From: baez@galaxy.ucr.edu Subject: E6 as determinant-preserving transformations Date: Mon, 14 May 2001 23:56:36 +0000 (UTC) Newsgroups: sci.math.research Summary: Actions of Lie group E6 on Jordan algebra of dimension 27 E6 has two different irreducible representations of dimension 27. We can think of these as representations on J and J*, where J is the Jordan algebra of 3x3 hermitian octonionic matrices. There is a cubic form det: J -> R which assigns to any such matrix its fairly obvious "determinant". This form is preserved by (a certain noncompact real form of) E6. In fact, in his paper "Lie groups in the foundations of geometry", Freudenthal says that E6 is precisely the group of linear transformations of J which preserve the determinant. However, he doesn't make this evident, and Adams in his book Lectures on Exceptional Lie Groups proves something a bit weaker, which has me puzzled. Namely: Adams proves that in addition to the cubic form det: J -> R there is a cubic form on the dual J*, which I will take the liberty of calling det*: J* -> R even though I see no way to concoct it from det without using extra structure - e.g. the inner product on J. Polarizing these cubic forms we get trilinear maps J x J x J -> R and J* x J* x J* -> R and dualizing these we get two "cross products" J x J -> J* and J* x J* -> J. Adams proves that the group of *pairs* of linear maps f: J -> J, g: J* -> J* preserving these two products is E6. In fact he also shows any pair preserving both products satisfies g = f*, so we may also state his result thus: E6 is the group of maps f: J -> J such that 1) f preserves det: J -> R and 2) f* preserves det*: J* -> R. So my question is: does condition 1) imply condition 2), and if so, why? PS - In fact points in the octonionic projective plane correspond to equivalence classes of nonzero elements p in J with cross product p x p = 0, where two elements are equivalent if they differ by a nonzero real multiple. Similarly, lines in the octonionic projective plane correspond to equivalence classes of nonzero elements L in J* with L x L = 0. To save time let me stop saying "equivalence classes". If p and p' are distinct points, the line containing them is p x p'. If L and L' are distinct lines, their point of intersection is L x L'. This makes it clear that any transformation preserving both cross products J x J -> J* and J* x J* -> J acts as a collineation of the octonionic projective plane. This ties in nicely with the description of E6 as the collineation group of this plane. PPS - There is a natural inner product on J which allows us to identify J with J*, identify det: J -> R with det*: J* -> R, and identify the two cross products J x J -> J* and J* x J* -> J. However, while this inner product is invariant under all automorphisms of J, it's not invariant under E6, which is why I avoided mentioning it above. PPPS - Hmm, I may have answered my own question. If f: J -> J preserves det, it preserves the cross product J x J -> J*, so it acts on points in the octonionic projective plane (since these are defined via p x p = 0). If we define a line to be an equivalence class of elements L in J* having the form p x q where p and q are distinct points, we see that f*: J* -> J* acts on lines. Since f(p) x f(q) = f*(p x q), we see that L is the line through p and q iff f*(L) is the line through f(p) and f(q). So f acts as a collineation of the octonionic projective plane, so f lies in E6, so f*: J* -> J* preserves det*. However, I'd be much happier if I could construct det* from det: then it would be utterly obvious that f: J -> J preserves det iff f*: J* -> J* preserves det*. ============================================================================== From: "R. Bryant" Subject: Re: E6 as determinant-preserving transformations Date: Tue, 15 May 2001 17:43:11 -0400 Newsgroups: sci.math.research baez@galaxy.ucr.edu wrote: > E6 has two different irreducible representations of dimension 27. > We can think of these as representations on J and J*, where J is > the Jordan algebra of 3x3 hermitian octonionic matrices. There is > a cubic form > > det: J -> R > > which assigns to any such matrix its fairly obvious "determinant". > This form is preserved by (a certain noncompact real form of) E6. > > In fact, in his paper "Lie groups in the foundations of geometry", > Freudenthal says that E6 is precisely the group of linear transformations > of J which preserve the determinant. However, he doesn't make this > evident, and Adams in his book Lectures on Exceptional Lie Groups > proves something a bit weaker, which has me puzzled. [ rest deleted -RLB] I may not be able to answer your question, but I do have a few comments that you might find helpful. First, it