Timeline for On morphisms of pure Hodge structures of decreasing weight
Current License: CC BY-SA 3.0
12 events
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| Aug 13, 2011 at 16:41 | history | edited | naf | CC BY-SA 3.0 |
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| Aug 13, 2011 at 16:33 | vote | accept | Hugo Chapdelaine | ||
| Aug 13, 2011 at 16:33 | comment | added | Hugo Chapdelaine | Ok Ulrich, I got it. I was too quick with your key remark: The key point is that $H^{p,q}$ as defined above is $0$ if $p+q>n$ but not if $p+q<n$. Thanks a lot for clearing out my confusion. | |
| Aug 13, 2011 at 14:23 | comment | added | naf | Since $f$ is defined over $\mathbb{Q}$ (so commutes with complex conjugation) one also gets $f(\bar{F}^qH) \subset \bar{F}^q H'$. Your counterexample doesn't apply here since the conclusion uses $n > n'$. | |
| Aug 13, 2011 at 14:08 | comment | added | Hugo Chapdelaine | You have $f(F^P H)\subseteq F^p H'$ but I don't quite see how you get from that $f(H^{p,q})\subseteq H'^{p,q}$. In fact my counterexample shows that it is wrong, isn't? | |
| Aug 13, 2011 at 14:06 | comment | added | Hugo Chapdelaine | But Ulrich I'm confused, you only seem to use the fact that $n\neq n'$ and moreover what about my counterexample | |
| Aug 13, 2011 at 14:01 | vote | accept | Hugo Chapdelaine | ||
| Aug 13, 2011 at 14:04 | |||||
| Aug 13, 2011 at 13:53 | vote | accept | Hugo Chapdelaine | ||
| Aug 13, 2011 at 14:00 | |||||
| Aug 13, 2011 at 13:49 | vote | accept | Hugo Chapdelaine | ||
| Aug 13, 2011 at 13:53 | |||||
| Aug 13, 2011 at 12:10 | comment | added | Donu Arapura | .... to accept. | |
| Aug 13, 2011 at 12:08 | comment | added | Donu Arapura | Yes, that's it. I somehow mislead myself into believing that filtered $\mathbb{Q}$-linear maps between pure Hodge structures are morphisms, but this only true if the weights are the same. Sometimes the obvious interpretation is the hardest. | |
| Aug 13, 2011 at 10:05 | history | answered | naf | CC BY-SA 3.0 |