\documentclass{article}$ \usepackage{amsmath,amssymb}$ \sloppy$ \begin{document}$ This output from the file \texttt{structcomplItex.tex}$ Computation of all complex structures on the real Lie Algebra$ USD {\mathcal{G}}_{6,3}.USD$ \par Case USD xi(1,6) = 0, xi(2,5) = 0.USD$ USD \\ xi(1,6):=0USD$ USD \\ xi(2,5):=0USD$ \smallskip \par $ Commutation relations for$ USD {\mathcal{G}}_{6,3}:USD\\$ USD[x(1),x(2)]=x(4)USD;$ USD[x(1),x(3)]=x(5)USD;$ USD[x(2),x(3)]=x(6)USD;$ \P$ Nonzero torsion$ \par$ Torsion equations to cancel (Latex output) : \\USD$ {1,2}|1\\xi(3,2)*xi(1,5) + xi(2,2)*xi(1,4) + xi(1,4)*xi(1,1)\\$ {1,2}|2\\ - xi(3,1)*xi(2,6) + xi(2,4)*xi(2,2) + xi(2,4)*xi(1,1)\\$ {1,2}|3\\ - xi(3,6)*xi(3,1) + xi(3,5)*xi(3,2) + xi(3,4)*xi(2,2) + xi(3,4)*xi(1,1 )\\$ {1,2}|4\\ - xi(4,6)*xi(3,1) + xi(4,5)*xi(3,2) + xi(4,4)*xi(2,2) + xi(4,4)*xi(1,1 ) - xi(2,2)*xi(1,1) + xi(2,1)*xi(1,2) + 1\\$ {1,2}|5\\ - xi(5,6)*xi(3,1) + xi(5,5)*xi(3,2) + xi(5,4)*xi(2,2) + xi(5,4)*xi(1,1 ) - xi(3,2)*xi(1,1) + xi(3,1)*xi(1,2)\\$ {1,2}|6\\ - xi(6,6)*xi(3,1) + xi(6,5)*xi(3,2) + xi(6,4)*xi(2,2) + xi(6,4)*xi(1,1 ) - xi(3,2)*xi(2,1) + xi(3,1)*xi(2,2)\\$ {1,3}|1\\xi(3,3)*xi(1,5) + xi(2,3)*xi(1,4) + xi(1,5)*xi(1,1)\\$ {1,3}|2\\xi(2,6)*xi(2,1) + xi(2,4)*xi(2,3)\\$ {1,3}|3\\xi(3,6)*xi(2,1) + xi(3,5)*xi(3,3) + xi(3,5)*xi(1,1) + xi(3,4)*xi(2,3)\\ $ {1,3}|4\\xi(4,6)*xi(2,1) + xi(4,5)*xi(3,3) + xi(4,5)*xi(1,1) + xi(4,4)*xi(2,3) - xi(2,3)*xi(1,1) + xi(2,1)*xi(1,3)\\$ {1,3}|5\\xi(5,6)*xi(2,1) + xi(5,5)*xi(3,3) + xi(5,5)*xi(1,1) + xi(5,4)*xi(2,3) - xi(3,3)*xi(1,1) + xi(3,1)*xi(1,3) + 1\\$ {1,3}|6\\xi(6,6)*xi(2,1) + xi(6,5)*xi(3,3) + xi(6,5)*xi(1,1) + xi(6,4)*xi(2,3) - xi(3,3)*xi(2,1) + xi(3,1)*xi(2,3)\\$ {1,4}|1\\xi(3,4)*xi(1,5) + xi(2,4)*xi(1,4)\\$ {1,4}|2\\xi(2,4)**2\\$ {1,4}|3\\xi(3,5)*xi(3,4) + xi(3,4)*xi(2,4)\\$ {1,4}|4\\xi(4,5)*xi(3,4) + xi(4,4)*xi(2,4) - xi(2,4)*xi(1,1) + xi(2,1)*xi(1,4)\\ $ {1,4}|5\\xi(5,5)*xi(3,4) + xi(5,4)*xi(2,4) - xi(3,4)*xi(1,1) + xi(3,1)*xi(1,4)\\ $ {1,4}|6\\xi(6,5)*xi(3,4) + xi(6,4)*xi(2,4) - xi(3,4)*xi(2,1) + xi(3,1)*xi(2,4)\\ $ {1,5}|1\\xi(3,5)*xi(1,5)\\$ {1,5}|3\\xi(3,5)**2\\$ {1,5}|4\\xi(4,5)*xi(3,5) + xi(2,1)*xi(1,5)\\$ {1,5}|5\\xi(5,5)*xi(3,5) - xi(3,5)*xi(1,1) + xi(3,1)*xi(1,5)\\$ {1,5}|6\\xi(6,5)*xi(3,5) - xi(3,5)*xi(2,1)\\$ {1,6}|1\\xi(3,6)*xi(1,5) + xi(2,6)*xi(1,4)\\$ {1,6}|2\\xi(2,6)*xi(2,4)\\$ {1,6}|3\\xi(3,6)*xi(3,5) + xi(3,4)*xi(2,6)\\$ {1,6}|4\\xi(4,5)*xi(3,6) + xi(4,4)*xi(2,6) - xi(2,6)*xi(1,1)\\$ {1,6}|5\\xi(5,5)*xi(3,6) + xi(5,4)*xi(2,6) - xi(3,6)*xi(1,1)\\$ {1,6}|6\\xi(6,5)*xi(3,6) + xi(6,4)*xi(2,6) - xi(3,6)*xi(2,1) + xi(3,1)*xi(2,6)\\ $ {2,3}|1\\xi(1,5)*xi(1,2) - xi(1,4)*xi(1,3)\\$ {2,3}|2\\xi(3,3)*xi(2,6) + xi(2,6)*xi(2,2) - xi(2,4)*xi(1,3)\\$ {2,3}|3\\xi(3,6)*xi(3,3) + xi(3,6)*xi(2,2) + xi(3,5)*xi(1,2) - xi(3,4)*xi(1,3)\\ $ {2,3}|4\\xi(4,6)*xi(3,3) + xi(4,6)*xi(2,2) + xi(4,5)*xi(1,2) - xi(4,4)*xi(1,3) - xi(2,3)*xi(1,2) + xi(2,2)*xi(1,3)\\$ {2,3}|5\\xi(5,6)*xi(3,3) + xi(5,6)*xi(2,2) + xi(5,5)*xi(1,2) - xi(5,4)*xi(1,3) - xi(3,3)*xi(1,2) + xi(3,2)*xi(1,3)\\$ {2,3}|6\\xi(6,6)*xi(3,3) + xi(6,6)*xi(2,2) + xi(6,5)*xi(1,2) - xi(6,4)*xi(1,3) - xi(3,3)*xi(2,2) + xi(3,2)*xi(2,3) + 1\\$ {2,4}|1\\ - xi(1,4)**2\\$ {2,4}|2\\xi(3,4)*xi(2,6) - xi(2,4)*xi(1,4)\\$ {2,4}|3\\xi(3,6)*xi(3,4) - xi(3,4)*xi(1,4)\\$ {2,4}|4\\xi(4,6)*xi(3,4) - xi(4,4)*xi(1,4) - xi(2,4)*xi(1,2) + xi(2,2)*xi(1,4)\\ $ {2,4}|5\\xi(5,6)*xi(3,4) - xi(5,4)*xi(1,4) - xi(3,4)*xi(1,2) + xi(3,2)*xi(1,4)\\ $ {2,4}|6\\xi(6,6)*xi(3,4) - xi(6,4)*xi(1,4) - xi(3,4)*xi(2,2) + xi(3,2)*xi(2,4)\\ $ {2,5}|1\\ - xi(1,5)*xi(1,4)\\$ {2,5}|2\\xi(3,5)*xi(2,6) - xi(2,4)*xi(1,5)\\$ {2,5}|3\\xi(3,6)*xi(3,5) - xi(3,4)*xi(1,5)\\$ {2,5}|4\\xi(4,6)*xi(3,5) - xi(4,4)*xi(1,5) + xi(2,2)*xi(1,5)\\$ {2,5}|5\\xi(5,6)*xi(3,5) - xi(5,4)*xi(1,5) - xi(3,5)*xi(1,2) + xi(3,2)*xi(1,5)\\ $ {2,5}|6\\xi(6,6)*xi(3,5) - xi(6,4)*xi(1,5) - xi(3,5)*xi(2,2)\\$ {2,6}|2\\xi(3,6)*xi(2,6)\\$ {2,6}|3\\xi(3,6)**2\\$ {2,6}|4\\xi(4,6)*xi(3,6) - xi(2,6)*xi(1,2)\\$ {2,6}|5\\xi(5,6)*xi(3,6) - xi(3,6)*xi(1,2)\\$ {2,6}|6\\xi(6,6)*xi(3,6) - xi(3,6)*xi(2,2) + xi(3,2)*xi(2,6)\\$ {3,4}|1\\ - xi(1,5)*xi(1,4)\\$ {3,4}|2\\ - xi(2,6)*xi(2,4)\\$ {3,4}|3\\ - xi(3,6)*xi(2,4) - xi(3,5)*xi(1,4)\\$ {3,4}|4\\ - xi(4,6)*xi(2,4) - xi(4,5)*xi(1,4) - xi(2,4)*xi(1,3) + xi(2,3)*xi(1,4 )\\$ {3,4}|5\\ - xi(5,6)*xi(2,4) - xi(5,5)*xi(1,4) - xi(3,4)*xi(1,3) + xi(3,3)*xi(1,4 )\\$ {3,4}|6\\ - xi(6,6)*xi(2,4) - xi(6,5)*xi(1,4) - xi(3,4)*xi(2,3) + xi(3,3)*xi(2,4 )\\$ {3,5}|1\\ - xi(1,5)**2\\$ {3,5}|3\\ - xi(3,5)*xi(1,5)\\$ {3,5}|4\\ - xi(4,5)*xi(1,5) + xi(2,3)*xi(1,5)\\$ {3,5}|5\\ - xi(5,5)*xi(1,5) - xi(3,5)*xi(1,3) + xi(3,3)*xi(1,5)\\$ {3,5}|6\\ - xi(6,5)*xi(1,5) - xi(3,5)*xi(2,3)\\$ {3,6}|2\\ - xi(2,6)**2\\$ {3,6}|3\\ - xi(3,6)*xi(2,6)\\$ {3,6}|4\\ - xi(4,6)*xi(2,6) - xi(2,6)*xi(1,3)\\$ {3,6}|5\\ - xi(5,6)*xi(2,6) - xi(3,6)*xi(1,3)\\$ {3,6}|6\\ - xi(6,6)*xi(2,6) - xi(3,6)*xi(2,3) + xi(3,3)*xi(2,6)\\$ {4,5}|4\\xi(2,4)*xi(1,5)\\$ {4,5}|5\\ - xi(3,5)*xi(1,4) + xi(3,4)*xi(1,5)\\$ {4,5}|6\\ - xi(3,5)*xi(2,4)\\$ {4,6}|4\\ - xi(2,6)*xi(1,4)\\$ {4,6}|5\\ - xi(3,6)*xi(1,4)\\$ {4,6}|6\\ - xi(3,6)*xi(2,4) + xi(3,4)*xi(2,6)\\$ {5,6}|4\\ - xi(2,6)*xi(1,5)\\$ {5,6}|5\\ - xi(3,6)*xi(1,5)\\$ {5,6}|6\\xi(3,5)*xi(2,6)\\$ USD$ \par Simultaneous resolution of the nonzero torsion equations and the matrix$ equation USD J^2 = -I USD in the case USD xi(1,6) = 0. USD$ \\ One first gets$ \\ from equation USD24|1USD :$ \\ USD xi(1,4):=0$ USD\\ and from equation USD26|3USD :$ \\USD xi(3,6):=0$ USD\\ and from equation USD35|1USD :$ \\USD xi(1,5):=0$ USD\\ and from equation USD36|2USD :$ \\USD xi(2,6):=0USD$ \\ and from equation USD14|2USD :$ \\USD xi(2,4):=0USD$ \\ and from equation USD15|3USD :$ \\USD xi(3,5):=0USD$ \par With these values, \textit{Reduce} computes again all equations.$ Then, one gets from equation USD13|3USD, as USD xi(3,4) \neq 0 USD :$ \\USD xi(2,3):=0USD$ \\ and from equation USD12|3USD :$ \\USD xi(2,2):= - xi(1,1)USD$ \\ and from equation USD14|4USD :$ \\USD xi(4,5):=0USD$ \\ and from equation USD14|5USD :$ \\USD xi(5,5):=xi(1,1)USD$ \\ and from equation USD14|6USD :$ \\USD xi(6,5):=xi(2,1)USD$ \\ and from equation USD24|4USD :$ \\USD xi(4,6):=0USD$ \\ and from equation USD24|5USD :$ \\USD xi(5,6):=xi(1,2)USD$ \\ and from equation USD24|6USD :$ \\USD xi(6,6):= - xi(1,1)USD$ \\ and from equation USD34|5USD :$ \\USD xi(1,3):=0USD$ \par With these values, \textit{Reduce} computes again all equations.