was E. Cartan (in his thesis) who first claimed that E6 can be defined as the stabilizer of a cubic form in 27 variables. He explicitly writes the cubic down, without mentioning the octonions or Jordan algebras or any of the other things that we usually quote. He doesn't give a proof, but a proof is not very hard to come by, given his very explicit formula Here it is: Take two 6-vectors x = (x_i) and y = (y_j) and a skew-symmetric 6x6 matrix z = (z^{ij}). Now consider the expression D = Pf(z) + z^{ij} x_i y_j (summation convention intended) (where Pf(z) means the Pfaffian of z). D is a cubic expression in 27 variables. Cartan says that the stabilizer of D is E6. (Actually, he is working over the complex field at this point. Only years later, when he classifies the real forms, does he give the corresponding cubics in 27 real or complex variables (which one depends on the real form; when the variables are complex, one must also require that the group preserve an Hermitian form or a real involution). Here is a sketch of the argument: First, note that D is invariant under SL(6) acting on x and y as elements of a vector space V of dimension 6 and on z as an element of the second exterior power of V*, since D is the essentially the sum of the (exterior algebra ) cube of z and the term z(x,y). This also makes is clear that D is invariant under SL(2) acting as unimodular changes of basis in x and y. Thus, D is invariant under SL(6) x SL(2). (Actually, you have to mod out by the subgroup (-I_6,-I_2) since this acts trivially. If you now break up gl(27) under the action of sl(6)+sl(2) in this way, you find that there are not very many pieces and you can rule out all but one fairly easily, that one being isomorphic to the tensor product of the third exterior power of V with the standard 2-dimenional representation of sl(2). Note that the nonzero weights of sl(2) and sl(6) plus the weights in this representation make up the root system of E6. By just checking an element in this last piece, you see that it does preserve D, so it follows that this 78-dimensional algebra is the tangent space to the stablizer of D. Thus, the identity component of stab(D) is E6. It remains to show that E6 = stab(D). Now, the E6 algebra obviously acts irreducibly on C^{27} since SL(6) x SL(2) itself preserves only one non-trivial spliiting. The commuting ring of the representation is therefore the scalars, but only the cube roots of unity preserve the cubic. Thus, the center of stab(D) is the cube roots of unity. Note that this center lies in E6 because it already lies in SL(6) x SL(2). It is known that an outer automorphism of E6 swtiches the two nontrivial cube roots of unity, so it follows that conjugation by an element of stab(D) is never an outer automorphism. Thus, an element of stab(D) can be written as an element of the center of stab(D) (which equals the center of E6) with an element of E6. Thus stab(D) = E6, as desired. By the way, note that the mapping (x,y,z) -> (-x,-y,z) lies in E6 and is the involution that corresponds to the symmetric pair (e6, sl(6)+sl(2)). Finally, when Cartan classifies the representation of E6, he notes that the two 27-dimensional representations are dual to each other (and not isomorphic, of course). There is a trivial, abstract nonsense argument that, because E6 preserves a cubic in its 'standard' 27-dimensional representation W, it must preserve one on its dual: The nontrivial pairing W x W* -> C induces a nondegenerate pairing S^3(W) x S^3(W*) --> C and the E6-invariant element D in S^3(W*) then defines (via its kernel) an E6-invariant hyperplane H_D in S^3(W). Since E6 is simple, there must be an E6-invariant line L_D in S^3(W) that is complementary to H_D, now let D* in L_D be the element that pairs with D to be 1. Then D* is an E6-invariant cubic form on W*. I hope that this helps, or is at least interesting. Yours, Robert Bryant ============================================================================== From: Linus Kramer Subject: Re: E6 as determinant-preserving transformations Date: Thu, 17 May 2001 11:13:56 +0200 Newsgroups: sci.math.research Freudenthal proves that the real Lie group E_6(-26) is precisely the group which preserves det in his paper Freudenthal, Oktaven, Ausnahmegruppen und Oktavengeometrie. Geom. Dedicata 19 (1985), no. 1, 7--63. It's a very nice and readable paper (written in german, though). You might also find the following interesting: Veldkamp, Freudenthal and the octonions. Nieuw Arch. Wisk. (4) 9 (1991), no. 2, 145--162. Regards, Linus Kramer -- Linus Kramer Mathematisches Institut Universitaet Wuerzburg Am Hubland 97074 Wuerzburg Germany E-mail: kramer@mathematik.uni-wuerzburg.de http://www.mathematik.uni-wuerzburg.de/~kramer