# Alternatives to Calabi Yau?

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## Summary:

Alternatives to Calabi Yau?

## Main Question or Discussion Point

Are there alternatives to Calabi Yau spaces describing dimensions in superstring theory? If yes, what are they? If no, why?

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PeterDonis
Mentor
2019 Award
Moderator's note: Moved thread to the Beyond the Standard Model forum.

StenEdeback
Demystifier
Gold Member
Yes, D-brane worlds.

StenEdeback
Thank you, Demystifier!
Sten E

Demystifier
haushofer
You mean by "describing dimensions" "compactifications to 4 spacetime dimensions?"

StenEdeback
Thank you haushofer!
Yes, that is a good way of putting it. I wonder if Calabi Yau are the only players in that game, or if there are other ways of handling all these extra dimensions.

Demystifier
Gold Member
As for compactification, there are also orbifolds.

StenEdeback
samalkhaiat
As for compactification, there are also orbifolds.
Calabi-Yau is an orbifold.

StenEdeback
Thank you Demystifier and samalkhaiat! I will have a look at orbifolds.

Demystifier
Gold Member
Calabi-Yau is an orbifold.
Yes. But some orbifolds are not manifolds, and obviously I meant those orbifolds.

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StenEdeback
samalkhaiat
I wonder if Calabi Yau are the only players in that game,
Yes, they seem so. Broadly speaking, compactification requires Ricci-flat compact (complex) space. Ricci-flatness and compactness are what CY spaces have. Of course, finding the right one is another story.

StenEdeback
Demystifier
Gold Member
compactification requires Ricci-flat
Why does compactification requires Ricci flatness? Does it depend on a requirement that some supersymmetry survives at low energies, or is there an argument that does not depend on supersymmetry?

samalkhaiat
Why does compactification requires Ricci flatness?
Consider $M^{D} = M^{d} \times K^{D-d}$. From Supergravity action you obtain the Einstein equation in $M^{D}$: $$R^{(D)}_{AB} = 0.$$ This implies, $$R^{(d)}_{\mu\nu} = 0, \ \mbox{&} \ R^{(D-k)}_{mn} = 0.$$
Does it depend on a requirement that some supersymmetry survives at low energies,
I would agree with that, if I was a phenomenologist. Mathematically, world-sheet supersymmetry means that there are Killing spinors, $\nabla \epsilon = 0$, on the target space of the critical dimension $M^{10}$. Then, one can easily show that $\nabla \epsilon = 0 \ \Rightarrow \ R^{(10)}_{AB} = 0$. So, if you consider the solution $M^{10} = M^{4} \times K^{6}$, then $\mbox{Ric}(K) = 0$.
is there an argument that does not depend on supersymmetry?
You can't avoid supersymmetry in superstrings. Where do the fermions come from?

Demystifier and StenEdeback
Demystifier
Gold Member
You can't avoid supersymmetry in superstrings. Where do the fermions come from?
Well, at least for academic purposes one can study compactification in 26-dimensional bosonic string theory. By replacing supergravity action with bosonic field action, your argument can be used to argue that we would need Calabi-Yau even then, am I right?

StenEdeback
samalkhaiat
No. In order for the compact internal space $K^{D-4}$ to be a CY space, it must admit a flat Kahler metric. In superstring, the Kahler metric comes for free and we can show that it is flat. This is not so in bosonic string theory. The pure Plyakov action of bosonic string is very boring. However, interesting things happen if you include the NLSM torsion potential (known as the Kalb-Ramond field $H_{AB}(X) = H_{[AB]}(X)$) in the action and assume that the metric $G_{AB}(X)$ and the torsion potential $H_{AB}(X)$ are independent of $X^{0}$.