I am studying the String Theory book by Becker, Becker, and Schwarz, and I decided to verify the Hodge diamonds for a CY3 and a CY4. These can be found on page 365 and they are eq.(9.14) and (9.16). It was very easy for me to derive the precise form of the Hodge diamond following what they mention in the book. A problem arose when I tried to repeat the computation for the case of a CY4.

Let me first state the problem/question. Why for a CY4 the Hodge number $h^{2,0}$ is equal to zero? This is not obvious to me from the relations they give in the book.

Allow me to show you how I have worked out the rest of the elements of the Hodge diamond.

Let me write here the properties the book gives for the Hodge numbers. For a Calabi-Yau n-fold we have that -these are eq.(9.10)-(9.12) in the book

$\begin{equation} \begin{split} h^{p,0} &= h^{n-p,0} \\ h^{p,q} &= h^{q,p} \\ h^{p,q} &= h^{n-q,n-p} \end{split} \end{equation}$

and we know that for a simply connected manifold $h^{1,0}=h^{0,1}=0$ and that a compact connected Kahler manifold has $h^{0,0}=1$.

From the first of the above properties, we have the following relations

$\begin{equation} \begin{split} &h^{4,3} = h^{3,4}, \qquad h^{4,2} = h^{2,4}, \qquad h^{4,1} = h^{1,4}, \qquad h^{4,0} = h^{0,4}, \qquad \qquad h^{3,2} = h^{2,3}, \qquad h^{3,1} = h^{1,3}, \\ &h^{3,0} = h^{0,3}, \qquad h^{2,1} = h^{1,2}, \qquad h^{2,0} = h^{0,2}, \qquad h^{1,0} = h^{0,1} \end{split} \end{equation}$

From the second property we have

$\begin{equation} h^{4,0} = h^{0,0} \qquad h^{3,0} = h^{1,0} \end{equation}$

And finally, from the third one we obtain

$\begin{equation} h^{4,4} = h^{0,0}, \qquad h^{4,3} = h^{1,0}, \qquad h^{4,2} = h^{2,0}, \qquad h^{4,1} = h^{1,0}, \qquad h^{3,3} = h^{1,1}, \qquad h^{3,2} = h^{2,1}. \end{equation}$

The undetermined $h^{2,2}$ is given by -see eq.(9.17)

$\begin{equation} h^{2,2} = 2 (22+2h^{1,1}+2h^{1,3}-h^{1,2}) \end{equation}$

If we impose all of the above to the Hodge diamond we get precisely what is shown in the book with the only difference that in the book they seem to have that $h^{2,0}=0$ which I cannot obtain.

  • 1
    $\begingroup$ Applying Hirzebruch Riemann-Roch for trivial bundle $\mathcal{O}_X$ and use the fact $c_1=0$ and expansion for Todd class, one obtains that $h^{2,0}+2=(3c_2\cdot c_2-c_4)/720$, but it is not clear to me how to compute $c_2\cdot c_2$. Hope someone can help. $\endgroup$
    – AG learner
    Apr 7, 2020 at 21:46
  • $\begingroup$ Thanks for the comment. Can you tell me what $c_1$ and $c_{2}$ are, please? $\endgroup$
    – user494842
    Apr 8, 2020 at 7:25
  • 1
    $\begingroup$ They are shorthands for first and second Chern classes of tangent bundle $T_X$. For computation involving Hirzebruch Riemann-Roch, see Salamon's paper. $\endgroup$
    – AG learner
    Apr 8, 2020 at 13:10
  • $\begingroup$ @AGlearner thank you for the suggestion $\endgroup$
    – user494842
    Apr 8, 2020 at 13:17

1 Answer 1


The answer depends on the definition of a CY variety that you are using.

If the definition is just the triviality of the canonical class and simply-connectedness then it is just not true that $h^{2,0}$ is zero. In fact, any hyperkahler fourfold is a counterexample (for instance, the Hilbert square of a K3-surface).

A more restrictive definition includes the assumption $h^{2,0} = 0$ (in fact, this is motivated by the Bogomolov Decomposition Theorem).

So, I guess, in your case the condition $h^{2,0} = 0$ holds by definition.

  • $\begingroup$ Thank you for your answer. The truth is that in the book it is not explicitly stated that $h^{2,0}=0$, however, it is the only way to obtain the Hodge diamond in the way that is presented in the book. Could you give me a good reference for the Bogomolov Decomposition Theorem or add a brief description? $\endgroup$
    – user494842
    Apr 8, 2020 at 9:11
  • 2
    $\begingroup$ The original reference is mathnet.ru/php/… $\endgroup$
    – Sasha
    Apr 8, 2020 at 9:16
  • $\begingroup$ Thanks a lot for the reference $\endgroup$
    – user494842
    Apr 8, 2020 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy