Timeline for Is there a compact complex manifold with $b_1(X)=b_2(X)=b_3(X)=b_4(X)=0$?
Current License: CC BY-SA 4.0
12 events
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Sep 8, 2022 at 21:59 | vote | accept | AmorFati | ||
Dec 31, 2021 at 14:34 | comment | added | YCor | @DonuArapura oops, indeed you're right. | |
Dec 31, 2021 at 13:27 | comment | added | Donu Arapura | @YCor Unless I'm mistaken, a compact Hermitian locally symmetric space would be Kähler, so $b_2$ would be positive. | |
Dec 31, 2021 at 13:09 | answer | added | Denis T | timeline score: 10 | |
Dec 31, 2021 at 12:46 | comment | added | Michael Albanese | There are infinitely many six-dimensional rational homology spheres which admit almost complex structures (and infinitely many which do not), but as far as I know, it is unknown whether any of them admit a complex structure. Aside from dimension two, rational homology spheres in every other dimension cannot even admit an almost complex structure. | |
Dec 31, 2021 at 9:44 | comment | added | YCor | I guess that an irreducible locally symmetric space has vanishing of Betti numbers below the rank (or so)? If so, irreducible locally symmetric Hermitian space of large enough rank would work. Hopefully somebody will confirm, as the literature looks quite arid to me. | |
Dec 31, 2021 at 4:52 | comment | added | abx | In dimension 3 an example would be a homology $\Bbb{Q}$-sphere. This is very close to ask whether $\Bbb{S}^6$ admits a complex structure... | |
Dec 31, 2021 at 2:21 | comment | added | AmorFati | @dhy Is there an example in complex dimension $3$? It doesn't seem to work for $\mathbb{S}^1 \times \mathbb{S}^5$. | |
Dec 31, 2021 at 2:06 | history | edited | AmorFati | CC BY-SA 4.0 |
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Dec 31, 2021 at 2:05 | comment | added | dhy | I think the answer to mathoverflow.net/questions/117098/… applies here as well. | |
Dec 31, 2021 at 1:55 | comment | added | Fernando Muro | Do you mean Betti numbers? The point seems to work. | |
Dec 31, 2021 at 0:29 | history | asked | AmorFati | CC BY-SA 4.0 |