Timeline for What is the cohomology of the $C^{\infty}$-Koszul complex on $\mathbb{C}^2$?
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| Dec 9, 2021 at 9:50 | answer | added | Richard Lärkäng | timeline score: 1 | |
| Dec 8, 2021 at 7:33 | comment | added | Vladimir Dotsenko | Indeed, @Kapil is absolutely right. Forget for a second about the rightmost term $\mathbb{C}$ (which is usually not included in the Koszul complex anyway). Then your second complex is obtained from your first complex by extending the ground ring from $\mathbb{C}$ to the ring of antiholomorphic functions on $\mathbb{C}^2$, which immediately shows that its homology is the latter ring. | |
| Dec 8, 2021 at 5:51 | comment | added | Kapil | As far as you are concerned there is no difference between $\mathbb{C}^2$ and $\mathbb{R}^4$. In both cases you would be calculating the complex valued functions on them and evaluating at the origin. Hence, the correct Koszul complex to compare with is the one for $\mathbb{R}^4$. This observation would probably give you the answer to your question. | |
| Dec 8, 2021 at 5:39 | history | asked | Zhaoting Wei | CC BY-SA 4.0 |