It's relatively straightforward to provide a coordinate-free definition of the symplectic form on a cotangent bundle; the usual way to do this is to construct the tautological 1-form $$\lambda(\xi) = \langle D\pi(\xi), \pi'(\xi)\rangle,$$ where $D\pi: TT^*M \to TM$ is the pushforward of the bundle projection $\pi: T^*M \to M$ and $\pi': TT^*M \to T^*M$ is a different bundle projection. Then one evaluates and gets a section of the dual bundle to $TT^*M$, i.e. an element of $\Omega^1(T^*M)$.

This is really elegant because we don't have to sit around and provide a coordinateful description of $\lambda$ and check that it makes sense on different charts - we can just get up and go do some symplectic geometry, once we define $\omega = d\lambda$. It also makes it obvious that $\omega$ is closed, as $d\omega = d^2\lambda = 0$.

It also makes it very straightforward to prove that $\alpha^*\lambda = \alpha$ for any $\alpha \in \Omega^1(M)$, thought of as a map $\alpha: M \to T^*M$. From this we can take derivatives to get a coordinate-free proof that $\lambda^*\omega = \omega$, and so we see easily that cotangent bundles are actually Liouville manifolds.

However, checking that $\omega$ is non-degenerate and hence a symplectic form seems to always be done in coordinates: $T^*M$ is locally trivial, so in a neighborhood we have the usual $\omega = \sum_i dp^i \wedge dq^i$ which is manifestly non-degenerate.

Is there a coordinate-free way to show that $\omega = d\lambda$ is non-degenerate?

One plausible route would be to give a coordinate-free definition of an $\omega$-tame (i.e., $\omega(v, Jv) > 0$) almost complex structure - from here, non-degeneracy is immediate. The standard complex structure on tangent bundles (the one that shows up in the symplectic geometry of Hamiltonian mechanics) should work fine - it's both $\omega$-tame and $\omega$-compatible - but I can't see how to define this intrinsically.


Your post has some 'soft-question' aspects which I shall first very briefly address.

  • What 'intrinsic' means is not so clear and frequently depends on one's understanding of the term. Several arguments could never explicitly mention a choice of coordinates, but could nevertheless introduce additional data (which are thus 'extrinsic' to the initial data, yet still 'intrinsically defined' on the manifold) whose existence is best established via the use of coordinates.

  • The reason why the nondegeneracy of $\mathrm{d}\lambda$ is almost always proved in coordinates could also have several soft (yet interesting) answers: besides its simplicity and the fact that 'cotangent coordinates' are prime examples of Darboux coordinates, the appearance of the exterior derivative makes it difficult to have only pointwise considerations (although nondegeneracy is a pointwise property) and thereby somewhat forces to introduce ad hoc quantities (such as extensions of covectors into differential 1-forms).

  • There is no 'canonical' almost complex structure on $T^*M$, so $\omega$-tameness of some (ad hoc) almost complex structure seems as difficult to establish than the nondegeneracy of $\omega$ itself.

The fact that $(T^*M, \mathrm{d}\lambda)$ is Liouville is an important clue towards perhaps 'the most intrinsic' proof that $\omega := \mathrm{d}\lambda$ is nondegenerate.

Step 1. Observe that for each differential 1-form $\alpha \in \Omega^1(M)$, the map $$T_{\alpha} : T^*M \to T^*M : \beta \mapsto \beta + \alpha_{\pi(\beta)}$$ is a symplectic diffeomorphism. This can be proved intrinsically, for instance via the method at the end of my answer to this post. The importance of this observation is that the maps $T_{\alpha}$ allow for a (canonical) 'parallel transport in the fibers of $T^*M$'. In particular, it is sufficient to prove that $\omega$ is nondegenerate along the zero section of $\pi : T^*M \to M$.

Step 2. We have a smooth $\mathbb{R}$-action on $T^*M$ given by dilations in the fibers: $A_r(\alpha) := r \cdot \alpha := e^{r}\alpha$ for $r \in \mathbb{R}$ and $\alpha \in T^*M$. We notice that $\pi \circ A_r = \pi$ and that $\pi' \circ (A_r)_* = A_r \circ \pi'$ for any $r$. Also denote $\eta$ the smooth vector field whose flow induces this action and observe that $\pi_* \eta = 0$ (so that it is tangent to the fibers of $T^*M$).

