Timeline for Almost Hermitian manifolds of constant curvature
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2017 at 6:13 | vote | accept | C.F.G | ||
| Sep 6, 2017 at 15:18 | answer | added | Robert Bryant | timeline score: 4 | |
| Sep 6, 2017 at 15:05 | comment | added | C.F.G | Can you write your counterexample as a Answer? Please accept my apology for this request and thank you for sharing your wisdom with me. | |
| Sep 6, 2017 at 14:51 | comment | added | Robert Bryant | I don't understand your comment or what you mean by 'not useful for conformal flatness'. The case $\dim = 5$ cannot happen, since we must be on an even dimensional manifold. The fact is that, in dimension $4$, your conditions only constrain the Weyl curvature of the underlying metric $g$, not the Ricci curvature, but they don't force the Weyl curvature to be zero; it's just required to lie in a certain $2$-dimensional subbundle that is defined by the $\mathrm{U}(2)$-structure given by $(g,\mathcal{J})$ on $M$. If you add that $g$ is Einstein, then, yes, you'll get constant curvature. | |
| Sep 6, 2017 at 14:42 | comment | added | C.F.G | Dear prof. Bryant, many thanks for your comment. May be the case dim $=4$ is a exception (such as Milnor exotic sphere! :) and I didn,t cheacked it. can be extend your conterexample to dim $=5$? That is surprising for me that the condition (2) is not useful for conformally flatness because the right hand side of (2) is $X_i$-free. | |
| Sep 6, 2017 at 14:17 | comment | added | Robert Bryant | The answer is 'no', at least when $2n=4$. In that case, your conditions only involve the Weyl curvature of the underlying metric $g$, so, in particular, if $g$ is conformally flat, then your conditions are satisfied, and there are many conformally flat metrics in dimension $4$ that do not have constant curvature. Then just choose any $g$-compatible almost complex struture $\mathcal{J}$, and you have a counterexample. Perhaps you meant to ask whether it can be deduced that $g$ is conformally flat? (This does not follow immediately since your conditions don't force the Weyl curvature to vanish.) | |
| Sep 6, 2017 at 7:46 | history | edited | C.F.G | CC BY-SA 3.0 |
added 12 characters in body
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| Sep 6, 2017 at 7:02 | history | asked | C.F.G | CC BY-SA 3.0 |