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It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

Postcript. Let us say that a Riemannian manifold $M$ has polynomial volume growth if there is a polynomial $p$ such that volume of any $r$-ball in $M$ has volume at most $p(r)$.

Evidently, if a manifold admits an embedding with uniformly thick tubular neighborhood into $\mathbb{R}^N$, then it has polynomial volume growth. I suspect that polynomial volume growth is also sufficient for $N\gg \deg p$.

If we assume a bit better regularity (say a bound on covariant derivatives of curvature tensor), then it can be proved. It is done by applying the result in "The intrinsic dimensionality of graphs" by Krauthgamer and Lee (thaks to aorq, see my question) together with the Nash embedding theorem.

It seems to be unknown if any complete Riemannian manifold with bounded geometry admits an isometric immersion with bounded normal curvatures. But once this problem is solved, the same argument could be used.

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

Postcript. Let us say that a Riemannian manifold $M$ has polynomial volume growth if there is a polynomial $p$ such that volume of any $r$-ball in $M$ has volume at most $p(r)$.

Evidently, if a manifold admits an embedding with uniformly thick tubular neighborhood into $\mathbb{R}^N$, then it has polynomial volume growth. I suspect that polynomial volume growth is also sufficient for $N\gg \deg p$.

If we assume a bit better regularity (say a bound on covariant derivatives of curvature tensor), then it can be proved. It is done by applying the result in "The intrinsic dimensionality of graphs" by Krauthgamer and Lee (thaks to aorq, see my question) together with the Nash embedding theorem.

It seems to be unknown if any complete Riemannian manifold with bounded geometry admits an isometric immersion with bounded normal curvatures. But once this problem is solved, the same argument could be used.

P.P.S. Now it is written Tubed embeddings.

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

Postcript. Let us say that a Riemannian manifold $M$ has polynomial volume growth if there is a polynomial $p$ such that volume of any $r$-ball in $M$ has volume at most $p(r)$.

Evidently, if a manifold admits an embedding into $\mathbb{R}^N$, then it has polynomial volume growth. I suspect that polynomial volume growth is also sufficient for $N\gg \deg p$.

If we assume a bit better regularity (say the covariant derivatives of its curvature tensor are in Hoelder class), then it can be proved. It is done by applying the result in "The intrinsic dimensionality of graphs" by Krauthgamer and Lee (thaks to aorq, see my question) together with the Nash embedding theorem.

It seems to be unknown if any complete Riemannian manifold with bounded geometry admits an isometric immersion with bounded normal curvatures. But once this problem is solved, the same argument can be used.

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

Postcript. Let us say that a Riemannian manifold $M$ has polynomial volume growth if there is a polynomial $p$ such that volume of any $r$-ball in $M$ has volume at most $p(r)$.

Evidently, if a manifold admits an embedding with uniformly thick tubular neighborhood into $\mathbb{R}^N$, then it has polynomial volume growth. I suspect that polynomial volume growth is also sufficient for $N\gg \deg p$.

If we assume a bit better regularity (say a bound on covariant derivatives of curvature tensor), then it can be proved. It is done by applying the result in "The intrinsic dimensionality of graphs" by Krauthgamer and Lee (thaks to aorq, see my question) together with the Nash embedding theorem.

It seems to be unknown if any complete Riemannian manifold with bounded geometry admits an isometric immersion with bounded normal curvatures. But once this problem is solved, the same argument could be used.

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

Postcript. Let us say that a Riemannian manifold $M$ has polynomial volume growth if there is a polynomial $p$ such that volume of any $r$-ball in $M$ has volume at most $p(r)$.

Evidently, if a manifold admits an embedding into $\mathbb{R}^N$, then it has polynomial volume growth. I suspect that polynomial volume growth is also sufficient for $N\gg \deg p$.

If we assume a bit better regularity (say the covariant derivatives of its curvature tensor are in Hoelder class), then it can be proved. It is done by applying the result in "The intrinsic dimensionality of graphs" by Krauthgamer and Lee (thaks to aorq, see my question) together with the Nash embedding theorem.

It seems to be unknown if any complete Riemannian manifold with bounded geometry admits an isometric immersion with bounded normal curvatures. But once this problem is solved, the same argument can be used.