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Jul 29 at 18:19 vote accept Akiva Weinberger
Jul 29 at 0:49 answer added Skip timeline score: 4
Jul 29 at 0:35 history edited Skip CC BY-SA 4.0
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Sep 9, 2018 at 11:16 comment added arsmath In general, a division algebra satisfies that the equations $xz = y$ and $z'x = y$ always has a solution for any nonzero $x$. For alternative algebras this implies the existence of a unit, but in general it doesn't.
Sep 9, 2018 at 11:11 comment added Akiva Weinberger @arsmath I don't have much experience with nonalternative algebras, but 1*a=a seemed pretty reasonable to me
Sep 9, 2018 at 11:10 comment added arsmath Why would it close up the search? "Unital" is not a natural property for non-alternative algebras.
Sep 9, 2018 at 9:53 history edited Akiva Weinberger CC BY-SA 4.0
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Sep 9, 2018 at 9:16 comment added Akiva Weinberger This looks like it answers it
Sep 9, 2018 at 9:02 comment added Akiva Weinberger I wonder if analyzing linearly independent vector fields on spheres will help us find, or prove the nonexistence of, such an algebra? (I know you can go the other way, that is, use algebras to find linearly independent vector fields on spheres.)
Sep 9, 2018 at 6:24 comment added Akiva Weinberger Also, Wikipedia doesn't seem to be consistent between unital and unitary.
Sep 9, 2018 at 6:23 comment added Akiva Weinberger @abx You know, I assumed nonassociative implies dimension 8, but I guess that might not actually be true. (Associative implies $\Bbb R$, $\Bbb C$, or $\Bbb H$, with or without the unital condition, at least according to Wikipedia.) If someone presented a unital dimension 4 division algebra other than $\Bbb H$ I'd accept it as an answer also.
Sep 9, 2018 at 6:18 comment added abx What about dimension 4? Is it known?
Sep 9, 2018 at 5:31 history edited Akiva Weinberger CC BY-SA 4.0
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Sep 9, 2018 at 5:26 history edited Akiva Weinberger CC BY-SA 4.0
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Sep 9, 2018 at 5:07 comment added Akiva Weinberger It occurs to me that I should have written some other dimension 2, dimension 4, and dimension 8 things in that last bullet point since $a*b=\overline{ab}$ generalizes to the quaternions as well.
Sep 9, 2018 at 5:01 history edited Akiva Weinberger CC BY-SA 4.0
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Sep 9, 2018 at 0:41 comment added John Wayland Bales This does not answer your question, but evidently there are examples over finite fields. tandfonline.com/doi/abs/10.1080/…
Sep 9, 2018 at 0:25 review First posts
Sep 9, 2018 at 0:26
Sep 9, 2018 at 0:23 history asked Akiva Weinberger CC BY-SA 4.0