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Feb 5, 2021 at 2:36 comment added Will Sawin Shouldn't the edge length be $2 \sqrt{6}$ here? (Or $\sqrt{6}$ if you normalize the hypercube edge length to $1$.)
Feb 3, 2021 at 13:55 comment added Richard Stanley In fact, a regular simplex of dimension $n$ can be inscribed in a hypercube of dimension $n$ if and only if $n+1$ is the order of a Hadamard matrix. If we also want every automorphism of the simplex to extend to an automorphism of the hypercube, then the $n=2^m-1$ result is correct.
Feb 3, 2021 at 11:22 answer added Adam P. Goucher timeline score: 7
Feb 3, 2021 at 8:24 history edited YCor CC BY-SA 4.0
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Feb 3, 2021 at 3:55 history edited Daniel Sebald CC BY-SA 4.0
Fixed error
Feb 3, 2021 at 3:55 comment added Daniel Sebald Later on in 4.5: “The proof of the ‘only if’ part of the statement of the last theorem may have seemed a bit unclear to the reader, and that is because it is false.”
Feb 3, 2021 at 3:11 comment added Tom Goodwillie You can't even have a triangle of vertices with side lengths $\sqrt 3$.
Feb 3, 2021 at 2:53 comment added Gerry Myerson "Theorem 4.5. A regular simplex of dimension n can be inscribed in a hypercube of dimension n if and only if $n=2^m −1$ for some m." $n=11$ is not of that form.
Feb 3, 2021 at 2:53 comment added Daniel Sebald Oops, you’re right.
Feb 3, 2021 at 2:52 history edited Daniel Sebald CC BY-SA 4.0
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Feb 3, 2021 at 2:39 comment added Gerry Myerson Wouldn't that hypercube have edges of length $2$, not $1$?
Feb 3, 2021 at 2:19 comment added Daniel Sebald Are you sure about that?
Feb 3, 2021 at 2:08 history edited Daniel Sebald CC BY-SA 4.0
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Feb 3, 2021 at 1:27 comment added Tom Goodwillie There is no such simplex.
Feb 3, 2021 at 0:47 history asked Daniel Sebald CC BY-SA 4.0