The Lie algebra $\mathfrak{e}_8$ has (at least) two ways to be written as a direct sum of $31$ Cartan subalgebras.

First, Thompson and Smith showed that the (compact or complex) Lie group $\mathrm{E}_8$ contains a maximal abelian subgroup which is finite of order $2^5$, with normalizer the nonsplit "Dempwolff" extension $2^5 \cdot \mathrm{SL}_5(\mathbb{F}_2)$. Being maximal abelian, this $2^5$ group has no fixed points in the adjoint representation $\mathfrak{e}_8$. The other $31$ characters of $2^5$ are permuted by $\mathrm{SL}_5(\mathbb{F}_2)$, and so the corresponding eigenspaces inside $\mathfrak{e}_8$ are all of the same dimension and decompose $\mathfrak{e}_8$ as a direct sum of $31$ eight-dimensional subspaces. In fact, each of these eight-dimensional eigenspaces is a Cartan in $\mathfrak{e}_8$, since the bracket must add eigenvalues. All together this writes $\mathfrak{e}_8$ as a direct sum of $31$ Cartans indexed by the nontrivial points in $\mathbb{F}_2^5$.

Second, $\mathrm{E}_8$ contains a maximal abelian subgroup of order $5^3$, which might have been discovered by Serre but I'm not sure. Its normalizer is the nonsplit extension $5^3 \cdot \mathrm{SL}_3(\mathbb{F}_5)$. Again since $5^3$ is maximal abelian it has no fixed points in $\mathfrak{e}_8$, and the other eigenspaces are permuted by $\mathrm{SL}_3(\mathbb{F}_5)$ and so of equal dimension $2 = 248 / (5^3-1)$. Take any line through the origin in $5^3$. The corresponding four two-dimensional eigenspaces all commute (although I do not have a one-line proof of this fact) and sum to a Cartan subalgebra. All together this writes $\mathfrak{e}_8$ as a direct sum of $31$ Cartans indexed by the points in the projective plane over $\mathbb{F}_5$.

These two direct sum decompositions are not conjugate: if they were, then $2^5$ would commute with (a conjugate of) $5^3$, but these subgroups are each maximal abelian. Are the related in some other way? Is there some exceptional combinatorics that encompasses and explains both decompositions simultaneously?

The Lie algebra $\mathfrak{e}_8$ has (at least) two ways to be written as a direct sum of $31$ Cartan subalgebras.

First, Thompson and Smith showed that the (compact or complex) Lie group $\mathrm{E}_8$ contains a maximal abelian subgroup which is finite of order $2^5$, with normalizer the nonsplit "Dempwolff" extension $2^5 \cdot \mathrm{SL}_5(\mathbb{F}_2)$. Being maximal abelian, this $2^5$ group has no fixed points in the adjoint representation $\mathfrak{e}_8$. The other $31$ characters of $2^5$ are permuted by $\mathrm{SL}_5(\mathbb{F}_2)$, and so the corresponding eigenspaces inside $\mathfrak{e}_8$ are all of the same dimension and decompose $\mathfrak{e}_8$ as a direct sum of $31$ eight-dimensional subspaces. In fact, each of these eight-dimensional eigenspaces is a Cartan in $\mathfrak{e}_8$, since the bracket must add eigenvalues. All together this writes $\mathfrak{e}_8$ as a direct sum of $31$ Cartans indexed by the nontrivial points in $\mathbb{F}_2^5$.

Second, $\mathrm{E}_8$ contains a maximal abelian subgroup of order $5^3$, which might have been discovered by Serre but I'm not sure. Its normalizer is the nonsplit extension $5^3 \cdot \mathrm{SL}_3(\mathbb{F}_5)$. Again since $5^3$ is maximal abelian it has no fixed points in $\mathfrak{e}_8$, and the other eigenspaces are permuted by $\mathrm{SL}_3(\mathbb{F}_5)$ and so of equal dimension $2 = 248 / (5^3-1)$. Take any line through the origin in $5^3$. The corresponding four two-dimensional eigenspaces all commute (although I do not have a one-line proof of this fact) and sum to a Cartan subalgebra. All together this writes $\mathfrak{e}_8$ as a direct sum of $31$ Cartans indexed by the points in the projective plane over $\mathbb{F}_5$.

These two direct sum decompositions are not conjugate: if they were, then $2^5$ would commute with (a conjugate of) $5^3$, but these subgroups are each maximal abelian. Are they related in some other way? Is there some exceptional combinatorics that encompasses and explains both decompositions simultaneously?