# Does the 3875-dimensional rep of $E_8$ have a solution to $x\star x=0$?

Consider the compact Lie group $$E_8$$. Its second-smallest fundamental representation is $$3875$$-dimensional and admits a symmetric invariant form, and so is real: $$E_8 \curvearrowright \mathbb{R}^{3875}$$. Furthermore, this irrep admits a unique (up to scale) $$E_8$$-invariant symmetric $$3$$-tensor, studied for example in Garibaldi and Guralnik, Simple groups stabilizing polynomials, 2015. Using this inner product, I can think of this invariant tensor as a commutative but non-associative multiplication $$\star : \mathbb{R}^{3875} \otimes \mathbb{R}^{3875} \to \mathbb{R}^{3875}$$.

Question: Does $$\mathbb{R}^{3875}$$ contain any nonzero vectors $$x$$ such that $$x \star x = 0$$ for this multiplication?

Let me mention two ways that one could try to answer this question. I wasn't able to carry either out to completion, and there might be other approaches.

First, pick a random vector $$y \in \mathbb{R}^{3875}$$, and consider the symmetric 2-tensor $$x_1 \otimes x_2 \mapsto \langle x_1 \star x_2, y\rangle$$, where of course $$\langle,\rangle$$ denotes the $$E_8$$-invariant inner product. This 2-tensor is the symmetric bilinear form corresponding to $$\| x\|^2 = \langle x \star x, y\rangle$$. Suppose that you had access to a multiplication table for $$\star$$. Then you could write down this inner product, and diagonalize it — diagonalizing an inner product is fast on the computer — and see if there are any null vectors. If you for some $$y$$ this inner product is definite, then there are no solutions, and if on the other hand a couple different $$y$$s have the same null vector, then probably there is a solution. However, I was unable to build a multiplication table for $$\star$$. Note that it would suffice to write down a set of generators for the $$\mathrm{Lie}(E_8)$$-action on $$\mathrm{Sym}^3(\mathbb{R}^{3875})$$, since finding a common eigenvalue is pretty fast, and for that, it would suffice to write down generators for the action on $$\mathbb{R}^{3875}$$, which is to say it would suffice to construct a crystal basis. But my computer timed out when I asked it to do that.

Second, consider the cubic function $$f(x) = \langle x \star x, x\rangle$$ corresponding to the symmetric $$3$$-tensor. A solution to $$x\star x = 0$$ is the same as a critical point of $$f$$. We may restrict to the unit sphere $$S = S^{3874} \subset \mathbb{R}^{3875}$$; then a solution to $$x\star x = 0$$ is the same as a critical point of $$f|_S$$ at which $$f$$ vanishes. One could hope that perhaps $$f$$ is a Morse–Bott function on $$S$$. It definitely is not Morse because it is $$E_8$$-invariant, and I expect but haven't proved that $$\mathrm{Lie}(E_8)$$ acts freely on $$S$$. Furthermore, one could hope that the critical points at which $$f$$ vanishes have the same number of attracting and repelling directions — $$1813 = (3874 - 248)/2$$ of each. Finally, one could hope that perhaps the cells in this Morse(–Bott) complex carry some symplectic or complex structure forcing them to be even-dimensional? This happens for example for flag manifolds. One should be careful a bit: $$E_8$$, and hence the groupoid $$S/E_8$$, has torsion in its homology, which is consistent with even-dimensional cells and freeness of the $$\mathrm{Lie}(E_8)$$-action only if the stabilizers are nontrivial finite groups. Conversely, perhaps $$S/E_8$$ has so much homology that there must be a degree-$$1813$$ critical point.

• Guessing that this real representation has a natural form over $\mathbf{Q}$, the given product $\star$ is probably also with rational coefficients, and then one can also ask whether $v\neq 0$ such that $v\ast v=0$ can be chosen with rational entries. – YCor Jun 24 '19 at 6:23
• Incidentally, there is an obvious analogous question with E8 replaced by the Monster: does the Griess algebra have a real nilpotent element? Going in the other direction, perhaps it’s worth warming up with small groups preserving unique algebra structures on smaller real representations. – Theo Johnson-Freyd Jun 24 '19 at 13:27

The basic representation of $$E_8$$ has character $$j(\tau)^{1/3} = q^{-1/3}(1+248q+4124q^2 + \cdots)$$, and the 4124 decomposes as $$1+248+3875$$. By Frenkel-Kac-Segal, the basic representation has an $$E_8$$-lattice vertex algebra structure. The tensor on the 3875-dimensional subspace of Virasoro-primary vectors comes from restricting and projecting the $$-\cdot_1-$$-product on the 4124-dimensional weight 2 subspace, so it suffices to produce a primary vector $$v$$ such that $$v_1v = 0$$.
The $$E_8$$-lattice vertex algebra has a grading by the $$E_8$$-lattice, and the weight 2 subspace has lattice-degree supported by the lattice vectors of norm at most 4. Let $$v$$ be a nonzero element of weight 2 that is homogeneous with respect to lattice-grading, and whose lattice-degree has norm 4. Then $$v$$ is primary, and $$v_1v = 0$$ because it has lattice-degree of norm 16. This construction works over the standard self-dual $$\mathbb{Z}$$-form as well as the $$\mathbb{R}$$-form you consider.
• The vector $v$ you describe has the following simpler description (pointed out to me by Skip Garibaldi): choose a highest weight vector; then its square is a weight vector, but of weight not among the weights of the representation. Indeed, the lattice grading you are using is precisely the weight grading. That grading works great for the complex form, of course, and for the split form, which I believe is what you mean by the standard $\mathbb{Z}$-form? But not the compact form. Indeed, a weight vector definitely is null for the invariant inner product, which is positive-definite in my case. – Theo Johnson-Freyd Jun 23 '19 at 22:09
• I'm a bit confused by the equation $4124 = 1 + 248 + 3875$. Since we obviously also have $1 = 1$ shouldn't it follow by analogy that $248 = 1 + 248$? But that is clearly false. Can you explain what is going on here? Thanks in advance! – Vincent Jan 13 '20 at 21:16
• @Vincent The equation is a decomposition of a representation into a direct sum of irreducible representations. The 248-dimensional irreducible representation of $E_8$ does not have a 1 dimensional summand. – S. Carnahan Jan 14 '20 at 4:55
• Ha, thank you. Now I think I see where my confusion is coming from. I was thinking of 'ordinary' $E_8$ rather than affine $E_8$. Thanks for your patience – Vincent Jan 15 '20 at 18:55