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Given an isometric embedding of the tangent bundle $TM$ of a Riemannian manifold with Sasaki metric into $ℝ^N$ with standard metric. My intuition tells me that the vector spaces $T_pM$ ($p ∈ M$) are mapped to affine sub spaces of $ℝ^N$ but cannot find a rigourous proof. Can you help?

Additional thoughts:

The existence of the isometric embedding is guaranteed by the Nash embedding theorem. It is unclear to me though how this embedding looks like (see my other question).

Here isometry does not mean that the distance function is preserved (that would be isometry in the metric space sense) but the Riemannian metric is preserved: $⟨dι_v A, dι_v B⟩_{ℝ^N} = ⟨A, B⟩_{G_s}$ for any $v ∈ TM$, $A, B ∈ T_vTM$, $ι$ is the isometric embedding.

The Sasaki metric restricted to $T_pM$ is what one would expect: the vertical part of $T_vTM$ ($v ∈ T_pM$) can be identified with $T_pM$ but answer my question one probably has to be very careful how to exploit this identification.

flukx
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  • Can someone fix the second link, please? I cannot figure out my markdown error. Thanks. – flukx May 15 '22 at 13:38
  • The second link was apparently broken by the HTML tag. – Arthur May 15 '22 at 13:43
  • @Arthur thanks! I thought I ruled that out but apparently not in the correct manner. My intention was to highlight that technically the first paragraph constitutes the entire question and the rest are additional thoughts I had that might or might not be helpful. – flukx May 15 '22 at 13:48
  • If I understand the notation in the linked post, the Sasaki metric of a circle is a flat cylinder, and an isometric embedding of a flat cylinder in Euclidean space certainly need not map tangent lines to Euclidean lines. (Dimension $4$ is enough.) Generally, a high-codimension Riemannian embedding is flexible, so offhand the answer appears to be "no," not just in this isolated example. – Andrew D. Hwang May 15 '22 at 13:59

1 Answers1

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$\newcommand{\Reals}{\mathbf{R}}$Every Riemannian manifold can be isometrically embedded in a sphere; for a Sasaki metric the fibres map to bounded sets, which are therefore not affine.

To flesh out the comment, IIRC, a flat cylinder may be isometrically embedded into a bounded subset of Euclidean four-space by choosing a bounded, complete plane spiral $C$, such as the polar graph $r = 2 + \tanh\theta$, and mapping $S^{1} \times \Reals$ to $S^{1} \times C$ by the identity in the first factor and an arclength parametrization in the second.

Added: I overlooked that the linked question asks about closed (i.e., compact) Riemannian manifolds. The linked answer does not assume closedness, however. For self-containedness: If $(M, g)$ is a Riemannian manifold, there exists an isometric embedding into some Euclidean space $\Reals^{N}$ by the Nash embedding theorem. The real line embeds isometrically in a flat torus $S^{1} \times S^{1}$ (using for example the curve below from the linked post), so $\Reals^{N}$ embeds isometrically in $(S^{1} \times S^{1})^{N} \subset \Reals^{4N}$. Composing gives an isometric embedding of $(M, g)$ in a torus (which incidentally embeds isometrically in a Euclidean sphere).

In particular, a Sasaki metric embeds isometrically into a bounded subset of some Euclidean space, so its fibres do not generally map to affine subsets.

In case it's useful for posterity, these curves can be chosen real-analytically, so the resulted bounded embedding preserves whatever real smoothness is possessed by the Nash embedding.

Embedding the line in a torus, wrapped

Andrew D. Hwang
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  • This is counter-intuitive but that's math. I am trying to understand the calculation in the example and will come back. – flukx May 15 '22 at 16:51
  • These are $C^1$ isometries. The story is totally different for $C^2$ isometries. You have to decide which ones you want to work with. Most likely, you want $C^2$ or smoother. – Deane May 15 '22 at 17:27
  • @Deane If I'm not mistaken the constructions here are $C^{\infty}$...? – Andrew D. Hwang May 15 '22 at 19:02
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    @AndrewD.Hwang, sorry. I glanced at your reference too quickly. But observe that the statement is for closed Riemannian manifolds only. – Deane May 15 '22 at 19:06