$ Then the only nonzero integrability equation left $ USD xi(1,2)*xi(2,1) + xi(1,1)**2 +1 = 0 USD$ makes sense only if USD xi(1,2) \neq 0 ,USD$ and then gives USD xi(2,1) :USD$ \\USD xi(2,1):= - (xi(1,1)**2 + 1)/xi(1,2)USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ The USD 3 \times 3 USD term in USD J^2 USD is: $ USD xi(3,4)*xi(4,3) + xi(3,3)**2. USD$ This makes sense only if USD xi(3,4) \neq 0 ,USD$ and then gives USD xi(4,3) :USD$ \\USD xi(4,3):= - (xi(3,3)**2 + 1)/xi(3,4)USD$ \\ From the USD 3 \times 4 USD term in USD J^2 USD one gets: $ \\USD xi(4,4):= - xi(3,3)USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ From the USD 3 \times 2 USD term in USD J^2 USD one gets : $ \\USD xi(4,2):=( - xi(3,3)*xi(3,2) + xi(3,2)*xi(1,1) - xi(3,1)*xi(1,2))/xi(3,4) USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ From the USD 3 \times 1 USD term in USD J^2 USD one gets : $ \\USD xi(3,2):=(xi(1,2)*(xi(4,1)*xi(3,4) + xi(3,3)*xi(3,1) + xi(3,1)*xi(1,1)))/( xi(1,1)**2 + 1)USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ From the USD 5 \times 4 USD term in USD J^2 USD one gets : $ \\USD xi(5,3):=( - xi(6,4)*xi(1,2) + xi(5,4)*xi(3,3) - xi(5,4)*xi(1,1))/xi(3,4) USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ From the USD 5 \times 3 USD term in USD J^2 USD one gets : $ \\USD xi(5,4):=(xi(1,2)*( - xi(6,4)*xi(3,3) - xi(6,4)*xi(1,1) + xi(6,3)*xi(3,4)) )/(xi(1,1)**2 + 1)USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ From the USD 5 \times 2 USD term in USD J^2 USD one gets : $ \\USD xi(5,1):=(xi(6,4)*xi(4,1)*xi(1,2) + xi(6,3)*xi(3,1)*xi(1,2) - xi(6,2)*xi(1 ,1)**2 - xi(6,2))/(xi(1,1)**2 + 1)USD$ \par \textit{Reduce} now computes the matrix USD J^2 USD,$ which must be equal to USD -I USD.$ From the USD 5 \times 1 USD term in USD J^2 USD one gets : $ \\USD xi(5,2):=(xi(1,2)*( - xi(6,4)*xi(4,1)*xi(3,4)*xi(3,3)*xi(1,2) + xi(6,4)*xi (4,1)*xi(3,4)*xi(1,2)*xi(1,1) - xi(6,4)*xi(3,3)**2*xi(3,1)*xi(1,2) - xi(6,4)*xi( 3,1)*xi(1,2) + xi(6,3)*xi(4,1)*xi(3,4)**2*xi(1,2) + xi(6,3)*xi(3,4)*xi(3,3)*xi(3 ,1)*xi(1,2) + xi(6,3)*xi(3,4)*xi(3,1)*xi(1,2)*xi(1,1) - 2*xi(6,2)*xi(3,4)*xi(1,1 )**3 - 2*xi(6,2)*xi(3,4)*xi(1,1) + xi(6,1)*xi(3,4)*xi(1,2)*xi(1,1)**2 + xi(6,1)* xi(3,4)*xi(1,2)))/(xi(3,4)*(xi(1,1)**4 + 2*xi(1,1)**2 + 1))USD$ \par Now the nonzero torsion equations left are :$ Torsion equations to cancel (Latex output) : USD$ USD$ \\ \P \\$ \par The matrix USD J USD is :\\$ USD J^1_1=xi(1,1);USD\\$ USD J^1_2=xi(1,2);USD\\$ USD J^1_3=0;USD\\$ USD J^1_4=0;USD\\$ USD J^1_5=0;USD\\$ USD J^1_6=0;USD\\$ USD J^2_1= - (xi(1,1)**2 + 1)/xi(1,2);USD\\$ USD J^2_2= - xi(1,1);USD\\$ USD J^2_3=0;USD\\$ USD J^2_4=0;USD\\$ USD J^2_5=0;USD\\$ USD J^2_6=0;USD\\$ USD J^3_1=xi(3,1);USD\\$ USD J^3_2=(xi(1,2)*(xi(4,1)*xi(3,4) + xi(3,3)*xi(3,1) + xi(3,1)*xi(1,1)))/(xi(1 ,1)**2 + 1);USD\\$ USD J^3_3=xi(3,3);USD\\$ USD J^3_4=xi(3,4);USD\\$ USD J^3_5=0;USD\\$ USD J^3_6=0;USD\\$ USD J^4_1=xi(4,1);USD\\$ USD J^4_2=(xi(1,2)*( - xi(4,1)*xi(3,4)*xi(3,3) + xi(4,1)*xi(3,4)*xi(1,1) - xi(3 ,3)**2*xi(3,1) - xi(3,1)))/(xi(3,4)*(xi(1,1)**2 + 1));USD\\$ USD J^4_3= - (xi(3,3)**2 + 1)/xi(3,4);USD\\$ USD J^4_4= - xi(3,3);USD\\$ USD J^4_5=0;USD\\$ USD J^4_6=0;USD\\$ USD J^5_1=(xi(6,4)*xi(4,1)*xi(1,2) + xi(6,3)*xi(3,1)*xi(1,2) - xi(6,2)*xi(1,1) **2 - xi(6,2))/(xi(1,1)**2 + 1);USD\\$ USD J^5_2=(xi(1,2)*( - xi(6,4)*xi(4,1)*xi(3,4)*xi(3,3)*xi(1,2) + xi(6,4)*xi(4,1 )*xi(3,4)*xi(1,2)*xi(1,1) - xi(6,4)*xi(3,3)**2*xi(3,1)*xi(1,2) - xi(6,4)*xi(3,1) *xi(1,2) + xi(6,3)*xi(4,1)*xi(3,4)**2*xi(1,2) + xi(6,3)*xi(3,4)*xi(3,3)*xi(3,1)* xi(1,2) + xi(6,3)*xi(3,4)*xi(3,1)*xi(1,2)*xi(1,1) - 