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The exceptional complex simple Lie algebra $F_4$ has an irreducible 26-dimensional representation $V$ with Dynkin label [0,0,0,1] in the usual ordering of the simple roots one can find, say, in Humphreys's book on Lie algebras and representation theory. In fact, $F_4$ can be defined as the Lie subalgebra of $\mathfrak{sl}(V)$ which preserves a symmetric inner product and a certain cubic form on $V$.

Now there are three different real forms of $F_4$ and my question is about what happens to $V$ when restricting to these real forms. The three real forms are the compact real form, the split real form and a third form. They can be distinguished by the 'index' of the Killing form; i.e., if the Killing form $\kappa(X,Y) = \operatorname{Tr} \operatorname{ad}_X \operatorname{ad}_Y$ has signature $(p,q)$, its index is $p-q$. I am most familiar with the compact real form, for which the Killing form is negative-definite, whence of index $-52$. The split real form has index $4$ and the third real form has index $-20$, and are denoted $F_4^4$ and $F_4^{-20}$, respectively.

I would like to know the following (pointers to the literature would also be greatly appreciated):

Questions

What is the type of $V$ under the different real forms? I know that for the compact real form it is real, but I would like to know also for $F_4^{-20}$ and $F_4^4$.

And if the type is real (as I suspect is the case), what is the signature of the invariant inner product on the underlying real representation $V_{\mathbb{R}}$?

Thanks in advance!

Edit

Based on Jim's answer below, the representations are of real type in all cases. From Bruce's answer, it would seem that for the split case $F_4^4$ the signature is (14,12).

In a rather convoluted calculation, I seem to find that for $F_4^{-20}$ the signature is (16,10), but I would like confirmation since I have seen at least one claim in the Physics literature (see last equation in §4) that it is (25,1).

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(TeXnical comment: use \operatorname rather than \mathrm in $\kappa(X,Y) = \dots$ to make the spacing work; currently "tr ad" runs together into "trad".) Nice question. –  Theo Johnson-Freyd May 4 '12 at 13:27
    
Thanks - I have followed your TeXnical suggestion. –  José Figueroa-O'Farrill May 4 '12 at 13:57
    
Can you settle the case $F_4^{-20}$ by restricting the 26 dimensional representation to the maximal compact subgroup? –  Bruce Westbury May 4 '12 at 16:32
    
Not really. The maximal compact of $F_4^{-20}$ is $\operatorname{Spin}(9)$, under which the 26-dimensional representation breaks up as a direct sum of the trivial 1-dimensional rep, the defining 9-dimensional rep and the 16-dimensional spinor rep. That reducible $\Spin(9)$-module admits invariant inner products of signatures $(25,1)$ and $(16,10)$, among others. –  José Figueroa-O'Farrill May 4 '12 at 17:04
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4 Answers

up vote 11 down vote accepted

I think the best way to see the signature of these quadratic forms is by using the formula from "A Classification Theorem for Albert Algebras" by R. Parimala, R. Sridharan, and Maneesh L. Thakur, Trans. AMS 350 #3, March 1998.

All forms of $F_4$ arise from Albert algebras. Over $R$, these are 27-dimensional algebras, whose unital automorphisms form groups of type $F_4$. They are classified, over fields of characteristic neither $2$ nor $3$, by cohomological invariants $f_3$ and $f_5$. These cohomological invariants determine 3-fold and 5-fold Pfister forms, $\phi_3$ and $\phi_5$ respectively.

The formula of P-S-T (above), or maybe originally due to Serre, is that for an Albert algebra $A$ over $k$, $$Q_A \perp \phi_3 \cong <2,2,2> \perp \phi_5.$$

Now there are only two Pfister forms over $R$ for $\phi_3$ and $\phi_5$. The signature of $\phi_3$ is either $(8,0)$ or $(4,4)$. Similarly, the signature of $\phi_5$ is either $(32,0)$ or $(16,16)$. The signature of $<2,2,2>$ is $(3,0)$. Hence the possibilities for the signature $(p,n)$ of $Q_A$ are: $$(p,n) + (8,0) = (3,0) + (32,0),$$ $$(p,n) + (8,0) = (3,0) + (16,16),$$ $$(p,n) + (4,4) = (3,0) + (32,0),$$ $$(p,n) + (4,4) = (3,0) + (16,16).$$

Only three cases are possible: $(p,n) = (27,0)$ or $(p,n) = (11,16)$ or $(p,n) = (15,12)$.

