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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 $h^{0,0} = h^{1,1} < h^{2,2} = h^{3,3} < h^{4,4} = h^{5,5} < h^{6,6}=\cdots h^{[\frac{n}{2}],[\frac{n}{2}]}$.

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$h^{1,1}=h^{2,2}<h^{3,3}=h^{4,4}<h^{5,5}=h^{6,6}<$ $h^{0,0}=h^{1,1}<h^{2,2}=h^{3,3}<h^{4,4}=h^{5,5}<h^{6,6}=\cdots.$ –  Ping May 23 '12 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 '12 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 '12 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 '12 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 '12 at 12:39
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1 Answer 1

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 $h^{1,1}>0$.

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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 '12 at 11:04
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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 '12 at 2:38
    
@YangMills: Thanks. I have modified my original question. –  Ping May 25 '12 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 '12 at 14:46
    
@YangMills: Of course. But obviously I want to know whether or not there exist examples for all $n$. –  Ping May 28 '12 at 0:59
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