# The universal property of the Liouville $1$-form

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?

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Is there also an "algebraic Liouville $1$-form" for (nice) schemes? –  Martin Brandenburg Oct 1 '13 at 8:10
@Martin Brandenburg Definition: a scheme $Y$ over a field $k$ is "nice" if $\Omega_k^{**}Y=\Omega_kY$. In this trivial case (that includes the smooth schemes), there is a Liouville $1$-form (see below) –  Sasha Anan'in Oct 3 '13 at 10:18

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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What is the arrow $T^*Y \to T^*X$ you used in the last line? –  Michael Bächtold Oct 1 '13 at 17:19
Your answer is probably the correct one if on replaces the functor $\Omega^1 Y$ with what algebraic geometres might denote with $\Omega^1_{Y/X}$ (in differential geometry I have heard this called the the module of one forms on $X$ relative to the map $\pi: Y\to X$. An element of it is a map which to every point $p\in Y$ associates a co-vector at $\pi(p)$.) One only needs to observe that the Loiuville form is actually en element of $\Omega^1_{T^*X/X}$ and the construction of $f:Y\to T^*X$ should be obvious. –  Michael Bächtold Oct 1 '13 at 17:57
@Martin Brandenburg If this is not my "common false belief" that $T^*Y\to T^*X$ does not exist in general, then you might include it in your list. The rest is very well done and in what follows I just complete the proof of the existence and uniqueness of $f$ (for a change, in the category of smooth algebraic varieties, thus killing two birds). –  Sasha Anan'in Oct 1 '13 at 18:41
@Martin Brandenburg First, I remind the definition of $T^*X$ in the case of smooth affine $X=\text{Spec}A$ over a field $k$ : $T^*X:=\text{Spec}A[\Omega_k^*A]$, where $\Omega_k^*A:=\text{Hom}_A(\Omega_kA,A)$, because $T^*X$ is glued from those affine ones for a general smooth $X$ (note that $\Omega_k^{**}A=\Omega_kA$ if $X$ is smooth because $\Omega_kA$ is a finite rank free $A$-module). The problem of existence and uniqueness is local in $Y$ and then in $X$. So, we assume both $X$ and $Y$ affine and the rest is obvious. Also, your isomorphism is natural in $X$, more universality. –  Sasha Anan'in Oct 1 '13 at 18:45
@MartinBrandenburg, given a smooth map $X\to Y$, I know how to construct its cotangent lift $T^\ast Y\to T^\ast X$, only when $X\to Y$ is a local diffeomorphism, but not in general. So we can work at least with the subcategory of local diffeomorphisms $X\to Y$. –  Giuseppe Tortorella Oct 2 '13 at 11:16