The question:
Let $S = \left(\begin{array}{cc} 0 & I_3 \\ I_3 & 0 \end{array}\right)$ and let $$L = \mathfrak{so}_6(\mathbb{C})= \{x \in \operatorname{End}(\mathbb{C}^6)\ |\ ^{\mathrm t}xS + Sx = 0 \}$$
1) Find a basis for $L$ and prove that $\dim L =15$ (Hint: write $x = \left(\begin{array}{cc} A & B \\ C & D \end{array}\right)$ in block form with $A,B,C,D$ blocks of order $3 \times 3$)
2) Show that every weight space $L_{\lambda}$ corresponding to a non-zero weigh $\lambda$ is 1-dimensional.
What is meant by $^{\mathrm t}xS + Sx = 0$ (what does the t stand for?) I will better understand what to do. What I know is that the $\mathfrak{so}(6)$ is rotations in 6-dimensional space about a point and the 6-D matrix has a determinant of 1 (don't know how that helps with the question though). Now for
1) I did a computation: $$Y= \left(\begin{array}{cc} A & B \\ C & D \end{array}\right)\left(\begin{array}{cc} 0 & I_3 \\ I_3 & 0 \end{array}\right) =\left(\begin{array}{cc} BI_3, AI_3 \\ DI_3, CI_3 \end{array}\right)Z= \left(\begin{array}{cc} 0 & I_3 \\ I_3 & 0 \end{array}\right)\left(\begin{array}{cc} A & B \\ C & D \end{array}\right) =\left(\begin{array}{cc} I_3C, I_3D \\ I_3A, I_3B \end{array}\right)$$ $$Y-Z = \left(\begin{array}{cc} BI_3-I_3C, AI_3 -I_3D\\ DI_3-I_3A, CI_3 - I_3B \end{array}\right)=0$$ Now if 0 is the 0-matrix we get equalities for example $BI_3=I_3C$ and $I_3B=CI_3$ thus $I_3 =C^{-1}I_3B = B^{-1}I_3C$ and so we have that $I_3^{2}=C^{-1}I_3BB^{-1}I_3C = C^{-1}I_{3}^{2}C$. Thus $$I_{3}C = CI_{3}$$ Now if $I_3$ is supposed to denote the 3 × 3 identity matrix we have $A=D$ and $B=C$. But now what? And furthermore how do they get a dimension of 15? I am guessing I might be way off track on this one.. $^{\mathrm t}xS + Sx = 0$ however starts to look similar to $[S,x]=0$ and I am wondering if the algebra I did above can be used to show that we are dealing with an equation of this form? - which leads us to part 2-
2) Now by definition, weights are elements of the dual space $H^*$ for $H$ an Abelian sub-algebra of $L$. For each $\alpha \in H^*$, let $$L_{\alpha} := \{x \in L := \{x \in L:[h,x] = \alpha(h)x\ \ \forall h \in H\}$$ One of these weight spaces is the zero weight spaces $$L_0 = \{z \in L:[h,z]=0\ \ \forall h \in H\}$$ Which of course starts looking pretty similar to $^{\mathrm t}xS + Sx = 0$ as noted above. Furthermore We can write the decomposition of $L$ into weight spaces for $H$ as $L = L_0 \oplus \bigoplus_{\alpha \in \Phi} L_{\alpha}$ so if we are indeed dealing with an abelian lie algebra (if $[S,x]=0$) then would we just have the disappearance of the terms $ \bigoplus_{\alpha \in \Phi} L_{\alpha}$.
I know I wrote a lot but am really trying to understand what is going on here.
Thanks for any insight.
Brian