2*xi(6,2)*xi(3,4)*xi(1,1)**3 - 2*xi(6,2)*xi(3,4)*xi(1,1) + xi(6,1)*xi(3,4)*xi(1,2)*xi(1,1)**2 + xi(6,1)*xi(3 ,4)*xi(1,2)))/(xi(3,4)*(xi(1,1)**4 + 2*xi(1,1)**2 + 1));USD\\$ USD J^5_3=(xi(1,2)*( - xi(6,4)*xi(3,3)**2 - xi(6,4) + xi(6,3)*xi(3,4)*xi(3,3) - xi(6,3)*xi(3,4)*xi(1,1)))/(xi(3,4)*(xi(1,1)**2 + 1));USD\\$ USD J^5_4=(xi(1,2)*( - xi(6,4)*xi(3,3) - xi(6,4)*xi(1,1) + xi(6,3)*xi(3,4)))/( xi(1,1)**2 + 1);USD\\$ USD J^5_5=xi(1,1);USD\\$ USD J^5_6=xi(1,2);USD\\$ USD J^6_1=xi(6,1);USD\\$ USD J^6_2=xi(6,2);USD\\$ USD J^6_3=xi(6,3);USD\\$ USD J^6_4=xi(6,4);USD\\$ USD J^6_5= - (xi(1,1)**2 + 1)/xi(1,2);USD\\$ USD J^6_6= - xi(1,1);USD\\$ USDUSD J^2 = \begin{pmatrix}$ -1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ -1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ -1&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ -1&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ -1\end{pmatrix}USDUSD$ \\$ det J:=1$ Trace J:=0$ \par Now we'll use equivalence by automorphisms$ of the form$ \\$ USDUSD \Phi = \begin{pmatrix}$ 1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 1&$ 0&$ 0&$ 0\\$ b(4,1)&$ b(4,2)&$ b(4,3)&$ 1&$ 0&$ 0\\$ b(5,1)&$ b(5,2)&$ b(5,3)&$ 0&$ 1&$ 0\\$ b(6,1)&$ b(6,2)&$ b(6,3)&$ 0&$ 0&$ 1\end{pmatrix}USDUSD$ USD det \Phi:=1USD$ \\ USD J2:=\Phi^{-1}*J*\Phi USD has entries :$ \\USD J2(1,1):=xi(1,1)USD\\$ \\USD J2(1,2):=xi(1,2)USD\\$ \\USD J2(1,3):=0USD\\$ \\USD J2(1,4):=0USD\\$ \\USD J2(1,5):=0USD\\$ \\USD J2(1,6):=0USD\\$ \\USD J2(2,1):= - (xi(1,1)**2 + 1)/xi(1,2)USD\\$ \\USD J2(2,2):= - xi(1,1)USD\\$ \\USD J2(2,3):=0USD\\$ \\USD J2(2,4):=0USD\\$ \\USD J2(2,5):=0USD\\$ \\USD J2(2,6):=0USD\\$ \\USD J2(3,1):=b(4,1)*xi(3,4) + xi(3,1)USD\\$ \\USD J2(3,2):=(b(4,2)*xi(3,4)*xi(1,1)**2 + b(4,2)*xi(3,4) + xi(4,1)*xi(3,4)*xi( 1,2) + xi(3,3)*xi(3,1)*xi(1,2) + xi(3,1)*xi(1,2)*xi(1,1))/(xi(1,1)**2 + 1)USD\\$ \\USD J2(3,3):=b(4,3)*xi(3,4) + xi(3,3)USD\\$ \\USD J2(3,4):=xi(3,4)USD\\$ \\USD J2(3,5):=0USD\\$ \\USD J2(3,6):=0USD\\$ \\USD J2(4,1):=( - b(4,3)*b(4,1)*xi(3,4)*xi(1,2) - b(4,3)*xi(3,1)*xi(1,2) + b(4, 2)*xi(1,1)**2 + b(4,2) - b(4,1)*xi(3,3)*xi(1,2) - b(4,1)*xi(1,2)*xi(1,1) + xi(4, 1)*xi(1,2))/xi(1,2)USD\\$ \\USD J2(4,2):=( - b(4,3)*b(4,2)*xi(3,4)**2*xi(1,1)**2 - b(4,3)*b(4,2)*xi(3,4)** 2 - b(4,3)*xi(4,1)*xi(3,4)**2*xi(1,2) - b(4,3)*xi(3,4)*xi(3,3)*xi(3,1)*xi(1,2) - b(4,3)*xi(3,4)*xi(3,1)*xi(1,2)*xi(1,1) - b(4,2)*xi(3,4)*xi(3,3)*xi(1,1)**2 - b( 4,2)*xi(3,4)*xi(3,3) + b(4,2)*xi(3,4)*xi(1,1)**3 + b(4,2)*xi(3,4)*xi(1,1) - b(4, 1)*xi(3,4)*xi(1,2)*xi(1,1)**2 - b(4,1)*xi(3,4)*xi(1,2) - xi(4,1)*xi(3,4)*xi(3,3) *xi(1,2) + xi(4,1)*xi(3,4)*xi(1,2)*xi(1,1) - xi(3,3)**2*xi(3,1)*xi(1,2) - xi(3,1 )*xi(1,2))/(xi(3,4)*(xi(1,1)**2 + 1))USD\\$ \\USD J2(4,3):=( - b(4,3)**2*xi(3,4)**2 - 2*b(4,3)*xi(3,4)*xi(3,3) - xi(3,3)**2 - 1)/xi(3,4)USD\\$ \\USD J2(4,4):= - (b(4,3)*xi(3,4) + xi(3,3))USD\\$ \\USD J2(4,5):=0USD\\$ \\USD J2(4,6):=0USD\\$ \\USD J2(5,1):=(b(6,1)*xi(1,2)**2*xi(1,1)**2 + b(6,1)*xi(1,2)**2 - b(5,3)*b(4,1) *xi(3,4)*xi(1,2)*xi(1,1)**2 - b(5,3)*b(4,1)*xi(3,4)*xi(1,2) - b(5,3)*xi(3,1)*xi( 1,2)*xi(1,1)**2 - b(5,3)*xi(3,1)*xi(1,2) + b(5,2)*xi(1,1)**4 + 2*b(5,2)*xi(1,1) **2 + b(5,2) - b(4,1)*xi(6,4)*xi(3,3)*xi(1,2)**2 - b(4,1)*xi(6,4)*xi(1,2)**2*xi( 