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Counting triangles in squares in 7 Dimensions

In \( \text{3D} \) it's possible to make a regular tetrahedron with integer coordinates that all lie on the vertices of a cube. Note that the analogous phenomena is not possible in \(2\) dimensions. That is, we cannot create an equilateral triangle with co-ordinates that all lie on a square.

In another problem we showed an analogue of this occurs in infinitely many higher dimensions. Ie It is possible to make a regular n-simplex who's vertices all lie on the vertices of a regular n-hypercube. Specifically, one can show that this happens for 7-simplices and 7-hypercubes. I have 2 questions that I have tried to answer but would like to see if anyone gets the same result. SO here are the questions.

  1. How many distinct 7-simplices that have vertices on a 7-hypercube all share at least one common vertex?

  2. How many distinct 7-simplices in total have vertices on a common 7-hypercube?

  3. What's the size of an orbit of a single 7-simplex in the set of all the 7-simplices under the group of the symmetries of the 7-hypercube?

I have tried to answer all of these questions but if you don't want to be influenced by these (as they may be wrong) I will put these in the comments.

Details and Assumptions:

  • A regular \(n\)-dimensional simplex in \( \mathbb{Z}^n \) has \(n+1\) vertices that are all an equal distance apart. (It's like an \(n\)-dimensional version of an equilateral triangle!)

  • Here is the Wikipedia article on hypercubes. (It's like an \(n\)-dimensional version of a square!)

Note by Roberto Nicolaides
1 month, 3 weeks ago

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Answers I got for each question were 5,80,16 respectively for questions 1,2 and 3. If I have time later and anyone asks I'll write down how I got these numbers. Roberto Nicolaides · 1 month, 3 weeks ago

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@Roberto Nicolaides For question 1, I'm getting an answer of more than 5. However, if it was assumed that all the simplexes would only intersect at 1 vertex, then the answer would be 5, which isn't true (I'm not sure if you made the same assumption as me). What I did was to consider the Hadamard Matrix of order 8. Up to equivalence, it is known that there is only 1 unique Hadamard Matrix for order 8. So to find all the 7-simplex, we only need to consider the rearrangement of the rows for the Hadamard Matrix given by Michael here, which is

\[{H}_{2}=\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}\]

From here, it is easy to find more than 5 solutions. Julian Poon · 1 month, 3 weeks ago

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@Julian Poon Julian, I think this is a wonderful observation! Thanks so much :)

Update : Found my mistake, I now make it our to be 30 sharing at least one common vertex and 480 simplices in total with 16 still being in a common orbit! Roberto Nicolaides · 1 month, 3 weeks ago

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@Roberto Nicolaides How did you get 30 though? As of right now the only method I see consists of tedious counting... Julian Poon · 1 month, 3 weeks ago

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@Julian Poon It was a bit of a tedious counting argument using the fact that every hadamard matrix is generated by 3 linear independents Row vectors. So it came down to counting the number of ways we can create these tripples. Let me know if you want me to try write this down explicitly :) I realise I divided by an unnessary factor of 3! = 6 before. The same question seems exceptionally interesting in 11 dimensions!! Roberto Nicolaides · 1 month, 3 weeks ago

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@Roberto Nicolaides I don't really understand. What construction did you use that requires the 3 row vectors? By the way, I tried to find more 7-simplexes with different lengths and proved that there aren't anymore apart from those described by Hadamard matrixes. Julian Poon · 1 month, 3 weeks ago

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@Julian Poon I did not explain myself well at all, apologies. Maybe I can find some time this weekend to write down what I did rigorously and explicitly.

Nice result checking for other lengths! I expect we can prove that they must all be of this type (eg a fixed amount of 1s in each row given the dimension) by looking at two different formulas for the hypervolume of an n-simplex!

I think this is a super interesting problem to look at it in general- in particular looking at the groups acting on the set of all the simplices in the case where n=11. I suspect one can derive an interesting construction of some of the sporadic simple groups by looking at these simplices in different dimensions!

Glad that you seem to find the problem interesting too! I think I once gave a talk on this and will have the slides somewhere - let me know if you want a copy! Roberto Nicolaides · 1 month, 3 weeks ago

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@Roberto Nicolaides H'okay, where can you send me? Julian Poon · 1 month, 3 weeks ago

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@Julian Poon I suppose I could send you an email via my university email or send you a copy via drop box? Is there a discrete way to send info such as emails via Brilliant - something like private messaging? I don't know if there is a way to do this all within Brilliant.

I will also aim to start writing a complete account of what I did, dotting the is and crossing the ts this weekend ( I think it'll be good to clarify some thoughts in my head also!) Roberto Nicolaides · 1 month, 3 weeks ago

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@Roberto Nicolaides In the case where n=11, there are also no 11-simplices that can't be expressed with Hadamard. As of now, I have yet to find a method to prove for all \(n=4k-1\). My bounding method only works for \(n<12\). I'll work on it during the weekend as that's when my exams end. Julian Poon · 1 month, 3 weeks ago

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@Julian Poon Interesting, I'd love to see this bounding method. I've dropped you an email so feel free to delete this comment if/when you receive it :) Roberto Nicolaides · 1 month, 2 weeks ago

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