Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Has anyone ever bothered to write down the 26-dimensional fundamental representation of $F_4$? I wouldn't mind looking at it. Is it in $\mathfrak{so}(26)$?

I'm familiar with the construction of the fundamental representation for $G_2$ where you can use use the fact that the groups is automorphism group of the octonions to put linear relations on $\mathfrak{so}(7)$. (Elements of ${\mathfrak{g}}_2$ are the derivations)

Answering the same question for $E_6$, $E_7$ or $E_8$ would also be welcome here.

share|improve this question
    
Perhaps I should clarify. My main question is really "what does the matrix look like?" What do you have to impose on the entries of the ${\mathfrak{so}}(26)$ matrix to be in the $F_4$ subalgebra? Ideally with a handy reference. –  Q.Q.J. Feb 26 '10 at 18:20

4 Answers 4

up vote 14 down vote accepted

As I understand the question, the OP would be happy to see a description of the lowest-dimension fundamental representation of $F_4$ (and perhaps $E_6$, $E_7$, $E_8$), and is happy with the description of $G_2$ acting on the space of trace-zero octonions.

Yes - people have "bothered to write down" representations of $F_4$ and the other exceptional groups in this spirit. Chevalley, Schafer, Albert, Jacobson, Freudenthal, and Tits are the names that come to mind first.

$F_4$ acts naturally as automorphisms of the 27-dimensional exceptional Jordan algebra $J_{3,O}$ -- this is the Jordan algebra of 3 by 3 Hermitian matrices with entries in the octonions (which octonion algebra you use depends on or determines the form of $F_4$). Since $F_4$ acts as algebra automorphisms, it preserves the unit element of this algebra, and preserves the trace form as a result. It follows that $F_4$ acts on the 26-dimensional trace-zero subspace of the 27-dimensional algebra. This is quite close, in spirit, to the example of $G_2$ acting on trace-zero octonions.

Also, $F_4$ acts on this 26-dimensional space, preserving the nondegenerate symmetric trace form: $$(X,Y) \rightarrow Tr(X \cdot Y).$$

$E_6$ also acts on the 27-dimensional Jordan algebra above, but not as algebra automorphisms. Instead, $E_6$ can be viewed as the linear automorphisms of this 27-dimensional space that preserve the cubic norm form (the "determinant" of a 3 by 3 Hermitian octonionic matrix). I believe this goes back to Chevalley and Schafer about 60 years ago.

$E_7$ acts naturally on a 56-dimensional space, studied by Freudenthal. This is the space of two-by-two matrices, with diagonal entries in the base field, and off-diagonal entries in the exceptional Jordan algebra mentioned above: $2+27+27 = 56$. $E_7$ can be viewed as the group of linear automorphisms of this 56-dimensional space preserving a quartic form, I believe.

The smallest irreducible representation of $E_8$ is the adjoint representation of $E_8$ on its own Lie algebra -- so you have to construct $E_8$ to represent it, in a sense.

A nice recent survey of related topics, and a source of other references is Baez's survey on the octonions.

share|improve this answer
    
Thanks Marty, some of these facts were new to me. Do you know if anyone who has carried through from the definition of algebra automorphisms of the exceptional Jordan algebra to explicit matrices has put the explicit result in print somewhere? This is what I intended by the phrase "bothering to write it down" –  Q.Q.J. Feb 27 '10 at 11:24
1  
Ooof! I don't know of anyone who would want to write down a 27 by 27 matrix, one for each of the 52 basis elements of the Lie algebra of type F_4! I have worked with this representation in the past, but always saving space by working with 3x3 octonionic Hermitian matrices, and certain subgroups of F_4 which act nicely (with easy-to-write formulas). –  Marty Feb 27 '10 at 11:45
    
It would be a bit crazy wouldn't it? Do you know if there is an easy way to identify the parabolic subalgebras in the 3x3 octonion matrix? –  Q.Q.J. Mar 1 '10 at 15:33
2  
The parabolic subgroups of $F_4$ can be identified with stabilizers of certain "flags" in $J_0$ (the trace-zero elements in the 27-dimensional Jordan algebra) -- this is the "metasymplectic geometry" which I think was described by Freudenthal in the 1950s. There are four maximal parabolics (up to conjugacy) in $F_4$. One such parabolic can be viewed as the stabilizer of a norm-zero line (for the cubic norm) in $J_0$. Other maximal parabolics are stabilizers of certain "lines", "planes", and "symplecta" in the projective space $PJ_0$. It's hard to find references! Try Freudenthal or Tits. –  Marty Mar 1 '10 at 19:19
    
Perhaps this article can be of some help: arxiv.org/abs/1006.3407 –  Vít Tuček Feb 4 '11 at 11:25

Answerlet: it's either in ${\mathfrak so}(26)$ or ${\mathfrak sp}(13)$, that is to say, it's either a complexification of a real representation or forgetful from a quaternionic representation. Indeed, this is true of every irrep of $F_4$. I don't instantly know which one occurs for your particular irrep.