1,1) + b(4,1)*xi(6,3)*xi(3,4)*xi(1,2)**2 + xi(6,4)*xi(4,1)*xi(1,2)**2 + xi(6,3)* xi(3,1)*xi(1,2)**2 - xi(6,2)*xi(1,2)*xi(1,1)**2 - xi(6,2)*xi(1,2))/(xi(1,2)*(xi( 1,1)**2 + 1))USD\\$ \\USD J2(5,2):=(b(6,2)*xi(3,4)*xi(1,2)*xi(1,1)**4 + 2*b(6,2)*xi(3,4)*xi(1,2)*xi( 1,1)**2 + b(6,2)*xi(3,4)*xi(1,2) - b(5,3)*b(4,2)*xi(3,4)**2*xi(1,1)**4 - 2*b(5,3 )*b(4,2)*xi(3,4)**2*xi(1,1)**2 - b(5,3)*b(4,2)*xi(3,4)**2 - b(5,3)*xi(4,1)*xi(3, 4)**2*xi(1,2)*xi(1,1)**2 - b(5,3)*xi(4,1)*xi(3,4)**2*xi(1,2) - b(5,3)*xi(3,4)*xi (3,3)*xi(3,1)*xi(1,2)*xi(1,1)**2 - b(5,3)*xi(3,4)*xi(3,3)*xi(3,1)*xi(1,2) - b(5, 3)*xi(3,4)*xi(3,1)*xi(1,2)*xi(1,1)**3 - b(5,3)*xi(3,4)*xi(3,1)*xi(1,2)*xi(1,1) + 2*b(5,2)*xi(3,4)*xi(1,1)**5 + 4*b(5,2)*xi(3,4)*xi(1,1)**3 + 2*b(5,2)*xi(3,4)*xi (1,1) - b(5,1)*xi(3,4)*xi(1,2)*xi(1,1)**4 - 2*b(5,1)*xi(3,4)*xi(1,2)*xi(1,1)**2 - b(5,1)*xi(3,4)*xi(1,2) - b(4,2)*xi(6,4)*xi(3,4)*xi(3,3)*xi(1,2)*xi(1,1)**2 - b (4,2)*xi(6,4)*xi(3,4)*xi(3,3)*xi(1,2) - b(4,2)*xi(6,4)*xi(3,4)*xi(1,2)*xi(1,1)** 3 - b(4,2)*xi(6,4)*xi(3,4)*xi(1,2)*xi(1,1) + b(4,2)*xi(6,3)*xi(3,4)**2*xi(1,2)* xi(1,1)**2 + b(4,2)*xi(6,3)*xi(3,4)**2*xi(1,2) - xi(6,4)*xi(4,1)*xi(3,4)*xi(3,3) *xi(1,2)**2 + xi(6,4)*xi(4,1)*xi(3,4)*xi(1,2)**2*xi(1,1) - xi(6,4)*xi(3,3)**2*xi (3,1)*xi(1,2)**2 - xi(6,4)*xi(3,1)*xi(1,2)**2 + xi(6,3)*xi(4,1)*xi(3,4)**2*xi(1, 2)**2 + xi(6,3)*xi(3,4)*xi(3,3)*xi(3,1)*xi(1,2)**2 + xi(6,3)*xi(3,4)*xi(3,1)*xi( 1,2)**2*xi(1,1) - 2*xi(6,2)*xi(3,4)*xi(1,2)*xi(1,1)**3 - 2*xi(6,2)*xi(3,4)*xi(1, 2)*xi(1,1) + xi(6,1)*xi(3,4)*xi(1,2)**2*xi(1,1)**2 + xi(6,1)*xi(3,4)*xi(1,2)**2) /(xi(3,4)*(xi(1,1)**4 + 2*xi(1,1)**2 + 1))USD\\$ \\USD J2(5,3):=(b(6,3)*xi(3,4)*xi(1,2)*xi(1,1)**2 + b(6,3)*xi(3,4)*xi(1,2) - b(5 ,3)*b(4,3)*xi(3,4)**2*xi(1,1)**2 - b(5,3)*b(4,3)*xi(3,4)**2 - b(5,3)*xi(3,4)*xi( 3,3)*xi(1,1)**2 - b(5,3)*xi(3,4)*xi(3,3) + b(5,3)*xi(3,4)*xi(1,1)**3 + b(5,3)*xi (3,4)*xi(1,1) - b(4,3)*xi(6,4)*xi(3,4)*xi(3,3)*xi(1,2) - b(4,3)*xi(6,4)*xi(3,4)* xi(1,2)*xi(1,1) + b(4,3)*xi(6,3)*xi(3,4)**2*xi(1,2) - xi(6,4)*xi(3,3)**2*xi(1,2) - xi(6,4)*xi(1,2) + xi(6,3)*xi(3,4)*xi(3,3)*xi(1,2) - xi(6,3)*xi(3,4)*xi(1,2)* xi(1,1))/(xi(3,4)*(xi(1,1)**2 + 1))USD\\$ \\USD J2(5,4):=( - b(5,3)*xi(3,4)*xi(1,1)**2 - b(5,3)*xi(3,4) - xi(6,4)*xi(3,3)* xi(1,2) - xi(6,4)*xi(1,2)*xi(1,1) + xi(6,3)*xi(3,4)*xi(1,2))/(xi(1,1)**2 + 1) USD\\$ \\USD J2(5,5):=xi(1,1)USD\\$ \\USD J2(5,6):=xi(1,2)USD\\$ \\USD J2(6,1):=( - b(6,3)*b(4,1)*xi(3,4)*xi(1,2) - b(6,3)*xi(3,1)*xi(1,2) + b(6, 2)*xi(1,1)**2 + b(6,2) - 2*b(6,1)*xi(1,2)*xi(1,1) - b(5,1)*xi(1,1)**2 - b(5,1) + b(4,1)*xi(6,4)*xi(1,2) + xi(6,1)*xi(1,2))/xi(1,2)USD\\$ \\USD J2(6,2):=( - b(6,3)*b(4,2)*xi(3,4)*xi(1,2)*xi(1,1)**2 - b(6,3)*b(4,2)*xi(3 ,4)*xi(1,2) - b(6,3)*xi(4,1)*xi(3,4)*xi(1,2)**2 - b(6,3)*xi(3,3)*xi(3,1)*xi(1,2) **2 - b(6,3)*xi(3,1)*xi(1,2)**2*xi(1,1) - b(6,1)*xi(1,2)**2*xi(1,1)**2 - b(6,1)* xi(1,2)**2 - b(5,2)*xi(1,1)**4 - 2*b(5,2)*xi(1,1)**2 - b(5,2) + b(4,2)*xi(6,4)* xi(1,2)*xi(1,1)**2 + b(4,2)*xi(6,4)*xi(1,2) + xi(6,2)*xi(1,2)*xi(1,1)**2 + xi(6, 2)*xi(1,2))/(xi(1,2)*(xi(1,1)**2 + 1))USD\\$ \\USD J2(6,3):=( - b(6,3)*b(4,3)*xi(3,4)*xi(1,2) - b(6,3)*xi(3,3)*xi(1,2) - b(6, 3)*xi(1,2)*xi(1,1) - b(5,3)*xi(1,1)**2 - b(5,3) + b(4,3)*xi(6,4)*xi(1,2) + xi(6, 3)*xi(1,2))/xi(1,2)USD\\$ \\USD J2(6,4):= - b(6,3)*xi(3,4) + xi(6,4)USD\\$ \\USD J2(6,5):= - (xi(1,1)**2 + 1)/xi(1,2)USD\\$ \\USD