[Math] The universal property of the Liouville $1$-form

ct.category-theorydg.differential-geometrysg.symplectic-geometry

I am not totally sure if this question is appropriate for MathOverflow, or if it more adeguate to MathStackexchange.
As usual any feedback is welcome.

Introduction

Given an arbitrary smooth manifold $Q$, on the cotangent bundle $T^\ast Q$ there exists a $1$-form $\lambda_Q$, which is variously known as the Liouville $1$-form, or the tautological $1$-form.

Local expression in fibered coordinate

For any local coordinate system $q_i$ on $Q$, let $(q_i,p_i)$ be the associate coordinate on $T^\ast Q$.
Then, locally, $\lambda_Q$ can be given by $$\lambda_Q=\sum_i p_i \cdot dq_i.\tag{$\star$}$$
These local descriptions can be correctly patched together to give a global $1$-form on $T^\ast Q$.

Intrinsic expression

For any $1$-form $\phi$ on $Q$, we have also $$\phi^\ast\lambda_Q=\phi,\tag{$\star \star$}$$ where in the left-hand side we are looking at $\phi$ as a section $\phi:Q\to T^\ast Q$ of the cotangent bundle $\tau_Q^\ast:T^\ast Q\to Q$.
Indeed this condition is enough to completely determine $\lambda_Q\in\Omega^1(T^\ast Q)$ as its unique solution.

Question

In some references (cfr. these lecture notes on page 8), I have found condition $(\star\star)$ referred to as the universal property of the Liouville $1$-form.
All the examples I know of mathematical objects characterized (up to isomorphisms) by a certain universal property, can be recast in the language of category theory, as universal objects of some category (cfr. for example here).
Now my question is:

The universal property $(\star ~ \star)$ of the Liouville $1$-form can be recast in the language of category theory? or otherwise, in what sense can it be called a universal property?

Best Answer

I'm not an expert in differential geometry, but it seems to me that the property of the Liouville $1$-form only talks about the given manifold and its tangent bundle, no other manifolds are involved, hence this isn't a universal property. But I think that we can generalize the property as follows:

Let $\mathsf{Mfd}/X$ denote the category of smooth manifolds $Y$ equipped with a smooth map $Y \to X$. Consider the functor $(\mathsf{Mfd}/X)^{\mathrm{op}} \to \mathsf{Vect}$ which maps $Y \to X$ to $\Omega^1(Y)$. Then I claim that this functor is represented by the Liouville $1$-form $(T^* X \to X,\lambda_X)$. This means: Given $Y \to X$ and $\omega \in \Omega^1(Y)$, there is a unique smooth map $f : Y \to T^* X$ over $X$ such that $f^* \lambda_X = \omega$.

In fact, one defines $f$ to be the composition $Y \xrightarrow{\omega} T^* Y \to T^* X$. Then $f^* \lambda_X = \omega^* \lambda_Y = \omega$.

EDIT: As mentioned in the comments, one has to take relative differential forms $\Omega^1(Y/X)$.

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