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Among simple Lie groups, $Spin(8)$ is the most symmetrical one in the sense that $Out(Spin(8))$ is the largest possible group. A description of this outer automorphism groups is as follows. $Spin(8)$ has 2 half-spinor representations and one vector representation (coming from standard representation of SO(8)) all of them of dimension 8. In general one can pre-compose a representation of a Lie group with an automorphism of the group to get another representation. In this special case $Out(Spin(8))$ permutes above three representations and this we get an isomorphism of $Out(Spin(8))$ and $S_3$.

My questions:

1- Is there any way to see an automorphism explicitly which interchanges the vector representation and a half-spinor representation? (It is easy to see there exists such an automorphism. But I'd like to have an explicit construction of such an automorphism.)

2- Apparently there is a 27-dimensionla Jordan algebra which has $Aut(Spin(8))$ as a subgroup of its automorphism group. Can anyone explain what is this Jordan algebra and how should I think about $Aut(Spin(8))$ acting on it?

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There's an answer that uses symmetries of the Dynkin diagram, en.wikipedia.org/wiki/File:Dynkin_diagram_D4.png in that the automorphisms of D4 (the relation to $S_3$ should be obvious) give rise to automorphisms of the Lie algebra $so(8)$, and then of $Spin(8)$, and these are outer automorphisms. –  David Roberts Dec 18 '12 at 0:20
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You ask for an explicit construction of these automorphisms. Unfortunately, there are a number of different ways to understand Spin(8) explicitly, and without knowing how you understand Spin(8) explicitly, we can't really figure out how to answer your question in a way that's useful to you. –  Alexander Woo Dec 18 '12 at 1:45
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Is the explicit formula using octonion multiplication acceptable as an explanation? –  Robert Bryant Dec 18 '12 at 3:23
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4 Answers 4

Here's my favorite way to answer your question. Hopefully the answer to Robert Bryant's question is "yes".

Let $A$ be the ring of octonions (the "nonsplit" octonions over ${\mathbb R}$); it comes with an involution $\alpha \mapsto \bar \alpha$, from which there is a trace $Tr(\alpha) = \alpha + \bar \alpha$ and a norm $N(\alpha) = \alpha \cdot \bar \alpha$. From this, we get a trilinear form $T: A \otimes A \otimes A \rightarrow {\mathbb R}$ given by $$T(\alpha, \beta, \gamma) = Tr( (\alpha \beta) \gamma) = Tr(\alpha (\beta \gamma)).$$ (Multiplication is nonassociative, but the traces work out to the same result.)

The group $Spin(8)$ can be constructed as the group of triples $(g_1, g_2, g_3) \in SO(A,N)^3$ of "rotation matrices" with respect to the norm quadratic form, such that for all $(\alpha, \beta, \gamma) \in A^3$, $$T(g_1 \alpha, g_2 \beta, g_3 \gamma) = T(\alpha, \beta, \gamma).$$

The full group of outer automorphisms is now almost clear -- cyclic permutations of $(g_1, g_2, g_3)$ give automorphisms of $Spin(8)$ defined above since $T(\alpha, \beta, \gamma) = T(\beta, \gamma, \alpha)$.

Edit below to reflect comments and correspondence with Daemi, and the comment of Bryant


The full $S_3$ action on $Spin(8)$ is obtained from cyclic permutations and the following slightly subtle action of transpositions. Let $C$ denote the main involution of $A$ (the one for which $Tr(\alpha) = \alpha + C(\alpha)$). For any $g \in SO(A,N)$, define $\bar g = C \circ g \circ C$; then $\bar g \in SO(A,N)$ as well.

The action of a transposition on $Spin(8)$ follows: for the transposition $(12)(3)$, the associated outer automorphism of $Spin(8)$ sends $(g_1, g_2, g_3)$ to $(\bar g_2, \bar g_1, g_3)$. The other transpositions act in the analogous ways.


The Jordan algebra is the exceptional Jordan algebra of 3x3 Hermitian matrices with octonion entries: $$J = \left\lbrace \left( \begin{array}{ccc} a & \alpha & \bar \beta \cr \bar \alpha & b & \gamma \cr \beta & \bar \gamma & c \\ \end{array} \right) : \alpha, \beta, \gamma \in A, a,b,c \in {\mathbb R} \right\rbrace.$$

The group $Spin(8)$ acts on the triple of octonions $(\alpha, \beta, \gamma)$ via the natural representation from above. It acts trivially on the real numbers $a,b,c$, and this gives an action of $Spin(8)$ on the exceptional Jordan algebra. The outer automorphism group $S_3$ acts by permuting $(a,b,c)$ and $(\alpha, \beta, \gamma)$ simultaneously. Together, these give an action of $S_3 \ltimes Spin(8)$ on the exceptional Jordan algebra.

Reference update:

The material above can be found in my paper on $D_4$ modular forms, Amer. J. of Math. 128 (2006), 849-898.

The construction of $Spin(8)$ (over ${\mathbb Z}$) follows from Proposition 4.8 of M.-A. Knus, R. Parimala, and R. Sridharan, "On Compositions and Triality," J. reine angew. Math., 457:45–70, 1994.

