Timeline for Why can we not always take a Kähler class to be in rational cohomology?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2015 at 9:43 | history | edited | diverietti | CC BY-SA 3.0 |
modified the last part following the comments of ACL
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| Mar 3, 2015 at 9:25 | comment | added | diverietti | @ArtiePrendergast-Smith: it was my pleasure! I didn't want to "steal" your answer, but at some point I saw that you weren't answering and so... Anyway, maybe I shall modify my $\mathbb Q^2$ in $\mathbb Q^3$, to avoid the problem pointed out by ACL. | |
| Mar 3, 2015 at 8:39 | comment | added | user5117 | @diverietti: thanks for writing out the details. I thought of doing so, then the point raised by ACL occurred to me. The minor amount of extra effort to fix this turned me off... | |
| Mar 2, 2015 at 20:31 | comment | added | diverietti | Ahhhhh ok! Now I see the point!! Yes, indeed! My exemple was somehow artificial just to explain what was going on... You are right! | |
| Mar 2, 2015 at 17:10 | comment | added | ACL | I was just claiming that one can think of having $V_{\mathbf R}^{1,1}$ to be the line spanned by $(1,\sqrt 2)$, but that that it does not correspond to a an actual Hodge structure on $\mathbf Q^2$. | |
| Mar 2, 2015 at 13:59 | comment | added | diverietti | Right, there is no pure Hodge structure of weight 2 with $\dim V^{1,1}_\mathbb R=1$. But I am still missing something in order to understand your comment above... | |
| Mar 2, 2015 at 12:57 | comment | added | ACL | @diverietti. If $\dim(V)=2$, there is no Hodge structure on $V$ such that $\dim(V^{1,1}_{\mathbf R})=1$, is there? | |
| Mar 2, 2015 at 9:50 | comment | added | diverietti | @ACL. Sorry but I am not sure I understand well your comment... The solution to what? And what do you mean about the $V^{01}$? Cheers, Simone. | |
| Mar 2, 2015 at 8:45 | comment | added | ACL | It is funny that this example be essentially the solution, but never appears for Hodge structures (what would $V^{0,1}$ be?). | |
| Mar 2, 2015 at 7:47 | history | edited | diverietti | CC BY-SA 3.0 |
slight change following YangMills comment.
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| Mar 2, 2015 at 7:42 | history | edited | diverietti | CC BY-SA 3.0 |
slight change following YangMills comment.
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| Mar 2, 2015 at 7:40 | comment | added | diverietti | Dear YangMills, I think this is a misunderstanding: I was speaking about a general dense subset, saying in parenthesis to keep in mind the case of $V_\mathbb Q$. But I admit that the way I wrote it can be misleading. I try to fix it. | |
| Mar 2, 2015 at 4:12 | comment | added | YangMills | "may very well have empty intersection" is not correct since the $0$ vector is always in this intersection | |
| Mar 1, 2015 at 21:22 | history | answered | diverietti | CC BY-SA 3.0 |