How to understand the symplectic 2-form $dp\wedge dq$?

I am reading V.I. Arnold's book "Mathematical methods of Classical mechanics". At the beginning of chapter 8 section B I can find the theorem (I limit the question to the $1$-dimensional case):

The cotangent bundle $T^*V$ has a natural sympletic structure. In local coordinates described above, this sympletic structure is given by the formula $\omega^2 = dp\wedge dq$

It is not clear what kind of tangent vector I should plug into this form. Certainly, amongst other stuff, something like $c\frac{\partial}{\partial q}$ (where $c$ some number). However, how do elements of the cotangent bundle tangent space look like?

It seems to be that Arnold considers the phase space as the cotangent bundle of an underlying manifold $V$ which is denoted by $T^*V$. According to Arnold the set $T^*V$ has a natural structure of a differentiable manifold of dimension $2n$ (as the phase space). A point of $T^*V$ is a 1-form on the tangent space to $V$ at some point $q$.

So if $\omega^2$ lives on such a manifold its vectors to operate on would be elements of $T(T^*V)$, and I really don't know how the base vectors of such tangent space look like. What is the tangent of an element of this space (for example, the tautological form $\omega^1 = p dq$)? For a simple point $q$ of $V$ it would be something like $c\frac{\partial}{\partial q}$, but what happens if the point I want my tangent vector touch is a section of a cotangent bundle?

I admit, this could be a question of Mathematics stack exchange, but as these forms $\omega^1$ and $\omega^2$ also appear in theoretical mechanics I post my question here.

• I'm not sure what you mean by "how the base vectors of such tangent space look like". The phase space is a manifold of dimension 2n, with canonical coordinates $(q^i,p_i)$. That the $p_i$ "originally" acted as 1-forms on the configuration space is of little importance, the vectors on the phase space are spanned, as usual, by the derivatives w.r.t. the coordinates, i.e. $q^i$ and $p_i$. – ACuriousMind Jan 3 '17 at 23:00

I'm not sure if I get the question correctly. If $V$ is a smooth manifold, then there exists a vector space $T_p^*V$, which is called the cotangent vector space @ point $p$ and is the dual space of $T_pV$ (the tangent space @ point $p$). Any element of $T^*_pV$ is a linear combination of its dual basis $$d_px^1,\cdots,d_p x^{\dim(V)}$$ where $d_p:C^\infty(V)\tilde{\to}T^*_pV,f\mapsto d_pf$ is a linear map called the gradient of $f$ @ $p$. One can generalise this, so that $$d:C^\infty(V)\tilde{\to}\Gamma(T^*V)$$ $$f\mapsto df$$ where $\Gamma(T^*V)$ is the set of all covector fields. Equivalently, one can write this as $$d:\Omega^0(V)\tilde{\to}\Omega^1(V)$$ where $\Omega^0(V)\equiv C^\infty(V)$ is the set of all 0-forms and $\Omega^1(V)\equiv\Gamma(T^*V)$ is the set of all 1-forms. Then, we have the definition of the wedge product, i.e $$\wedge:\Omega^n(V)\times\Omega^m(V)\tilde{\to}\Omega^{n+m}(V)$$ $$(\omega,\sigma)\mapsto \omega\wedge\sigma$$ and so, if $p,q\in C^\infty(V)$, then $dp,dq\in\Gamma(T^*V)$, but that means $dp,dq\in\Omega^1(V)$ with $dp\wedge dq\in\Omega^2(V)$, where (I'm not going to give the general definition) $$dp\wedge dq=dp\otimes dq-dq\otimes dp$$ The question still remains. What does this holy monster eat for dinner. Well, by definition I believe it eats 2 vector fields $X_1,X_2\in\Gamma(TV)$, since $$\Omega^2(V):=\{\omega:\Gamma(TV)\times\Gamma(TV)\tilde{\to}C^\infty(V)\}$$ Basically, I will give the general definition of the wedge product for extra help, where $$(\omega\wedge\sigma)(X_1,\cdots,X_{n+m})=\frac{1}{n!m!}\sum_{\pi\in Perm(n+m)}sign(\pi)(\omega\otimes\sigma)(X_{\pi(1)},\cdots, X_{\pi(n+m)})$$ with $\omega,\sigma$ defined above. In simpler words, for each permutation between the vector fields just change the sign. If you don't remember how a vector field is strictly defined, then let me remind you it is a smooth section $\sigma$ of the tangent bundle $TV$, for which $\pi\circ\sigma=e$, where $\pi:TV\to V$ is the smooth projection map and $e=id_V$ the identity on $V$. Because this definition confuses me, I prefer to think that the vector field is just a field, whose evaluation at a point gives a vector. I hope I helped a bit, but then again I'm not completely sure if this answers your question.
• Thank you for the long answer. My real problem is to imagine (in analogy to a tangent space $T_p M$ of a point p in a manifold $M$) an element of the tangent space $T(T^*_p M)$ to a point x of $T_p^*M$ which is according to your explanation is (at least very similar) to an element of the dual space of $T_p^*M$, i.e. $T_p M$. That's seems to be rather strange. But in the meantime I've thought, coordinates are needed on $T_p^*M$, once I've them, via Darboux's theorem I can set up $dp \wedge dq$ and plug in the corresponding tangent vectors (in the corresponding coordinate system). – Frederic Thomas Jan 6 '17 at 16:37
• The point is that Arnold only explains Darboux theorem at the end of chapter 8 and not at the place where he defines $dp \wedge dp$. That creates most of the confusion. – Frederic Thomas Jan 6 '17 at 16:37
Yes, that is correct you can evaluate on sections of $T(T^*V)$ so vector fields on cotangent bundle total space. In the case you have described you can write as combinations of $\frac{\partial}{\partial q}$ and $\frac{\partial}{\partial p}$ over some functions. So for example the 1-form Liouville $\lambda = p dq$ we may try on $\frac{\partial}{\partial q}$ to get $p$ since they are dual bases by construction.