I've been searching the literature for a direct definition of the group $E_8$ (over a general field, but even a definition of just one incarnation would be great). I knew (from talking to people) that there's probably nothing available, but I'm confused about one point.

By "direct definition", I mean something other than a definition of $E_8$ as a group of automorphisms of its own Lie algebra.

Something promising is the "octo-octonionic projective space" $(\mathbb{O} \otimes \mathbb{O} )\mathbb{P}^2$ -- the group of isometries of the latter is meant to be a form of $E_8$. In his paper on the octonions, John Baez mentions this, but warns that $(\mathbb{O} \otimes \mathbb{O} )\mathbb{P}^2$ can only be defined in terms of $E_8$, so this is circular, and he adds "alas, nobody seems to know how to define [it] without first defining $E_8$. Thus this group remains a bit enigmatic."

The existence of the book Geometry of Lie groups by Boris Rosenfeld confuses me. In it, he claims to construct the plane, calling it $(\mathbb{O} \otimes \mathbb{O} ) {\simeq \atop S^2}$ (I cannot even reproduce it well in Latex). See Theorem 7.16 in particular.

The problem is that each object, in this book, is claimed to be definable "by direct analogy" with some other object, itself usually not quite defined in full, and so on. I'm having an awful lot of trouble reading Rosenfeld's book. In the end (see 7.7.3) he claims that everything can be carried out over a finite field, yielding a definition of $E_8(q)$. But I cannot find the details anywhere -- the thing is, I don't even know if I'm reading a survey, or a complete treatment with proofs that escape me.

Is anybody on MO familiar with Rosenfeld's book? Or is there an alternative reference for this mysterious "octo-octonionic plane"?

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    I heard rumours that Rosenfeld's boook fell short of actually defining $(\mathbb O\otimes \mathbb O)\mathbb P^2$. – André Henriques Aug 20 at 17:25
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    I know two basic ways to "define" algebraic objects, e.g. groups. One is "inductive": to give generators and relations for the group, building it out of smaller/simpler pieces. This is what Andre Henriques does below. The other is "coinductive": to define the group as the automorphisms of some object. That approach is the one mentioned by Scott Carnahan. I think you are asking for that approach. If so, Scott's answer is perhaps "the best", in the sense that the most direct construction (root lattice) $\to$ (Lie group) passes through vertex algebras. – Theo Johnson-Freyd Aug 21 at 1:25
  • Eschenburg and Hosseini (2013; pdf) have some discussion of the “Rosenfeld planes” $\mathbb{(O\otimes L)P}^2$ for $\mathbb{L\in\{R,C,H,O\}}$. Also related: mathoverflow.net/questions/99736/… – Francois Ziegler Aug 21 at 22:46
  • I like many answers, do I have to accept one? – Pierre Aug 22 at 8:19

Here's an easy, direct definition of $E_8$.

The compact Lie group $E_8$ is the colimit in the category of topological groups of the following diagram of groups $$ {\scriptstyle\begin{matrix} &SU(2)&\\[-1mm] &\downarrow&\\[-1mm] &SU(3)&\\[-1mm] &\uparrow&\\[-1mm] SU(2)\to SU(3) \leftarrow SU(2)&\!\!\!\!\!\to SU(3) \leftarrow SU(2)\to SU(3) \leftarrow\!\!\!& SU(2)\to SU(3) \leftarrow SU(2)\to SU(3) \leftarrow SU(2)\to SU(3) \leftarrow SU(2)\\ \end{matrix}}, $$ modulo the normal subgroup $N$ generated by commutators of non-adjacent $SU(2)$'s.

Namely: $$ E_8=\mathrm{colim}\left(\scriptstyle\begin{matrix} SU(2)&\!\!\!\!\!\to SU(3) \leftarrow SU(2)\to SU(3) \leftarrow\!\!\!& SU(2)&&SU(2)\\ \downarrow&\downarrow&\downarrow&&\downarrow\\ SU(3)&SU(3)&SU(3)&&SU(3)\\ \uparrow&\uparrow&\uparrow&&\uparrow\\ SU(2)&SU(2)&SU(2)&\!\!\!\!\!\to SU(3) \leftarrow\!\!\!& SU(2) \end{matrix}\right)/N. $$

Here, whenever we see a subdiagram $SU(2)\to SU(3) \leftarrow SU(2)$, the two maps are given by $\big(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big)\mapsto \big(\begin{smallmatrix}a&b&0\\c&d&0\\0&0&1\end{smallmatrix}\big)$ and $\big(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big)\mapsto \big(\begin{smallmatrix}1&0&0\\0&a&b\\0&c&d\end{smallmatrix}\big)$.

