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Given any $n\geq 2$, is there an example of $n$-dimensional compact Kahler manifold such that its Hodge numbers satisfy $h^{1,1} = h^{2,2} < h^{3,3} = h^{4,4} < h^{5,5} = h^{6,6} < \cdots h^{[\frac{n}{2}],[\frac{n}{2}]}$.

Note that 2-plane complex Grassmannian's Hodge numbers satisfy $h^{0,0} = h^{1,1} < h^{2,2} = h^{3,3} < h^{4,4} = h^{5,5} < h^{6,6}=\cdots h^{[\frac{n}{2}],[\frac{n}{2}]}$.

edited May 24 at 11:07

asked May 23 2012 at 11:11

Ping
280●2●10

	$h^{1,1}=h^{2,2}<h^{3,3}=h^{4,4}<h^{5,5}=h^{6,6}<$ $h^{0,0}=h^{1,1}<h^{2,2}=h^{3,3}<h^{4,4}=h^{5,5}<h^{6,6}=\cdots.$ – Ping May 23 2012 at 11:12
	So strange! I don't why in the main text the two above formulae cannot be exhibited regularly and completely. So I give the formulae in the above comments. Sorry. – Ping May 23 2012 at 11:15
	Sorry. The correct two formulas is as follows. $$h^{1,1}=h^{2,2}<h^{3,3}=h^{4,4}<h^{5,5}=h^{6,6}<\cdots.$$ $$h^{0,0}=h^{1,1}<h^{2,2}=h^{3,3}<h^{4,4}=h^{5,5}<h^{6,6}=\cdots.$$ – Ping May 23 2012 at 11:18
	The formulae don't display because you've only got one '$' symbol. You should have one at the beginning and one at the end of the formulae. – Michael Albanese May 23 2012 at 11:29
	@Michael Albanese I am sure I used the correct symbols as this is not my first time to ask questions. But I really do not know why the expressions are as above.:-) – Ping May 23 2012 at 12:39

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YangMills · Answer 1 · 2012-05-31 07:53:53Z UTC

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EDIT: this answer refers to a previous version of the question.

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 Kähler manifold has $h^{1,1}>0$.

edited May 31 at 7:53

answered May 24 at 10:20

YangMills
3,340●10●16

Sorry. You misunderstand my meaning. The sequence in my question should stop at $h^{[\frac{n}{2}],[\frac{n}{2}]}$. Otherwise this question is clearly makes no sense. – Ping May 24 at 11:04

I know, my answer was meant to show you that you should put a bit more effort in writing your questions correctly! – YangMills May 25 at 2:38

@YangMills: Thanks. I have modified my original question. – Ping May 25 at 4:33

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Your modified question is trivial for $n<6$, since in that case $\mathbb{CP}^n$ works. – YangMills May 26 at 14:46

@YangMills: Of course. But obviously I want to know whether or not there exist examples for all $n$. – Ping May 28 at 0:59

the existence of compact Kahler manifolds satisfying some Hodge numbers' restrictions

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