# Root system for $\mathfrak{so}(6)$ algebra [Solved]

Let us consider, for example, $$\mathfrak{so}(6)$$, with fifteen generators defined as (note: it sometimes has an $$i$$ in front)

$$\left(M_{ij}\right)_{ab}=-\left(\delta_{ia}\delta_{jb}-\delta_{ib}\delta_{ja}\right),\qquad i,j,a,b=1,\dots,6\,,\tag{1}$$

whose algebra is

$$\left[M_{ij},M_{k\ell}\right]=-\left(\delta_{i\ell}M_{jk}+\delta_{jk}M_{i\ell}-\delta_{ik}M_{j\ell}-\delta_{j\ell}M_{ik}\right).\tag{2}$$

It is usual (actually, I dont know if this is the unique option, that's why I said usual) to consider $$M_{12}$$, $$M_{34}$$ and $$M_{56}$$ ($$H_i=M_{(2i-1)(2i)}$$, in general) as the Cartan generators. The positive simple roots in this case are found to be (page 84):

$$E_{\alpha_1}=M_{35}-i M_{36}+i M_{45}+M_{46},\\ E_{\alpha_2}=M_{13}-i M_{14}+i M_{23}+M_{24},\\ E_{\alpha_3}=M_{13}-i M_{14}-i M_{23}-M_{24}, \tag{3}$$

My question is if it is possible to have another set of Cartan generators, e.g. $$M_{16}$$, $$M_{25}$$ and $$M_{34}$$ or a combination of them. How could I get this second set of "Cartans" from the first one? Does the transformation also apply for the roots?

My second question is about finding the roots in terms of the generators $$M_{ij}$$ for the case of $$\mathfrak{so}(6)$$ whose generators are written in such a way they show the nine generators of $$\mathfrak{u}(3)$$. The $$\mathfrak{u}(3)$$ subalgebra of $$\mathfrak{so}(6)$$ is composed by those generators of $$\mathfrak{so}(6)$$ that are invariant under $$J=\begin{pmatrix} 0 & I_3 \\ -I_3 & 0\end{pmatrix}$$. Thus, those of $$\mathfrak{so}(6)/\mathfrak{u}(3)$$ are

$$K_1=\frac{1}{2}\left(M_{12}-M_{45}\right),\quad K_2=\frac{1}{2}\left(M_{15}-M_{24}\right),\quad K_3=\frac{1}{2}\left(M_{13}-M_{46}\right),\\ K_4=\frac{1}{2}\left(M_{16}-M_{34}\right),\quad K_5=\frac{1}{2}\left(M_{26}-M_{35}\right),\quad K_6=\frac{1}{2}\left(M_{23}-M_{56}\right), \tag{4}$$

and those of $$\mathfrak{u}(3)$$,

$$T_1=\frac{1}{2}\left(M_{12}+M_{45}\right),\quad T_2=\frac{1}{2}\left(M_{15}+M_{24}\right),\quad T_3=\frac{1}{2}\left(M_{13}+M_{46}\right),\\ T_4=\frac{1}{2}\left(M_{16}+M_{34}\right),\quad T_5=\frac{1}{2}\left(M_{26}+M_{35}\right),\quad T_6=\frac{1}{2}\left(M_{23}+M_{56}\right),\\ T_7=M_{14},\quad T_8=M_{25},\quad T_9=M_{36}. \tag{5}$$

Notice that the first Cartans, $$M_{12}$$, $$M_{34}$$ and $$M_{56}$$, and the second ones, $$M_{16}$$, $$M_{25}$$ and $$M_{34}$$, appear mixed in this representation. In particular, for my purposes, I need the latter set as being the Cartans or included in them.

Summarizing: what is the operation to "rotate" the Cartan generators and roots to another set of them? And, in general, given a set of generators, how do I get the new Cartan generators and roots?

I know that for $$\mathfrak{so}(6)$$ the Cartan matrix and subalgebra are

$$A_{ij}=\begin{pmatrix} 2 & -1 & -1\\ -1 & 2 & 0 \\ -1 & 0 & 2 \end{pmatrix},\quad \left[H_i,E_j\right]=A_{ij}E_j,\quad \left[E_i,E_{-j}\right]=\delta_{ij}H_j \tag{6}$$

but I dont know how to implement an efficient way to "extract" the roots, now, in terms of $$K_i$$ and $$T_i$$.

• Also, you are aware that $\mathfrak{so}(6)$ is a simple Lie algebra, hence that any $\mathfrak{u}_3$ you find in it is a subalgebra but not an ideal (and almost certainly not unique), hence the expression "$\mathfrak{so}(6)/\mathfrak{u}(3)$" does not even define a Lie algebra, and that $\oplus$ decomposition in your title and question body cannot be a direct sum of Lie algebras? Oct 6, 2020 at 16:22
• I feel like most of my questions and comments have not been addressed, in any case I do not quite understand what you want. But you know that e.g. for any $A \in SO(6)$ (the group), conjugating $M \mapsto A^{-1}MA$ is an automorphism of the Lie algebra, hence I assume e.g. $A^{-1}M_{12}A, A^{-1}M_{34}A, A^{-1}M_{56}A$ will be a new set of "Cartans" (whatever you mean by that, please clarify that still). Oct 7, 2020 at 21:38
• I guess that the $H_i$ are supposed to be parts of a Cartan-Weyl or Chevalley basis (en.wikipedia.org/wiki/Chevalley_basis), i.e. in particular $\alpha_i(H_i)=2$ for the three simple roots $\alpha_i$. If one demands enough normalisation, such are unique at least in one given Cartan subalgebra; but as said before, there are tons of Cartan subalgebras in this Lie algebra (actually, any three linearly independent elements span one). Oct 8, 2020 at 17:17
• Now, if $\mathfrak h$ is a Cartan subalgebra of a Lie algebra $\mathfrak g$, and $\alpha$ is an automorphism of $\mathfrak g$, then $\alpha(\mathfrak h)$ is another CSA. (Over $\mathbb C$, the converse holds i.e. all CSA's are conjugate to each other via automorphisms; I don't know what exactly could be said in this case.) Now I described standard automorphisms of $\mathfrak{so}(6)$ in my comment above. Just take some nice elements of the group $SO(6)$ and conjugate your $H_i$ with them, then you get some nice new choices. Oct 8, 2020 at 17:26
• @TorstenSchoeneberg Juste made that. It solved my problem. Indeed, there were lots of choices. I picked the one I needed. Thanks Oct 8, 2020 at 17:35