If you want the complex Lie group $E_8$, use $SL(2)$'s and $SL(3)$'s instead of $SU(2)$'s and $SU(3)$'s.

If you want the group of $k$-points of the algebraic group $E_8$, use $SL(2,k)$'s and $SL(3,k)$'s ($k$ an arbitrary ring). [Edit: this doesn't work for arbitrary rings — see the comments below]

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    What are the maps in the one spot, where $SU(2)$ goes into 3 $SU(3)$'s? – Donu Arapura Aug 20 at 18:13
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    Doesn't the version for rings imply that $G(E_8, R)$ is generated by its $A_2$ subsystem subgroups? This is not true in general, but only holds for some specific rings, say, satisfying $\dim(Max(R))\leqslant 2$. – Andrei Smolensky Aug 20 at 18:27
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    If this is "easy", no wonder I don't get it. – Monty Harder Aug 20 at 22:04
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    Reference, please. – Vít Tuček Aug 20 at 22:19
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    @Vincent It does for simply-laced diagrams, because on the level of elementary generators all the defining relations are inside $A_2$ subsystems. This is most easily seen by looking at the associated Steinberg group, see "Steinberg groups as amalgams" by Allcock or "On the centrality of $K_2$ for Chevalley groups of tyle $E_\ell$" by Sinchuk. – Andrei Smolensky Aug 21 at 14:16

The algebraic group $E_8$ is the group of automorphisms of the $E_8$ lattice vertex algebra, by Frenkel-Kac and Segal. This vertex algebra has a self-dual integral form, so the construction works over arbitrary commutative rings. To construct the vertex algebra, one only needs the rank 8 unimodular $E_8$ lattice, not the 248-dimensional $E_8$ Lie algebra.

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    Great answer! It'll take me some time to fully appreciate it, but I think this was the type of thing I was after. I'll probably accept this answer soon, but since so many specialists seem to be around, let me ask even more: does anybody know how to construct the building $E_8(K)$ directly? Yesterday I would have bet that nobody did, but I keep being surprised! – Pierre Aug 21 at 8:02
  • Note that the 1-product on the degree one elements of that vertex algebra is nothing else than the Lie algebra of $E_8$. So this answer is not very far away from saying that $E_8$ is the automorphism group of the $E_8$ Lie algebra. – André Henriques Aug 23 at 20:05
  • @André: my understanding of these algebras is very shallow at the moment, but it seems that the construction of the algebra from the lattice is relatively direct. I have in mind the "magic square" of Tits and others, where the construction of the Lie algebra E8 is... well, not to my taste. You're right, though. I should have changed my question to mention that nice constructions of the Lie algebra are acceptable answers... – Pierre Aug 24 at 8:52
  • The construction of the $E_8$ Lie algebra from the $E_8$ root system is also very direct. Take the 8dim Cartan $\oplus$ a 1dim root space for every one of the 248 roots. Then define the Lie bracket in "the only reasonable way possible". Now, there's a subtlety: "the only reasonable way possible" involves annoying $\pm 1$'s. Things become clear when you understand that these $\pm 1$'s are equivalent to a $\mathbb Z/2$-central extension of the $E_8$ root lattice. Now, the construction of the $E_8$ vertex algebra from the $E_8$ root lattice also uses that same $\mathbb Z/2$-central extension... – André Henriques Aug 24 at 15:27
  • @AndréHenriques The vertex operators "generated by" the weight 1 space give an integrable action of the $E_8$ formal group over $\mathbb{Z}$, not just the Lie algebra. – S. Carnahan Aug 24 at 17:31

Recently Lusztig gave a much simpler definition of $E_8$ (and all the simple Lie algebras/groups) that avoids the usual sign issue with the standard Chevalley/Serre construction. See Lusztig - On conjugacy classes in the group $E_8$ and Geck - On the construction of semisimple Lie algebras and Chevalley groups. I think this construction of Lusztig's is not as well known as it should be.

  • Is it related to the (vertex agebra motivated) construction described around p. 140 of the following notes of a course by Richard Borcherds ? people.brandeis.edu/~syzygy/LieGroups.pdf – BS. Aug 21 at 7:28
  • @BS: I'm not sure about that. I think Lusztig's construction is coming from the theory of canonical bases (not vertex algebras). – Sam Hopkins Aug 21 at 12:05

Here's one of my favorite definitions, that gives a construction of (many forms) of $E_8$ as well as a plethora of other exceptional groups. I hope it's ok to construct the Lie algebras here, and let the groups be their automorphism groups.