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I took the liberty of editing the answer to get the braces and matrix to work. The solution was to use \lbrace and \rbrace instead of \{ and \}, respectively; and also to use \cr instead of \\. –  José Figueroa-O'Farrill Dec 18 '12 at 11:05
    
@Marty: Very nice. Setting aside the Jordan algebra aspect (i.e., focusing just on the first question), this answer works using any octonion algebra over any $\mathbf{Z}[1/2]$-algebra: see the section "The group Spin and triality" in the Book of Involutions. Probably somewhere in the discussion of Jordan algebras in the Book of Involutions the 2nd question is addressed more generally (at least away from characteristics 2 and 3). –  user29720 Dec 18 '12 at 11:53
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One thing that may make it a bit eaiser to see how triality works is to note that $\mathrm{Tr}(\alpha\beta)= \mathrm{Tr}(\beta\alpha)$, which is what explains the action of the symmetric group. Also, if you want detailed proofs of the above statements, I recommend the book Spinors and Calibrations by F. Reese Harvey, who does an excellent job of explaining all of this (and the geometry of $\mathrm{F}_4$ and $\mathrm{Spin}(9)$ as well). –  Robert Bryant Dec 18 '12 at 16:12
    
Since kreck brought it up, it works over Z too, using Coxeter's maximal order in the octonions. Thanks also to Figueroa-O'Farrill for fixing the Tex! –  Marty Dec 18 '12 at 16:18
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For more algebraic flavour and construction over a general field see the book Octonions, Jordan Algebras, and Exceptional Groups by Tonny Springer. –  Vít Tuček Jan 1 '13 at 1:31
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In addition to the above answers involving spinors and/or octonions, you might be interested in Cartan's original construction of the triality automorphisms, which is very explicit and takes just a couple of pages in his beautiful little paper Le principe de dualité et la théorie des groups simples et semi-simples (Bull. Sc. Math 49 (1925), 361–374). The description is in Section 5 of that article, mainly on pages 368 and 369, though you'll probably be interested in the very geometric construction that he makes to interpret the outer automorphisms in the concluding sections 6 and 7 of that paper.

The main difference from what has been said in answer so far is that, instead of constructing spinors (which, of course, he already knew how to do at that point), Cartan works on the centerless simple group $G = \mathrm{SO}(8)/\lbrace\pm\mathrm{I}\rbrace$ (which he calls the 'adjoint group' of $\mathrm{SO}(8)$), since the triality automorphisms actually do act on $G$ and not just on its Lie algebra. He then uses a clever choice of notation to write down an explicit action of $S_3$, the symmetric group on $3$ letters, on the Lie algebra of $G$ in such a way that these automorphisms are Lie algebra automorphisms that preserve the obvious maximal torus and yet are outer because they don't come from the Weyl group.

Another great source, of course, is Chevalley's beautiful little book The algebraic theory of spinors (1954), the last chapter of which is all about triality. He takes pains to do everything over general fields as well, so it's quite a useful treatment.

Finally, Cartan also wrote about the tie of triality with $\mathrm{F}_4$ acting irreducibly on the $26$-dimensional space of traceless $3$-by-$3$ Hermitian octonian matrices in Section V of his paper Sur des familles remarquables d'hypersurfaces isoparamétriques dans les espaces sphériques (Math. Zeitschrift 45 (1939), 335–367). The relevant passage is on pages 354 and 355, where he explains the construction.

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I don't have time to look into the mathematics, but you should take a look at the classic little paper $Spin(8)$, triality, $F_4$, and all that, by Frank Adams. It is in Superspace and supergravity, Cambridge University Press 1981. (Also in Volume II of Adams' collected works). It starts with the identification of the automorphism group of $Spin(8)$ with the symmetric group $\Sigma_3$ and it has the construction of the exceptional Jordan algebra of dimension 27.

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@Marty: Thanks for your response. I wasn't aware of existence of such construction of $Spin(8)$ and it makes clear to me some of the properties of it. I just believe action of outomorphism group is not by just permuting $g_1$, $g_2$, and $g_3$, but involves taking some conjugation as well. For example: $$(g_1,g_2,g_3) \to (\bar{g}_1,\bar{g}_3,\bar{g}_2)$$ is an automorphism in your description of $Spin(8)$, while: $$(g_1,g_2,g_3) \to (g_1,g_3,g_2)$$ is not. Am I right or I'm making a stupid mistake.

This construction gives a nice lift of $Out(Spin(8))$ to $Aut(Spin(8))$. I would like to have such a lift for which fixed points of its action on $Spin(8)$ is just identity element. But the lift from this construction doesn't have this property. Do you think can I change this lift by some inner automorphisms that at the end of the day I get my desired lifting?

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I'm not sure if continuing questions are supposed to be given as answers... but here are a few comments. First, I don't know what you would mean by conjugating $g_1$, $g_2$, and $g_3$. They are real matrices. The fixed points of the triality automorphism on $Spin(8)$ form the subgroup $G_2$. Changing the lift should just conjugate the $G_2$ within the $Spin(8). –  Marty Dec 18 '12 at 21:42
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@Marty: Daemi is not being clear in his notation. What he means by $\bar g_i$ is actually $C\circ g_i\circ C$ where $C:\mathbb{O}\to\mathbb{O}$ is octonionic conjugation. One must do something like this because, although $\mathrm{Tr}\bigl(\alpha(\beta\gamma)\bigr)=\mathrm{Tr}\bigl((\alpha\beta)\gamma‌​\bigr)=\mathrm{Tr}\bigl(\gamma(\alpha\beta)\bigr)$, you don't have $\mathrm{Tr}\bigl(\alpha(\beta\gamma)\bigr)=\mathrm{Tr}\bigl(\gamma(\beta\alpha)‌​\bigr)$. Instead, $\mathrm{Tr}\bigl(\alpha(\beta\gamma)\bigr)=\mathrm{Tr}\bigl(\bar\gamma(\bar\bet‌​a\bar\alpha)\bigr)$, etc. –  Robert Bryant Dec 21 '12 at 21:00
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