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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
    Commented Sep 24, 2018 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
    Commented Sep 24, 2018 at 5:55
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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
    Commented Sep 24, 2018 at 7:53
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    $\begingroup$ The isometric embedding and the introduction of $TM$ are probably a bit of red herring here. I think what you are asking is this: Given a Riemannian manifold $(N,h)$ and a Riemannian submanifold $(M,g)$, the musical isomorphisms allows you to define a a smooth map $\sigma: T^*M \to T^*N$. Your question is whether that the pull-back by $\sigma$ of the canonical symplectic structure on $T^*N$ is equal to the canonical symplectic structure on $T^*M$. Or, in other words, you are asking whether the embedding $\sigma$ makes $T^*M$ a symplectic submanifold of $T^*N$. $\endgroup$
    – Willie Wong
    Commented Sep 26, 2018 at 4:46
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    $\begingroup$ I hope I am not mistaken, but I think this then follows from the fact that embedded Riemannian submanifolds have normal tubular neighborhoods. This I think implies you have local coordinates $(x^1, \ldots, x^n)$ of $N$, inducing coordinates $(x^1, \ldots, x^n, y^1, \ldots, y^n)$ of $T^*N$, such that the embedding $\sigma$ sends $T^*M$ (locally) to the set $\{x^{m+1}= \cdots = x^{n} = y^{m+1} = \cdots =y^{n} = 0\}$. $\endgroup$
    – Willie Wong
    Commented Sep 26, 2018 at 5:16

2 Answers 2

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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
    Commented Sep 25, 2018 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
    Commented Sep 25, 2018 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
    Commented Sep 26, 2018 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
    Commented Sep 26, 2018 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
    Commented Sep 26, 2018 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
    Commented Sep 24, 2018 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
    Commented Sep 24, 2018 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
    Commented Sep 24, 2018 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
    Commented Sep 25, 2018 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
    Commented Sep 25, 2018 at 23:29

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