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I apologize in advance if this question has an obvious answer.

Let $(M,g)$ be a Riemannian manifold. Then the tangent bundle $TM$ carries a natural symplectic structure $\omega_g$. In fact $\omega_g$ is the pull back of the canonical symplectic structure of the cotangent bundle via the obvious diffeomorphism between $TM$ and $T^* M$ which is defined by the inner product $g$.

The standard structure of $T\mathbb{R}^n=\mathbb{R}^n \times \mathbb{R}^n$ is denoted by $\omega$.

For every Riemannian manifold $(M,g)$, is there an isometric embedding $j$ of $M$ in some $\mathbb{R}^n$ such that $j^*(\omega)=\omega_g?$

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    $\begingroup$ For the question as stated, I would have thought that the "naturality" implies $j^{\ast} \omega = \omega_{j^{\ast} \delta}$. Here $\delta$ is the Euclidean metric and $\omega = \omega_{\delta}$. Then the solution is simply that there exists an isometric embedding into some $\mathbb{R}^n$ since then $g = j^{\ast} \delta$. $\endgroup$ – Paul Bryan Sep 24 '18 at 1:15
  • $\begingroup$ @PaulBryan may be I am missing some thing but the problem is that the symplectic structure originally comes from cotangent bundle but the "cotangent bundle" is a contravariant (and not covariant) functor. Does realy your argument work? $\endgroup$ – Ali Taghavi Sep 24 '18 at 5:55
  • $\begingroup$ I don't know if it works. It was jus a first guess. To be sure, can't you just check directly from the definitions? I suppose the answer depends on exactly what "natural" means here. Often in this sort of context where some tensor/form in DG is constructed from another, natural means "in some canonical way" but this may not be necessarily be the case here. I guess you construct $\omega_g$ by some lift of $g$ to $TM$ by splitting $TTM$ into $VM$ and $TM$ via the Levi Civita connection and the use a metric contraction of the canonical sympletic form on $T^{\ast}M$. Chase it through now and see. $\endgroup$ – Paul Bryan Sep 24 '18 at 6:19
  • $\begingroup$ @PaulBryan $\omega_g$ is constructed as follows: the metric gives a dieffeomorphism between $TM$ and $T^* M$, on the other hand $T^* M$ has a canonical symplectic structure (independent of any Riemannian metric). We pull back this canonical symplectic structure via the above mentioned diffeomorphism. The resulting structure is denoted by $\omega_g$. With such construction, we have the naturality property you mentioned when we have DIFFEOMORPHISM isometry but I am not sure it works for isometric embedding(not necessarilly surjective). $\endgroup$ – Ali Taghavi Sep 24 '18 at 7:35
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    $\begingroup$ I see. Still seems plausible to me, perhaps naively. I'm in transit so I don't have time to check it properly right now sorry. $\endgroup$ – Paul Bryan Sep 24 '18 at 7:53
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This is always the case (using "naturality" as Paul Bryan suggested in the comments).

Let $f: M \to N$ be a smooth map between Riemannian manifolds $(M,g)$ and $(N, h)$. Let $g^\flat: TM \to T^*M$ denote the musical isomorphism induced by $g$ (and similarly $h^\flat$). Now, $f$ is an isometry if and only if $h_{f(m)} (T_m f (X), T_m f (\cdot)) = g_m (X, \cdot)$ holds for all $X \in T_m M$. In other words, $g^\flat (X) = f^* (h^\flat (T_m f(X)))$ or shorter $$g^\flat = T^* f \circ h^\flat \circ Tf,$$ where $T^*f: T^* N \to T^* M$ is the cotangent lift of $f$ (often called a point transformation). Using the fact that cotangent lifts are always symplectic maps, we have $$\omega_g = (g^\flat)^* \omega_M = (Tf)^* (h^\flat)^* \omega_N = (Tf)^* \omega_h,$$ where $\omega_M$ and $\omega_N$ denote the canonical symplectic forms on $T^* M$ and $T^* N$, respectively.

