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I've been trying to study differential geometry on the context of physics but somethings are really cloudy and I can't figure out a correct and fluid interpretation on some things from the textbooks alone. So, my questions:

1) How can I understand what a cotangent bundle is intuitively? That is, to be convinced that it is the space of linear functionals on the tangent bundle by some geometric insight.

2) How can I see that a differential form (say, a 2-form $\omega$) is a map from some manifold to its cotangent bundle? Or that it can be viewed as a section of it?

3) Also, here is how I see understand a pullback: let $\varphi:M \to N$ be a map between manifolds and $\omega$ (which I don't know how to define since my second question is exactly regarding if $\omega(Y)\in T^{ *}N$ for $Y\in N$) a differential r-form on $N$. The pullback $\varphi^{*} \omega$ is a map which takes elements of $M$ to the image of the r-form correspondent to the image (regarding $\varphi$) of the original element? That is, takes an element $X$ of $M$ to $\omega(\varphi (X))$. Is it correct?

Please correct me if any of the statements is not correct. Thanks!

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    If you want "to be convinced that it is the space of linear functionals on the tangent bundle," you must have some *other* definition of the cotangent bundle in mind. What is that other definition? Otherwise the question makes no sense: that's just the definition. It's like asking to be convinced that a topological space is a set together with a subset of the powerset that contains the empty set and the whole set and is closed under infinite unions and finite intersections. – 2017-03-01
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    @symplectomorphic I see! I guess that what I meant (or should have, then) is what it means geometrically (if it means something, particularly) to be the space of linear functionals on the tangent bundle. Thanks for the comment! – 2017-03-01

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  1. The cotangent bundle is not the "space of linear functionals on $TM$" rather than it is a bundle over $M$ whose fiber at a point $p \in M$ consists of linear functionals on the tangent space $T_p M$. This is usually just the definition of the cotangent bundle so I don't see how one should be convinced this is the case unless the definition of the cotangent bundle given is not what I refer to.
  2. A differential $k$-form on $M$ can be thought of as an object $\omega$ that for each $p \in M$ gives you a alternating multilinear map $\omega|_{p} \colon T_p M \times \dots \times T_p M \rightarrow \mathbb{R}$ (where there are $k$ copies of $T_p M$ on the domain) and satisfies some regularity conditions. In other words, for each $p \in M$ we want to choose a $k$-multilinear alternating map on $T_pM$ (note how at each $p \in M$ this is a different vector space!) "which varies smoothly" with $p$. One way to define such a form rigorously is to construct a bundle $\Lambda^k(T^{*}M) \rightarrow M$ (which includes specifying the topology, the smooth structure and the vector bundle structure) whose fiber at each $p \in M$ is the space of $k$-multilinear alternating maps on $T_p M$. Then a $k$-form is just a smooth section $\omega \colon M \rightarrow \Lambda^k(T^{*}M)$ of the bundle. For each $p \in M$, $\omega|_{p}$ (which more properly should be written as $\omega(p)$ because it is the value of a function at a point $p$ but the $\omega|_{p}$ notation makes things appear less cluttered later) "chooses" a $k$-multilinear alternating map on $T_pM$ and the "smooth" part makes such that the choices vary smoothly with $p$.
  3. I don't understand your sentence but let me describe the pullback operation rigorously. Given a $k$-form $\omega$ on $N$ and a smooth map $f \colon M \rightarrow N$, we want to define a $k$-form on $M$. That is, for each $p \in M$, we need to specify a $k$-multilinear alternating map on $T_p M$. This is done by the formula: $$ (f^{*} \omega)|_{p}(X_1,\dots,X_k) := \omega|_{f(p)}(df|_p(X_1), \dots, df|_p(X_k)). $$ Namely, we evaluate the $k$-multilinear map $\omega|_{f(p)}$ defined on $T_{f(p)}M$ with the pushforwards under the differential $df|_p$ of the tangent vectors $X_1,\dots,X_k$.
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    Thanks for the answer! So, regarding 1) as I said on another comment, I see! I guess that what I meant (or should have, then) is what it means geometrically (if it means something, particularly) to be the space of linear functionals on the tangent bundle. 2) So, seeing a k-form as something whose codomain is a cotangent bundle comes from looking at how it varies regarding the (continuous) change of the vector space (the tangent bundle to each point on the manifold) where it is defined? – 2017-03-01
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    Well, regarding $(1)$, I think it is easier to start with the geometric meaning of $TM$ first. In physics, this is often called the phase space. You know that to determine the motion of a particle uniquely in (say) $\mathbb{R}^3$ under the influence of various forces, you need to specify the particle's position and velocity at some time. By moving from $\mathbb{R}^3$ to $T(\mathbb{R}^3) = \mathbb{R}^3 \times \mathbb{R}^3$ (where the first three coordinates are the position coordinates and the last three coordinates are the velocity coordinates) you can just specify an initial point in – 2017-03-01
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    $T(\mathbb{R}^3)$ and then the trajectory of the point (**in $T(\mathbb{R}^3)$**, so you track down both position and velocity) will be determined uniquely. Now think of the same situation, only now the particle is constrained to lie on some manifold (say the sphere). You can imagine a small ball connected a rigid rod which is fixed to the origin at one side but can rotate (like a "two-dimensional" pendulum). To determine the trajectory of the ball, you need to specify a unique point in $T(S^2)$ - the initial position and the initial velocity which is now a tangent vector to the sphere. – 2017-03-01
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    The geometric picture of $T^{*}M$ is harder to explain but is related to the Lagrangian v.s Hamiltonian point of view of analytical mechanics and, roughly speaking, tracking a point in $T^{*}M$ amounts to tracking its generalized position and momentum instead of position and velocity. I didn't understand your second point. The codomain of a $k$-form is not the cotangent bundle, only of a $1$-form. This is because to each point $p \in M$ you associate a linear functional on $T_pM$ just like a vector field is a function from $M$ to the tangent bundle (to each point you associate a tangent vector – 2017-03-01
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    I didn't understand. Isn't it general that the codomain of a k-form is the cotangent bundle of the manifold where it is defined? Is it the alternating multilinearity condition that makes this not to be the case? And what does it means to say that the codomain is a _section_ of the cotangent bundle? Also, how is a vector field a function from $M$ to its tangent bundle? I understand that this is all very simple, but my knowledge on topology and differentiable manifolds is almost nonexisting, I've been (probably wrongfully) trying to study through an easypath towards the physics applications. – 2017-03-01
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    A vector field is a function from $M$ to the tangent bundle, not the cotangent bundle. I think it is best you read about it in some textbook, it's too much to explain in the comments. Are you using some textbook to study? What is your mathematical background? – 2017-03-01
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    I'm studying some lecture notes right now (was going through Fecko's _Differential geometry and Lie groups for physicists_, but soon decided by a less complete approach). I'm familiar mostly with multivariable calculus, analytic geometry, tensor analysis and some abstract algebra. – 2017-03-01
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    I can recommend reading the relevant parts about differential forms and vector bundles in "Gauge Fields, Knots and Gravity" by John Baez. They are well-written and might help you with intuition and basic definitions. – 2017-03-01