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http -> https (the question was bumped anyway)
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Martin Sleziak
Martin Sleziak
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I can't comment on Noah's answer, so:

The reason for the $C^1$ condition is that Nash's $C^1$ embedding theoremNash's $C^1$ embedding theorem says that any Riemannian $k$-manifold with a short embedding into $\mathbb{R}^n$ has an isometric $C^1$ embedding into $\mathbb{R}^n$ for any $n > k$. In particular, there is an isometric $C^1$ embedding of the hyperbolic plane into $\mathbb{R}^3$. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

I can't comment on Noah's answer, so:

The reason for the $C^1$ condition is that Nash's $C^1$ embedding theorem says that any Riemannian $k$-manifold with a short embedding into $\mathbb{R}^n$ has an isometric $C^1$ embedding into $\mathbb{R}^n$ for any $n > k$. In particular, there is an isometric $C^1$ embedding of the hyperbolic plane into $\mathbb{R}^3$. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

I can't comment on Noah's answer, so:

The reason for the $C^1$ condition is that Nash's $C^1$ embedding theorem says that any Riemannian $k$-manifold with a short embedding into $\mathbb{R}^n$ has an isometric $C^1$ embedding into $\mathbb{R}^n$ for any $n > k$. In particular, there is an isometric $C^1$ embedding of the hyperbolic plane into $\mathbb{R}^3$. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

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Ben McKay
Ben McKay
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I can't comment on Noah's answer, so:

The reason for the C^1$C^1$ condition is that Nash's C1$C^1$ embedding theorem says that any Riemannian k$k$-manifold with a short embedding into Rn$\mathbb{R}^n$ has an isometric C1$C^1$ embedding into Rn$\mathbb{R}^n$ for any n > k$n > k$. In particular, there is an isometric C1$C^1$ embedding of the hyperbolic plane into R3$\mathbb{R}^3$. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

I can't comment on Noah's answer, so:

The reason for the C^1 condition is that Nash's C1 embedding theorem says that any Riemannian k-manifold with a short embedding into Rn has an isometric C1 embedding into Rn for any n > k. In particular, there is an isometric C1 embedding of the hyperbolic plane into R3. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

I can't comment on Noah's answer, so:

The reason for the $C^1$ condition is that Nash's $C^1$ embedding theorem says that any Riemannian $k$-manifold with a short embedding into $\mathbb{R}^n$ has an isometric $C^1$ embedding into $\mathbb{R}^n$ for any $n > k$. In particular, there is an isometric $C^1$ embedding of the hyperbolic plane into $\mathbb{R}^3$. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

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Tom Church
Tom Church
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I can't comment on Noah's answer, so:

The reason for the C^1 condition is that Nash's C1 embedding theorem says that any Riemannian k-manifold with a short embedding into Rn has an isometric C1 embedding into Rn for any n > k. In particular, there is an isometric C1 embedding of the hyperbolic plane into R3. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.