The Unapologetic Mathematician

Mathematics for the interested outsider

Coordinate Vectors Span Tangent Spaces

Given a point p in an n-dimensional manifold M, we have the vector space \mathcal{T}_pM of tangent vectors at p. Given a coordinate patch (U,x) around p, we’ve constructed n coordinate vectors at p, and shown that they’re linearly independent in \mathcal{T}_pM. I say that they also span the space, and thus constitute a basis.

To see this, we’ll need a couple lemmas. First off, if f is constant in a neighborhood of p, then v(f)=0 for any tangent vector v. Indeed, since all that matters is the germ of f, we may as well assume that f is the constant function with value c. By linearity we know that v(c)=cv(1). But now since 1=1\cdot1 we use the derivation property to find

\displaystyle v(1)=v(1\cdot1)=v(1)1(p)+1(p)v(1)=2v(1)

and so we conclude that v(1)=v(c)=0.

In a slightly more technical vein, let U be a “star-shaped” neighborhood of 0\in\mathbb{R}^n. That is, not only does U contain 0 itself, but for every point x\in U it contains the whole segment of points tx for 0\leq t\leq1. An open ball, for example, is star-shaped, so you can just think of that to be a little simpler.

Anyway, given such a U and a differentiable function f on it we can find n functions \psi_i with \psi_i(0)=D_if(0), and such that we can write

\displaystyle f=f(0)+\sum\limits_{i=1}^nu^i\psi_i

where u^i is the ith component function.

If we pick a point x\in U we can parameterize the segment c(t)=tx, and set \phi=f\circ c to get a function on the unit interval \phi:[0,1]\to\mathbb{R}. This function is clearly differentiable, and we can calculate

\displaystyle\phi'(t)=\sum\limits_{i=1}^n\frac{\partial f}{\partial x^i}\frac{dc^i}{dt}=\sum\limits_{i=1}^nD_if(tx)x^i

using the multivariable chain rule. We find


We can thus find the desired functions \psi_i by setting


Now if we have a differentiable function f defined on a neighborhood U of a point p\in M, we can find a coordinate patch (U,x) — possibly by shrinking U — with x(p)=0 and x(U) star-shaped. Then we can apply the previous lemma to f\circ x^{-1} to get

\displaystyle f\circ x^{-1}=f(p)+\sum\limits_{i=1}^nu^i\psi_i

with \psi_i(0)=\left[\frac{\partial}{\partial x^i}(p)\right](f). Moving the coordinate map to the other side we find

\displaystyle f=f(p)+\sum\limits_{i=1}^nx^i\left(\psi_i\circ x\right)

Now we can hit this with a tangent vector v

\displaystyle\begin{aligned}v(f)&=v\left(f(p)+\sum\limits_{i=1}^nx^i\left(\psi_i\circ x\right)\right)\\&=v\left(f(p)\right)+\sum\limits_{i=1}^nv\left(x^i\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^n\left(v(x^i)\left[\psi_i\circ x\right](p)+x^i(p)v\left(\psi_i\circ x\right)\right)\\&=\sum\limits_{i=1}^nv(x^i)\psi_i(0)\\&=\sum\limits_{i=1}^nv(x^i)\left[\frac{\partial}{\partial x^i}(p)\right](f)\end{aligned}

where we have used linearity, the derivation property, and the first lemma above. Thus we can write

\displaystyle v=\sum\limits_{i=1}^nv(x^i)\frac{\partial}{\partial x^i}(p)

and the coordinate vectors span the space of tangent vectors at p.

As a consequence, we conclude that \mathcal{T}_pM always has dimension n — exactly the same dimension as the manifold itself. And this is exactly what we should expect; if M is n-dimensional, then in some sense there are n independent directions to move in near any point p, and these “directions to move” are the core of our geometric notion of a tangent vector. Ironically, if we start from a more geometric definition of tangent vectors, it’s actually somewhat harder to establish this fact, which is partly why we’re starting with the more algebraic definition.

March 31, 2011 Posted by | Differential Topology, Topology | 8 Comments