Lie algebra $\mathfrak{so}(9)$ as a subalgebra of $\mathfrak{f}_4$

Question

What are the explicit Lie algebra mono-morphisms from $\mathfrak{so}(9)$ into the exceptional Lie algebra $\mathfrak{f}_4$ and from $\mathfrak{f}_4$ to $\mathfrak{e}_6$? Are these explicitly described in any book?

Answer 1

One way to describe the embedding $Spin_9\to F_4$ is as follows. $F_4$ is the group of automorphisms of the Jordan algebra $\mathcal J$ of traceless (or, equivalently, matrices with any given constant (real) trace, since an automorphism must fix the the real multiples of the identity and preserve the trace) Hermitian $3\times 3$ matrices with Cayley numbers (octonions) as coefficients. The algebra of Cayley numbers is a non-associative normed division algebra over $\mathbf R$ of dimension $8$. It follows that $\mathcal J$ is a real vector space of dimension $3\cdot 8 + 2 = 26$ ($3-1=2$ for the diagonal entries, which are real and add up to zero). This gives a $26$-dimensional representation $\rho$ of $F_4$ which is irreducible ($\omega_4$ is the notation of Bourbaki) and in fact the lowest dimensional non-trivial representation of $F_4$. The structure of Jordan algebra is given by $a\cdot b = \frac12(ab+ba)$ for $a$, $b\in\mathcal J$ (on the right hand side we have the usual multiplication of matrices, but recall that the Cayley algebra is neither associative, nor commutative). Now let $$p=\left(\begin{array}{ccc}1&0&0;\\ 0&0&0; \\ 0&0&0\end{array}\right)\in\mathcal J.$$ Then the isotropy subgroup of $F_4$ at $p$ is isomorphic to $Spin_9$. The orbit through $p$ is thus $F_4/Spin_9$, namely, the Cayley projective plane.

The representation $\mathcal J$ is in some sense what you need to get $E_6$ from $F_4$. In fact, $\mathfrak e_6=\mathfrak f_4 + \mathcal J$ as vector spaces, where the Lie algebra structure is as follows: for $X$, $Y\in \mathfrak f_4$, $a$, $b\in \mathcal J$ we have that $[X,Y]$ is the bracket in $\mathfrak f_4$, $[X,a]=\rho(X)a$ is the action $\rho$ of $\mathfrak f_4$ on $\mathcal J$ and $[a,b]\in\mathfrak f_4$ is defined by $\langle X,[a,b]\rangle = \langle \rho(X)a,b\rangle$ for all $X\in \mathfrak f_4$ and $\langle X,Y\rangle = Kill_{\mathfrak f_4}(X,Y) + trace(\rho(X)\rho(Y))$. This is in fact a standard construction, taking into account that $F_4$ is a symmetric subgroup of $E_6$ and $E_6/F_4$ is a symmetric space.

There are other things but at this point I'd refer you to the very nice book "Lectures on Exceptional Lie Groups", J. F. Adams Edited by Zafer Mahmud and Mamoru Mimura, 1996, Chicago Lectures in Mathematics.

Answer 2

I'm not sure what kind of information is being asked for in the question, but Claudio has offered a coherent viewpoint based on some explicit Lie algebra theory. It's also possible to work on the level of internal structure, given by Dynkin (and extended Dynkin) diagrams, etc. Chapter X of the standard book by Helgason Differential Geometry, Lie Groups and Symmetric Spaces (partly influenced by work of Kac) draws together a lot of the classification-related material you might need.

Embeddings of the Lie algebra of type $B_4$ into the Lie algebra of type $F_4$ occur naturally in terms of the extended Dynkin diagram for $F_4$ (page 503 of Helgason, for example). Here you see the diagram of $B_4$: for a fixed Cartan subalgebra of the bigger algebra, you get simple roots for the subalgebra sharing this Cartan subalgebra by taking the two long simple roots and adjacent short simple root together with the negative of the highest (long) root of $F_4$ which occurs as the extra node in the extended diagram. What you have here is not a Levi subalgebra of a standard parabolic in $F_4$ but rather a pseudo-Levi subalgebra.

