I am following “String Theory and M-Theory” by Becker, Becker, and Schwarz and I am currently studying chapter 9. I have a question - or better yet a point of confusion - regarding the derivation of the massless four-dimensional spectra when considering the IIA/B compactified on a ${\rm CY}_3$. Some context follows in order to present my query in a complete way.

We are considering a compactification of the IIA/B of the form $\mathcal{M}_{10} = \mathcal{M}_{4} \times {\rm CY}_3$. Since the spacetime metric is a sum of a four-dimensional and a six-dimensional piece the Laplacian also assumes the same form, namely $\Delta_{10} = \Delta_{4} + \Delta_{6}$. Then, the number of massless modes in the four-dimensional space is given by the number of the zero modes of the six-dimensional Laplacian. The number of these zero-modes of interest is given by the Betti number.

Let me be a bit more explicit and consider the ten-dimensional two-form field (this is the example considered on page 386). In what follows I am sticking to the notation in the book, that is the splitting of the indices is $M=(\mu,m)$. The two-form field can be written as

$\begin{equation} B_{MN} = B_{\mu \nu} \oplus B_{\mu n} \oplus B_{m n} \end{equation}$

Counting and interpretation.

From the four-dimensional point of view, the first term of the above relation is a two-form, the second one is a gauge field (one-form) and the final term is just a scalar (zero-form). From the six-dimensional point of view, the first term is a zero-form and the associated Betti number is the $b_0$. In the CY$_3$ case, we have $b_0=1$. The second term is a one-form in the six-dimensional picture and thus the related Betti number is $b_1=0$. The final term is a two-form and we have $b_2=h^{1,1}$.

Therefore, the number of massless states in the four-dimensional theory is one massless two-form, no massless gauge field and $h^{1,1}$ scalars.

The above situation is an example that I understand.

My question:

My confusion lies when the authors consider the CY$_3$ compactification of type IIA/B theories. Let's just take IIB on $\mathcal{M}_{4} \times $ CY$3$ to be concrete. Exercise 9.13 from the book on page 403 is precisely that. Let me present one case that confuses me. Consider the ${\rm SU}(3)$ covariant splitting of the indices - following the book - $M=(\mu,i,\overline{i})$. The metric is decomposed as

$\begin{equation} G_{MN} = G_{\mu \nu} \oplus G_{ij} \oplus G_{i \bar{\jmath}} \end{equation}$

In the result of the exercise, it is stated that the first of the above is associated to $1$ which is the $b_0$ and makes sense, the third is related to the $h^{1,1}$ which is the result of $b_1$ and also makes sense but the term $G_{ij}$ is said to be related to $h^{2,1}$. The only Betti number on a CY$_3$ related to that Hodge number is $b_{3}$. This is precisely what does not make any sense to me. The term $G_{ij}$ has two indices on CY$_3$ and is thus a two-form so I was expecting that we would seek the $b_2$ number. Of course, I have similar questions with the indices in the rest of the $p$-forms in that exercise; I just wanted to give a simple example.

Can someone explain what I am missing or misunderstood?

I am certain that the book has no typo as the result that is presented has a nice interpretation in the context of mirror symmetry. To be precise, one can check that under the change $h^{1,1} \leftrightarrow h^{2,1}$ the vector and hypermultiplets in the resulting four-dimensional theories get interchanged.

Edit: After the answer by ACuriousMind

I do not disagree with the suggestion about the $C_{\mu i j \overline{k}}$ part of the four-form and the vector multiplet. It has three indices on the $CY_3$ and that's fine. If you look though the gravity multiplet, you also see $C_{\mu i j k}$, which has three indices on the Calabi-Yau and I was expecting that it would be associated with $b_3$ and not $b_0$. I know that $b_3 = 2(1+h^{1,2})$, but it is not clear why $C_{\mu i j k}$ is associated with the $1$ and $C_{\mu i j \overline{k}}$ is related to $h^{1,2}$. Can you please add a comment regarding that point as well?


1 Answer 1


First, note that the "easy" way to see the relevance of $h^{2,1}$ for the vector multiplet is to just look at the $C_{\mu i j \bar{k}}$ vector part of it.

Second, the metric zero modes are special because of the metric moduli and you cannot treat them like the zero modes of any other field. It is not a coincidence that BBS discuss the structure of the metric moduli space before giving you these exercises. $h^{2,1}$ are also the complex structure moduli and the zero modes associated with $G_{ij}$ are these, so the structure of the multiplet containing both $C_{\mu i j \bar{k}}$ and $G_{ij}$ works out.

  • $\begingroup$ Maybe I chose a bad example. Thanks for the explanation. Allow me to pose a closely related question below. Please see the edited OP. $\endgroup$
    – user172341
    Apr 11, 2020 at 10:59
  • 1
    $\begingroup$ @Konstantinos You should probably re-read what the $h^{i,j}$ mean and how they relate to the split complex indices. The $C_{\mu ijk}$ is a $(3,0)$-form on the CY-fold and hence has zero modes $h^{3,0} = 1$. The $C_{\mu ij\bar{k}}$ is a $(2,1)$-form and hence has zero modes $h^{2,1}$. The $b_k$ are the sum of all $h^{i,j}$ with $i + j = k$ and are not what you should be thinking about here. $\endgroup$
    – ACuriousMind
    Apr 11, 2020 at 11:05
  • $\begingroup$ Ok. Thanks for that. My misunderstanding was the following. I thought I had to only consider the independent Hodge numbers, the ones that compose the Hodge diamond. Now, the whole construction is very clear. Thanks again $\endgroup$
    – user172341
    Apr 11, 2020 at 11:06

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