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Added warning that some of my comments are wrong.
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Sam Nead
Sam Nead
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Is Hubert and Schmidt's assertion wrong?

No, they are correct. They allow squares where the sidelength is not equal to one.

If you like, we can say that the two translation surfaces $(X, \omega)$ and $(X, r\omega)$ (for $r$ positive and real) are "scalar multiples" of each other. Then under any definitions, the scalar multiples of the square-tiled surfaces are dense in the space of translation surfaces.

EDIT: There is some nonsense (due to me) in the comments below. Caveat emptor!

Is Hubert and Schmidt's assertion wrong?

No, they are correct. They allow squares where the sidelength is not equal to one.

If you like, we can say that the two translation surfaces $(X, \omega)$ and $(X, r\omega)$ (for $r$ positive and real) are "scalar multiples" of each other. Then under any definitions, the scalar multiples of the square-tiled surfaces are dense in the space of translation surfaces.

Is Hubert and Schmidt's assertion wrong?

No, they are correct. They allow squares where the sidelength is not equal to one.

If you like, we can say that the two translation surfaces $(X, \omega)$ and $(X, r\omega)$ (for $r$ positive and real) are "scalar multiples" of each other. Then under any definitions, the scalar multiples of the square-tiled surfaces are dense in the space of translation surfaces.

EDIT: There is some nonsense (due to me) in the comments below. Caveat emptor!

Source Link
Sam Nead
Sam Nead
  • 28.2k
  • 5
  • 72
  • 131

Is Hubert and Schmidt's assertion wrong?

No, they are correct. They allow squares where the sidelength is not equal to one.

If you like, we can say that the two translation surfaces $(X, \omega)$ and $(X, r\omega)$ (for $r$ positive and real) are "scalar multiples" of each other. Then under any definitions, the scalar multiples of the square-tiled surfaces are dense in the space of translation surfaces.