Timeline for Groups becoming algebraic groups
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2012 at 11:01 | comment | added | user28172 | @Yves: Ah yes, I had misunderstood the intended meaning of "or" when Jeremy wrote "(of projections to $X$ or $Y$)". | |
| Nov 18, 2012 at 10:32 | comment | added | YCor | @nosr: obviously you require the map to be a morphism in both $x$- and $y$-fibers. | |
| Nov 18, 2012 at 10:25 | vote | accept | Jérémy Blanc | ||
| Nov 18, 2012 at 2:21 | comment | added | Andreas Blass | Since there was some discussion about the extent to which the group structure is needed, the following might be relevant. If $k$ is a countably infinite field, then there is a function $f:k\times k\to k$ that is not a polynomial even though, for each fixed value of $x$ in $k$, $f(x,y)$ is a polynomial function of $y$ and, for each fixed $y\in k$, $f(x,y)$ is a polynomial function of $x$. (There is no such $f$ when $k$ is uncountable.) | |
| Nov 17, 2012 at 23:42 | comment | added | user28172 | @Jeremy: I don't know what such a result for $X \times Y \rightarrow Z$ could be, since we could use a morphism $f_x: Y \rightarrow Z$ on the $x$-fiber that has horribly non-algebraic dependence on $x$. I've never heard of a fiberwise criterion for a map of $k$-points to be a morphism. | |
| Nov 17, 2012 at 23:39 | answer | added | user28172 | timeline score: 12 | |
| Nov 17, 2012 at 22:20 | comment | added | YCor | @Jeremy well, I only thought of the question in terms of the more general question whether a fiberwise morphism function on a product is a morphism. I'd be surprised that the question with additional hypotheses (associativity etc) would change a lot. | |
| Nov 17, 2012 at 21:42 | comment | added | Jérémy Blanc | I mean a map $X\times Y\to Z$ where the restriction on each fibre (of projections to $X$ or $Y$) is a morphism "should be" a morphism. I have heard such kind of result but if it is true it should be harder to prove than my "simple" question. | |
| Nov 17, 2012 at 21:04 | comment | added | Jérémy Blanc | The motivation is that I have really abstract examples that I want to be algebraic group and which satisfy the conditions. By the way, I think that the result is true, because the left-multiplication is a morphism, and the map $G\times G\to G$ is given by morphisms in each fibre so should be a morphism, but the last part is really harder without group actions (and I've just heard it, without seeing a proof). | |
| Nov 17, 2012 at 19:36 | comment | added | anon | What's the motivation for the question? Even if the answer were yes (which I doubt), it wouldn't make it easier to verify that a variety is an algebraic group. | |
| Nov 17, 2012 at 19:25 | comment | added | user28172 | @Mikhail: In char. $p$ such near-misses (losing (3)) abound. Let $H$ be a linear algebraic group and $H'$ an infinitesimal non-normal closed subgroup scheme of $H$. The quotient scheme $G = H'\backslash H$ is a smooth affine variety and the natural map $H(k) \rightarrow G(k)$ is bijective. If $G(k)$ is thereby equipped with the group structure of $H(k)$ then this is non-algebraic (as otherwise the quotient map $H \rightarrow G$ would be a homomorphism of algebraic groups and so its schematic kernel $H'$ would be normal), and (1) and (2) hold but (3) fails. | |
| Nov 17, 2012 at 19:02 | comment | added | Mikhail Bondarko | The multplicative group of quaternions (presented as $\mathbb{C}\times \mathbb{C} \setminus \{0\}$ yields a non-algebraic example of your (1)+(2); yet it fails (3). | |
| Nov 17, 2012 at 18:08 | comment | added | Jérémy Blanc | As I am talking about an algebraic variety, I meaned the (closed) points. I hope it is clearer now. | |
| Nov 17, 2012 at 18:07 | history | edited | Jérémy Blanc | CC BY-SA 3.0 |
added 112 characters in body
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| Nov 17, 2012 at 18:03 | comment | added | Will Sawin | What set are you using to be the elements of the group? | |
| Nov 17, 2012 at 18:01 | comment | added | YCor | you might specify if you have in mind characteristic zero or arbitrary characteristic. | |
| Nov 17, 2012 at 17:33 | history | asked | Jérémy Blanc | CC BY-SA 3.0 |