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


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.

April 13, 2011 Posted by | Differential Topology, Topology | 5 Comments