Why is cotangent more canonical than tangent? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:37:43Z http://mathoverflow.net/feeds/question/17325 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent Why is cotangent more canonical than tangent? Eric Zaslow 2010-03-06T22:40:59Z 2010-10-24T19:28:48Z <p>You don't need a metric to define the differential of a function, and the cotangent bundle carries a canonical one-form.</p> <p>But you do need a metric to define the gradient, and the tangent bundle does not have a canonical vector field.</p> <p>These are not difficult truths, but still... why the preference toward "co"?</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/17330#17330 Answer by Orbicular for Why is cotangent more canonical than tangent? Orbicular 2010-03-06T23:24:19Z 2010-03-06T23:24:19Z <p>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).<br> 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).<br> 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!<br> 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).</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/17333#17333 Answer by Deane Yang for Why is cotangent more canonical than tangent? Deane Yang 2010-03-06T23:51:39Z 2010-03-07T03:49:12Z <p>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.</p> <p>I hate to ruin such a nice terse answer, but I might as well describe how <em>I</em> 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:</p> <p>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.</p> <p>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.</p> <p>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).</p> <p>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.</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/17337#17337 Answer by algori for Why is cotangent more canonical than tangent? algori 2010-03-07T00:13:46Z 2010-03-07T00:13:46Z <p>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.</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/17355#17355 Answer by Terry Tao for Why is cotangent more canonical than tangent? Terry Tao 2010-03-07T04:08:02Z 2010-03-07T04:08:02Z <p>If you want to differentiate functions <em>from</em> a manifold to (say) the real line R, then you want to use the cotangent bundle on the manifold.</p> <p>If instead you want to to differentiate functions <em>to</em> the manifold from the real line (i.e. parameterised curves), then you want to use the tangent bundle on the manifold.</p> <p>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.</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/32206#32206 Answer by Matheus for Why is cotangent more canonical than tangent? Matheus 2010-07-16T17:31:32Z 2010-07-16T17:31:32Z <p>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.</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/32239#32239 Answer by Richard Borcherds for Why is cotangent more canonical than tangent? Richard Borcherds 2010-07-16T22:35:52Z 2010-07-16T22:44:22Z <p>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. </p> <p>(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.)</p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/32289#32289 Answer by Michael for Why is cotangent more canonical than tangent? Michael 2010-07-17T15:09:56Z 2010-07-17T15:12:38Z <p>In my opinion there is no asymmetry between the tangent an cotangent bundle.</p> <p>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 <code>$$E=\sum \dot{x}_i \partial_{\dot{x}_i}$$</code> (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 $\pi\colon T^*M\to M$.</p> <p>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.</p> <p>It's more of a yin &amp; yang. </p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/34054#34054 Answer by Peter Dalakov for Why is cotangent more canonical than tangent? Peter Dalakov 2010-07-31T20:01:18Z 2010-07-31T20:01:18Z <p>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 $V\oplus V$. </p> http://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent/43420#43420 Answer by Sina for Why is cotangent more canonical than tangent? Sina 2010-10-24T19:28:48Z 2010-10-24T19:28:48Z <p>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. </p> <p>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. </p>