Aren't the edges of a $d$ dimensional Tetrahedron always equi-angular?

For equiangular lines, we have Matousek's proof that the number of such lines in $$d$$ dimensional space can't be more than $$\frac{d(d+1)}{2}$$. We then have this AEIS sequence that tells us the actual number of equiangular lines. For four dimensions, the number of equiangular lines possible is $$6$$.

Now, I believe I've found a way to configure $$\frac{d(d+1)}{2}$$ equiangular lines in $$d$$ dimensions; the edges of the $$d$$-cell. And this contradicts the AEIS sequence. There must be something wrong in my logic and I need help finding what it is.

Let's start with $$2$$ dimensions. We have an equilateral triangle (ABC) in this space. And any pair of lines of this triangle are at $$60^{\circ}$$ to each other. One can even translate all of them to the origin and they will remain equiangular (if we consider only the acute or only obtuse angles for any pair of lines).

Now, consider $$3$$ dimensions. Take the center of mass of this equilateral triangle (D) and put a point there. Move this point in a way that it is orthogonal to the plane of the triangle in the third dimension. By symmetry, DA, DB and DC are equiangular with each other. Also, the angle between DA and any of the edges of the triangle (AB, BC or AC) is the same by symmetry. As we move D away from this plane, we get a Tetrahedron at some point. This happens when the angles between DA, DB and DC with the plane of triangle ABC becomes $$60^{\circ}$$. The Tetrahedron has four vertices and they are all connected, so $${4 \choose 2}$$ edges, which form a system of equiangular lines. So, we get $$6$$ equiangular lines and this is consistent with the AEIS sequence.

But now, we can continue doing this as we increase the dimensionality of the space. We can take the Tetrahedron in $$3$$ dimensions and a point $$E$$ at its center of mass, moving it along the fourth dimension orthogonal to the 3-d space of the Tetrahedron. This will give us the $$5$$-cell. And per arguments before, all the edges of this object should be equiangular. And there are $$\frac{4(4+1)}{2}=10$$ edges, meeting the upper bound of Matousek's proof but contradicting the AEIS sequence which says there should be only $$6$$.

• per arguments before, all the edges of this object should be equiangular does not sound convincing at all. Can you elaborate (perhaps, compute the exact coordinates of $E$, and the corresponding angles)? Commented Aug 8, 2021 at 22:10
• We can also use symmetry arguments. Rotate the 5-cell so that the base Tetrahedron is ABCE and D is the point extending into the fourth dimension. Now, ABCE should be identical to ABCD (since all Tetrahedral faces are identical). But let me explicitly find the coordinates and compute the angles. Commented Aug 8, 2021 at 22:41

I found the resolution to the paradox and it is very, very disappointing. The Tetrahedron has let me down. Unlike the equilateral triangle, its edges don't form a system of equiangular lines. Let the base triangle be ABC and the vertex at the top be D. AB and DC are actually orthogonal ($$90^{\circ}$$) and not at $$60^{\circ}$$ to each other while AB and BC are at $$60^{\circ}$$.