Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

1 Answer 1

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 $\Omega^1(Y/X)$.

share|improve this answer
2  
What is the arrow $T^*Y \to T^*X$ you used in the last line? –  Michael Bächtold Oct 1 '13 at 17:19
2  
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
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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