As $F_4$ acts on the orthogonal complement of the identity, and the identity has positive norm, the possible signatures for the 26-dimensional rep of $F_4$ are: $$(26,0), (10,16), (14,12).$$

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Beautiful! Thanks a lot. This is the sort of confirmation I was hoping for. –  José Figueroa-O'Farrill May 4 '12 at 20:05
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Actually, this answer is already in Élie Cartan's 1914 classification of the real forms Les groupes réels simples, finis et continus. He shows there, in the final pages, that there are three real forms of $F_4$, all appearing in $SL(26,\mathbb{R})$: $F^4_{4}$ (the split form) preserves an inner product of type $(14,12)$, $F^{-20}_4$ preserves an inner product of type $(10,16)$, and $F^{-52}_4$ (the compact form) preserves an inner product of type $(26,0)$. –  Robert Bryant May 9 '12 at 21:50
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The answer to the first question (as you expect) is that the representation is real in each case. I'm less sure how to answer the second question explicitly. But among the many possible older sources in the mathematics or physics literature there is one worth mentioning:

J. Tits, Tabellen zu den einfachen Lie Gruppen und ihren Darstellungen, Lect. Notes in Math. 40 (1967), Springer-Verlag

This early volume of the series (in large page form soon abandoned) is drawn from lectures Tits gave in Bonn in 1966. He summarizes a lot of basic material concisely (in 53 typed pages) with some indications as to proofs and earlier references. Especially relevant for your purpose is Section 11 on representations of the real forms, together with the detailed tables for simple types including the three real forms of $F_4$ (pages 43-44). But note that his numbering convention for this Dynkin diagram is the reverse of the usual Bourbaki convention which I followed.

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All the forms of $F_4$ can be defined as automorphism groups of some Jordan algebra of three by three matrices with entries in octonions / split octonions / complexified octonions. These algebras are all of dimension 27 over the appropriate field and the subspaces of trace-free matrices are the irreducible 26-dimensional representations of the various forms of $F_4$. The invariant quadratic form is $A\mapsto \mathrm{Tr}(A^2)$. (And the invariant cubic form is $A\mapsto \mathrm{det}(A)$.)

The group $F_4^{-20}$ is according to Yokota (but I guess that one can dig this up also out of the work of Veldkamp or Springer) the automorphism group of the real Jordan algebra $J(1,2,\mathbb{O}) = \{X\in \mathrm{M}(3,\mathbb{O}\otimes_\mathbb{R}\mathbb{C}) \, |\, I_1 \overline{X}^tI_1 = X \}$ where $I_1 = \mathrm{diag}(-1,1,1)$. Now the computation of the signature of $A\mapsto \mathrm{Tr}(A^2)$ is a matter of simple calculation.

The other two real cases $F_4^{-52}$, $F_4^4$ follow similarly since they are the automorphism groups of $J(3,\mathbb{O}) = \{ X\in M(3,\mathbb{O})\,|\, X^t =X \}$ and $J(3,\mathbb{O}) = \{ X\in M(3,\mathbb{O}')\,|\, X^t=X \}$ respectively.

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+1 for the reference to Yokota's paper. I had missed that! And indeed, this seems to settle it nicely as well. –  José Figueroa-O'Farrill May 9 '12 at 20:42
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The paper http://arxiv.org/abs/math/0203010 discusses various constructions of real forms of the Freudenthal magic square. For $F_4$ they obtain $F_4^4$ and $F_4^{-52}$ but not $F_4^{-20}$. I had a quick scan but did not see the 26-dimensional representation explicitly but it must be there even if only implicitly.

This paper has been published at:

MR2020553 (2005b:17017)
Barton, C. H. ;  Sudbery, A.
Magic squares and matrix models of Lie algebras.
Adv. Math.  180  (2003),  no. 2, 596--647.
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Thanks. I believe that the magic square only gives those two real forms, because there are only two "real" forms of the octonions: the usual ones and the so-called split octonions. The usual octonions give the compact real form of $F_4$, whereas in a what is surely only a linguistic coincidence, the split octonions give the split real form of $F_4$. –  José Figueroa-O'Farrill May 4 '12 at 15:52
    
And, yes, the 26 dimensional representation is there implicitly. The Jordan algebra of which F_4 is the automorphism algebra is the one of $3\times 3$ traceless, hermitian octonionic matrices, and that's a real vector space of dimension 26. –  José Figueroa-O'Farrill May 4 '12 at 15:55
    
I think it might be a linguistic convenience. A complex composition algebra has two real forms which are called the compact real form and the split real form. –  Bruce Westbury May 4 '12 at 16:10
    
Then the naive candidate for the inner product is $(A,B)=tr(AB)$. This should then give the signature. –  Bruce Westbury May 4 '12 at 16:13
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Each entry in the square depends on an unordered pair of composition algebras so you might expect three real forms. –  Bruce Westbury May 4 '12 at 16:30
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