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From Analysis and Algebra on Differentiable Manifolds, first edition, exercise 2.6.4., question 1 (slightly edited for this post):

Let $\vartheta$ be the canonical 1-form on the cotangent bundle $T^* M$ over a $C^\infty$ n-manifold $M$. Prove that $d\vartheta$ is a 2-form $\Omega$ on $T^* M$ such that the vertical bundle of the natural projection $p:T^*M\rightarrow M$ is a Lagrangian foliation; that is, the fibres of $p$ are totally isotropic submanifolds.

From previous exercises the coordinate system on $T^*M$ is of the form $(q^1,..,q^n,p_1,...,p_n)$, and to prove the assertion, the authors use the coordinates expression of $\Omega=dp_i\wedge dq^i$ and note that the tangent space to the fibres of $p$ is locally spanned by $\frac{\partial}{\partial p_i}$.

So my questions are:

  1. How is $\Omega$ defined on the submanifold ? Is it restricted in some specific way, and what would the coordinate expression of this restriction be ?
  2. If not, is there a meaning for the expression $\frac{\partial}{\partial q^i}$ on the fibres as submanifolds ? Because on a given fibre, all $q^i$ are constant, so you can't apply derivatives along $q^i$ to functions defined on those submanifolds. And, ultimately this would give me the meaning of $dq^i$ and how to apply it to prove the non degenerate nature of $\Omega$ on the fibres.
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Are you sure about the result that should be proved? What about $2d\nu$? – Daan Michiels Feb 5 '13 at 19:10
The result to prove is the non degenerate nature of the 2 form on the fibres, that is certain. They do not use the 2 form on any vectors, but the definition of a non degenerate 2 form use its action on pairs of vectors, hence my question. – vkubicki Feb 7 '13 at 12:15
Could you please post the complete exercise 2.6.4? You only wrote part of the exercise (points (2) and (3) are missing), and what is to be proved now ("that $d\nu$ is the only...") is wrong. – Daan Michiels Feb 7 '13 at 12:48
The post has been edited to include the verbatim formulation of question 1. Question 1 should be answered independently of question 2 and 3, which introduce new objects. – vkubicki Feb 8 '13 at 1:59
My point is that question (1) should not be answered independently of questions (2) and (3). They are three conditions that belong together. The exercise is: prove dat $d\theta$ is the only 2-form $\Omega$ on $T^*M$ such that (1), (2) and (3) are all true. Indeed, $d\theta$ is not the only 2-form satisfying condition (1), since $2d\theta$ also does. – Daan Michiels Feb 8 '13 at 10:02
up vote 3 down vote accepted

This answer is rather verbose. I hope I didn't overdo it.

  1. Yes, $\Omega$ gets restricted. Suppose $x\in T^*M$ is some point in the cotangent bundle and write $F$ for the fibre that contains $x$. Then our job is to define the restriction of $\Omega$ to $F$, which we will for clarity call $\Omega'$. If $v,w\in T_xF$ are two tangent vectors to the fibre, then we should find a formula to define $\Omega'(v,w)$. Note that $v,w$ can also be regarded as elements of $T_x(T^*M)$ (we identify them with their pushforward under the inclusion map of $F$ into $T^*M$, if you wish). This means we can just set $$ \Omega'(v,w) = \Omega(v,w) ,$$ which is the reason why $\Omega'$ is usually just called $\Omega$.

    Coordinate expression. On the fibre $F$ we have coordinates $p_i$ (no coordinates $q^i$ anymore on the fibre). To find an expression for the restricted version $\Omega'$ of $\Omega$ in these coordinates, it suffices to calculate the values $$ \Omega'\left( \frac{\partial}{\partial p_i}, \frac{\partial}{\partial p_j} \right) $$ for all $i$ and $j$ (actually, calculating these for $i<j$ is enough by antisymmetry). To calculate these, note that $\frac{\partial}{\partial p_i}$ as a tangent vector to the fibre corresponds to $\frac{\partial}{\partial p_i}$ as a tangent vector to $T^*M$. This gives us $$ \Omega'\left( \frac{\partial}{\partial p_i}, \frac{\partial}{\partial p_j} \right) = \Omega\left( \frac{\partial}{\partial p_i}, \frac{\partial}{\partial p_j} \right) = 0 .$$ In this equation, the arguments to $\Omega'$ should be regarded as tangent vectors to $F$, whereas the arguments to $\Omega$ should be regarded as tangent vectors to $T^*M$. We have shown that $\Omega'=0$ because it returns $0$ for any pair of basis vectors. In fact, showing that $\Omega'=0$ is exactly showing that $F$ is isotropic.

  2. As far as I'm aware, $\frac{\partial}{\partial q^i}$ does not have a meaning on the fibres as submanifolds. Note, however, that $dq^i$ most definitely does have a meaning on the fibres as submanifolds (even though this meaning turns out to be zero). In fact, there are two equivalent ways of giving meaning to $dq^i$.

    First approach. We can use the same procedure we used for $\Omega$ and restrict $dq^i$ to a 1-form on the fibre $F$. In other words, here is how the resulting 1-form would work: given a vector $v$ tangent to $F$, we can regard it as a tangent vector to $T^*M$ and then feed it to the original $dq^i$ (a 1-form on $T^*M$) to produce a number.

    Second approach. We can regard $q^i$ as a smooth function on the fibre $F$ (by restriction) and then calculate its differential $dq^i$ (strictly speaking $d({q^i}_{\mid F})$) which is a 1-form on $F$.

    Equivalence of the two approaches To show that the two approaches are equivalent, we'll make the following observation: if $f$ is a smooth function on $M$ and $N$ is a submanifold of $M$, then the 1-form on $N$ obtained by differentiation $f_{\mid N}$ equals the restriction of $df$ to $N$. Indeed, suppose $v$ is a tangent vector to $N$. Take a smooth curve $\gamma$ in $N$ which has speed $v$ at time $t=0$. The two forms are equal because both take the value $$ \frac{d}{dt}_{\mid t=0}\left( f(\gamma(t)) \right) $$ on $v$ (check this as an exercise).

    The case at hand. I claimed that in your question, $dq^i$ is actually just the 1-form $0$. Indeed, $q^i$ is constant on $F$ (by definition of fibre), so that its differential is zero.

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So to put it another way, the restriction of $\Omega$ to the fibres is the image of $\Omega$ by the pullback of the inclusion map $i:F\rightarrow T^* M$ ? – vkubicki Feb 10 '13 at 7:43
In fact, I misunderstood the definition of an isotropic submanifold, I thought that the restriction of the symplectic form had to be non degenerate too. – vkubicki Feb 10 '13 at 7:54

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