Timeline for When does a finite morphism induce isomorphism on cohomology?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 14, 2012 at 15:26 | history | edited | Karl Schwede | CC BY-SA 3.0 |
Fixed typo
|
| Sep 14, 2012 at 14:33 | vote | accept | Paul Graaf | ||
| Sep 14, 2012 at 12:41 | history | edited | Karl Schwede | CC BY-SA 3.0 |
added 40 characters in body
|
| Sep 14, 2012 at 12:22 | history | edited | Karl Schwede | CC BY-SA 3.0 |
added 1177 characters in body
|
| Sep 14, 2012 at 11:33 | comment | added | Karl Schwede | Definitely you can't do the trick I did above. There is never a splitting of $O_X \to O_{\widetilde{X}}$ and the trace is worthless. Let me think about normality briefly and then edit my answer. | |
| Sep 14, 2012 at 4:29 | comment | added | Paul Graaf | @Karl Scwede: Thanks for the answer. I see that the map in question need not be surjective. I had the normalization map $\bar{X} \to X$ in mind. Can we achieve injectivity if $X$ is not normal? Can the trace map be defined in non-normal situations? | |
| Sep 12, 2012 at 18:22 | comment | added | Karl Schwede | Indeed you are right. | |
| Sep 12, 2012 at 17:08 | comment | added | Will Sawin | Using the flat base change theorem, we can formalize your "I don't think one should expect that $H^n(X,\mathcal O_X) \to H^n(Y,\mathcal O_Y)$. Because $\mathbb C$ is flat over $\mathbb R$, $Y\to X$ is flat base change, so $H^n(Y,\mathcal O_Y)=H^n(X,\matcal O_X) \otimes_{\mathbb R} \mathbb C$, and the map between them is the obvious one. It is indeed never surjective when $H^n(X,\mathcal O_X)\neq 0$. | |
| Sep 12, 2012 at 16:57 | history | edited | Karl Schwede | CC BY-SA 3.0 |
added 251 characters in body; added 22 characters in body
|
| Sep 12, 2012 at 16:34 | history | answered | Karl Schwede | CC BY-SA 3.0 |