The situation I am describing is a $10D$ heterotic string theory which is compactified on a Calabi-Yau to get a $N=1$, $4D$ effective theory. It is mentioned in Ashoke Sen's notes on string compactification that the difference between the number of complex structure moduli and the Kahler moduli, i.e. $(h_{2,1}-h_{1,1})$ is the number of generations of the standard model. What is the exact argument for this?


1 Answer 1


In general in such heterotic compactifications on a Calabi-Yau manifold $X$, you have a structure group $G$ whose commutant is the effective low-energy structure group $H$, and you get effective supermultiplets transforming in representations $(R_i,r_i)$ of $G\times H$ by decomposing the adjoint of the heterotic $E_8$ into these irreducible representations of $G\times H$.

It turns out that the multiplicity $n_{r_i}$ of the representation $r_i$ is given by the dimension $h^1(X,V_{R_i})$ of the first cohomology $H^1(X,V_{R_i})$ of the bundle $V_{R_i}$ associated to the representation $R_i$ that occurs together with $r_i$, and the number of effective generations of particles is the difference between the number of generations $n_{r_i}$ and the number of anti-generations $n_{\bar{r}_i}$. Furthermore, $V_{\bar{r}_i} = V_{r_i}^\ast$, i.e. the anti-generation bundle is the dual of the generation bundle, so $h^1(X,V_{\bar{R}_i}) = h^2(X,V_{R_i})$, and the number of effective generations is $h^2(X,V) - h^1(X,V)$, which, incidentally, is the Euler characteristic of $V$ if $h^0$ and $h^3$ vanish, as they often do.

In the case you are asking about, the $V_{R_i}$ is simply taken to be the tangent bundle of the Calabi-Yau, so the $h^i(X,V)$ are simply the standard cohomologies associated to $X$. Therefore, the number of effective generations is $h^2 - h^1$.

See chapters 5.6 through 5.8 of "Heterotic and M-theory Compactifications for String Phenomenology" and the references therein for the full argument.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.