Begin with a field $k$ of characteristic zero. Let $B$ and $C$ be composition algebras over $k$. I.e., $B$ and $C$ are $k$-algebras, isomorphic to $\bar k$ or $(\bar k \times \bar k)$ or $M_2(\bar k)$ or $O_{\bar k}$ (split octonion algebra over $\bar k$) after passing to an algebraic closure $\bar k / k$. For the split $E_8$, take $B = C = O_k$.

Let $A = B \otimes_k C$, viewed as a $k$-algebra with involution, typically nonassociative. For $E_8$, $A = O_k \otimes_k O_k$. This is an important example of a "structurable algebra," defined and studied by Bruce Allison.

Allison, B. N., Tensor products of composition algebras, Albert forms and some exceptional simple Lie algebras, Trans. Am. Math. Soc. 306, No. 2, 667-695 (1988). ZBL0649.17006.

Now I'll review the construction of a Lie algebra from $A$, following section 1 of the above reference. If $x \in A$, define $L_x \in End_k(A)$ by $L_x(y) = xy$. Define $R_x \in End_k(A)$ by $R_x(y) = yx$. Define $V_{x,y} \in End_k(A)$ by $$V_{x,y}(z) = (x \bar y)z + (z \bar y)x - (z \bar x)y.$$

Let ${\mathfrak g}_0$ be the $k$-subspace of $End_k(A)$ generated by the endomorphisms $V_{x,y}$ for all $x,y \in A$. A key identity (for all structurable algebras) implies that ${\mathfrak g}_0$ is not just a $k$-subspace... it's a Lie algebra!

Lie ${\mathfrak g}_{\pm 1} = A$. Let ${\mathfrak g}_{\pm 2} = A_0 = \{ a \in A : a + \bar a = 0 \}$. Define $${\mathfrak g} = {\mathfrak g}_{-2} \oplus {\mathfrak g}_{-1} \oplus {\mathfrak g}_{0} \oplus {\mathfrak g}_1 \oplus {\mathfrak g}_2.$$ Allison defines a 5-graded Lie algebra structure on this direct sum, directly from the algebra-with-involution structure on $A$. We've already seen the Lie bracket on ${\mathfrak g}_0$. The bracket $[X,Y]$ with $X \in {\mathfrak g}_0$ and $Y \in {\mathfrak g}_1$ is the obvious one. For the rest, I'll refer to Section 1 of Allison's paper, since it's a bit tedious (and takes up a half-page).

Anyways, for $A = O \otimes O$, the 64-dimensional structurable algebra, this gives an explicit construction of a Lie algebra of type $E_8$. In general, one gets a nice construction of 5-graded Lie algebras... it's very helpful for understanding "Heisenberg parabolics" for example.

  • Thanks, very interesting, too. But I have to say, all the constructions of the Lie algebra of $E_8$ using direct sums of vector spaces and ad hoc brackets (such as that in F. Adams's book on exceptional groups) are what I'm trying to a avoid... – Pierre Aug 22 at 7:46
  • Well -- everyone's got their own taste. But it's not really "ad hoc". Constructions via Jordan algebras lead to a deep understanding of Lie algebras with minuscule parabolics. Constructions (like above) via structurable algebras lead to a deep understanding of Lie algebras with 2-step parabolics. Since all simple Lie algebras have a natural 2-step parabolic (Heisenberg parabolic), this is important in practice! – Marty Aug 22 at 23:40

The most direct definition is probably as (the identity component of) the stabilizer of a tensor. More precisely, in The octic $E_8$ invariant by Cederwall and Palmkvist an explicit symmetric 8-form $f$ on $\mathbb{R}^{248}$ is constructed, having the compact $E_8$ (times $\{\pm1\}$) as the stabilizer in $GL(248, \mathbb{R})$. It is shown, moreover, in Simple groups stabilizing polynomials by Garibaldi and Guralnick, that any $E_8$-invariant 8-form that is not a scalar multiple of the fourth power of the Killing form $\kappa$ will do, and that the stabilizer in $GL(248, \mathbb{C})$ is the complex form (times $\mu_8$). The invariant 8-forms can be produced from arbitrary 8-forms by averaging with respect to the Haar measure (and almost all of them are not proportional to $\kappa^4$).

There seems to be a construction of $\mathfrak{e}_8$ using some sort of geometrical objects in Configurations of lines and models of Lie algebras by Laurent Manivel.

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