The Unapologetic Mathematician

Mathematics for the interested outsider

Cotangent Vectors, Differentials, and the Cotangent Bundle

There’s another construct in differential topology and geometry that isn’t quite so obvious as a tangent vector, but which is every bit as useful: a cotangent vector. A cotangent vector \lambda at a point p\in M is just an element of the dual space to \mathcal{T}_pM, which we write as \mathcal{T}^*_pM.

We actually have a really nice example of cotangent vectors already: a gadget that takes a tangent vector at p and gives back a number. It’s the differential, which when given a vector returns the directional derivative in that direction. And we can generalize that right away.

Indeed, if f is a smooth germ at p, then we have a linear functional v\mapsto v(f) defined for all tangent vectors v\in\mathcal{T}_pM. We will call this functional the differential of f at p, and write \left[df(p)\right](v)=v(f).

If we have local coordinates (U,x) at p, then each coordinate function x^i is a smooth function, which has differential dx^i(p). These actually furnish the dual basis to the coordinate vectors \frac{\partial}{\partial x^i}(p). Indeed, we calculate

\displaystyle\begin{aligned}\left[dx^i(p)\right]\left(\frac{\partial}{\partial x^j}(p)\right)&=\left[\frac{\partial}{\partial x^j}\right](x^i)\\&=\left[D_j(u^i\circ x\circ x^{-1})\right](x(p))\\&=\delta_j^i\end{aligned}

That is, evaluating the coordinate differential dx^i(p) on the coordinate vector \frac{\partial}{\partial x^j}(p) gives the value 1 if i=j and 0 otherwise.

Of course, the dx^j(p) define a basis of \mathcal{T}^*_pM at every point p\in U, just like the \frac{\partial}{\partial x^j}(p) define a basis of \mathcal{T}_pM at every point p\in U. This was exactly what we needed to compare vectors — at least to some extent — at points within a local coordinate patch, and let us define the tangent bundle as a 2n-dimensional manifold.

In exactly the same way, we can define the cotangent bundle \mathcal{T}^*M. Given the coordinate patch (U,x) we define a coordinate patch covering all the cotangent spaces \mathcal{T}^*_pM with p\in U. The coordinate map is defined on a cotangent vector \lambda\in\mathcal{T}^*_pM by

\displaystyle\tilde{x}(\lambda)=\left(x^1(p),\dots,x^n(p),\lambda\left(\frac{\partial}{\partial x^1}(p)\right),\dots,\lambda\left(\frac{\partial}{\partial x^n}(p)\right)\right)

Everything else in the construction of the cotangent bundle proceeds exactly as it did for the tangent bundle, but we’re missing one thing: how to translate from one basis of coordinate differentials to another.

So, let’s say x and y are two coordinate maps at p, defining coordinate differentials dx^i(p) and dy^j(p). How are these two bases related? We can calculate this by applying dy^j(p) to \frac{\partial}{\partial x^j}(p):

\displaystyle\begin{aligned}\left[dy^j(p)\right]\left(\frac{\partial}{\partial x^i}(p)\right)&=\left[\frac{\partial}{\partial x^i}\right](y^j)\\&=\left[D_i(u^j\circ y\circ x^{-1})\right](x(p))\\&=J_i^j(p)\end{aligned}

where J_i^j(p) are the components of the Jacobian matrix of the transition function y\circ x^{-1}. What does this mean? Well, consider the linear functional

\displaystyle\sum\limits_iJ_i^j(p)dx^i(p)

This has the same values on each of the \frac{\partial}{\partial x^i}(p) as dy^j does, and we conclude that they are, in fact, the same cotangent vector:

\displaystyle dy^j(p)=\sum\limits_iJ_i^j(p)dx^i(p)

On the other hand, recall that

\displaystyle\frac{\partial}{\partial x^i}(p)=\sum\limits_jJ_i^j(p)\frac{\partial}{\partial y^j}(p)

That is, we use the Jacobian of one transition function to transform from the dx^i(p) basis to the dy^j(p) basis of \mathcal{T}^*_pM, but the transpose of the same Jacobian to transform from the \frac{\partial}{\partial x^i}(p) basis to the \frac{\partial}{\partial y^j}(p) basis of \mathcal{T}_pM. And this is actually just as we expect, since the transpose is actually the adjoint transformation, which automatically connects the dual spaces.

About these ads

April 13, 2011 - Posted by | Differential Topology, Topology

3 Comments »

  1. […] patch on , we get a basis of for each . Then, just as we did with the tangent bundle and the cotangent bundle we can come up with a coordinate patch “induced by ” on each of our new bundles. In […]

    Pingback by Tensor Bundles « The Unapologetic Mathematician | July 6, 2011 | Reply

  2. […] Similarly, we pass from the -coordinate basis to the -coordinate basis of by using another Jacobian: […]

    Pingback by Change of Variables for Tensor Fields « The Unapologetic Mathematician | July 8, 2011 | Reply

  3. […] of the coordinate transformation from one patch to the other. Indeed, we use the Jacobian to change bases on the cotangent bundle, and transforming between these top forms amounts to taking the determinant of the transformation […]

    Pingback by Compatible Orientations « The Unapologetic Mathematician | August 29, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 393 other followers

%d bloggers like this: