the existence of compact Kahler manifolds satisfying some Hodge numbers' restrictions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:54:23Z http://mathoverflow.net/feeds/question/97745 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97745/the-existence-of-compact-kahler-manifolds-satisfying-some-hodge-numbers-restrict the existence of compact Kahler manifolds satisfying some Hodge numbers' restrictions Ping 2012-05-23T11:11:12Z 2013-06-18T07:14:55Z <p>Given any $n\geq 2$, is there an example of $n$-dimensional compact Kahler manifold such that its Hodge numbers satisfy $h^{1,1} = h^{2,2} &lt; h^{3,3} = h^{4,4} &lt; h^{5,5} = h^{6,6} &lt; \cdots h^{[\frac{n}{2}],[\frac{n}{2}]}$.</p> <p>Note that 2-plane complex Grassmannian's Hodge numbers satisfy $h^{0,0} = h^{1,1} &lt; h^{2,2} = h^{3,3} &lt; h^{4,4} = h^{5,5} &lt; h^{6,6}=\cdots h^{[\frac{n}{2}],[\frac{n}{2}]}$.</p> http://mathoverflow.net/questions/97745/the-existence-of-compact-kahler-manifolds-satisfying-some-hodge-numbers-restrict/97829#97829 Answer by YangMills for the existence of compact Kahler manifolds satisfying some Hodge numbers' restrictions YangMills 2012-05-24T10:20:16Z 2012-05-31T07:53:53Z <p>EDIT: this answer refers to a previous version of the question.</p> <p>Already for $n=3$ the answer is no. Indeed, $h^{3,3}=1$ so by your condition $h^{1,1}=h^{2,2}=0$ but a compact K&auml;hler manifold has $h^{1,1}>0$.</p>