This question is related to this one and also with another of my questions(see Addendum 2) and it asks for the coordinate-free description of the canonical 2-form $\omega$ defined in $T^*M$.

It is well-known that $\omega$ is (avoiding signs) the exterior derivative of the Liouville (also known as the tautological 1-form) $\theta$ defined as the section

$$ \theta:(p,\alpha)\in T^*M \longmapsto \theta_{(p,\alpha)}\in T^*_{(p,\alpha)}T^*M, $$

where $\theta_{(p,\alpha)}$ acts on vectors $X\in T_{(p,\alpha)}T^*M$ as

$$ \theta_{(p,\alpha)}(X)=\alpha\Big(T_{(p,\alpha)}\pi_{T^*M}X\Big), $$

being $T_{(p,\alpha)}\pi_{T^*M}$ the differential or tangent map of $\pi_{T^*M}:T^*M\rightarrow M$ at the point $(p,\alpha)$.

The problem of this description is that $\theta$ is defined pointwise, so we cannot apply the usual coordinate-free dexcription of the exterior derivative

$$ d\alpha(X,Y)=X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y]) $$

because our $\theta$ is not described in that way.

Thus my question is: How can I compute $\omega=d\theta$? The answer for the first question I have linked suggests it is necessary to extend the covectors (point 2 of the first considerations), altoguh I don't exactly how to do this. So I'm asking for the coordinate description of $\omega$ or also just for a reference where the above formula for $d$ has been satudied also pintwise, explianing the extensions and proivint the final result is independent of them.

Thanks in advance.


When I ask for a method to compute $d\theta$ I mean a coordinate-free way of computing the 2-form. I know the computation can be done considering local coordinates (Wiki has this computation for example) but I want a coordinate-free description.


2 Answers 2


I begin by giving my answer to your question, which will involve ad hoc elements. Afterwards, I reflect upon the question "how to compute $\omega = d\theta$ in a coordinate-free fashion?", as it appears to me to be a rather vague question and also a rather unfortunate requirement.

Let $p : T^*M \to M$ denote the canonical projection, $p' : T(T^*M) \to T^*M$ denote the canonical projection and $Tp = p_* : T(T^*M) \to TM$ denote the differential of $p$. Then for $X \in T(T^*M)$, $\theta(X) := \langle p'(X) , Tp(X) \rangle$ where $\langle - , - \rangle$ is the duality product between $T^*_mM$ and $T_mM$ for any $m \in M$.

Given any smooth function $f : N \to Q$ between manifolds and given a $p$-form $\alpha$ on $Q$, it is well-known that $f^*(d\alpha) = d(f^*\alpha)$. We are going to apply this fact to different choices of $f : N \to T^*M$ with $\alpha = \theta$.

Let's first prove that the fibers of $p$ are isotropic submanifolds of the 2-form $\omega = d\theta$. Given such a fiber $T^*_m M$, there is the inclusions $\iota : T^*_mM \subset T^*M$, hence $\iota^*\omega = d(\iota^*\theta)$. But for $Y \in T(T^*_mM)$, $$(\iota^*\theta)(Y) = \theta(\iota_* Y) = \langle p'(\iota_*Y), p_*\iota_* Y \rangle = \langle p'(\iota_*Y), 0 \rangle = 0,$$ hence $d(\iota^*\theta) = 0$.

Given a nonzero $X \in T_mM$, there exists $\beta \in T^*_mM$ such that $\beta(X) = -1$ Observe that $\beta$ can be extended into a 1-form defined in a neighborhood of $m$ (and in fact on the whole of $M$); this 1-form can be identified with a section $s_{\beta} : M \to T^*M$ of $p$. Given $\gamma \in T^*_mM$, there is an element of $T_{\gamma}T^*_mM$ which is rather canonically identified with $\beta$, which we will denote $Y$.

Denoting $x,y$ the coordinates on $\mathbb{R}^2$, consider an immersion (this exists) $f : \mathbb{R}^2 \to T^*M$ with the following properties: $f(0,0) = \gamma$, $f_{*(0,0)}(\partial_y) = Y$ and $f_{*(0,0)}(\partial_x) = X'$ is a lift of $X$ i.e. $p_*X' = X$, which we extend in a neighborhood. Observe that $0 = f_* 0 = f_*[\partial_x, \partial_y] = [f_*\partial_x, f_* \partial_y]$. Then morally we compute $$ \begin{align} (f^*\omega)(\partial_x,\partial_y) &= \omega( X', Y) = X'(\theta(Y)) - Y(\theta(X')) - \theta([X', Y]) \\ &= X'(0) - Y(\gamma(X)) - \theta(0) = - \frac{d}{dt}[(\gamma + t \beta)(X)] = -\beta(X) = 1 \, . \end{align} $$ This seems to prove that $\omega$ is nondegenerate at every point $\gamma \in T^*M$, and thus that it is therefore a symplectic form.

But this computation is not correct, as we need to extend $Y$ and $X'$ to be defined in a neighborhood of $\gamma$ in order to differentiate as we did. It is however sufficient for this computation that $f_*(\partial_y)$ be everywhere tangent to the fibers of $p$ (ideally chosen so that $f(0,t) = \gamma + t \beta$) and that, along the fiber $T^*_mM$, $f_{*}(\partial_x)$ be lifts of $X$. I leave to the reader the task of proving that such a $f$ exists...

The passage $Y(\gamma(x)) = \frac{d}{dt}[(\gamma + t \beta)(X)]$ might be harder to understand, except perhaps when $\gamma = 0$; it is due to the fact that $T^*M$ being a vector bundle, there is a canonical connection along each fibers of $p$. Nevertheless, I will rapidly sketch how we can reduce to the case $\gamma = 0$.

We now notice a following fundamental property of $\theta$. Given a 1-form $\beta$ on $M$, we can identify it with a section $s_{\beta} : M \to T^*M$ to $p$. For $X \in TM$, we compute $$ (s_{\beta}^*\theta)(X) = \theta(s_{\beta \, *}X) = \langle p'(s_{\beta \, *}X) , p_*s_{\beta \, *}X \rangle = \langle \beta , X \rangle = \beta(X) , $$ hence $s_{\beta}^*\theta = \beta$. Incidentally, $s_{\beta}^*\omega = d\beta$ and taking $\beta = 0$, we see that the 0-section is a Lagrangian submanifold for $\omega$.

Using this, we can show that whenever $\beta$ is a closed 1-form, the map $\Psi_{\beta} : T^*M \to T^*M : \gamma \mapsto \gamma + s_{\beta}(p(\gamma))$ is a (pre)symplectic map in the sense that it preserves $\theta$ and hence $\omega$ (this is not so easy to establish in a coordinate-free approach). Hence, taking $\beta$ to be a (closed) extension of $-\gamma$, the use of this map allows to compute $\omega$ only along the 0-section.

I now reflect upon the question "how to compute $\omega = d\theta$ in a coordinate-free fashion?"

First, by the very definition of a manifold, one could argue that there is no "coordinate-free" approach to differential geometry/topology, but at best a "non-manifestly coordinate-based" approach. Indeed, many of the arguments one could come with rely on constructions which are ultimately established by working in coordinates. When facing a new problem, it is unclear whether it could easily be solved using only already established constructions; perhaps a new construction would be more appropriate, and the way to achieve this construction might involve coordinates explicitly.

Secondly, what does "compute" mean? In the present context, my take on this question would be to describe $\omega$ in somewhat more explicit and intelligible terms. On the one hand, 'explicit' and 'intelligible' are vague qualificatives; the sought-after description should probably serve some definite purpose, for instance be fit for subsequent inquiry. On the other hand, the 'more intelligible terms' are probably ad hoc structures, objects with which we are more acquainted, such as coordinates; it seems arbitrary to snob the use of coordinates in favor of other referents. In that sense, the usual coordinate description of $\omega$ recalled in Camillo Arosemena-Serrato's answer is relevant, as it is intelligible and useful for further studies; incidentally and most interestingly, it shows that the canonical coordinates on $T^*M$ induced by any coordinates system on $M$ are Darboux coordinates. (By the way, this computation of $\omega$ is not independent of the coordinates on $T^*M$...)

Regarding the use of the formula $d\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y])$ to $\alpha = \theta$, it is not a problem that $\theta$ be defined pointwise; any explicit description of a function on a set is a pointwise description, which does not prevent us from performing calculus on the function.

It is difficult to evaluate $d \alpha$ in a coordinate-free way, for it requires us to compare the values of $\alpha$ in neighboring points, which is difficult without some identification of cotangent spaces at neighboring points. Such an identification is provided by a coordinates system; it is also provided by a connection (but the existence of connection is best established by the use of coordinates...). In any case, the identification is an ad hoc structure to the problem.

The formula however provides a coordinate-free translation of the problem. Instead of having to compute $d\alpha$, then evaluate it at a point $p$ on vectors $X,Y \in T_pQ$, one rather extends $X$ and $Y$ to vector fields near $p$ (the extensions are ultimately irrelevant ad hoc structures) and compute the righthand side of the equation. In order for this technique to be manageable, one would be best to choose extensions $X$ and $Y$ suited to the problem (i.e. for the specific $Q$ and $\alpha$). In the situation at hand, one sees that the extensions provided by the coordinate vector fields $\partial/\partial x^j$ and $\partial/\partial \xi_k$ are very well suited to the computation of $d\theta$ through the righthand side of the equation.

