Viewing exceptional Lie algebras via the classical ones

Question

I've been trying to understand the exceptional Lie algebras through the classical ones that I am more familiar with. In particular I wanted to get a handle on the root spaces and most discussions that I've read focus on the compact case (e.g. Baez's approach via the octonions and the magic square constructions) while I am more interested in the complex/split story.

To make it more clear what I mean, a choice of Cartan subalgebra in $\mathfrak{sl}(V)$ is equivalent to a choice of decomposition of $V$ into lines $L_1 \oplus \cdots \oplus L_n$ (The Cartan subalgebra being the diagonal guys with respect to this decomposition). Then the root spaces are all the $L_i^* \otimes L_j = \hom(L_i,L_j) \leq \mathfrak{sl}(V)$. We can also do this for $\mathfrak{so}(V)$ and $\mathfrak{sp}(V)$ by choosing decompositions into isotropic lines and identifying them with $\bigwedge^2V$ and $S^2V$.

For $\mathfrak{e}_7$ we can make use of the above by taking a $\mathfrak{sl}_8$ subalgebra. In particular we can choose the Cartan subalgebra to be a Cartan subalgebra of this subalgebra. Let $V$ be 8-dimensional and $\mathfrak{sl}_8 = \mathfrak{sl}(V)$. Then there is a complement to $\mathfrak{sl}_8$ in $\mathfrak{e}_7$ which is $\mathrm{ad}(\mathfrak{sl}_8)$ invariant and it is isomorphic to $\bigwedge^4V$ as a $\mathfrak{sl}_8$ representation. So $\mathfrak{e}_7 = \mathfrak{sl}(V) \oplus \bigwedge^4V$ (not direct as a sum of Lie algebras) and we can see its root system this way. Again the Cartan subalgebra defines a decomposition $V = L_1 \oplus \cdots\oplus L_8$. The root spaces in the $\mathfrak{sl}(V)$ part are just as before and the root spaces in $\bigwedge^4V$ are all the $L_i \wedge L_j \wedge L_k \wedge L_l$.

I know it is possible to do this for $\mathfrak{e}_8$ as well viewing it as a copy of $\mathfrak{so}_{16}$ together with one of the spin representations of $\mathfrak{so}_{16}$. Is there a way to see the other exceptional Lie algebras in this manner? I don't think $\mathfrak{e}_6$ contains a simple subalgebra with the same rank as it but it does have subalgebras like $\mathfrak{sl}_3 \oplus\mathfrak{sl}_3 \oplus\mathfrak{sl}_3 $ and $\mathfrak{sl}_6 \oplus\mathfrak{sl}_2 $.

Are there any good references on this style of approach to the exceptional Lie algebras or a good way to see what the decompositions and root systems would be?

Edit: More recently, I have found some good notes on constructions of this form in J. Adams's "Lectures on exceptional Lie groups"

Answer 1

Élie Cartan himself, recognized and used the following description of $\mathfrak{e}_6$: Let $V$ be a vector space of dimension $6$ and let $W$ be a vector space of dimension $2$. Then there is a vector space splitting $$ \mathfrak{e}_6 = \mathfrak{sl}(V)\oplus\mathfrak{sl}(W)\oplus \bigl(\Lambda^3(V)\otimes W\bigr) $$ and, moreover, $\mathfrak{e}_6$ is the Lie algebra of linear transformations of $A = \Lambda^2(V^*)\oplus (V\otimes W)$ that preserve a certain cubic form on $A$ that is invariant under the obvious representation of $\mathrm{SL}(V)\times\mathrm{SL}(W)$ on $A$. (This $A$ is one of the two lowest dimensional linear representations of $\mathfrak{e}_6$, the other is its dual.)

I don't know that Élie Cartan ever noticed this, or wrote about it, but I think there is a description along the following lines: (maybe in Freudenthal or Dynkin): Let $V_1$, $V_2$, and $V_3$ be three vector spaces of dimension $3$. Then there is a decomposition $$ \mathfrak{e}_6 = \mathfrak{sl}(V_1)\oplus\mathfrak{sl}(V_2)\oplus \mathfrak{sl}(V_3)\oplus (V_1\otimes V_2\otimes V_3)\oplus (V^*_1\otimes V^*_2\otimes V^*_3) $$ as a module over the group $\mathrm{SL}(V_1)\times\mathrm{SL}(V_2)\times \mathrm{SL}(V_3)$. I'll try to find the reference.