What is a specific example of a 6D Calabi-Yau manifold? are there simple ones like a 6-torus, $T^6=(S^1)^6$ , $S^3\times T^3$, or similar structures with products of Spheres and Torus?
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The 6-torus, obtained by identifying the boundaries of a parallelpiped of $\mathbb{R}^6$ (or, more properly, $\mathbb{C}^3$), is a Calabi-Yau manifold. Note that one characteristic of a Calabi-Yau manifold is that it is Ricci-flat, which may impose restrictions on whether you can have metrics of the form $S^n \times T^n$; certainly the "naïve" metric on $S^n$ is not Ricci-flat, but as far as I know it is an open question whether there exists a flat metric on $S^n$ for $n \geq 4$.
You may find this review paper useful, though I wrote it as a graduate student almost 20 years ago and there are probably inaccuracies & infelicities. In addition, there may have been discoveries in the intervening years that are not reflected there. Caveat lector.
answered May 24, 2022 at 14:25
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$\begingroup$ I am skeptical on whether we can have any physical mapping like "identifying" opposite edges of a parallelopiped, in reality people talk about circles when they are asked how can they identify two opposite edges of a square, but then you can never talk about the space being flat i guess. When you realize the "identification of opposite edges" as a circle, it is introducing some curvature. $\endgroup$ Commented May 25, 2022 at 4:38
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$\begingroup$ what I believe is that a torus is never flat - please correct me if I am wrong. $\endgroup$ Commented May 25, 2022 at 4:39
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$\begingroup$ @EdenZane: a torus does have to have extrinsic curvature if it's embedded in a higher-dimensional Euclidean space. But it is possible to mathematically define a torus via these sorts of identifications without reference to a higher-dimension space in which it's embedded, and in this case it has no intrinsic curvature (and it's meaningless to talk about extrinsic curvature). If you want more details on this, I would encourage you to ask it separately (probably over at Mathematics rather than here.) $\endgroup$ Commented May 25, 2022 at 11:39