Proof: $F_4$ has no Dynkin diagram symmetries, so the automorphism of the dominant Weyl chamber taking an irrep to its dual must be trivial. So $V \cong V^* $. Then the $F_4$-invariant inner product on $V$ lives either in $Sym^2 V$ or $Alt^2 V$, whence the dichotomy.

share|improve this answer
    
Clearly the compact real form is in the orthogonal group. By the Weyl unitary trick, this is the case also for the other real forms. –  José Figueroa-O'Farrill Feb 26 '10 at 12:16
2  
My previous comment is ambiguous and if taken literally it's wrong. I don't mean that the noncompact real forms are in $\mathfrak{so}(26)$, but that the invariant bilinear product is still symmetric! –  José Figueroa-O'Farrill Feb 26 '10 at 12:25

Here I quote some GAP code for generating F4 Lie algebra as matrices 27 x 27. It is possible to obtain them in dimension 26 but then you need to use Sqrt(2) or Sqrt(3). In dimension 27 it is more nice.

Matrices L1..L7 are 8x8 matrices of left multiplication by imaginary unit octonions e1..e7. R1..R7 are right multiplications by unit octonions. If you have troubles to obtain such please let me know.

I have also obtained Lie algebra of E6 as complex matrices 27x27 and Lie algebra E7 as quaternion matrices 28x28. I tried to obtain some nice way E8 Lie algebra in shape 31*8 dim matrices but no luck. It was in 2008 when I worked on these scripts. I have used Freudenthal, Tits, Vinberg papers when doing this.

Regards, Marek


v18 := BlockMatrix([[1,1,-R1], [2,2,-L1], [3,3,L1+R1]], 4,4); v28 := BlockMatrix([[1,1,-R2], [2,2,-L2], [3,3,L2+R2]], 4,4); v38 := BlockMatrix([[1,1,-R3], [2,2,-L3], [3,3,L3+R3]], 4,4); v48 := BlockMatrix([[1,1,-R4], [2,2,-L4], [3,3,L4+R4]], 4,4); v58 := BlockMatrix([[1,1,-R5], [2,2,-L5], [3,3,L5+R5]], 4,4); v68 := BlockMatrix([[1,1,-R6], [2,2,-L6], [3,3,L6+R6]], 4,4); v78 := BlockMatrix([[1,1,-R7], [2,2,-L7], [3,3,L7+R7]], 4,4);

S:= DiagonalMat([1,-1,-1,-1,-1,-1,-1,-1]);

Build now elliptic version of F4. The name is from that exp(t*[[0,-1], [2,0]]) is ellipse. Really exp(t*[[0,-1], [2,0]]/Sqrt(2)) is ellipse.

p1 := [[0, -1, 1]]; p2 := [[-1, 1, 0]];

n := NullMat(8,8); v:= n+p1; vt := -2*TransposedMat(v);

This a0 really corresponds to ad([[0,-1,0],[1,0,0],[0,0,0]]) derivation on h3O. a0 := BlockMatrix([[1,2,S], [2,1,-S], [3,4, v], [4,3, vt]], 4,4);

w := n+p2; wt := -2*TransposedMat(w);

b0 := BlockMatrix([[2,3,S], [3,2,-S], [1,4, w], [4,1, wt]], 4,4); mats3 := [ v18, v28, v38, v48, v58,v68,v78, a0, b0];; ms:=List(mats3, x->x{[1..27]}{[1..27]}); f4_e := LieAlgebra( Rationals, ms);

share|improve this answer

While there have been many people who have done this, the first person to do so was Élie Cartan, who wrote down a basis for the matrix algebra ${\frak{f}}_4\subset{\frak{so}}(26)$ in his 1894 thesis Sur la structure des groupes de transformations finis et continus. See pp 144--147 for the explicit formulae over $\mathbb{C}$ or $\mathbb{R}$ (for the split form). In a later paper, Les groupes réels simples finis et continus (1914), he explicitly exhibited (conjugate linear) involutions on $\mathbb{C}^{26}$ whose fixed subgroups would yield the three real forms of $F_4$, see pp. 343--352.

He also gives explicit formulae for $E_6$, $E_7$ and $E_8$ in their lowest dimensional representations in the above two papers.

Apropos the OP's comment My main question is really "what does the matrix look like?" What do you have to impose on the entries of the so(26) matrix to be in the F4 subalgebra?: It seems to me that you are asking how the linear transformations in the various algebras are characterized. So, for example, one might say that ${\frak{so}}(n)$ is characterized by being the skewsymmetric $n$-by-$n$ matrices or, alternatively, you might say that these are the linear transformations that, to first order, preserve the standard (positive definite) inner product on $\mathbb{R}^n$.

Along these lines, Cartan again gives the answers for $F_4$, $E_6$, and $E_7$ in his 1894 thesis: He shows that ${\frak{f}}_4\subset{\frak{so}}(26)$ is characterized as the Lie algebra of the stabilizer of a quadratic form and a certain projective cone of dimension $16$ in $26$ dimensions. (Later, he realized that it could be characterized as the stabilizer of a homogeneous cubic polynomial in $26$ variables.) He also described ${\frak{e}}_6\subset{\frak{sl}}(27)$ as the stabilizer of a homogeneous cubic polynomial in $27$ variables, and ${\frak{e}}_7\subset{\frak{sp}}(28)$ as the stabilizer of a homogeneous quartic polynomial in $56$ variables (and a symplectic form, though, for the Lie algebra, this is not necessary).

Finally, although Cartan does not seem to have noticed this, it turns out that ${\frak{e}}_8\subset{\frak{so}}(248)$ can be characterized in ${\frak{gl}}\bigl({\frak{e}}_8\bigr)$ as the stabilizer of the Cartan $3$-form of the algebra ${\frak{e}}_8$ itself.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.