On the one hand, for fixed $r \in \mathbb{M}$ and $\xi \in TT^*M$ we thus compute $$\begin{align} (A_r^* \lambda)(\xi) &= \lambda((A_r)_* \xi) = \langle \pi'(\xi), \pi_*(A_r)_*\xi \rangle = \langle \pi'((A_r)_*\xi), (\pi \circ A_r)_*\xi \rangle \\ &= \langle A_r(\pi'(\xi)), \pi_*\xi \rangle = \langle e^r(\pi'(\xi)), \pi_*\xi \rangle = e^r\lambda(\xi) \, . \end{align}$$ Consequently $\mathcal{L}_{\eta}\lambda = \lim_{r \to 0} \frac{A_r^* \lambda - \lambda}{r} = \lim_{r \to 0} \frac{e^r - 1}{r} \lambda = \lambda$. On the other hand, by Cartan's formula, $\mathcal{L}_{\eta}\lambda = \eta \lrcorner \, \mathrm{d}\lambda + \mathrm{d}(\eta \lrcorner \, \lambda) = \eta \lrcorner \, \omega + \mathrm{d}(\lambda(\eta))$. Since $\eta$ is parallel to the fibers, i.e. $\pi_*\eta$ is the zero vector field on $M$, we have $\lambda(\eta) = 0$.

Hence $\lambda = \eta \lrcorner \, \omega$, so $\eta$ is the Liouville vector field (associated to $\lambda$). Since the fibers of $T^*M$ are vector spaces, we can canonically identify tangent spaces to them to themselves. Under that correspondence, for $\beta \in T^*M$ we have $\eta_{\beta} = \beta$.

Step 3. Let $\beta \in T^*M \setminus 0_M$, where $0_M : M \to T^*M$ denotes the zero section of $\pi$. Then $\eta_{\beta} = \beta$. It is possible to extend $\beta$ into a differential 1-form $\tilde{\beta}$. Via step 1, we can consider $(T_{-\tilde{\beta}})_*(\eta_{\beta})$ which is the result of the parallel transport of $\eta_{\beta} = \beta$ at $\beta$ to a vector (which we still denote $\beta$) at the zero section (via a symplectic diffeomorphism).

Worded differently, this argument shows that any vector $\xi \in T_{0_M}T^*M$ which is tangent to the fibers of $\pi$ can be identified via a symplectomorphism to a value of $\eta$ elsewhere in the same fiber (namely at $\beta = \pi'(\xi)$).

So for such a vector, $\xi \lrcorner \, \omega_{0_M} = \eta \lrcorner \, \omega_{\beta} = \beta$. If this is nonzero, then there is a vector $\zeta$ to $M \simeq 0_M$ such that $\omega_{0_M}(\xi, \zeta) = \beta(\zeta) \neq 0$.

Conversely, given a nonzero $\zeta \in T(0_M) \simeq TM$, there is a 1-form $\beta$ such that $\beta(\zeta) \neq 0$. Setting $\xi := (T_{-\tilde{\beta}})_*(\eta_{\beta})$ we get $\omega_{0_M}(\xi, \zeta) \neq 0$.

Since $T_{0_M}T^*M \simeq T^*M \oplus T(0_M)$, it follows that $\omega$ is nondegenerate along $0_M$, hence everywhere according to step 1.

The fact that a $\omega$-compatible almost complex structure $J$ induces a Riemannian metric $g_J$ on $T^*M$ gives a clue towards another possible proof which I will only briefly sketch.

Let $g$ be a Riemannian metric on $M$. (This is an ad hoc quantity, but from there, there are 'distinguished' constructions available (some people would speak of 'canonical constructions', but others would refrain from doing so because of the modern categorical meaning given to 'canonical' which is not met here).) It induces a bundle isomorphism $g_{\flat} : TM \to T^*M$, but also the 'Levi-Civita' Riemannian metric $\tilde{g}$ on $TM$ (so that the $\tilde{g}$-orthogonal to the tangent spaces to the fibers of $TM$ defines the Levi-Civita connection). With this connection at hand, we can decompose $TTM \simeq V \oplus H = TM \oplus TM$ (into vertical and horizontal distributions) and hence define a $\tilde{g}$-compatible almost complex structure which intertwines $H$ and $V$. Using $g_{\flat}$ to push these structures forward on $T^*M$ (or to pullback $\omega$ on $TM$), we can compare these structures to one another and prove that they are compatible.

  • $\begingroup$ Dear @JordanPayette: Your answer is complitely subestimated in my opinion. It answers OP question and provides lot of information, not only regarding symplectic geometry but also Kähler manifolds. I have asked a question here asking for the coordinate-free description of the canonical 2-form in $T^*M$. As you say, it is needed to consider some extensions. I would like you to have a look at my question. Thanks. $\endgroup$
    – Dog_69
    May 3 '19 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.