J2(6,6):= - xi(1,1)USD\\$ From these formulas, one has that$ USD xi(1,2),xi(3,4) USD are invariant under the action of USD \Phi USD$ USD J2^1_2 = xi(1,2)J2^3_4 =xi(3,4)USD\\$ Take the following values$ \\ USD b(4,1):=( - xi(3,1))/xi(3,4)USD$ in order to get USD J2^3_1=0 USD and $ \\ USD b(4,3):=( - xi(3,3))/xi(3,4)USD$ in order to get USD J2^3_3=0 USD and $ \\ USD b(6,3):=xi(6,4)/xi(3,4)USD$ in order to get USD J2^6_4=0 USD and $ \\ USD b(4,2):= - (xi(1,2)*(xi(4,1)*xi(3,4) + xi(3,3)*xi(3,1) + xi(3,1)*xi(1,1)) )/(xi(3,4)*(xi(1,1)**2 + 1))USD$ in order to get USD J2^4_1=0 USD and $ \\ USD b(6,2):=(2*b(6,1)*xi(3,4)*xi(1,2)*xi(1,1) + b(5,1)*xi(3,4)*xi(1,1)**2 + b (5,1)*xi(3,4) + xi(6,4)*xi(3,1)*xi(1,2) - xi(6,1)*xi(3,4)*xi(1,2))/(xi(3,4)*(xi( 1,1)**2 + 1))USD$ in order to get USD J2^6_1=0 USD and $ \\ USD b(6,1):=( - b(5,2)*xi(3,4)*xi(1,1)**4 - 2*b(5,2)*xi(3,4)*xi(1,1)**2 - b(5 ,2)*xi(3,4) - xi(6,4)*xi(4,1)*xi(3,4)*xi(1,2)**2 - xi(6,4)*xi(3,3)*xi(3,1)*xi(1, 2)**2 - xi(6,4)*xi(3,1)*xi(1,2)**2*xi(1,1) + xi(6,2)*xi(3,4)*xi(1,2)*xi(1,1)**2 + xi(6,2)*xi(3,4)*xi(1,2))/(xi(3,4)*xi(1,2)**2*(xi(1,1)**2 + 1))USD$ in order to get USD J2^6_2=0 USD and $ \\ USD b(5,3):=(xi(1,2)*( - xi(6,4)*xi(3,3) - xi(6,4)*xi(1,1) + xi(6,3)*xi(3,4)) )/(xi(3,4)*(xi(1,1)**2 + 1))USD$ in order to get USD J2^6_3=0 .USD $ \\ The matrix USD \Phi USD is with these values$ \\$ USD \Phi^1_1=1;USD\\$ USD \Phi^1_2=0;USD\\$ USD \Phi^1_3=0;USD\\$ USD \Phi^1_4=0;USD\\$ USD \Phi^1_5=0;USD\\$ USD \Phi^1_6=0;USD\\$ USD \Phi^2_1=0;USD\\$ USD \Phi^2_2=1;USD\\$ USD \Phi^2_3=0;USD\\$ USD \Phi^2_4=0;USD\\$ USD \Phi^2_5=0;USD\\$ USD \Phi^2_6=0;USD\\$ USD \Phi^3_1=0;USD\\$ USD \Phi^3_2=0;USD\\$ USD \Phi^3_3=1;USD\\$ USD \Phi^3_4=0;USD\\$ USD \Phi^3_5=0;USD\\$ USD \Phi^3_6=0;USD\\$ USD \Phi^4_1=( - xi(3,1))/xi(3,4);USD\\$ USD \Phi^4_2= - (xi(1,2)*(xi(4,1)*xi(3,4) + xi(3,3)*xi(3,1) + xi(3,1)*xi(1,1))) /(xi(3,4)*(xi(1,1)**2 + 1));USD\\$ USD \Phi^4_3=( - xi(3,3))/xi(3,4);USD\\$ USD \Phi^4_4=1;USD\\$ USD \Phi^4_5=0;USD\\$ USD \Phi^4_6=0;USD\\$ USD \Phi^5_1=b(5,1);USD\\$ USD \Phi^5_2=b(5,2);USD\\$ USD \Phi^5_3=(xi(1,2)*( - xi(6,4)*xi(3,3) - xi(6,4)*xi(1,1) + xi(6,3)*xi(3,4))) /(xi(3,4)*(xi(1,1)**2 + 1));USD\\$ USD \Phi^5_4=0;USD\\$ USD \Phi^5_5=1;USD\\$ USD \Phi^5_6=0;USD\\$ USD \Phi^6_1=( - b(5,2)*xi(3,4)*xi(1,1)**4 - 2*b(5,2)*xi(3,4)*xi(1,1)**2 - b(5, 2)*xi(3,4) - xi(6,4)*xi(4,1)*xi(3,4)*xi(1,2)**2 - xi(6,4)*xi(3,3)*xi(3,1)*xi(1,2 )**2 - xi(6,4)*xi(3,1)*xi(1,2)**2*xi(1,1) + xi(6,2)*xi(3,4)*xi(1,2)*xi(1,1)**2 + xi(6,2)*xi(3,4)*xi(1,2))/(xi(3,4)*xi(1,2)**2*(xi(1,1)**2 + 1));USD\\$ USD \Phi^6_2=( - 2*b(5,2)*xi(3,4)*xi(1,1)**5 - 4*b(5,2)*xi(3,4)*xi(1,1)**3 - 2* b(5,2)*xi(3,4)*xi(1,1) + b(5,1)*xi(3,4)*xi(1,2)*xi(1,1)**4 + 2*b(5,1)*xi(3,4)*xi (1,2)*xi(1,1)**2 + b(5,1)*xi(3,4)*xi(1,2) - 2*xi(6,4)*xi(4,1)*xi(3,4)*xi(1,2)**2 *xi(1,1) - 2*xi(6,4)*xi(3,3)*xi(3,1)*xi(1,2)**2*xi(1,1) - xi(6,4)*xi(3,1)*xi(1,2 )**2*xi(1,1)**2 + xi(6,4)*xi(3,1)*xi(1,2)**2 + 2*xi(6,2)*xi(3,4)*xi(1,2)*xi(1,1) **3 + 2*xi(6,2)*xi(3,4)*xi(1,2)*xi(1,1) - xi(6,1)*xi(3,4)*xi(1,2)**2*xi(1,1)**2 - xi(6,1)*xi(3,4)*xi(1,2)**2)/(xi(3,4)*xi(1,2)*(xi(1,1)**4 + 2*xi(1,1)**2 + 1)) ;USD\\$ USD \Phi^6_3=xi(6,4)/xi(3,4);USD\\$ USD \Phi^6_4=0;USD\\$ USD \Phi^6_5=0;USD\\$ USD \Phi^6_6=1;USD\\$ Then one gets for USDJ2USD \\$ USD J2^1_1=xi(1,1);USD\\$ USD