonions $\mathbb O(2)$. The conjugacy class of size 255 consist 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 $\mathbb O(q)$ ?