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You don't need a metric to define the differential of a function, and the cotangent bundle carries a canonical one-form.

But you do need a metric to define the gradient, and the tangent bundle does not have a canonical vector field.

These are not difficult truths, but still... why the preference toward "co"?

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    $\begingroup$ I think the question, as written, is somewhat misleading. The tangent and cotangent bundles are equally canonical. They come for free with the differentiable structure. Another thing which comes for free with the differentiable structure are the smooth functions, which are arrows from the objects. Hence it is not surprising that contravariant functors are preferred. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Mar 6, 2010 at 23:15
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    $\begingroup$ If you construct them as the tangent and cotangent sheaf, the tangent sheaf is easier to construct as it is nothing more than the sheaf defined by $T(U):=Der_{\mathbb{R}}(\mathcal{O}_{\mathcal{M}}(U),\mathcal{O}_{\mathcal{M}}(U))$. The cotangent sheaf is just constructed by the universal property of the module of differentials. $\endgroup$
    – Harry Gindi
    Commented Mar 7, 2010 at 0:03
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    $\begingroup$ In fpqc's formula the ring O_M(U) must be replaced with the restriction sheaf (O_M)|_U, just like definition of Hom sheaf. Otherwise it doesn't even make sense as a presheaf. As for the cotangent sheaf, a definition is no harder: (Delta_M)*(I/I^2) on M akin to algebraic case. Strictly speaking, modules of differentials in the algebraic sense are not logically relevant at this step since M x M is not built via tensor products (but the theories behave similarly, and for completed stalk calculations for analytic M the link with modules of continuous differentials is very useful) $\endgroup$
    – BCnrd
    Commented Mar 7, 2010 at 14:46
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    $\begingroup$ The word "canonical" seems to be an unintended pun (cf canonical transformations in symplectic geometry and all that). $\endgroup$
    – Victor Protsak
    Commented Jul 17, 2010 at 20:20

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If you want to differentiate functions from a manifold to (say) the real line R, then you want to use the cotangent bundle on the manifold.

If instead you want to to differentiate functions to the manifold from the real line (i.e. parameterised curves), then you want to use the tangent bundle on the manifold.

So the preference comes from whether you want to use the manifold as the domain or as the range of the functions one is differentiating.

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    $\begingroup$ Situation is opposite: If I want to differentiate a function on manifold at a point, then I need a tangent vector at that point. $\endgroup$
    – Petya
    Commented Mar 7, 2010 at 4:13
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    $\begingroup$ No, Terry's answer is right (see also the first paragraph of Deane's answer). A derivative is a linear approximation, so the derivative of a function $\mathbb{R}^n \to \mathbb{R}$ is a linear form on $\mathbb{R}^n$. In the case of a manifold this means you get an element of $T^*M$, which of course turns out to be the differential. $\endgroup$
    – Matt Noonan
    Commented Mar 7, 2010 at 4:51
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    $\begingroup$ This nicely captures the duality between the two, but are these answers really honest? Do y'all really think of these two structures with the democracy professed in these answers? I don't (I understand them, but I don't) and I have to twist my mind a bit to find the structure dual to the canonical form. Maybe it's just me. (Sorry, this isn't really a mathematics question.) $\endgroup$
    – Eric Zaslow
    Commented Mar 7, 2010 at 4:52
  • $\begingroup$ Democracy? Not really. In my answer, I definitely view the tangent bundle as being more fundamental, and the cotangent bundle as arising naturally from differentiating functions. On the other hand, the canonical 1-form and its covariant derivative make the cotangent bundle in many ways much more interesting to study for its own sake than the tangent bundle. $\endgroup$
    – Deane Yang
    Commented Mar 7, 2010 at 5:06
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    $\begingroup$ @Deane Yang: I suppose you mean "exterior differential" instead of covariant derivative! $\endgroup$
    – Orbicular
    Commented Mar 7, 2010 at 9:07
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Neither is more canonical than the other. The tangent bundle of $M$ represents the set of all possible derivatives of maps $R \rightarrow M$, and the cotangent bundle of $M$ represents the set of all possible derivatives of maps $M \rightarrow R$. They are dual to each other.

I hate to ruin such a nice terse answer, but I might as well describe how I think of the tangent and cotangent bundles. I have a personal prejudice for using only freshman calculus and basic linear algebra as much as possible and avoiding multivariable calculus. So here's the way I see things:

The idea is to build everything using only linear algebra and the definition of the derivative of a real-valued function of one real variable.

So for me you have to first define the tangent bundle as the set of all possible velocity vectors of parameterized curves in the given manifold. The first observation is that if you fix a point in the manifold, the set of all possible velocity vectors based at that point has a natural vector space structure.

Next, given a real-valued function on a manifold, you want to define its derivative. Well, if all you have is the 1-variable derivative, then the only thing you can do is to compose the function with a parameterized curve. Then you observe the following: The value of the derivative at a point actually depends only on the velocity vector of the curve at that point and is a linear function of the velocity vector. Therefore, the set of all possible derivatives of a real-valued function is naturally dual to the tangent bundle (viewed as the set of all possible velocity vectors). That's the cotangent bundle (the set of all possible derivatives of real-valued functions on the manifold).

