Finding subgroups of E6 (and other Lie groups) using the Dynkin diagram

Question

According to the wikipedia article on $E6$, the group has subgroups isomorphic to $SU(3) \times SU(3) \times SU(3)$, $SU(6) \times SU(2)$, and $SO(10) \times U(1)$, and it is claimed these can be read off from its Dynkin diagram. I can see that if we delete some edges from the Dynkin diagram of $E6$, we can find the Dynkin diagrams of these three groups, however, the presence of an edge indicates the corresponding generators do not commute. So while this argument seems valid for the case of $SO(10) \times U(1)$, in the other cases I do not see why the various non-abelian factors commute. How is one able to see these subgroups using the Dynkin diagram?

Answer 1

Edit: I first talked about subgroups in my answer. One should rather talk about sub-algebras and then switch to the group afterward. To address the question of the group you need to focus on root lattices, see the edit at the end. I keep the notation of the wikipedia article.

I don't think it suffices to "delete an edge". I use here the reference Borel-de Siebenthal's article "Les sous-groupes fermés de rang maximum des groupes de Lie clos", Commentary mathematici Helvetici, Vol 23, 1949, pp 200-222 (Warning it is in French). There are two kinds of maximal reductive sub-algebra of maximal rank we are considering here: those which are semi-simple and those which are not.

The non-semi-simple ones (but maximal reductive of maximal rank anyway) are constructed by simply deleting an edge. Since you seem not to be worried about them, I won't.

To construct the other kind of maximal sub-algebras of maximal rank, i.e. maximal semi-simple sub-algebras of maximal rank, you should follow these steps:

• Choose a Cartan sub-algebra of your simple algebra and an order for the root system. Set $\{\alpha_1,\dots,\alpha_r\}$ be the associated simple roots.

• Find $\alpha_0$ the smallest root in the root system. Then I claim that for any $k\in \{1,\dots,r\}$, $\{\alpha_0,\dots,\alpha_r\}\setminus \{\alpha_k\}$ generates a system of roots.

• Each maximal semi-simple sub-algebra of maximal rank has a root system constructed as above. Actually, any system of roots associated to a semi-simple sub-algebra of maximal rank (not necessarily maximal) can be constructed applying a finite number of times this construction to the original root system.

This is how you construct certain classes of maximal sub-algebras of simple Lie algebras (warning: not all of the system constructed like this are associated to proper and maximal semi-simple sub-algebras). Now you can read off the construction right from the Dynkin diagram.

• First draw the Dynkin diagram of your simple Lie algebra.

• Draw the $\alpha_0$ on the Dynkin diagram.

• Delete from the Dynkin diagram one of the $\alpha_k$ for $1\leq k\leq r$ and read the Dynkin diagram of a semi-simple sub-algebra of maximal rank (up to some exception it is maximal).

For example with $E_6$:

• I use the numeration given in the Wikipedia article that you have highlighted. $$\begin{array}{ccccccccc}1&-&2&-&3&-&4&-&5\\&&&&|&&&&\\&&&&6&&&&\end{array}$$

• If my calculations are correct (well it can also directly be read on page 219 in the reference I have given), the smallest root is given by:

$$-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4-\alpha_5-2\alpha_6 $$ and, once again if my calculations are correct, the root on the Dynkin diagram goes like this: $$\begin{array}{ccccccccc}1&-&2&-&3&-&4&-&5\\&&&&|&&&&\\&&&&6&&&&\\&&&&|&&&&\\&&&&0&&&&\end{array}$$

• For instance, $SU(3)\times SU(3)\times SU(3)$ is obtained by taking out the root $3$, because when you delete $3$ and the root associated to it, you are left with: $$\begin{array}{ccccccccc}1&-&2& & & &4&-&5\\&&&& &&&&\\&&&&6&&&&\\&&&&|&&&&\\&&&&0&&&&\end{array}$$

• If you take out $6$, you are left with: $$\begin{array}{ccccccccc}1&-&2&-&3&-&4&-&5\\&&&&&&&&\\&&&&&&&&\\&&&&&&&&\\&&&&0&&&&\end{array}$$ and this is $SU(6)\times SU(2)$ and you get the same thing by taking out $2$ or $4$.

• If you take out $1$ or $5$, you recover the root system of $E_6$ (so you don't have a proper semi-simple sub-algebra).

Edit: (following Jason De Vito's comment) For the record, with what I have done here, we only identify the isogeneous class of the semi-simple maximal subgroup. To identify the group, you only need to compute its center. Like I have said in the comment, to be complete, one should understand the quotient group :$$\Lambda_W/\Lambda_k $$ where $\Lambda_W$ is the lattice of fundamental weights of $E_6$ and $\Lambda_k$ is the lattice generated by $\{\alpha_0,\dots,\alpha_r\}\setminus \{\alpha_k\}$. Now I make this computation for the $SU(3)\times SU(3)\times SU(3)$ case, i.e. $k=3$. I denote $\Lambda_R$ the root lattice of $E_6$ (i.e. the lattice generated by $\alpha_1,\dots,\alpha_6$).

• First we know that (e.g. by computing the inverse of the Dynkin matrix) $\Lambda_W/\Lambda_R$ is of order $3$.

• As a result it suffices to compute $\Lambda_R/\Lambda_3$. It is clearly generated by the image of $\alpha_3$. Since $3\alpha_3$ is clearly in $\Lambda_3$, we see that $\Lambda_R/\Lambda_3$ is either of order $3$ or $1$. If it were of order $1$ then $\Lambda_R=\Lambda_3$ but this is impossible (we already know that they are not the same root system). Thus we have $[\Lambda_W:\Lambda_3]$ is of order $9$.

• In particular the center of the semi-simple subgroup of $E_6$ whose Lie algebra is the one of $SU(3)\times SU(3)\times SU(3)$, is not of order $27$ and hence not isomorphic to the group $SU(3)\times SU(3)\times SU(3)$.