J2^1_2=xi(1,2);USD\\$ USD J2^1_3=0;USD\\$ USD J2^1_4=0;USD\\$ USD J2^1_5=0;USD\\$ USD J2^1_6=0;USD\\$ USD J2^2_1= - (xi(1,1)**2 + 1)/xi(1,2);USD\\$ USD J2^2_2= - xi(1,1);USD\\$ USD J2^2_3=0;USD\\$ USD J2^2_4=0;USD\\$ USD J2^2_5=0;USD\\$ USD J2^2_6=0;USD\\$ USD J2^3_1=0;USD\\$ USD J2^3_2=0;USD\\$ USD J2^3_3=0;USD\\$ USD J2^3_4=xi(3,4);USD\\$ USD J2^3_5=0;USD\\$ USD J2^3_6=0;USD\\$ USD J2^4_1=0;USD\\$ USD J2^4_2=0;USD\\$ USD J2^4_3=( - 1)/xi(3,4);USD\\$ USD J2^4_4=0;USD\\$ USD J2^4_5=0;USD\\$ USD J2^4_6=0;USD\\$ USD J2^5_1=0;USD\\$ USD J2^5_2=0;USD\\$ USD J2^5_3=0;USD\\$ USD J2^5_4=0;USD\\$ USD J2^5_5=xi(1,1);USD\\$ USD J2^5_6=xi(1,2);USD\\$ USD J2^6_1=0;USD\\$ USD J2^6_2=0;USD\\$ USD J2^6_3=0;USD\\$ USD J2^6_4=0;USD\\$ USD J2^6_5= - (xi(1,1)**2 + 1)/xi(1,2);USD\\$ USD J2^6_6= - xi(1,1);USD\\$ \\$ USDUSD J2 = \begin{pmatrix}$ xi(1,1)&$ xi(1,2)&$ 0&$ 0&$ 0&$ 0\\$ - (xi(1,1)**2 + 1)/xi(1,2)&$ - xi(1,1)&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ xi(3,4)&$ 0&$ 0\\$ 0&$ 0&$ ( - 1)/xi(3,4)&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(1,1)&$ xi(1,2)\\$ 0&$ 0&$ 0&$ 0&$ - (xi(1,1)**2 + 1)/xi(1,2)&$ - xi(1,1)\end{pmatrix}USDUSD$ Hence we are in the case where$ USD xi(3,1)=xi(3,3)=xi(4,1)=xi(6,1)=xi(6,2)=xi(6,3)=xi(6,4)= 0 USD: \\$ \\$ USDUSD J = \begin{pmatrix}$ xi(1,1)&$ xi(1,2)&$ 0&$ 0&$ 0&$ 0\\$ - (xi(1,1)**2 + 1)/xi(1,2)&$ - xi(1,1)&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ xi(3,4)&$ 0&$ 0\\$ 0&$ 0&$ ( - 1)/xi(3,4)&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(1,1)&$ xi(1,2)\\$ 0&$ 0&$ 0&$ 0&$ - (xi(1,1)**2 + 1)/xi(1,2)&$ - xi(1,1)\end{pmatrix}USDUSD$ \par Now, to get USD J2(3,4) = 1, USDwe'll use equivalence by the automorphism$ \\$ USDUSD \Psi_1 = \begin{pmatrix}$ 1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ xi(3,4)&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(3,4)&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ xi(3,4)\end{pmatrix}USDUSD$ Then USD J2 = \Psi_1^{-1}J\Psi_1 USD is the matrix$ \\$ USDUSD J2 = \begin{pmatrix}$ xi(1,1)&$ xi(1,2)&$ 0&$ 0&$ 0&$ 0\\$ - (xi(1,1)**2 + 1)/xi(1,2)&$ - xi(1,1)&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(1,1)&$ xi(1,2)\\$ 0&$ 0&$ 0&$ 0&$ - (xi(1,1)**2 + 1)/xi(1,2)&$ - xi(1,1)\end{pmatrix}USDUSD$ \par Then, to get USD J2(1,2) = 1, USD we'll use equivalence by the automorphism $ \\$ USDUSD \Psi_2 = \begin{pmatrix}$ xi(1,2)&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ xi(1,2)&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ xi(1,2)&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(1,2)**2&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ xi(1,2)\end{pmatrix}USDUSD$ Then USD NJ2 = \Psi_2^{-1} J2 \Psi_2 USD is the matrix$ \\$ USDUSD NJ2 = \begin{pmatrix}$ xi(1,1)&$ 1&$ 0&$ 0&$ 0&$ 0\\$ - (xi(1,1)**2 + 1)&$ - xi(1,1)&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(1,1)&$ 1\\$ 0&$ 0&$ 0&$ 0&$ - (xi(1,1)**2 + 1)&$ - xi(1,1)\end{pmatrix}USDUSD$ \par Finally, to get USD J2(1,1) = 0, USD$ we'll use equivalence by the automorphism$ \\$ USDUSD \Psi_3 = \begin{pmatrix}$ 1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ - xi(1,1)&$ 1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 1&$ 0\\$ 0&$ 0&$ 0&$ 0&$ - xi(1,1)&$ 1\end{pmatrix}USDUSD$ Then USD NNJ2 = \Psi_3^{-1} NJ2 \Psi_3 USD is the matrix$ \\$ USDUSD NNJ2 = \begin{pmatrix}$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ -1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ 1\\$ 0&$ 0&$ 0&$ 0&$ -1&$ 0\end{pmatrix}USDUSD$ \par Introduce the automorphism USD \Psi = \Psi_1 \Psi_2 \Psi_3 :USD$ \\$ USDUSD \Psi = \begin{pmatrix}$ xi(1,2)&$ 0&$ 0&$ 0&$ 0&$ 0\\$ - xi(1,1)&$ 1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ xi(3,4)*xi(1,2)&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ xi(1,2)&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ xi(3,4)*xi(1,2)**2&$ 0\\$ 0&$ 0&$ 0&$ 0&$ - xi(3,4)*xi(1,2)*xi(1,1)&$ xi(3,4)*xi(1,2)\end{pmatrix}USDUSD$ \\$ USD det \Psi:=xi(3,4)**3*xi(1,2)**6USD$ Then USD J2 = \Psi^{-1}J\Psi USD is the matrix$ \\$ USDUSD J2 = \begin{pmatrix}$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ -1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ 1\\$ 0&$ 0&$ 0&$ 0&$ -1&$ 0\end{pmatrix}USDUSD$ We've thus got into the case of a complex structure USD J USD where$ USD xi(3,1)=xi(3,3)=xi(4,1)=xi(6,1)=xi(6,2)=xi(6,3)=xi(6,4)= 0 USD: \\$ xi(1,1) = 0, USD and USD xi(1,2) = xi(3,4) = 1 .USD $ \\$ USDUSD J = \begin{pmatrix}$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ -1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ 1\\$ 0&$ 0&$ 0&$ 0&$ -1&$ 0\end{pmatrix}USDUSD$ Hence, we have that$ any integrable complex structure with USD xi(1,6) = 0 , xi(2,5) = 0 USD$ is equivalent to the structure defined by $ USDUSD J = \begin{pmatrix}$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ -1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 1&$ 0&$ 0\\$ 0&$ 0&$ -1&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ 1\\$ 0&$ 0&$ 0&$ 0&$ -1&$ 0\end{pmatrix}USDUSD$ Recall the matrix USDJ_1 USD$ USDUSD J_1 = \begin{pmatrix}$ 0&$ 0&$ 0&$ 0&$ 0&$ 1\\$ 0&$ 0&$ 1&$ 0&$ 0&$ 0\\$ 0&$ -1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 1&$ 0\\$ 0&$ 0&$ 0&$ -1&$ 0&$ 0\\$ -1&$ 0&$ 0&$ 0&$ 0&$ 0\end{pmatrix}USDUSD$ Observe now that USD J USD is equivalent to USD J_1 USD by the automorphism$ USDUSD M = \begin{pmatrix}$ 0&$ 1&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 1&$ 0&$ 0&$ 0\\$ 1&$ 0&$ 0&$ 0&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ 0&$ 1\\$ 0&$ 0&$ 0&$ -1&$ 0&$ 0\\$ 0&$ 0&$ 0&$ 0&$ -1&$ 0\end{pmatrix}USDUSD$ \textit{i.e. } USD M^{-1}J M = J_1 USD, since$ %USD M J_1 - J M = mat((0,0,0,0,0,0),(0,0,0,0,0,0),(0,0,0,0,0,0),(0,0,0,0,0,0),(0,0,0,0,0,0),(0,0,0 ,0,0,0))$ USD M J_1 - J M =0.USD$ \par Hence, any integrable complex structure with $ USD xi(1,6) =0, xi(2,5) \neq 0 USD$ is equivalent to the structure defined by USD J_1 USD$ \\ check of torsion$ \\Torsion equations to cancel (Latex output) : \\USD$ USD$ Zero torsion$ \par Commutation relations of USD \mathfrak{m} : USD$ USD[\tilde{x}_1,\tilde{x}_3]=tildex_5USD;$ USD[\tilde{x}_1,\tilde{x}_4]=tildex_6USD;$ USD[\tilde{x}_2,\tilde{x}_3]=tildex_6USD;$ USD[\tilde{x}_2,\tilde{x}_4]= - tildex_5USD;$ \P$ \par Now we check if the condition USD [Jx,y]= J[x,y] USD is satisfied$ USD\forall x,y \in {\mathcal{G}}_{6,3},USD$ \textit{i.e.} if USD{\mathcal{G}}_{6,3}USD$ is a \textit{complex} algebra.$ \\USD J[x_j,x_k] \neq [Jx_j,x_k] USD in the following cases{{{1,1},x(4)}, {{1,2}, - x(3)}, {{2,1},x(3)}, {{2,2},x(4)}, {{3,1}, - x(6)}, {{3,2},x(5)}, {{4,1}, - x(5)}, {{4,2}, - x(6)}}$ \end{document}$