This for me is a nice coherent story that I can tell (and remember) without using any mathematical symbols at all but also one whose details can be fleshed out in a straightforward manner.

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    $\begingroup$ The question is presumably about deeper geometric meaning and not about how we come to the definitions. This answer is just explaining the well-known definition. The power of a notion is seen really when we go beyond the most basic applications and most basic case. For example going beyond smooth case to nonsmooth varieties (where tangent cone is often better than tangent bundle), noncommutative geometry, derived geometry (cotangent complex with all its applications...)... $\endgroup$
    – Zoran Skoda
    Commented Mar 8, 2010 at 16:07
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    $\begingroup$ @Zoran: I suppose you could be right. I simply gave my usual simple-minded view of the world (which more than suffices for anything I use the tangent or cotangent bundles for). Any chance you could elaborate on what you mean by "deeper geometric meaning"? I'm always interested in learning more about my blind spots. $\endgroup$
    – Deane Yang
    Commented Mar 8, 2010 at 21:06
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    $\begingroup$ @Zoran: I understand your point of view, but I think the issue is a difference in what different people think of as "geometric". For myself, the term means things to do with distances, angles, volumes, curves, surfaces and so on, so Deane's answer is very appealing to me. I think that for the purposes of deep understanding, motivating a definition from a simple case is complementary to seeing how it generalizes. $\endgroup$
    – Gordon Craig
    Commented Mar 9, 2010 at 22:52
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    $\begingroup$ This was Saunders Maclane's favorite point of view: you get tangents and differentials and their pairing all at the same time. I think Deane is referring to this. That is, for each p in M consider curves x(t) w/ x(0) = p, and for f defined in open U containing p, we have a pairing <x,f> = f'(0). A tangent vector is an equivalence class of such x where x ~ y if <x,-> = <y,-> and a cotangent is an equivalence class of f where f ~ g if <-,f> = <-,g>. $\endgroup$
    – Robert Bruner
    Commented Oct 4, 2014 at 3:17
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This is a triviality, but still: there is a pullback of a differential form, but in general no push-forward of a vector field. As a consequence, one gets e.g. for any smooth map $f:X\to Y$ of smooth manifolds a map of sheaves $f^{-1}\Omega^{\bullet}_Y\to\Omega^{\bullet}_X$; similar maps exist in the complex analytic and algebraic cases.

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    $\begingroup$ We can push forward vectors but not vector fields: given a vector field in the source, we have as many vectors over a point in the target as there are the preimages of the point. In particular, if the mapping is not injective, there may be two or more, if the mapping is not surjective, there may be none at all. $\endgroup$
    – algori
    Commented Mar 7, 2010 at 3:26
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    $\begingroup$ Here is a "philosophical" explanation. If f:M' ---> M is a map of manifolds and E' and E are vector bundles on M' and M resp. then for open U in M have E(U) ---> (fE)(f^{-1}(U)) since "pull back" sections: h_E:E ---> f_*fE on M, makes sense for any sheaf in role of E but "masks" intervention of non-bundle pushforward. For sheaf map f*(f_E') --> E' there's no way to hide pushforward. Total derivative is TM' --> f*(TM) over M', wrong direction to relate with h_{TM}, so must dualize: f*(TM) --> TM' composes on sections with h_{TM}, yields pullback on 1-forms. $\endgroup$
    – BCnrd
    Commented Mar 7, 2010 at 14:30
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    $\begingroup$ I think that the broken symmetry goes all the way back to the asymmetric definition of a function: a relation on $X \times Y$ such that there is only one $y$ related to each $x$. In the case of one-to-one functions, we can pull and push vectors and covectors equally well. In the case of general relations (ie, many-to-many "functions"), we generally can't push or pull anything. $\endgroup$
    – Matt Noonan
    Commented Mar 7, 2010 at 14:52
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    $\begingroup$ Deane, what is your definition of "push forward a smooth object"? I think smoothness is a red herring; even for set-theoretic vector fields (i.e., no smoothness or even continuity condition) or set-theoretic functions, there is no meaningful notion of "push forward" in the absence of some way to "integrate along fibers" (like a trace map for a finite covering space, or something more elaborate in case of a proper submersion, etc.) $\endgroup$
    – BCnrd
    Commented Mar 7, 2010 at 17:59
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    $\begingroup$ @Brian: Yes, you're completely right. We almost always pull back rather than push forward because we usually can't push forward at all. $\endgroup$
    – Deane Yang
    Commented Mar 7, 2010 at 21:09
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If you look at manifolds with singularities, as in algebraic geometry, then the cotangent sheaf seems easier to construct. In terms of modules over coordinate rings, the cotangent sheaf just corresponds to the universal module M with a derivation d from the coordinate ring R to M. For the module of the tangent sheaf, I cant think of anything easier than just taking the dual of M.

