Timeline for Alternative construction of the first Chern class map
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2016 at 3:47 | comment | added | Shubhodip Mondal | I am sorry, you are right. I should really mean $I^{\bullet +1}$. | |
| Jun 26, 2016 at 2:00 | comment | added | nfdc23 | You need to be careful about signs: since the complex $I^{\bullet}[1]$ differs from $I^{\bullet+1}$ by negating every other differential, the map $K^{\bullet} \rightarrow I^{\bullet}[1]$ is "off" from the analogous $K^{\bullet} \rightarrow I^{\bullet+1}$ by negating the map in every other degree. This then negates every other map in homology. Hence, your formulation as the separate question is really sensitive to whether you use $I^{\bullet}[1]$ or $I^{\bullet+1}$, and it seems you really want to be using the latter for the motivating context. Please think this through. | |
| Jun 25, 2016 at 8:46 | comment | added | Shubhodip Mondal | I posted this as a separate question mathoverflow.net/questions/243023/…. Please have a look. | |
| Jun 25, 2016 at 7:05 | comment | added | Shubhodip Mondal | (To show that it extends, we just need to keep in mind that $A''$ lands inside $\text{ker} (I^1 \to I^2)$.) | |
| Jun 25, 2016 at 6:47 | comment | added | Shubhodip Mondal | Thanks. I had thought of the following way to generalize this and get hold of the other connecting maps. Let $K^\bullet$ be a resolution for $A''$. We have a map $A'' \to I^1$ as you described. Now we can extend it to a map $K^0 \to I^1$. Then the whole thing gets extended, i.e., we get a map $K^\bullet \to I[1]^\bullet$, where $[.]$ denotes the shift. Now we can apply $F$ and get maps $R^i F (A'') \to R^{i+1} F (A')$. Do you know how they compare with the connecting maps? | |
| Jun 24, 2016 at 17:09 | comment | added | nfdc23 | Apply $F$ to a choice (unique up to homotopy) of map from the resolution $A \rightarrow A''$ (in degrees 0 and 1) of $A'$ to the resolution $I^{\bullet}$ of $A'$. That gives a map $F(A'') \rightarrow F(I^1)$ which by naturality of $F$ lands inside $\ker(F(I^1)\rightarrow F(I^2))$ and so induces a map $F(A'') \rightarrow H^1(F(I^{\bullet}))$. The sign discrepancy is an exercise in definition-chasing; I don't have a reference offhand (but it is basically a special case of the anti-commutativity of the lower-right square in the big diagram in Exercise 10.2.6 in Weibel's "Homological Algebra"). | |
| Jun 24, 2016 at 5:03 | comment | added | Shubhodip Mondal | I am sorry, I don't see how to obtain the map $F(A'') \to H^1( F(I^\bullet))$. Would you kindly elaborate? Also can you give me a reference? | |
| Jun 23, 2016 at 22:52 | comment | added | nfdc23 | I didn't see how to use the non-exact diagram of resolutions. The sign issue in degree 1 is a general fact of homological algebra: if $0 \rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ is a short exact sequence in an abelian category with enough injectives, $F$ is a left-exact additive functor to another abelian category, and $A'\rightarrow I^{\bullet}$ is an injective resolution then the connecting map $F(A'') \rightarrow R^1F(A)$ is negative of $F(A'')\rightarrow H^1(F(I^{\bullet}))$ arising from map from $A\rightarrow A''\rightarrow 0$ to $I^{\bullet}$ as resolutions of $A'$. | |
| Jun 23, 2016 at 17:38 | comment | added | Shubhodip Mondal | typo: $0 \to \mathscr O_X \to \mathcal A ^{0, \bullet}$ and $0 \to \Omega_X^1 \to \mathcal A ^{1, \bullet}$ | |
| Jun 23, 2016 at 17:28 | comment | added | Shubhodip Mondal | Some ideas: We have resolutions $0 \to \mathbf C \to \mathcal A_{\mathbf C}^{\bullet}$, $0 \to \mathscr O_X \mathcal A^{0, \bullet}$, $0 \to \mathcal A^{1, \bullet}$. Where in the resolutions the maps are given by $d, - \overline{\partial}$ and $ \overline{\partial}$. Now objects in the resolution can be connected by $\mathcal A ^k \to \mathcal A^{0,k} \to \mathcal A^{1,k}$, where the first map is given by projection and the second map is given by $\partial$. this gives us a commutative diagram. Can we get the boundary maps from this? (Unfrtunately the map of complexes is not exact) | |
| Jun 23, 2016 at 16:56 | comment | added | Shubhodip Mondal | I was naively wondering that all the connecting maps agree with the hodge theoritic inclusions! | |
| Jun 23, 2016 at 16:46 | comment | added | Shubhodip Mondal | Yes your elaboration is exactly what I had in mind. Thanks for writing it down in detail. So you are saying the zero degree connecting map is negative of Hodge theoritic inclusion right? Can you give a reference or show how it is done? Maybe that will help me to get some idea about the next connecting map! | |
| S Jun 23, 2016 at 16:14 | history | answered | nfdc23 | CC BY-SA 3.0 | |
| S Jun 23, 2016 at 16:14 | history | made wiki | Post Made Community Wiki by nfdc23 |