Timeline for On morphisms of pure Hodge structures of decreasing weight
Current License: CC BY-SA 3.0
28 events
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| S Apr 13, 2018 at 9:44 | history | suggested | random123 | CC BY-SA 3.0 |
made small changes so that the symbols with the prime as superscript appear correctly.
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| Apr 13, 2018 at 8:31 | review | Suggested edits | |||
| S Apr 13, 2018 at 9:44 | |||||
| Aug 13, 2011 at 16:33 | vote | accept | Hugo Chapdelaine | ||
| Aug 13, 2011 at 14:01 | vote | accept | Hugo Chapdelaine | ||
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| Aug 13, 2011 at 13:53 | vote | accept | Hugo Chapdelaine | ||
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| Aug 13, 2011 at 13:49 | vote | accept | Hugo Chapdelaine | ||
| Aug 13, 2011 at 13:53 | |||||
| Aug 13, 2011 at 10:05 | answer | added | naf | timeline score: 2 | |
| Aug 13, 2011 at 9:44 | comment | added | naf | I think your interpretation of the Scholie is correct: $f$ is NOT a morphism of Hodge structures but just a linear map of the underlying rational vector spaces which, when based changed to $\mathbb{C}$, preserves $F$. (For pure Hodge structures of a fixed weight $n$ this is a morphism of Hodge structures.) | |
| Aug 12, 2011 at 20:55 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
edited body; edited tags
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| Aug 12, 2011 at 18:49 | comment | added | Donu Arapura | I hadn't read that before. Scholie 5.1 is baffling to me as well, since a few lines up, Deligne makes a stronger statement "si $f:H\to H'$ un morphisme de structures de Hodge mixte pures de poids différents, alors $f$ torsion." If you figure it out, let us know. | |
| Aug 12, 2011 at 17:46 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 12, 2011 at 17:28 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 12, 2011 at 17:11 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 12, 2011 at 16:48 | comment | added | Hugo Chapdelaine | So in my question I'm only asking for a morphism of Hodge structure. My point is that a map which respects the torus action seems to be stronger than a map which only respects the filtration. | |
| Aug 12, 2011 at 16:46 | comment | added | Hugo Chapdelaine | Well, I never talked about mixed Hodge structure but only pure Hodge structure so there is no Galois action and therefore no weight filtration | |
| Aug 12, 2011 at 16:32 | comment | added | naf | @Hugo: I think you are right. I was confused by your use of the word morphism. There are no morphisms of (mixed) Hodge structures between pure Hodge structures of different weights but what you define is not a morphism (in the sense of Deligne) since there is no condition on the weight filtration. | |
| Aug 12, 2011 at 16:20 | comment | added | Hugo Chapdelaine | Well I'll try to resolve my confusion, any way thanks Donu. | |
| Aug 12, 2011 at 16:06 | comment | added | Donu Arapura | I claim that there are no nonzero morphisms (= $\mathbb{Q}$-linear filtered maps) from $\mathbb{Q}$ and the Tate twist $\mathbb{Q}(-1)$ (your example). But I'm afraid I'm going to have to leave this to you. | |
| Aug 12, 2011 at 15:52 | comment | added | Hugo Chapdelaine | So going back to my example we find that $F^k H_{C}=C$ if $k\leq 0$ and $0$ if $k>0$. Similarly we find that $H_{\mathbf{C}}'=\mathbf{C}$ if $k\leq 1$ and $0$ if $k>1$. Therefore for all $k$ we have that $F^k H_{\mathbf{C}}\subseteq F^k H_{\mathbf{C}}'$. | |
| Aug 12, 2011 at 15:48 | comment | added | Donu Arapura | Perhaps I understand the confusion, but I'm still not quite sure. A morphism defined over $\mathbb{Q}$, and in particular over $\mathbb{R}$, preserving the Hodge filtration $F$, must also preserve the conjugate $\bar F$. But $H^{pq}= F^p\cap \bar F^{q}$. So we are back to the my earlier point. The key point is the pair $(F,\bar F)$ are what Deligne calls opposed filtrations, and this leads to special features. Does this help? | |
| Aug 12, 2011 at 15:14 | comment | added | Hugo Chapdelaine | So you see I'm not asking that my map respects the torus action but simply that it respects the Hodge filtration which is somehow weaker | |
| Aug 12, 2011 at 15:14 | comment | added | Donu Arapura | The lemma (although not the proof) of my earlier comment is on page 25 of "Théorie de Hodge II". | |
| Aug 12, 2011 at 15:11 | comment | added | Hugo Chapdelaine | Well may be I'm confused but take $H_{Q}=Q$ and $H_{Q}'=Q$ with the identity map $\iota: H_Q\rightarrow H_{Q'}$. If you place $H_Q$ in degree $(0,0)$ and $H_{Q}'$ in degree $(1,1)$ then this respects the Hodge filtration, isn't ? | |
| Aug 12, 2011 at 15:08 | comment | added | Hugo Chapdelaine | My comment was meant to @Ulrich | |
| Aug 12, 2011 at 15:05 | comment | added | Hugo Chapdelaine | Well I have a counter-example to what you said take $H_{\mathbf{C}}=\mathbf{C}$ place in degree $(0,0)$ and take $H_{\mathbf{C}}=\mathbf{C}$ place in degree $(1,1)$. | |
| Aug 12, 2011 at 15:03 | comment | added | Donu Arapura | It's easier if you think that a weight HS has a bigrading $H\otimes \C$ with $\bar H^{pq}=H^{qp}$ and $p+q=n$. Since morphisms must preserve this, they would vanish for different weights. There is a lemma to be proved about the equivalence of this notion with representations of Deligne's torus, but it's not difficult. | |
| Aug 12, 2011 at 14:30 | comment | added | naf | One only needs to assume that $n \neq n'$... | |
| Aug 12, 2011 at 14:26 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |