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Fixed some typos
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edit approved Oct 1, 2015 at 16:17
K. P
edit approved Oct 1, 2015 at 16:17
K. P
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hallo,

iI have the following question. iI read about Bryant's theorem which saissays that: any real-analytic 3-dimensional Riemannian manifold $(Y,g)$ with real-analytic metric $g$ can be isometrically embedded as a special Lagrangian submanifold of some Calabi-Yau manifold $(X, \Omega, \omega)$. My question is: does this result also hold in dimensions greater than 3? Or is there any possibility to establish this? TanksThanks in advance.

Mira

hallo,

i have the following question. i read about Bryant's theorem which sais that: any real-analytic 3-dimensional Riemannian manifold $(Y,g)$ with real-analytic metric $g$ can be isometrically embedded as a special Lagrangian submanifold of some Calabi-Yau manifold $(X, \Omega, \omega)$. My question is: does this result also hold in dimensions greater than 3? Or is there any possibility to establish this? Tanks in advance.

Mira

I have the following question. I read about Bryant's theorem which says that: any real-analytic 3-dimensional Riemannian manifold $(Y,g)$ with real-analytic metric $g$ can be isometrically embedded as a special Lagrangian submanifold of some Calabi-Yau manifold $(X, \Omega, \omega)$. My question is: does this result also hold in dimensions greater than 3? Or is there any possibility to establish this? Thanks in advance.

Mira

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gary
gary
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Theorem of Bryant in higher dimensions

hallo,

i have the following question. i read about Bryant's theorem which sais that: any real-analytic 3-dimensional Riemannian manifold $(Y,g)$ with real-analytic metric $g$ can be isometrically embedded as a special Lagrangian submanifold of some Calabi-Yau manifold $(X, \Omega, \omega)$. My question is: does this result also hold in dimensions greater than 3? Or is there any possibility to establish this? Tanks in advance.

Mira