Timeline for Cohomology class of the intersection of two hypersurfaces
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2014 at 8:37 | vote | accept | Alex Gavrilov | ||
| Jan 29, 2014 at 11:13 | history | edited | Alex Gavrilov | CC BY-SA 3.0 |
added 73 characters in body
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| Jan 29, 2014 at 11:09 | answer | added | Alex Gavrilov | timeline score: 0 | |
| Jan 28, 2014 at 8:41 | comment | added | Alex Gavrilov | Looks like you are right. I was under impression that the components have the same class, because they are symmetric under the symmetry group of $X$. But under close examination, this group isn't connected. My bad. | |
| Jan 28, 2014 at 8:05 | comment | added | Alex Gavrilov | If you think that $p$ and $q$ are the classes of components of $Y\cap Z$, then I do not believe this. | |
| Jan 28, 2014 at 8:02 | comment | added | Alex Gavrilov | abx, what do you mean? With my notation, $[Y]=[Z]=h$, because these are hyperplane sections. | |
| Jan 27, 2014 at 18:06 | comment | added | Allen Knutson | A more general statement is, if $X$ smooth complete contains $Y,Z$ where $Y$ is a hypersurface containing no component of $Z$, then $[Y\cap Z] = [Y]\smile[Z]$. | |
| Jan 27, 2014 at 17:04 | comment | added | abx | This is well known. A 4-dimensional smooth quadric $X$ has $H^2(X,\Bbb{Z})=\Bbb{Z}.h$ and $H^4(X,\Bbb{Z})=\Bbb{Z}.p+\Bbb{Z}.q$, with $p^2=q^2=1$, $p.q=0$ and $h^2=p+q$. With your notation, $[Y]=p$ and $[Z]=q$. | |
| Jan 27, 2014 at 16:11 | comment | added | Alex Gavrilov | Why do you think so? | |
| Jan 27, 2014 at 14:17 | comment | added | abx | Of course it is $[Y]\smile [Z]$! This is clear in your example. Try again! | |
| Jan 27, 2014 at 13:41 | history | asked | Alex Gavrilov | CC BY-SA 3.0 |