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.
