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 $i$ th 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

$\displaystyle\begin{aligned}f(x)-f(0)&=\phi(1)-\phi(0)\\&=\int\limits_0^1\phi'(t)\,dt\\&=\sum\limits_{i=1}^nx^i\int\limits_0^1D_if(tx)\,dt\end{aligned}$

We can thus find the desired functions $\psi_i$ by setting

$\displaystyle\psi_i(x)=\int\limits_0^1D_if(tx)\,dt$

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.

8 Comments »

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Just after “multivariable chain rule”, f(p) – f(0) should be f(x) – f(0).

p lives in the manifold rather than R^n.

Comment by Rory Molinari | July 6, 2011 | Reply
Hi, I’m learning differential geometry and run into this blog, I must say you did a great job. But I have a question: “What does \circ means?”

Comment by guo wei | October 16, 2012 | Reply
It sounds like for some reason the TeX is not being rendered in your browser. \circ is the TeX code to generate the composition symbol $\circ$ .

Comment by John Armstrong | October 16, 2012 | Reply

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