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Consider a regular n-simplex (the n-dimensional generalisation of a triangle/tetrahedron).

A triangle will tile the plane in a triangular pattern.

In 4, 8 and 24 dimensions. Can we tile the volume with n-simplices?

And will the vertices of those simplices give the lattices $F_4$, $E_8$ and Leech lattice in turn?

i.e. will a 5-cell tile 4D space where the vertices form the lattice $F_4$. If not what shape does?

(These are the sphere packing solutions in those dimensions). But are they also tiling of n-simplices?

Edit: I just read that in fact a 24-cell can tessellate 4D space in a $F_4$ arrangement

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The point symmetry of $F_4$ is "o3o4o3o" (typewriter friendly notation of Dynkin symbol). Polytopes representing that symmetry can be obtained by extending either nodes into true (non-zero) edges, e.g. the 24-cell with Coxeter-Dynkin symbol "x3o4o3o".

That point symmetry is extendable into a euclidean space symmetry "o3o4o3o3o". A hyper-honeycomb representing that symmetry then is be obtained quite similarily, e.g. the already mentioned packing of 24-cells with symbol "x3o4o3o3o".

Just as the cells of the 24-cell are octahedra, as can be given symbolically by "x3o4o .", so the cells of that tetracomb (4D hyper-honeycomb) are indeed 24-cells, as can be seen symbolically "x3o4o3o .". And just as the vertex figure of the 24-cell is the cube, as symbolically is obtained accordingly by ". x4o3o", so alike the vertex figure of that tetracomb is deduced from ". x4o3o3o" as the tesseract. The tesseract is known to show up 8 cubical cells, therefore you'd get accordingly eight 24-cells per vertex in that tetracomb.

Alternatively you also could have considered the tetracomb "o3o4o3o3x" of the same euclidean space group instead. The cells here are ". o4o3o3x", also known as 16-cells. The vertex figure of that tetracomb would have been "o3o4o3x .", i.e. the 24-cell. As it is known that the 24-cell has 24 cells, accordingly this tetracomb will show up twenty-four 16-cells per vertex. - Infact, those too regular tetracombs are dual to each other.

Quite similarily the point group of $E_8$ is "o3o3o3o3o3o3o *e3o" (where the "*e" part is just the typewriter friendly inline representation of the tridental Dynkin symbol: the link adjoined to "*e" is meant to revisit the e-th so far already mentioned node). Polytopes representing that symmetry can be obtained by extending either nodes into true (non-zero) edges, e.g. the 8D Gosset polytopes $4_{21}$ with Coxeter-Dynkin symbol "x3o3o3o3o3o3o *e3o", $2_{41}$ with symbol "o3o3o3o3o3o3x *e3o" or $1_{42}" with symbol "o3o3o3o3o3o3o *e3x".

This point group further can be extended similarily into the euclidean space group "o3o3o3o3o3o3o3o *f3o". Thus the according octacombs (8D hyper-honeycomb) would be either $5_{21}$ or "x3o3o3o3o3o3o3o *f3o", $2_{51}$ or "o3o3o3o3o3o3o3x *f3o", resp. $1_{52}$ or "o3o3o3o3o3o3o3o *f3x".

The cells of $5_{21}$ then are both, "x3o3o3o3o3o3o3o *f ." (or 8D simplices), 17280 per vertex, and "x3o3o3o3o3o3o . *f3o" (or 8D cross-polytopes), 2160 per vertex.

The cells of $2_{51}$ are both, ". o3o3o3o3o3o3x *f3o" (Gosset polytopes $2_{41}$), 16 per vertex, and "o3o3o3o3o3o3o3x *f ." (or 8D simplices), 128 per vertex.

And finally the cells of $1_{52}$ then are both, ". o3o3o3o3o3o3o *f3x" (Gosset polytopes $1_{42}$), 9 per vertex, and "o3o3o3o3o3o3o . *f3x" (or 8D hemi-hypercubes), 9 per vertex as well.

Obviously all 3 of those octacombs therefore no longer are regular, as this would require just a single cell type at least.

(24D finally is beyond my grasp. I never managed to wrap my mind around that Leech lattice.)

But for sure, in any dimension you would also have the hypercube "x4o3...3o", which allows to fill the according all-space regularily, as is shown by the hyper-honeycomb "x4o3...3o4o".

OTOH, the general $n$D simplex is encoded by the Coxeter-Dynkin symbol "x3o3...3o", i.e. it belongs to the point group $A_n$ or "o3o3...3o". And that group can indeed be continued into the euclidean space group "o3o3...3o3o3*a" (closed loop Dynkin symbol). Thus there surely are according hyper-honeycombs, obtained by extending either nodes into true (non-zero) edges, e.g. "x3o3...3o3o3*a". Sadly that one (and any other choice alike) will not be regular. This is simply because the cells here would be multiple again. Infact those are "x . o3...3o3o3*a" (simplex), "x3o . o3o3...3o3o3*a" (rectified simplex), "x3o3o . o3...3o3o3*a" (birectified simplex), ..., "x3o3...3o . o3*a" (rectified simplex again), "x3o3...3o3o . *a" (simplex again).

For instance the $A_2$ group "o3o" is being extended into "o3o3o3*a", giving rise to the tiling "x3o3o*a", which shows up the tiles "x3o . *a" (triangle) and "x . o3*a" (triangle again). Thus this one is regular.

But already $A_3$ would give rise to the extended (euclidean) space group "o3o3o3o*a". For that would provide the honeycomb "x3o3o3o$*$a", and that one would have for cells the types "x3o3o . *a" (tetrahedron), "x3o . o3*a" (octahedron), and "x . o3o3*a" (tetrahedron again).

--- rk

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The angle of an equilateral triangle is a sixth of $2\pi$, so we can fit six of them round a point. Thus it is unsurprising that one can tile the plane with equilateral triangles.

The dihedral angle of the regular tetrahedron is $\cos^{-1}(1/3)$. That is not a rational multiple of $2\pi$, so one cannot fit them round an edge. Thus a space packing of regular tetrahedra is impossible.

In $n\ge4$ dimensions, the same problem happens. The dihedral angle of the regular $n$-simplex is $\cos^{-1}(1/n)$, not a rational multiple of $2\pi$, and again a tiling with regular simplices is impossible.

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  • $\begingroup$ Maybe the simplices are not regular. $\endgroup$ – Oscar Lanzi Feb 6 at 2:52

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