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  • $\begingroup$ Thank you for your answer. Just a question: When f is not surjective how can one define "The cotangent lift", that is $T^*f$? $\endgroup$ – Ali Taghavi Sep 25 '18 at 23:33
  • $\begingroup$ So I stil think that the naturality property mentioned by @Paul is not obvious(at least to me). am I missing something? $\endgroup$ – Ali Taghavi Sep 25 '18 at 23:38
  • $\begingroup$ @WillieWong yes I know the definition of the dual map(In its linear algebraic setting. But I mean that it is possible that $f$ is not surjective. So the domain of $T^* f$ is not the whole cotangent bundle $T^*N$. $\endgroup$ – Ali Taghavi Sep 26 '18 at 4:04
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    $\begingroup$ @AliTaghavi: ah, I see your point. But isn't that moot in this instance? You are looking at it acting on $h^\flat \circ Tf$ which would definitely have base point in the image of $f$. // Ah, I think I see your objection: you are wondering about "naturality" in this case. $\endgroup$ – Willie Wong Sep 26 '18 at 4:18
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    $\begingroup$ This is why one usually defines natural operations only in the context of diffeomorphisms. But here, loosely speaking, naturality means that everything transforms as one would expect it to be the case. $\endgroup$ – Tobias Diez Sep 26 '18 at 10:03
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It seems here we have an answer: https://www.sciencedirect.com/science/article/pii/S0926224502000670

This is how I would approach the question:

The question is equivalent to the existence of some immersion $f : M \to \mathbb{R}^N$ such that if $\omega_{\mathbb{R}}$ is the sympletic form of $\mathbb{R}^N$ and $\omega$ is the sympletic form of $M$, one has $f^{\ast}(\omega_{\mathbb{R}}) = \omega.$

Note that since $d\omega_{\mathbb{R}} = 0$ and $\mathbb{R}^N$ is simply connected, there exists $\theta \in \Omega^1(\mathbb{R}^N)$ such that $\omega_{\mathbb{R}} = d\theta.$ Therefore, $$f^{\ast}(d\theta) = \omega.$$ Therefore, $d(f^{\ast}\theta) = \omega.$ This implies a necessary condition is that $\omega$ is exact. If $H_{dR}^2(M) = 0$, then $\omega = d\tilde \theta$, for $\tilde \theta \in \Omega^1(M)$. This implies that:

$$d(f^*(\theta)) = d\tilde \theta.$$ Therefore, $$f^{\ast}(\theta) - \tilde \theta \in H_{dR}^1(M).$$ If we assume that this is zero, $f^{\ast}{\theta} = \tilde \theta$. Therefore, $$\tilde \theta (X) = \theta(df(X)), ~\forall X \in TM.$$ Note that $\theta = \langle Z,\cdot\rangle,$ for some vector field $Z \in T\mathbb{R}^N$. Therefore, $$\theta(X) = \langle df(X),Z\rangle,$$ and this equations suggests an equation for the isommetric immersion.

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    $\begingroup$ OP’s 2-form $\omega_g$ is on $TM$ (not $M$) and is always exact. $\endgroup$ – Francois Ziegler Sep 24 '18 at 3:52
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    $\begingroup$ @FrancoisZiegler Yes the 2 forms on TM are exact. So there is no matter of exactness. The matter is whether the "Naturality" mentioned by Paul is realy the case? What do you think about that?To be honnest I think that the problem is nontrivial. $\endgroup$ – Ali Taghavi Sep 24 '18 at 19:33
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    $\begingroup$ You are both right, I misread the questions at first, so exactness is not a problem, but I endorse I am pretty sure that this is a not trivial question, this is why I have suggested one approach to at least look at an immersion equation. Also, on the paper I've sent you it discuss the problem on a more general frame, considering embbedings of symplectic space in to other symplectic spaces, not necessarily $\mathbb{R}^N.$ $\endgroup$ – L.F. Cavenaghi Sep 24 '18 at 20:51
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    $\begingroup$ @L.F.Cavenaghi The point is that the natural symplectic structure comes from the cotangent bundle. On the other hand when $N$ is a submanifold of $M$, there is no a canonical embedding of $T^* N$ in $T^*M$ such that the embedding would be independemt of any metric. $\endgroup$ – Ali Taghavi Sep 25 '18 at 7:40
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    $\begingroup$ Of course your answer and the linked paper is interesting. Thanks for your attention to my question. You answer is not useless as i gave my +1. $\endgroup$ – Ali Taghavi Sep 25 '18 at 23:29

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