Timeline for Universal identities on cubic surfaces or hypersurfaces
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 25, 2017 at 16:13 | comment | added | Gro-Tsen | @YCor Concerning your question of whether $w_1\circ w_2$ can be universally undefined without $w_1$ being universally equal to $w_2$, my intuition is no (the two are equivalent): I have no proof, but a handwavy "explanation" is that the only other possibility is that $w_1$ and $w_2$ are "universally on a line", and I don't think it's possible to universally select a line on a cubic surface without breaking the symmetries of the lines. (I have to admit, there are a lot of things I'm confused about, though.) | |
| Nov 25, 2017 at 11:20 | comment | added | Gro-Tsen | I should add that $v∘((u∘v')∘w)=v'∘((u∘v)∘w)$ does not hold universally even though it does when $u,v,v',w$ are in the same plane (the case $w=v∘v'$ is the identity I mentioned). This shows, at least, that cubic hypersurfaces do not get all the identities from the law $(x,y)\mapsto -x-y$ from an Abelian group. | |
| Nov 25, 2017 at 3:23 | history | edited | Gro-Tsen | CC BY-SA 3.0 |
replace incorrect identity by a correct one (noted by YCor in comments)
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| Nov 25, 2017 at 3:14 | comment | added | Gro-Tsen | @YCor In fact, the identity $(u∘v)∘(u∘v')=u∘(v∘v')$ was a mistake of mine (I was confused by the previously mentioned question and the fact that in an elliptic curve we take a point $u$ of order $3$ for origin). The correct first nontrivial identity I meant to write is $v∘((u∘v')∘(v∘v'))=v'∘((u∘v)∘(v∘v'))$ satisfied by $(x,y)\mapsto -x-y$ on an Abelian group. (I will fix my question.) | |
| Nov 24, 2017 at 15:36 | comment | added | YCor | Another question is about listing universally undefined elements. I'm wondering if it follows naively from listing universal identities. Namely, is the following true (I don't conjecture anything): "For all $w_1,w_2$ well-defined, either they are universally equal, or $w_1\circ w_2$ is well-defined."? (by "well-defined" I mean "not universally undefined", but I don't like double negations). In any case, it doesn't seem tautologically true, because $(w_1,w_2)$ might fall in a proper subvariety that is not the diagonal. | |
| Nov 24, 2017 at 11:02 | comment | added | YCor | Is the identity $(uv)(uv')=u(vv')$ obvious? | |
| Nov 23, 2017 at 15:51 | comment | added | YCor | About your "note": to keep the universal algebra point of view you probably want to define the quotient of the free magma on $\Sigma\cup\{\bot\}$ by the relations $\bot\circ x=x\circ\bot =\bot$ for all $x$ and $w=\bot$ whenever $w$ is universally undefined. Or you don't want to include the first relations, but the 2-sided ideal generated by $\bot$ becomes quite messy and I'm not sure you want to keep track of it. | |
| Nov 23, 2017 at 15:45 | comment | added | YCor | Unlikely, but in principle the answer could depend on the characteristic of the field: it seems that you have one definition per characteristic (unless you mean "for every alg. closed field and hypersurface") | |
| Nov 23, 2017 at 15:41 | comment | added | Gro-Tsen | Hmpf… Just after posting, I realized that my question was essentially a duplicate of this one (unanswered). I still feel that I stated it more precisely (and perhaps too tediously), but if someone wants to close it as a duplicate, I won't feel offended. | |
| Nov 23, 2017 at 15:36 | history | asked | Gro-Tsen | CC BY-SA 3.0 |