Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a Calabi-Yau 4-fold, i.e., a connected 4-dimensional compact Kahler manifold with $K_{X} \cong \mathscr{O}_{X}$ and $h^{i} (X,O_{X} )= 0$ for $0 \lt i \lt 4$. Given a general 4-dimensional Weil torus $T$, one has a Hodge class contained in $H^{2,2}( X )$. I would like to ask whether the following equality holds:

$$ Hg^{2} ( X)_{prim}= Hg^{2}\left ( T \right )_{prim}$$

where $Hg^{2}\left ( X \right )_{prim}$ is a primitive Hodge class, and $Hg^2 ( \star )$ is defined as $ H^4 ( \star ,\mathbb{Q} )\cap H^{2,2} ( \star )$ for $\star = X,T$. If the equality is faulty, I would like to know why. If Hodge class is not unique, how do we get the classification of Hodge classes?

share|improve this question
1  
I tried to fix the English, but I don't know enough about Weil tori to tell if I preserved the meaning of the second sentence. –  S. Carnahan Sep 25 '12 at 9:03
    
Carnahan,thank you to fix English grammar fault,About the definition of Weil tori,see Voisin's papers and search in internet. –  喻yuwei Sep 25 '12 at 10:18
    
It might be useful to provide a link: math.jussieu.fr/~voisin/Articlesweb/takagifinal.pdf I don't think this is true. To see this, take $X$ to be product of to K3 surfaces with Picard number 20 and compute the two sides. (I'd do it, but I have to run.) –  Donu Arapura Sep 25 '12 at 13:39
1  
I don't understand the meaning of the sentence "Given a general 4-dimensional Weil torus T, one has a Hodge class contained in $H^{2,2}(X)$". What does $T$ have to do with $X$? Can you please explain it, or correct it if wrong? –  YangMills Sep 25 '12 at 16:26
    
Explain :the subspace$X^{'}$ of a Weil torus$X$ consists of the Hodge classes,and belongs to$H^{4}\left ( X,\mathbb{Q} \right )$ ,by Hodge decomposition,it contained in $H^{2,2}\left ( X \right )$ ,If it is not Hodge classes,maybe contained in$H^{1,3}\left ( X \right )$ ,$H^{3,1}\left ( X \right )$ ,$H^{0,4}\left ( X \right )$ or$H^{4,0}\left ( X \right )$ .Of course,the Hodge classes of$T$ whether exist in $X$ ,I hope have proof or counterexample .Prove its existence is a very diffficult. –  喻yuwei Sep 26 '12 at 2:53
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.