I'm aware that mathematically speaking, Calabi-Yau manifolds are complex manifolds with vanishing first Chern number. However from a physics point of view, Calabi-Yau manifolds are related to the solution of Einstein's field equation in vacuum environment (i.e., with vanishing stress–energy tensor). Since Einstein's field equation is on a 4-dimensional real manifold, why Calabi-Yau manifolds are complex? Is there a "real version" of Calabi-Yau manifold?
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$\begingroup$ Superficially, "algebraic geometry" works best over algebraically-closed fields, and $\mathbb C$ is the obvious choice when the underlying "arithmetical analysis" involves $\mathbb R$. Is this the sort of thing you're asking about? $\endgroup$– paul garrettCommented Oct 26, 2017 at 23:45
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1$\begingroup$ Here is my understanding of the story: if you are doing 4-dimensional GR, then indeed the relevant manifold is not a Calabi-Yau but an Einstein Lorentzian manifold (I might not have the terminology quite right.) However if you are doing some form of 10d supergravity/string theory the 6d manifold you are compactifying on gets forced to be a Calabi-Yau manifold for some reason. I really have no understanding of the physics so take this with a large grain of salt. $\endgroup$– dhyCommented Oct 27, 2017 at 0:02
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4$\begingroup$ Since you seem to be looking for physics motivation, the easiest answer is holonomy. If you compactify superstring theory on a manifold with SU(3) holonomy, then a good bit of the supersymmetry is preserved (1/4 of it). Turns out then only manifolds with SU(3) holonomy are Calabi-Yau 3-folds. Also note every complex n-manifold does have an underlying smooth 2n-manifold. $\endgroup$– JJJCommented Oct 27, 2017 at 0:04
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3$\begingroup$ To be a little more precise, the physics of supersymmetry requires a covariantly constant spinor. This implies the SU(3) holonomy, which implies the complex structure. $\endgroup$– Aaron BergmanCommented Oct 27, 2017 at 1:32
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1$\begingroup$ My answer to another MO question might be of interest: mathoverflow.net/a/43603/394 $\endgroup$– José Figueroa-O'FarrillCommented Oct 27, 2017 at 10:05
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2 Answers
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I think that one possible answer is that a Calabi-Yau manifold is a Riemannian manifold $M$ with $SU(n)$ Riemannian holonomy, where $2n=\dim_\mathbb R M$.
Such a manifold is then necessarily complex, and the Riemannian metric is the real part of a Kähler metric which has zero Ricci curvature. Since the Ricci form in complex geometry is always a representative of the first Chern class of the manifold, what you ask follows.
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2$\begingroup$ Shouldn't that be $n = \frac{1}{2} \dim_{\mathbb{R}} M$? $\endgroup$– VincentCommented Oct 27, 2017 at 10:23
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$\begingroup$ Yes, it should! :) I'll fix it. $\endgroup$ Commented Oct 27, 2017 at 10:37
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$\begingroup$ Thank you for the answer! Is there any intuitive reason for a $SU(n)$ holonomy group instead of a real one? $\endgroup$ Commented Oct 28, 2017 at 13:15
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$\begingroup$ I am sorry Xige, not sure to understand correctly your question. The Lie group $SU(n)$ is indeed a real Lie group! $\endgroup$ Commented Oct 29, 2017 at 9:49
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$\begingroup$ Oh I forgot to add... Compare to $SO(n)$, $SU(n)$is "complex" in the sense that it's a subgroup a $GL(n,\mathbb{C})$. So why don't people use $SO(n)$ for holonomy groups? $\endgroup$ Commented Nov 1, 2017 at 20:36
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You might consider other real forms of $SU(n)$; a Riemannian manifold with holonomy in such a real form will then arise as extra dimensions in string theories with reduced holonomy and a parallel spinor. The Riemannian manifold need not be complex. But the limit as you make the extra dimensions small will not be Lorentzian, unless the extra dimensions are Riemannian with a parallel spinor. To have a reduced dimension of 4 (for standard general relativity) and a total space dimension of 10 (for type A or type B string theory), you need a 6 dimensional Riemannian manifold of extra dimensions, with a parallel spinor. These conditions give you a Calabi-Yau manifold. If you ask for 7 extra dimensions, you get a G2 holonomy manifold, so extra dimensions are not always complex. There are compact Riemannian Einstein 6-manifolds beside Calabi-Yaus, but not with a parallel spinor.
answered Oct 27, 2017 at 10:03
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1$\begingroup$ Can a mare mortal understand the importance of parallel spinors here? $\endgroup$ Commented Oct 27, 2017 at 16:29
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$\begingroup$ Thank you for the answer! Since I don't know much about string theory, what does "parallel spinor" mean? Is it related to some special structure of corresponding spin group? $\endgroup$ Commented Oct 28, 2017 at 13:01
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$\begingroup$ Lucky for you, the term "parallel spinor" is purely differential geometry, and requires no physics to understand. It refers to a section of a spinor bundle invariant under parallel transport of a spin connection. See Lawson and Michelsohn, Spin Geometry, Princeton U. Press, 1990 for complete details of spinors and their associated connections. $\endgroup$ Commented Oct 28, 2017 at 13:40
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$\begingroup$ your first sentence is hard to understand. Real forms arise as extra dimensions? Real forms are Lie groups and dimensions are, well, dimensions. $\endgroup$– user141498Commented Jun 9, 2019 at 8:17