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fixed mistake
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André Henriques
André Henriques
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Let $\Sigma$ be your surface.
I'll construct a new surface $\tilde \Sigma$ as follows. Take $\Sigma$ apart into individual triangles, and reglue them with a twist at each edge (the adjacency graphs for the faces of $\Sigma$, and for the faces of $\tilde\Sigma$ are the same). Note that the surfaces $\Sigma$ and $\tilde\Sigma$ could have different genus, and that $\tilde\Sigma$ could be non-orientable.

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the universal cover of the new surface $\tilde\Sigma$ into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$.
That map is well defined up to an overall translation.

Let $\Sigma$ be your surface.
I'll construct a new surface $\tilde \Sigma$ as follows. Take $\Sigma$ apart into individual triangles, and reglue them with a twist at each edge (the adjacency graphs for the faces of $\Sigma$, and for the faces of $\tilde\Sigma$ are the same). Note that the surfaces $\Sigma$ and $\tilde\Sigma$ could have different genus, and that $\tilde\Sigma$ could be non-orientable.

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the new surface $\tilde\Sigma$ into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$.
That map is well defined up to an overall translation.

Let $\Sigma$ be your surface.
I'll construct a new surface $\tilde \Sigma$ as follows. Take $\Sigma$ apart into individual triangles, and reglue them with a twist at each edge (the adjacency graphs for the faces of $\Sigma$, and for the faces of $\tilde\Sigma$ are the same). Note that the surfaces $\Sigma$ and $\tilde\Sigma$ could have different genus, and that $\tilde\Sigma$ could be non-orientable.

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the universal cover of the new surface $\tilde\Sigma$ into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$.
That map is well defined up to an overall translation.

added 400 characters in body
Source Link
André Henriques
André Henriques
  • 43.2k
  • 5
  • 130
  • 264

Let $\Sigma$ be your surface.
I'll construct a new surface $\tilde \Sigma$ as follows. Take $\Sigma$ apart into individual triangles, and reglue them with a twist at each edge (the adjacency graphs for the faces of $\Sigma$, and for the faces of $\tilde\Sigma$ are the same). Note that the surfaces $\Sigma$ and $\tilde\Sigma$ could have different genus, and that $\tilde\Sigma$ could be non-orientable.

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the new surface $\tilde\Sigma$ into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$. 
That map is well defined up to an overall translation (assuming the surface is connected).

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the surface into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$. That map is well defined up to an overall translation (assuming the surface is connected).

Let $\Sigma$ be your surface.
I'll construct a new surface $\tilde \Sigma$ as follows. Take $\Sigma$ apart into individual triangles, and reglue them with a twist at each edge (the adjacency graphs for the faces of $\Sigma$, and for the faces of $\tilde\Sigma$ are the same). Note that the surfaces $\Sigma$ and $\tilde\Sigma$ could have different genus, and that $\tilde\Sigma$ could be non-orientable.

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the new surface $\tilde\Sigma$ into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$. 
That map is well defined up to an overall translation.

Source Link
André Henriques
André Henriques
  • 43.2k
  • 5
  • 130
  • 264

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the surface into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$. That map is well defined up to an overall translation (assuming the surface is connected).