The Unapologetic Mathematician

The Tangent Bundle

So far we’ve talked about tangent spaces one at a time. For each $p\in M$ we get a tangent space $\mathcal{T}_pM$ at $p$. But things get really interesting when we start to sew them all together.

So, let’s take the (disjoint) union of all the tangent spaces $\mathcal{T}_pM$ at once. It’s important here that we never identify any of these with each other; a tangent vector at $p$ is never the same as another tangent vector at $q$. So far this is just a bunch of $n$-dimensional vector spaces parameterized by the $n$-dimensional manifold $M$, but I say that we can actually give this whole set the structure of a $2n$-dimensional smooth manifold!

Indeed, all we need is to give a collection of patches covering the whole space, and we can do this starting from any atlas on $M$. Given an open coordinate patch $(U,x)$ in $M$, we will make a coordinate patch that covers all of the $\mathcal{T}_pM$ with $p\in U$. Given a $v\in\mathcal{T}_pM$ we need to come up with a point $\bar{x}(v)\in\mathbb{R}^{2n}$. We write

$\displaystyle\bar{x}(v)=\left(x^1(p),\dots,x^n(p),v(x^1),v(x^2),\dots,v(x^n)\right)$

As $v$ varies within $\mathcal{T}_pM$, the first $n$ components of the image — $x(p)$ — stay fixed, and the rest of them give a linear isomorphism from $\mathcal{T}_pM$ to $\mathbb{R}^n$.

On the other hand, how do things change as we vary the base point $p$? This, it turns out, is where the interesting stuff is happening. Normally we wouldn’t be able to tell anything about how $\mathcal{T}_pM$ and $\mathcal{T}_qM$ are related for $p\neq q$, but within a single coordinate patch we can use the coordinate map $x$ to define coordinate vectors at every single point $p\in U$, and this lets us compare vectors at different points by comparing their components with respect to these coordinate vector bases.

The catch is, this only works for points in the same coordinate patch, and different coordinate patches give us different ways of comparing tangent vectors at nearby points. So we can’t really say much about them, but it’s enough to define a coordinate patch.

So we have to check about where patches overlap. If we have $(U,x)$ and $(V,y)$ as two patches on $M$ then we already know that $y\circ x^{-1}$ is a smooth transition function. This handles the smoothness of the first $n$ components of the transition function $\bar{y}\circ\bar{x}^{-1}$. For the other components, we know that the transition function is a linear isomorphism, which is clearly smooth.

We call this manifold the “tangent bundle” of $M$ and write $\mathcal{T}M$. It comes equipped with a map $\pi:\mathcal{T}M\to M$, which I say is smooth. Indeed, we just need to check this on a single pair of coordinate patches. We can pick any patch $(U,x)$ on $M$ and the corresponding patch on $\mathcal{T}M$. Then given $(a,b)\in x(U)\times\mathbb{R}^n$ we get a vector $\bar{x}^{-1}(a,b)=v\in\mathcal{T}_{x^{-1}(a)}M$ by writing

$\displaystyle v=\sum\limits_{i=1}^nb^i\left(\frac{\partial}{\partial x^i}(x^{-1}(a))\right)$

Our projection then sends $v$ to its base-point $x^{-1}(a)$, which $x$ sends to $a$. That is, if we write our projection $\pi$ out in terms of coordinates $\bar{x}$ on $\mathcal{T}M$ and $x$ on $M$, it’s just the projection $x(U)\times\mathbb{R}^n\to x(U)$, which is obviously smooth.

April 4, 2011