(Added later: in particular you can construct the tangent sheaf from the cotanent sheaf, but there seems no easy way to go in the other direction, suggesting that the cotangent sheaf is more basic.)

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In my opinion there is no asymmetry between the tangent an cotangent bundle.

For example the universal one-form on the cotangent space you mention. There is an analog on the tangent space, namely the Euler field (I think it is called like that), which in local coordinates $(x_1,\ldots,x_n,\dot{x}_1,\ldots,\dot{x}_n)$ on $TM$ is given by $$ E=\sum \dot{x}_i \partial_{\dot{x}_i} $$ (the infinitesimal generator of homotheties). Actually it's better to think of it as a relative vector field along the projection $\pi:TM \to M$ (i.e. as section in the pull back of the bundle $TM$ to $TM$). Then it has the universal property that if you restrict $E$ to a section $s\colon M\to TM$ you get back $s$. Just like in the cotangent picture. To make the analogy more complete observe that the universal one form on $T^* M$ can also be seen as a relative form along the map $\pi\colon T^*M\to M$.

Algori mentions that there is a pullback of forms but no push forward of fields. That's on the level of sections of these vector bundles. But if you look at the level of the total spaces of them you find that there is a "pushforward" $Tf\colon TX\to TY$ (the tangent map) but no analog map between $T^*X$ and $T^*Y$ in either direction. So that balances the apparent asymmetry.

It's more of a yin & yang.

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There is an ``asymmetry'' already at the linear algebra level: given a (fin.dim.) vector space $V$, you get a canonical indefinite pairing and a symplectic form on $V\oplus V^\vee$, while you do not get these on $V\oplus V$.

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Just as a side remark, when dealing with geodesic flows in dynamical systems, it is more natural to think about these flows living inside the cotangent bundle because they become automatically symplectic, which is a nice (and "rigid") structure to play with: Lyapunov exponents are symmetric with respect to the origin, etc.

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Well, the cotangent bundle is not "more natural" than the tangent bundle (but this obviously depends on what you mean by "natural"!). It simply has "more canonical structure" associated to it - namely the Liouville one-form that you mentioned. This comes from the fact that the cotangent bundle is dual to the tangent bundle. Hence you can take a tangent vector to the total space of the cotangent bundle, hit it with the differential and get a vector on the manifold you started out with - this vector can now be paired with the basepoint of the tangent vector you started out with, giving you essentially the Liouville (or canonical one-form).
In contrast, there is no natural way to pair vectors (unless chosing an inner product); furthermore taking the cotangent of the projection reverses the direction of the mapping (it is "contravariant" in the sense of category theory).
Moreover there are also interesting things about the tangent bundle that the cotangent bundle does not have - vectors! These vectors can be integrated to flows (at least locally) and allowing you to deal with dynamical systems on the manifold! This is not possible for forms!
But of course, differentiating a function gives you a form (this can be checked, for instance, by looking at the transformation behaviour under chart changes). More generally the (principal) symbol of a linear partial differential operator is a (homogeneous) function on the cotangent bundle. That's also an important point (and causes the cotangent bundle to pop up in analysing PDE).

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The canonicity comes in terms of the symplectic structure. On cotangent bundle there is a canonical symplectic form that is independent of the hamiltonian where as on tangent bundle the symplectic form is dependent to the lagrangian.

Or put in another words because the cotangent bundle itself is already built from some 1-forms attached to each points, it is easy to build a symplectic from the cotangent bundle itself while since the tangent bundle is made of tangent vectors attached to each point, you need the help of a function to build a symplectic two form on it.

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Though I am late for this answer by almost 7 years, but still want to give a try. I have the following picture in mind.

$$\text{Cotangent Bundles}\leftrightarrow \text{Pull-backs}\leftrightarrow \text{Differentials}$$ $$\text{Tangent Bundles}\leftrightarrow \text{Push-forward}\leftrightarrow \text{Tangent Vectors}$$

Why the preference toward "co"?

One approach that comes to my mind is from Warner. Let $M$ be a manifold and $(U,\phi)$ and $(V,\psi)$ are two coordinate systems about $m$. If $\phi=(x_1,x_2,...,x_n)$ and $\psi=(y_1,y_2,...y_n)$, then we note that $$\frac{\partial}{\partial y_j}\Biggr|_m =\sum_{i=1}^n\frac{\partial x_i}{\partial y_j}\Biggr|_m \frac{\partial}{\partial x_i}\Biggr|_m $$

If we have $x_1=y_1$, then $\frac{\partial}{\partial y_1}\Bigr|_m \neq \frac{\partial}{\partial x_1}\Bigr|_m $ but $dy_1=dx_1$. This is because $\frac{\partial}{\partial x_i}\Bigr|_m $ depends on $\phi$ and not only on $x_i$. In this sense, differentials are more natural and hence cotangent bundles.

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