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"?
Best Answer
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]