From the internal viewpoint of the adjoint representation, this construction is entirely explicit. Moreover, beyond the arbitrary choice of a common Cartan subalgebra, the finitely many automorphisms of the bigger or the smaller Lie algebra stabilizing this Cartan subalgebra seem to provide the only variation in the embedding. Naturally you might prefer to realize all of this in terms of Jordan algebras and the like, but it's already visible in the abstract Lie algebras.

For the comparison of Lie algebras of types $F_4$ and $E_6$, probably the most natural Lie algebraic embedding is the "folding" suggested by Johannes. This too is treated in Helgason's chapter and in the book by Kac, as well as in algebraic group classifications. Here you realize the smaller Lie algebra as the fixed points in the bigger one of a natural diagram automorphism which identifies certain simple roots of $E_6$. Again I don't see more than one way to do this abstractly, but you can always apply some compatible automorphisms of the Lie algebras.

P.S. Concerning the question of which field you work over, it makes no difference if the field is real or complex (in fact, the Lie algebras involved come from a Chevalley form over the integers). What matters is that the Lie algebras be split, since otherwise you get into further questions about non-split real forms and their relationship.

Answer 3

There is a systematic construction of the exceptional Lie algebras that uses tensor products of composition algebras as the starting point. This is nicely explained in "Triality, exceptional Lie algebras, and Deligne dimension formulas" by Landsberg/Manivel (where you can also find references to the original sources of this observation). This gives constructions

$${\mathfrak f}_4 \quad = \quad ({\mathbb O} \otimes {\mathbb R}) \oplus ({\mathbb O} \otimes {\mathbb R}) \oplus ({\mathbb O} \otimes {\mathbb R}) \quad \oplus \quad t({\mathbb O}) \times t({\mathbb R})$$

$${\mathfrak e}_6 \quad = \quad ({\mathbb O} \otimes {\mathbb C}) \oplus ({\mathbb O} \otimes {\mathbb C}) \oplus ({\mathbb O} \otimes {\mathbb C}) \quad \oplus \quad t({\mathbb O}) \times t({\mathbb C})$$

Here $t({\mathbb A})$ denotes the triality algebra of $\mathbb A$, so $t({\mathbb R}) = 0$, $t({\mathbb C}) = {\mathbb R}^2$, $t({\mathbb O}) = {\mathfrak s \mathfrak o}_8$.

In this setting the ${\mathfrak s \mathfrak o}_9\subset {\mathfrak f}_4$ is given by ${\mathfrak s\mathfrak o}_9 = ({\mathbb O} \otimes {\mathbb R}) \oplus ( t({\mathbb O}) \times t({\mathbb R}))$ and the inclusion ${\mathfrak f}_4 \subset {\mathfrak e}_6$ is induced by ${\mathbb O} \otimes {\mathbb R} \subset {\mathbb O} \otimes {\mathbb C}$.

Answer 4

Another point of view. One can construct very explicitely $f_{4}$ by beginning with $so(9)$. Let $J_{ij}$ be a basis of $so(9)$. Consider the 16-dimensional spin representation of $so(9)$ with a basis $Q^{a}$ : one has $[J_{ij},Q^{a}]= \gamma_{ab,ij}Q_{b}$. Then consider the direct sum of $so(9)$ and of its spin representation. We want to construct a Lie algebra structure on this vector space. For $[J_{ij}, J_{kl}]$ one takes what we have in $so(9)$, and one set $[J_{ij},Q^{a}]= \gamma_{ab, ij}Q_{b}$ and $[Q^{a}, Q^{b}] = \gamma_{ab, ij} J_{ij}$.

There is some computation to do in order to prove that Jacobi identity is verified. The Lie algebra obtained is of dimension 36+16=52 and is $f_{4}$. By construction, $so(9)$ is a Lie subalgebra of $f_{4}$.

The same construction with $so(16)$ and its 128-dimensional spin representation gives a Lie algebra of dimension 120+128=248 which is $e_{8}$. The natural inclusion of $so(9)$ in $so(16)$ induces an inclusion of $f_{4}$ in $e_{8}$. In fact $e_{8}$ contains naturally $e_{6}$ : $e_{6}$ is the centralisator of $so(10) \times su(3) \subset so(10) \times su(4)=so(10) \times so(6) \subset so(16) \subset e_{8}$. With this explicit description, it is easy to see that $f_{4}$ is in fact contained in $e_{6}$.