Timeline for Why Use Hypercohomology When Defining the de Rham Cohomology of a Smooth Scheme over $k$?
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17 events
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| Apr 16, 2020 at 6:05 | history | edited | Praphulla Koushik | CC BY-SA 4.0 |
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| Apr 27, 2018 at 13:56 | vote | accept | Alekos Robotis | ||
| Apr 27, 2018 at 12:21 | answer | added | Donu Arapura | timeline score: 18 | |
| Apr 27, 2018 at 10:03 | comment | added | Denis Nardin | Long story short: you want the Mayer-Vietoris sequence. In fact choosing the value on affines plus the requirement of having a MV les forces you more or less to choose hypercohomology (which is just another version of sheavification for complexes). | |
| Apr 27, 2018 at 8:48 | review | Close votes | |||
| Apr 29, 2018 at 10:21 | |||||
| Apr 27, 2018 at 1:56 | comment | added | gdb | However, it is worth mentioning that the naive definition of de Rham cohomology does give you the right answer in the affine case because each term $\Omega^i_{X/k}$ is coherent. Thus the algebraic de Rham complex is a complex of $\Gamma$-acyclic sheaves. In this case you can use spectral sequence of a filtered complex to argue that $\mathrm H^i_{dR}(X)=\mathrm H^i_{dR-naive}(X)$ provided that $X$ is an affine scheme. | |
| Apr 27, 2018 at 1:51 | comment | added | gdb | There is no chance to have a result of this sort for the naive definition of the algebraic de Rham cohomology. For instance, by the very definition $\mathrm H^i_{dR-naive}(X)=0$ for $i>\operatorname{dim} X$ but we know (by the comparison of smooth de Rham cohomology with topological cohomology) that $\mathrm H^{2d}_{dR-\mathcal C^{\infty}}(X^{an})=\mathbf C$ for a smooth proper $\mathbf C$-scheme of dimension $d$. Also, you can see by hands that the naive de Rham complex doesn't give you the correct answer for $\mathbf P^1_{\mathbf C}$ (it never gives you cor. ans. for proper sm. schemes) | |
| Apr 27, 2018 at 1:45 | comment | added | gdb | First of all, as pointed out in other comments it doesn't make sense to consider 'usual cohomology' because the result won't be a vector space. You are probably asking why we define de Rham cohomology as hypercohomology of the de Rham complex rather than homology of the complex of its global sections. The main reason is that we want this invariant to be useful. For example, we want the algebraic de Rham cohomology of a smooth finite type $\mathbf C$ scheme to coincide with ($\mathcal C^{\infty}$ or holomorphic) de Rham cohomology of the associated analytic space. | |
| Apr 27, 2018 at 1:36 | comment | added | Mohan | The DeRham cohomolgy is a $k$-vector space and not a sheaf of $k$-vector spaces. | |
| Apr 27, 2018 at 0:19 | comment | added | Alekos Robotis | So we want to use the hypercohomology so that the result is a $k-$vector space, rather than a sheaf of $k-$vector spaces? | |
| Apr 27, 2018 at 0:15 | comment | added | Mohan | Again, complex as usual means what? One interpretation is to take the homology of the complex as it stands, then you get sheaves as the result, which is not what one is looking for. The complex is exact as sheaves of $k$-vector spaces in the analytic category. | |
| Apr 26, 2018 at 22:30 | history | edited | Alekos Robotis | CC BY-SA 3.0 |
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| Apr 26, 2018 at 22:26 | comment | added | Alekos Robotis | Perhaps ordinary does not add anything to the statement. I just mean why "bother" with the hypercohomology. What problem would one encounter by defining the de Rham cohomology of $X/k$ by the cohomology of the complex as usual. | |
| Apr 26, 2018 at 22:24 | comment | added | Mohan | What exactly do you mean by ordinary cohomology? | |
| Apr 26, 2018 at 22:19 | history | edited | Alekos Robotis | CC BY-SA 3.0 |
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| Apr 26, 2018 at 22:15 | review | First posts | |||
| Apr 27, 2018 at 0:12 | |||||
| Apr 26, 2018 at 22:13 | history | asked | Alekos Robotis | CC BY-SA 3.0 |