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## Main Question or Discussion Point

Hello,

Is the law for matrix multiplication in SL(2,C) the same as usual ? I try to solve the equation [tex]A_{k}k_{0}A_{k}^{\dagger}=k[/tex] where [tex]k_0[/tex] corresponds to the unit vector [tex]\{0,0,1\}[/tex] and [tex]k[/tex] is an arbitrary vector, i.e.:

[tex]k0=

\left( \begin{array}{cc}

2 & 0 \\

0 & 0 \\

\end{array} \right)

[/tex]

[tex]k=

\left( \begin{array}{cc}

1+n_3 & n_- \\

n_+ & 1-n_3 \\

\end{array} \right)

[/tex]

If I try to solve for

[tex]A_k=

\left( \begin{array}{cc}

a & b \\

c & d \\

\end{array} \right)

[/tex]

this gives (where [tex]a*[/tex] is the conjugate of [tex]a[/tex]):

[tex]A_{k}k_{0}A_{k}^{\dagger}=

\left( \begin{array}{cc}

2aa* & 2ac* \\

2ca* & 2cc* \\

\end{array} \right)

[/tex]

So this gives conditions on [tex]\{a,c\}[/tex] but can [tex]\{b,c\}[/tex] be arbitrary ? How do I solve this equation and obtain the expression of [tex]A_k[/tex] involving only [tex]n_+, n_-[/tex] and [tex]n_3[/tex] ?

Thanks a lot for your help!

Is the law for matrix multiplication in SL(2,C) the same as usual ? I try to solve the equation [tex]A_{k}k_{0}A_{k}^{\dagger}=k[/tex] where [tex]k_0[/tex] corresponds to the unit vector [tex]\{0,0,1\}[/tex] and [tex]k[/tex] is an arbitrary vector, i.e.:

[tex]k0=

\left( \begin{array}{cc}

2 & 0 \\

0 & 0 \\

\end{array} \right)

[/tex]

[tex]k=

\left( \begin{array}{cc}

1+n_3 & n_- \\

n_+ & 1-n_3 \\

\end{array} \right)

[/tex]

If I try to solve for

[tex]A_k=

\left( \begin{array}{cc}

a & b \\

c & d \\

\end{array} \right)

[/tex]

this gives (where [tex]a*[/tex] is the conjugate of [tex]a[/tex]):

[tex]A_{k}k_{0}A_{k}^{\dagger}=

\left( \begin{array}{cc}

2aa* & 2ac* \\

2ca* & 2cc* \\

\end{array} \right)

[/tex]

So this gives conditions on [tex]\{a,c\}[/tex] but can [tex]\{b,c\}[/tex] be arbitrary ? How do I solve this equation and obtain the expression of [tex]A_k[/tex] involving only [tex]n_+, n_-[/tex] and [tex]n_3[/tex] ?

Thanks a lot for your help!