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Sep 18, 2019 at 20:14 vote accept M. Winter
Sep 17, 2019 at 18:07 history edited M. Winter CC BY-SA 4.0
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Sep 17, 2019 at 8:16 answer added M. Winter timeline score: 4
Sep 17, 2019 at 5:19 history edited M. Winter CC BY-SA 4.0
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Sep 17, 2019 at 5:13 comment added M. Winter @Bullet51 I think the cartesian product of a $d_1$-cube and a $d_2$-cube is a $(d_1+d_2)$-cube. You are right about the latter two examples. I already mentioned these in my comment, but I should edit them into my post.
Sep 16, 2019 at 23:30 comment added LeechLattice What about Cartesian products of $d$-cubes, rhombic dodecahedrons and rhombic triacontahedrons?
Sep 16, 2019 at 20:31 comment added M. Winter @IlyaBogdanov You are absolutely right. And also the rhombic triacontahedron. I have overlooked these. So I guess my question now is whether there are any more?
Sep 16, 2019 at 20:27 comment added Ilya Bogdanov Rhombic dodecahedron?
Sep 16, 2019 at 17:26 history asked M. Winter CC BY-SA 4.0