  • $\begingroup$ First of all, let me thank you for your answer. I really appreciate the effort you have done typing such a complete and elaborate answer. Thanks. Now, regarding the expression ''coordinate-free description'', I think most people working in differential topology what it means: we want an expression for which coordinates are not seen explictly; for forms and tensors in general this alternative way usually is simply the charcaterisation of such objects by their action on vector fields and 1-forms. $\endgroup$
    – Dog_69
    May 7, 2019 at 22:01
  • $\begingroup$ At this point of the conversation let me point that I said $\omega$ is independent of the choice of coordinates in the sense that, under diffeomorphisms $\Phi$, $\omega'=\Phi_*\omega$ (Remark 2). $\endgroup$
    – Dog_69
    May 7, 2019 at 22:01
  • $\begingroup$ Finally, and regarding the proof, I have few questions. The first one is about the canonical connections defined on each fibre of $p'$. I guess it consists simply in using the vector space structure, but I would like you to add some details. And -the other is about the proof of the fact that the zero section is Lagrangian. Isn't this a consequence of the fact that the fibres of $p$ are isotropic? And the last one, when you talk about prolongating vector fields, if you chose the such vector fields, wouldn't it mean the result would be the expression of $\omega$ in coordinates? Thanks again. $\endgroup$
    – Dog_69
    May 7, 2019 at 22:42
  • $\begingroup$ @Dog_69 That is how I understood your use of coordinate-free description. Still, one faces the problem of distinguishing/describing the vectors on which $\omega$ acts. In a Darboux chart $(x_1, y_1, x_2, y_2)$ the vectors $\partial_{y_1}$ and $\partial_{y_2} + \partial_{x_1}$ span a space transverse to the Lagrangian space $Span(\partial_{x_1}, \partial_{x_2})$ but is not itself Lagrangian; how would we detect that without the Darboux chart? By the way, this example proves that the reason why the 0-section is Lagrangian is not because the fibers are. $\endgroup$ May 8, 2019 at 1:49
  • 1
    $\begingroup$ @Dog_69 It is sufficient to compute $\omega$ between vectors based at $p$ with $p$ belonging to the zero section. The vectors themselves are not required to be tangent to the zero section (if we were to impose this requirement, then since the zero section is Lagrangian, the answer would always be zero.) $\endgroup$ May 15, 2019 at 16:25

Let us see first how the $\theta$ you define can be expressed using local coordinates $(U,x_1,\ldots,x_n)$ of $M$. If $\xi\in T_x^*M$, then $\xi=\sum_{i=1}^n\xi_idx_i$ for some $\xi_i\in\mathbb R$, and thus we get coordinates on the open set $T^*U$ of $T^*M$ by setting $T^\ast U\ni(x,\xi)\mapsto (x_1,\ldots,x_n,\xi_1,\ldots,\xi_n)\in\mathbb R^{2n}$. See Ana Cannas' notes.

Let us express the $1$-form you define in terms of these coordinates. $\pi$ when restricted to $T^*U$ can be seen as $(x,\xi)\mapsto x$. Thus, $T_{(x,\lambda)}\pi$ is the projection $\sum_{i=1}^na_i\frac{\partial}{\partial x_i}+\sum_{i=1}^nb_i\frac{\partial}{\partial \xi_i}\mapsto \sum_{i=1}^na_i\frac{\partial}{\partial x_i}$. Thus the definition you gave of $\theta$ implies that if $\xi=\sum_{i=1}^n\xi_idx_i\in T_x^*U$, then $\theta_{(x,\xi)}=\sum_{i=1}^n\xi_idx_i$, where in the last equality we consider $\sum_{i=1}^n\xi_idx_i$ as a $1$-form of $T_{(x,\alpha)}^\ast(T^\ast U)$ using the coordinates of $T^*U$ given above. This gives the definition of the Lioville form using coordinates. This does not depend on the coordinates.

From this description we get the expresion of the canonical symplectic form of $T^*M$ by $\sum_{i=1}^ndx_i\wedge d\xi_i$, by noticing that in $T^*U$ we have $d\theta=d((x,\xi)\mapsto \xi)=\sum_{i=1}^nd\xi_i\wedge dx_i=-\sum_{i=1}^ndx_i\wedge d\xi_i$. This definition of this $2$-form does not depend on the coordinates as $\theta$ is independent of coordinates.

  • $\begingroup$ Your answer is actually independent of the coordinates you set, but it's not the coordinate-free description I am looking for. $\endgroup$
    – Dog_69
    May 4, 2019 at 9:26
  • $\begingroup$ @Dog_69 I showed your defintion could is equivalent to the one using coordinates to obtain the relation $\omega=-d\theta$. $\endgroup$ May 5, 2019 at 2:29
  • 1
    $\begingroup$ Yes but that is not what I am looking for. I asked for a coordinate-free expression for $d\theta$ and your answer uses coordinates. I insist, coordinate-free and independent of the choice of coordinates it is not the same. I have edited my question to make this point clearer. $\endgroup$
    – Dog_69
    May 5, 2019 at 5:42

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